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ARTICLES2023 January66 1 1 10.1007/s11425-016-5135-4 February 8, 2023 May 5,
2023 , 2023 ) The unitary subgroups of group algebras of a class of finite
$2$-groups with derived subgroup of order $2$

<EMAIL_ADDRESS><EMAIL_ADDRESS>
Yulei Wang

Yulei Wang, Heguo Liu

# The unitary subgroups of group algebras of a class of finite $2$-groups with
derived subgroup of order $2$

Yulei Wang  Heguo Liu  Department of Mathematics, Henan University of
Technology, Zhengzhou 450001, China  Department of Mathematics, Hainan
University, Haikou 570228, China

###### Abstract

Let $p$ be a prime and $F$ be a finite field of characteristic $p$. Suppose
that $FG$ is the group algebra of the finite $p$-group $G$ over the field $F$.
Let $V(FG)$ denote the group of normalized units in $FG$ and let $V_{*}(FG)$
denote the unitary subgroup of $V(FG)$. If $p$ is odd, then the order of
$V_{*}(FG)$ is $|F|^{(|G|-1)/2}$. However, the case when $p=2$ still is open.
In this paper, the order of $V_{*}(FG)$ is computed when $G$ is a nonabelian
$2$-group given by a central extension of the form

$1\longrightarrow\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}\longrightarrow
G\longrightarrow\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}\longrightarrow
1$

and $G^{\prime}\cong\mathbb{Z}_{2}$, $n,m\geq 1$. Further, a conjecture is
confirmed, namely, the order of $V_{*}(FG)$ can be divisible by
$|F|^{\frac{1}{2}(|G|+|\Omega_{1}(G)|)-1}$, where $\Omega_{1}(G)=\\{g\in G\ |\
g^{2}=1\\}$.

###### keywords:

normalized unit, unitary subgroup, inner abelian $p$-group, central extension

20C05,20D15

##  1 Introduction

In this paper, $p$ always is a prime, $F$ is a finite field of characteristic
$p$ and $G$ is a finite $p$-group.

For an integral ring $Z$, let $U(Z)$ and $U(ZG)$ be the multiplicative group
of $Z$ and the integral group ring $ZG$, respectively. Suppose that $f$ is a
homomorphism of the group $G$ into $U(Z)$, we define an anti-automorphism of
the ring $ZG$: $x^{f}=\sum\limits_{g\in G}\alpha_{g}f(g)g^{-1}\in ZG$, where
$x=\sum\limits_{g\in G}\alpha_{g}g\in ZG$. An element $u\in U(ZG)$ is said to
be $f$-unitary if the inverse element $u^{-1}$ coincides with the element
$u^{f}$ or $-u^{f}$. Obviously, all $f$-unitary elements of the group $U(ZG)$
form a subgroup, which is denoted by $U_{f}(ZG)$. The interest in $U_{f}(ZG)$
arouse from algebraic topology and unitary $K$-theory in [1, 14]. The study
and description about $U_{f}(ZG)$ in certain cases is known as Novikov’s
problem.

In particular, $f$ is trivial, namely, $f(g)=1$ for all $g\in G$. At this
time, we denote this trivial homomorphism $f$ by $*$. Let $V(FG)$ be the group
of normalized units in $FG$, namely

$V(FG)=\left\\{\sum\limits_{g\in G}\alpha_{g}g\in FG\ |\ \sum\limits_{g\in
G}\alpha_{g}=1\right\\}.$

Obviously, $V(FG)$ is a subgroup of $U(FG)$. An element $x\in V(FG)$ is called
unitary (also called unitary normalized unit) if $x^{*}=x^{-1}$. We denote by
$V_{*}(FG)$ the subgroup of all unitary elements of $V(FG)$. The problem of
the description of invariants of $V_{*}(FG)$ was raised by Novikov, and Serre
has showed that there is a relation between the self-dual normal basis of the
finite Galois extension $L$ over $F$ with Galois group $G$ and the unitary
subgroup of the group algebra $FG$ in [16], where $char(F)=2$.

The order of $V_{*}(FG)$ is equal to $|F|^{(|G|-1)/2}$ when $char(F)>2$ in
[9]. Thus we only consider how to compute order of $V_{*}(FG)$ when $p=2$. In
fact, it is particularly challenging to compute the order of $V_{*}(FG)$ when
$p=2$ as Balogh put in [2]. At this time, there is an interesting conjecture
in [7], namely, the order of $V_{*}(FG)$ is divisible by
$F^{\frac{1}{2}(|G|+|\Omega_{1}(G)|)-1}$, where $\Omega_{1}(G)=\\{g\in G\ |\
g^{2}=1\\}$.

For finite abelian $p$-groups, the relative results of $V_{*}(FG)$ have been
obtained in [3, 4, 7, 8]. However, when $G$ is nonabelian, a few facts about
$V_{*}(FG)$ can be known. For several special classes of groups, some results
can only be obtained, see [4, 5, 10, 11, 12, 17]. In particular, the orders of
unitary subgroups of group algebras of a class of extraspecial $2$-groups, a
dihedral group, a quaternion group have been determined in [9]. In [2], Balogh
gave the order of unitary subgroups of group algebras of finite $2$-groups
which satisfies special conditions. Blackburn determined the isomorphism types
of groups of prime power order with derived subgroup of prime order in [6],
but we need further determine the more accurate structure in order to compute
the unitary subgroups of group algebras of the class of $2$-groups. In [17],
we studied the unitary normalized units of a nonabelian $2$-group with a
derived subgroup of order $2$, which is given by a central extension of a
cyclic group by an elementary abelian $2$-group. Now we will consider a
nonabelian $2$-group $G$ given by a central extension of the form

$1\longrightarrow\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}\longrightarrow
G\longrightarrow\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}\longrightarrow
1$

and $G^{\prime}\cong\mathbb{Z}_{2}$, $m,n\geq 1$.

First, we will give some fundamental conclusions in the second section. And
then we give the accurate structure of the above group $G$ according to the
inner abelian finite $2$-groups and central product in the third section. In
the forth section, we will determine the unitary subgroups of the group
algebra $FG$, where $char(F)=2$. Moreover, a conjecture about the order of
$V_{*}(FG)$ is confirmed.

##  2 Preliminaries

To avoid confusion, we will explain some notations. Let $G[2^{i}]$ denote the
subgroup $\langle g\in G\ |\ g^{2^{i}}=1\rangle$ and let $G^{2^{i}}$ denote
the subgroup $\langle g^{2^{i}}\ |\ g\in G\rangle$. Let $G^{(k)}$ denote the
direct product of $k$ groups $G^{\prime}s$. For $x=\sum\limits_{g\in
G}\alpha_{g}g\in FG$, the support $\\{g\in G\ |\ \alpha_{g}\neq 0\\}$ of $x$
is denoted by $supp(x)$. For any subset $S$ of $G$, we define
$\widehat{S}:=\sum\limits_{g\in S}g$. For an element $c$ of $G$, we denote by
$\Omega_{c}(G)$ the set $\\{g\in G\ |\ g^{2}=c\\}$. We denote by $d(G)$ the
number of the minimal generated elements of finite $p$-group $G$. The
combinatorial number is denoted by $\binom{n}{i}$ , namely,
$\binom{n}{i}=\frac{n!}{(n-i)!i!}$. Let

$\gamma_{1}(k):=\binom{k}{0}3^{k}+\binom{k}{2}3^{k-2}+\cdots+\binom{k}{l}3^{k-l}=2^{2k-1}+2^{k-1},$

where $l=k$ if $k$ is even and $l=k-1$ if $k$ is odd. Let

$\gamma_{2}(k):=\binom{k}{1}3^{k-1}+\binom{k}{3}3^{k-3}+\cdots+\binom{k}{l}3^{k-l}=2^{2k-1}-2^{k-1},$

where $l=k-1$ if $k$ is even and $l=k$ if $k$ is odd. Other notations used are
standard (as in [13, 15]).

######  Definition 2.1.

A group $G$ is a central product of the normal subgroups $H$ and $K$ if
$G=HK$, $[H,K]=1$, and denoted by $H\mathbin{\mathsf{Y}}K$. The central
product of $k$ groups $H^{\prime}s$ is denoted by $H^{\mathbin{\mathsf{Y}}k}$.

######  Definition 2.2.

A finite $p$-group $G$ is called inner abelian if $G$ is nonabelian and every
proper subgroup of $G$ is abelian.

######  Lemma 2.3 ([7]).

Let $G$ be a finite abelian $2$-group and $F$ a finite field of characteristic
$2$. Then

$|V_{*}(FG)|=|G^{2}[2]|\cdot|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1}.$

######  Lemma 2.4 ([9]).

Let $F$ be a finite field of characteristic $2$.

(i) If $G$ is a dihedral group of order $2^{n+1}$, then
$|V_{*}(FG)|=|F|^{3\cdot 2^{n-1}}$.

(ii) If $G$ is a quaternion group of order $2^{n+1}$, then
$|V_{*}(FG)|=4|F|^{2^{n}}$.

######  Lemma 2.5 ([2]).

Let $F$ be a finite field of characteristic $2$. Let $G=K\times E$, where
$K=\langle a,b\ |\ a^{4}=b^{4}=1,[a,b]=a^{2}\rangle$ and $E$ be a finite
elementary abelian $2$-group. Then
$|V_{*}(FG)|=4|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1}$.

In [12], Kaur and Khan studied the unit group $U(F(G\times A))$ and unitary
unit group $U_{*}(F(G\times A))$ of the group algebra $F(G\times A)$ of the
direct product of an arbitrary finite group $G$ and a finite elementary
abelian $2$-group $A$ over a field $F$ with characteristic $2$. Similarly, we
may obtain the results of $V_{*}(F(G\times A))$.

######  Lemma 2.6.

Let $F$ be a finite field of characteristic $2$. Suppose that $G$ is a finite
$2$-group and $A$ is an elementary abelian $2$-group of order $2^{k}$. Then
the unitary normalized unit subgroup $V_{*}(F(G\times A))$ is semidirect
product of the group $W^{*}$ and the unitary normalized unit subgroup
$V_{*}(FG)$, where $W^{*}=(\cdots((A^{*}_{k}\rtimes A^{*}_{k-1})\rtimes
A^{*}_{k-2}\rtimes\cdots)\rtimes A^{*}_{1}$ such that each $A^{*}_{i}$ is an
elementary abelian $2$-group of order $|F|^{2^{i-2}(|G|+|\Omega_{1}(G)|)}$ and
the order of $V_{*}(F(G\times A))$ is
$|V_{*}(FG)|\cdot|F|^{\frac{1}{2}(|G|+|\Omega_{1}(G)|)(|A|-1)}$. Further, if
$|V_{*}(FG)|=l|F|^{\frac{1}{2}(|G|+|\Omega_{1}(G)|)-1}$ for some nature number
$l$, then $|V_{*}(F(G\times A))|=l|F|^{\frac{1}{2}(|G\times
A|+|\Omega_{1}(G\times A)|)-1}$.

######  Proof 2.7.

Let $A=\langle a_{1}\rangle\times\langle a_{2}\rangle\times\cdots\times\langle
a_{k}\rangle$. Write $G_{k}:=G\times A=G_{k-1}\times\langle a_{k}\rangle$ and
$G_{0}:=G$, where $G_{k-1}=G\times\langle
a_{1}\rangle\times\cdots\times\langle a_{k-1}\rangle$. Let $\theta$ be the
projection of $G_{k}$ onto $G_{k-1}$. Assume that
$\theta^{\prime}:FG_{k}\rightarrow FG_{k-1}$ is an algebra homomorphism over
$F$ which is a linear extension of the projection $\theta$. Obviously, the
kernel $\mathrm{Ker}\theta^{\prime}$ of homomorphism is an ideal of $FG_{k}$
generated by $a_{k}-1$. Under the map $\theta^{\prime}$, the image of a
unitary normalized unit in $V(FG_{k})$ is a unitary normalized unit in
$V(FG_{k-1})$. Further, since $\theta$ is an epimorphism fixing $G_{k-1}$, the
restricted map $\theta^{\prime}|_{V_{*}(F(G_{k}))}$ is an epimorphism from
$V_{*}(FG_{k})$ onto $V_{*}(FG_{k-1})$, and its kernel
$A^{*}_{k}=(1+\mathrm{Ker}\theta^{\prime})\bigcap V_{*}(FG_{k})$ is an
elementary abelian group. From this, for arbitrary $x\in A^{*}_{k}$, we have
$x^{*}=x$. Suppose that $x=1+\sum\limits_{g\in G_{k-1}}\alpha_{g}g(a_{k}-1)$,
then

$x=1+\sum\limits_{g\in\Omega_{1}(G_{k-1})}\alpha_{g}g(a_{k}-1)+\sum\limits_{g\in
I}\alpha_{g}(g+g^{-1})(a_{k}-1),$

where $I$ is a subset of $G_{k-1}\setminus\Omega_{1}(G_{k-1})$ such that if
$g\in I$ then $g^{-1}\notin I$. Therefore, the cardinality of $I$ is
$2^{k-2}(|G|-|\Omega_{1}(G)|)$. From this, the order of $A^{*}_{k}$ is
$|F|^{2^{k-2}(|G|+|\Omega_{1}(G)|)}$.

Since there is an inclusion map $i:V_{*}(FG_{k-1})\rightarrow V_{*}(FG_{k})$,
we have $V_{*}(FG_{k})=A^{*}_{k}\rtimes V_{*}(FG_{k-1})$. Assume that
$A^{*}_{i}=(1+\mathrm{Ker}\theta^{\prime})\bigcap V_{*}(FG_{i})$, then by
induction one can prove the lemma.

If $|V_{*}(FG)|=l|F|^{\frac{1}{2}(|G|+|\Omega_{1}(G)|-1)}$ for some nature
number $l$, then

$\begin{aligned} |V_{*}(F(G\times
A))|&=l|F|^{\frac{1}{2}(|G|+|\Omega_{1}(G)|)-1}|F|^{\frac{1}{2}(|G|+|\Omega_{1}(G)|)(|A|-1)}\\\
&=l|F|^{\frac{1}{2}(|G||A|+|\Omega_{1}(G)||A|)-1}\\\
&=l|F|^{\frac{1}{2}(|G\times A|+|\Omega_{1}(G\times A)|)-1}.\end{aligned}$

Note that the central product of $Q_{8}$ and $Q_{8}$ is isomorphic to the
central product of $D_{8}$ and $Q_{8}$. From this, the structure of an
extraspecial $p$-group is as follows.

######  Lemma 2.8 ([15]).

An extraspecial $p$-group is a central product of $n$ nonabelian subgroups of
order $p^{3}$ and has order $p^{2n+1}$. If $p=2$, $G$ is a central product of
$D_{8}^{\prime}s$ or a central product of $D_{8}^{\prime}s$ and a single
$Q_{8}$.

For an inner abelian finite $p$-group, we have the following results.

######  Lemma 2.9 ([18]).

Let $G$ be a finite $p$-group, then the following properties are equivalent:

(i) $G$ is an inner abelian group;

(ii) $d(G)=2$ and $|G^{\prime}|=p$.

(iii) $d(G)=2$ and $\zeta G=Frat(G)$.

######  Lemma 2.10 ([18]).

Let $G$ be an inner abelian finite $p$-group, then $G$ is one of the following
types:

(i) $Q_{8}$, the quaternion group;

(ii) $M_{p}(n,m)=\langle a,b\ |\
a^{p^{n}}=b^{p^{m}}=1,a^{b}=a^{1+p^{n-1}}\rangle$, $n\geq 2,m\geq 1$;

(iii) $M_{p}(n,m,1)=\langle a,b,c\ |\
a^{p^{n}}=b^{p^{m}}=c^{p}=1,[a,b]=c,[c,a]=[c,b]=1\rangle$, $n\geq m\geq 1$.

##  3  Isomorphism types of the group

Let $G$ be a nonabelian $2$-group given by a central extension of the form

$1\longrightarrow N\longrightarrow
G\longrightarrow\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}\longrightarrow
1$

and $G^{\prime}=\langle c\rangle\cong\mathbb{Z}_{2}$,
$N\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$, $n,m\geq 1$. Without loss
of generality, we suppose $n\geq m\geq 1$ in the following sections.

Note that $G^{\prime}\leq\mathrm{Frat}\ G\leq N\leq\zeta G$. Hence $G/\zeta G$
is an elementary abelian $2$-group. Further, we have the following lemma.

######  Lemma 3.1.

(1) For any two elements $\bar{x}=x\zeta G$ and $\bar{y}=y\zeta G$ of $G/\zeta
G$, write $[x,y]=c^{r}$ ($0\leq r\leq 1$) and $f(\bar{x},\bar{y})=r$, then
$G/\zeta G$ becomes a nondegenerate symplectic space over a field $F$ with $2$
elements.

(2) $G=E\mathbin{\mathsf{Y}}\zeta G$, where $E$ is a central product of some
inner abelian groups. Furthermore, these inner abelian groups have the
isomorphism classes: $Q_{8}$, $M_{2}(u,v)$ and $M_{2}(w,1,1)$, where
$u,v,w\leq n+1$.

######  Proof 3.2.

(1) Obviously $f$ is well-defined. For $x$, $y$, $x_{i}$, $y_{i}\in G$,
$i=1,2$, we have $[x_{1}x_{2},y]=[x_{1},y][x_{2},y]$ and
$[x,y_{1}y_{2}]=[x,y_{1}][x,y_{2}]$, thus $f$ is bilinear. Since $[x,x]=1$ and
$[x,y]=[y,x]^{-1}$, $f(\bar{x},\bar{x})=0$ and
$f(\bar{x},\bar{y})=-f(\bar{y},\bar{x})$, so $G/\zeta G$ is a symplectic space
over $F$. If $f(\bar{x},\bar{y})=0$ for all $y\in G$, then $[x,y]=1$, thus
$x\in\zeta G$. It follows that the symplectic space $G/\zeta G$ is
nondegenerate.

(2) From (1), we may assume that the dimension of symplectic space $G/\zeta G$
is $2k$, and $\\{\bar{x}_{1},\ldots,\bar{x}_{k},\bar{y}_{1},\ldots,\\\
\bar{y}_{k}\\}$ is a basis of $G/\zeta G$, where $\bar{x}_{i}=x_{i}\zeta G$
and $\bar{y}_{i}=y_{i}\zeta G$ for $i=1,\ 2,\ \ldots,\ k$, satisfying:
$f(\bar{x}_{i},\bar{y}_{i})=1$, that is, $[x_{i},y_{i}]=c$; For $i\neq j$,
$f(\bar{x}_{i},\bar{x}_{j})=f(\bar{y}_{i},\bar{y}_{j})=f(\bar{x}_{i},\bar{y}_{j})=0$,
that is, $[x_{i},x_{j}]=[y_{i},y_{j}]=[x_{i},y_{j}]=1$.

Let $G_{i}:=\langle x_{i},y_{i}\rangle$, then for $i\neq j$, we have that
$[G_{i},G_{j}]=1$. Obviously $d(G_{i})=2$ and $|G_{i}^{\prime}|=2$, also by
Lemma 2.9, we have $G_{i}$ is inner abelian. Note that $G_{i}^{2}\leq N$.
According to Lemma 2.10, the isomorphism classes of $G_{i}$ are $Q_{8}$,
$M_{2}(u,v)$ and $M_{2}(w,1,1)$, where $u,v,w\leq n+1$.

Let $r:=d(\zeta G)$. Suppose that

$\displaystyle\zeta G=\langle z_{1}\rangle\times\langle
z_{2}\rangle\times\cdots\times\langle
z_{r}\rangle\cong\mathbb{Z}_{2^{n_{1}}}\times\mathbb{Z}_{2^{n_{2}}}\times\cdots\times\mathbb{Z}_{2^{n_{r}}},n_{1}\geq
n_{2}\geq\cdots\geq n_{r}\geq 1.$ (1)

Since $(\zeta G)^{2}\leq N$, we have $n_{i}=1$ for $3\leq i\leq r$ and
$\langle z_{1}^{2}\rangle\times\langle z_{2}^{2}\rangle=\mathrm{Frat\zeta
G}\leq N$. If $n_{2}\geq 2$, then

$\displaystyle N\bigcap\left(\langle z_{3}\rangle\times\cdots\times\langle
z_{r}\rangle\right)=1,$ (2)

otherwise, $|N[2]|\geq 2^{3}$, a contradiction. If $n_{2}=1$, then $m$ must be
$1$ by $N\leq\zeta G$. At this time, by adjusting the parameters
$z_{1},z_{2},\ldots,z_{r}$, we similarly may obtain (2). Further we may obtain
the following lemma.

######  Lemma 3.3.

(1) $n\leq n_{1}\leq n+1$ and $m\leq n_{2}\leq m+1$.

(2) The isomorphism classes of $\zeta G$ are as follows:

(i)
$A_{1}\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$.
In this case, $N=\langle z_{1}\rangle\times\langle z_{2}\rangle$.

(ii)
$A_{2}\cong\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$.
In this case, $N=\langle z_{1}^{2}\rangle\times\langle z_{2}\rangle$.

(iii)
$A_{3}\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$.
In this case, $N=\langle z_{1}\rangle\times\langle z_{2}^{2}\rangle$.

(iv)
$A_{4}\cong\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$.
In this case, $N=\langle z_{1}^{2}\rangle\times\langle z_{2}^{2}\rangle$.

######  Proof 3.4.

(1) Since $N\leq\zeta G$, $n=\mathrm{Exp}N\leq\mathrm{Exp}(\zeta G)=n_{1}$.
Also since $(\zeta G)^{2}\leq N$, $n_{1}-1\leq n$. From this, we have $n\leq
n_{1}\leq n+1$.

If $n_{2}\leq m-1$, then $N\leq\zeta G$ implies that
$\mathbb{Z}_{2^{n-m+1}}\times\mathbb{Z}_{2}\cong N^{2^{m-1}}\leq(\zeta
G)^{2^{m-1}}\cong\mathbb{Z}_{2^{n_{1}-m+1}}$, a contradiction. Hence
$n_{2}\geq m$.

Suppose that $n_{1}=n+1$. Note that
$\mathbb{Z}_{2^{n_{1}-1}}\times\mathbb{Z}_{2^{n_{2}-1}}\cong(\zeta G)^{2}\leq
N$. If $n_{2}\geq m+2$, then $|N|\geq 2^{n+m+1}$, a contradiction. From this,
we have $m\leq n_{2}\leq m+1$. Suppose that $n_{1}=n$. If $n_{2}=m+2$, then
$(\zeta G)^{2}\cong\mathbb{Z}_{2^{n-1}}\times\mathbb{Z}_{2^{m+1}}\cong
N\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$. From this, we have $n-1=m$
and $n_{1}=n=m+1<n_{2}$, a contradiction. If $n_{2}\geq m+3$, then
$2^{n+m+1}\leq|(\zeta G)^{2}|\leq|N|=2^{n+m}$, a contradiction. In a word, (1)
is true.

(2) We will next distinguish the isomorphism classes of $\zeta G$. Let
$N=\langle a\rangle\times\langle b\rangle$, where $|a|=2^{n}$ and $|b|=2^{m}$.

(i) Assume that $n_{1}=n$ and $n_{2}=m$. By (2), we obtain $\zeta G=\langle
a,b,z_{3},\ldots,z_{r}\rangle$, which is the isomorphism class $A_{1}$ of
$\zeta G$.

(ii) Assume that $n_{1}=n+1$ and $n_{2}=m$. We may let $z_{1}^{2}=a^{i}b^{j}$
since $\langle z_{1}^{2}\rangle\leq N$. If $n>m$, then $i$ and $2$ are coprime
since $|z_{1}^{2}|=2^{n}=|a|>|b|$. In this case, we have $N=\langle
a^{i}b^{j}\rangle\times\langle b\rangle=\langle z_{1}^{2}\rangle\times\langle
b\rangle$ and $\zeta G=\langle z_{1},b,z_{3},\ldots,z_{r}\rangle$ by (2),
which is the structure $A_{2}$ of $\zeta G$.

Suppose $n=m$. Obviously, one of $i$ and $j$ must be coprime to $2$. If $i$ is
coprime to $2$, then we have the structure $A_{2}$ of $\zeta G$ which is
similar to the case $n>m$. If $j$ is coprime to $2$, then $N=\langle
a^{i}b^{j}\rangle\times\langle a\rangle=\langle z_{1}^{2}\rangle\times\langle
a\rangle$ and $\zeta G=\langle z_{1},a,z_{3},\ldots,z_{r}\rangle$ by (2) which
is the structure $A_{2}$ of $\zeta G$.

(iii) Assume that $n_{1}=n$ and $n_{2}=m+1$. Not to cause confusion, we may
similarly let $z_{1}^{2}=a^{i}b^{j}$ and $z_{2}^{2}=a^{u}b^{v}$. Obviously,
$i$ must be divisible by $2$ and let $i=2i_{1}$.

If $v$ is coprime to $2$, then $N=\langle a\rangle\times\langle
z_{2}^{2}\rangle$ and $\zeta G=\langle a,z_{2},z_{3},\ldots,z_{r}\rangle$ by
(2), which is the isomorphism class $A_{3}$ of $\zeta G$.

Suppose that $v$ is divisible by $2$ and let $v=2v_{1}$. Obviously
$z_{2}^{2^{m}}=a^{2^{m-1}u}$ is of order $2$. Since $n=n_{1}\geq n_{2}=m+1$,
we have $n>m$. If $j$ is divisible by $2$, then
$z_{1}^{2^{n-1}}=a^{2^{n-1}i_{1}}$ of order 2 is equal to $a^{2^{m-1}u}$, a
contradiction. It follows that $(j,2)=1$. Further we have that $n=m+1$,
otherwise, $\langle z_{1}\rangle\bigcap\langle z_{2}\rangle=\langle
a^{2^{m-1}u}\rangle\neq 1$, a contradiction. Hence $(a^{i}b^{j})^{2^{m-1}}$ is
an element with order $2$ of $\langle z_{1}\rangle$. From this, we may obtain
$N=\langle a\rangle\times\langle a^{i}b^{j}\rangle=\langle
a\rangle\times\langle z_{1}^{2}\rangle$ and $\zeta G=\langle
a,z_{1},z_{3},\ldots,z_{r}\rangle$, which is the isomorphism class $A_{3}$ of
$\zeta G$ by adjusting the parameters $z_{1}$ and $z_{2}$.

(iv) Assume that $n_{1}=n+1$ and $n_{2}=m+1$. Since $(\zeta G)^{2}=\langle
z_{1}^{2}\rangle\times\langle z_{2}^{2}\rangle$ has order $2^{n+m}$, we have
$N=(\zeta G)^{2}$. In this case, we may take $a=z_{1}^{2}$ and $b=z_{2}^{2}$,
which is the isomorphism class $A_{4}$ of $\zeta G$.

According to Lemma 3.1, we know that $G$ is the central product of $E$ and
$\zeta G$. Further, suppose that $E$ is the central product of
$G_{1},G_{2},\ldots,G_{k}$, where the isomorphism classes of $G_{i}$
($i=1,2,\ldots,k$) are $Q_{8}$, $M_{2}(u,v)$ and $M_{2}(w,1,1)$, where
$u,v,w\leq n+1$ as in Lemma 3.1. Next we will determine the isomorphism
classes of $G$ according to the types of $\zeta G$ in Lemma 3.3.

###  3.1  The isomorphism type $A_{1}$ of $\zeta G$

In the section, let

$\zeta G=\langle z_{1}\rangle\times\langle
z_{2}\rangle\times\cdots\times\langle
z_{r}\rangle\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2},N=\langle
z_{1}\rangle\times\langle z_{2}\rangle.$

Obviously, $c\in N[2]=\langle z_{1}^{2^{n-1}}\rangle\times\langle
z_{2}^{2^{m-1}}\rangle$. If $c=z_{1}^{2^{n-1}}\cdot z_{2}^{2^{m-1}}$, then we
may rewrite $N=\langle z_{1}\rangle\times\langle z_{1}^{2^{n-m}}z_{2}\rangle$.
At this time, $c=(z_{1}^{2^{n-m}}z_{2})^{2^{m-1}}$. Without loss of
generality, we may always suppose $c\in\langle z_{1}\rangle$ or $c\in\langle
z_{2}\rangle$.

For every factor $G_{i}$ of the central product of $E$, we first determine the
types of $G_{i}\mathbin{\mathsf{Y}}N$. Note that $\zeta G_{i}\leq N$ in the
central product $G_{i}\mathbin{\mathsf{Y}}N$. For convenience, the notations
as $M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$,
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$,
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}M_{2}(n+1,1)$ and so on, imply the
intersections of the factors of the central products are the group $\langle
c\rangle$ in the following parts.

######  Lemma 3.5.

Suppose $n\geq m\geq 1$ and $c\in\langle z_{1}\rangle$.

(i) If $G_{i}\cong M_{2}(n+1,m+1)$ , then $G_{i}\mathbin{\mathsf{Y}}N=G_{i}$.
When $n>m$, $G_{i}\ncong M_{2}(m+1,n+1)$.

(ii) If $G_{i}\cong M_{2}(n+1,v)$, then $1\leq v\leq m+1$. When $v\leq m$,
$G_{i}\mathbin{\mathsf{Y}}N\cong M_{2}(n+1,1)\times\mathbb{Z}_{2^{m}}$.

(iii) If $G_{i}\cong M_{2}(u,n+1)$, then $2\leq u\leq m+1$ and $n=m$. When
$u<m+1$, $G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}$.

(iv) If $G_{i}\cong M_{2}(u,v)$, where $2\leq u<n+1$ and $1\leq v<n+1$, then
$G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$ or
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$.

(v) If $G_{i}\cong M_{2}(n+1,1,1)$, then $n=m$ and
$G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}$.

(vi) If $G_{i}\cong M_{2}(w,1,1)$, where $1\leq w<n+1$, then
$G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$ or
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$.

######  Proof 3.6.

(i) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{n+1}}=y_{i}^{2^{m+1}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{n}}\rangle\cong
M_{2}(n+1,m+1).$

Since $\zeta G_{i}=\langle x_{i}^{2}\rangle\times\langle
y_{i}^{2}\rangle\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$, we have
$N=\zeta G_{i}$ and $G_{i}\mathbin{\mathsf{Y}}N=G_{i}$.

If

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{m+1}}=y_{i}^{2^{n+1}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{m}}\rangle\cong
M_{2}(m+1,n+1),$

then $G_{i}^{\prime}=\langle x_{i}^{2^{m}}\rangle$. Thus
$x_{i}^{2^{m}}=c=z_{1}^{2^{n-1}}$. Since $\zeta G_{i}=\langle
x_{i}^{2}\rangle\times\langle y_{i}^{2}\rangle\leq N$, we have $N=\zeta G_{i}$
by comparing their orders. But, the case when $n>m$ implies that

$\langle c\rangle=N^{2^{n-1}}=(\zeta G_{i})^{2^{n-1}}=\langle
y_{i}^{2^{n}}\rangle,$

which is a contradiction. Hence $G_{i}\ncong M_{2}(m+1,n+1)$.

(ii) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{n+1}}=y_{i}^{2^{v}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{n}}\rangle\cong
M_{2}(n+1,v),$

then $G_{i}^{\prime}=\langle x_{i}^{2^{n}}\rangle$. Thus
$x_{i}^{2^{n}}=c=z_{1}^{2^{n-1}}$. Since $\zeta G_{i}=\langle
x_{i}^{2}\rangle\times\langle y_{i}^{2}\rangle\leq N$, we have
$2^{n+v-1}\leq|N|=2^{n+m}$. It follows $v-1\leq m$.

Suppose $v\leq m$. Let $x_{i}^{2}=z_{1}^{l}z_{2}^{j}$ and
$y_{i}^{2}=z_{1}^{s}z_{2}^{t}$, where $1\leq l,s\leq 2^{n},1\leq j,t\leq
2^{m}$. If $l$ is divisible by $2$, then
$c=x_{i}^{2^{n}}=z_{1}^{2^{n-1}l}z_{2}^{2^{n-1}j}=z_{2}^{2^{n-1}j}$, which is
a contradiction. From this, $l$ must be coprime to $2$. Since
$1=y_{i}^{2^{v}}=z_{1}^{2^{v-1}s}z_{2}^{2^{v-1}t}$, we have $s$ and $t$ are
divisible by $2^{n-v+1}$ and $2^{m-v+1}$, respectively. Let $s=2s_{1}$ and
$t=2t_{1}$. Hence $y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{1}}$ is of order $2$. At this
time, we may replace $y_{i}$ by $y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{1}}$.

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i},y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{1}},z_{2}\rangle=\langle
x_{i},y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{1}}\rangle\times\langle z_{2}\rangle\cong
M_{2}(n+1,1)\times\mathbb{Z}_{2^{m}}.$

(iii) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{u}}=y_{i}^{2^{n+1}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{u-1}}\rangle\cong
M_{2}(u,n+1).$

Since $\zeta G_{i}=\langle x_{i}^{2}\rangle\times\langle y_{i}^{2}\rangle\leq
N$, we have $2^{u-1+n}\leq|N|=2^{n+m}$. It follows $u-1\leq m$. Let
$x_{i}^{2}=z_{1}^{l}z_{2}^{j}$ and $y_{i}^{2}=z_{1}^{s}z_{2}^{t}$, where
$1\leq l,s\leq 2^{n},1\leq j,t\leq 2^{m}$.

If $n>m$, then $1\neq
y_{i}^{2^{n}}=z_{1}^{2^{n-1}s}z_{2}^{2^{n-1}t}=z_{1}^{2^{n-1}s}=c=x_{i}^{2^{u-1}}$,
which is a contradiction. Hence we obtain $n=m$. Suppose that $u<m+1$. Since
$1=x_{i}^{2^{u}}=z_{1}^{2^{u-1}l}z_{2}^{2^{u-1}j}$, we have both $l$ and $j$
are divisible by $2^{m-u+1}$. Let $l=2l_{1}$ and $j=2j_{1}$. If $t$ is
divisible by $2$, then $1\neq
y_{i}^{2^{n}}=z_{1}^{2^{n-1}s}z_{2}^{2^{n-1}t}=z_{1}^{2^{n-1}s}=c$, which is a
contradiction. Thus we have $t$ is coprime to $2$. In a word, we may replace
$x_{i}$ by $x_{i}z_{1}^{-l_{1}}z_{2}^{-j_{1}}$ and

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}}z_{2}^{-j_{1}},y_{i},z_{1}\rangle=\langle
x_{i}z_{1}^{-l_{1}}z_{2}^{-j_{1}},y_{i}\rangle\mathbin{\mathsf{Y}}\langle
z_{1}\rangle\cong M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}.$

(iv) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{u}}=y_{i}^{2^{v}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{u-1}}\rangle\cong
M_{2}(u,v),$

where $u,v\leq n$. Let $x_{i}^{2}=z_{1}^{l}z_{2}^{j}$ and
$y_{i}^{2}=z_{1}^{s}z_{2}^{t}$, where $1\leq l,s\leq 2^{n},1\leq j,t\leq
2^{m}$. Obviously, both $l$ and $s$ are divisible by $2$. Let $l=2l_{1}$ and
$s=2s_{1}$.

If $(j,2)=1=(t,2)$, then there exist $j_{1}$ and $t_{1}$ such that
$jj_{1}\equiv 1\pmod{2^{m}}$ and $tt_{1}\equiv 1\pmod{2^{m}}$. Hence
$x_{i}^{2j_{1}}=z_{1}^{2l_{1}j_{1}}z_{2}$ and
$y_{i}^{2t_{1}}=z_{1}^{2s_{1}t_{1}}z_{2}$. Note that $n\geq u\geq 2$ and
$c=z_{1}^{2^{n-1}}$. From this, we have
$(x_{i}^{j_{1}}y_{i}^{-t_{1}}z_{1}^{s_{1}t_{1}-l_{1}j_{1}+2^{n-2}})^{2}=1$. It
follows that

$G_{i}\mathbin{\mathsf{Y}}N=\langle
y_{i}z_{1}^{-s_{1}},x_{i}^{j_{1}}y_{i}^{-t_{1}}z_{1}^{s_{1}t_{1}-l_{1}j_{1}+2^{n-2}},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}.$

Suppose that $(j,2)=1$ and $2|t$. Let $t=2t_{2}$. Since the orders of
$x_{i}z_{1}^{-l_{1}}$ and $y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{2}}$ are $2^{m+1}$
and $2$, respectively, we have

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}},y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{2}},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}.$

Similarly, we may obtain the same result for the case when $2|j$ and
$(t,2)=1$.

Suppose that $2|j$ and $2|t$. Let $j=2j_{2}$ and $t=2t_{3}$. We may obtain

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}}z_{2}^{-j_{2}},y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{3}},z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}.$

(v) Let

$G_{i}=\langle x_{i},y_{i},c\ |\
x_{i}^{2^{n+1}}=y_{i}^{2}=c^{2}=1,[x_{i},y_{i}]=c,[x_{i},c]=[y_{i},c]=1\rangle\cong
M_{2}(n+1,1,1).$

Let $x_{i}^{2}=z_{1}^{l}z_{2}^{j}$, where $1\leq l\leq 2^{n},1\leq j\leq
2^{m}$. If $n>m$, then
$x_{i}^{2^{n}}=z_{1}^{2^{n-1}l}z_{2}^{2^{n-1}j}=z_{1}^{2^{n-1}l}$, which is a
contradiction. Hence $n=m$. At this time, if both $l$ and $j$ can be divisible
by $2$, then the order of $z_{1}^{l}z_{2}^{j}$ is less than $2^{m}$, which is
impossible. If $j$ can be divisible by $2$ and $(l,2)=1$, then we have
$x_{i}^{2^{m}}=z_{1}^{2^{m-1}l}z_{2}^{2^{m-1}j}=c$, which is also impossible.
From this, we have $(j,2)=1$. It follows that

$G_{i}\mathbin{\mathsf{Y}}N=\langle x_{i},y_{i},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}.$

(vi) Let

$G_{i}=\langle x_{i},y_{i},c\ |\
x_{i}^{2^{w}}=y_{i}^{2}=c^{2}=1,[x_{i},y_{i}]=c,[x_{i},c]=[y_{i},c]=1\rangle\cong
M_{2}(w,1,1),$

where $1\leq w<n+1$. Let $x_{i}^{2}=z_{1}^{l}z_{2}^{j}$, where $1\leq l\leq
2^{n},1\leq j\leq 2^{m}$. Since
$1=x_{i}^{2^{w}}=z_{1}^{2^{w-1}l}z_{2}^{2^{w-1}j}$, we have $l$ can be
divisible by $2$. Let $l=2l_{1}$.

If $(j,2)=1$, then $G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}},y_{i},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}.$

Suppose $j$ can be divisible by $2$ and let $j=2j_{1}$. Thus

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z^{-l_{1}}z_{2}^{-j_{1}},y_{i},z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}.$

######  Lemma 3.7.

Suppose $n\geq m\geq 1$ and $c\in\langle z_{2}\rangle$.

(i) If $G_{i}\cong M_{2}(m+1,n+1)$ , then $G_{i}\mathbin{\mathsf{Y}}N=G_{i}$.
When $n>m$, $G_{i}\ncong M_{2}(n+1,m+1)$.

(ii) If $G_{i}\cong M_{2}(n+1,v)$, where $1\leq v\leq m+1$, then $n=m$. When
$1\leq v\leq m$, $G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{m}}$.

(iii) If $G_{i}\cong M_{2}(u,n+1)$, where $2\leq u\leq m$, then
$G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}$.

(iv) If $G_{i}\cong M_{2}(u,v)$, where $2\leq u<n+1$ and $1\leq v<n+1$, then

$G_{i}\mathbin{\mathsf{Y}}N\cong\left\\{\begin{aligned}
&D_{8}\times\mathbb{Z}_{2^{n}}\ or\ Q_{8}\times\mathbb{Z}_{2^{n}},\ \ m=1.\\\
&M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}\ \mathrm{or}\
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}},\ \
m>1.\end{aligned}\right.$

(v) If $G_{i}\cong M_{2}(n+1,1,1)$, then $G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}.$

(vi) If $G_{i}\cong M_{2}(w,1,1)$, where $1\leq w<n+1$, then
$G_{i}\mathbin{\mathsf{Y}}N\cong M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}$ or
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}$.

######  Proof 3.8.

(i) - (iii) may be obtained similar to (i) -(iii) of Lemma 3.5.

(iv) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{u}}=y_{i}^{2^{v}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{u-1}}\rangle\cong
M_{2}(u,v),$

where $u,v\leq n$. Let $x_{i}^{2}=z_{1}^{l}z_{2}^{j}$ and
$y_{i}^{2}=z_{1}^{s}z_{2}^{t}$, where $1\leq l,s\leq 2^{n},1\leq j,t\leq
2^{m}$. Obviously, both $l$ and $s$ are divisible by $2$. Let $l=2l_{1}$ and
$s=2s_{1}$.

If $(j,2)=1=(t,2)$, then there exist $j_{1}$ and $t_{1}$ such that
$jj_{1}\equiv 1\pmod{2^{m}}$ and $tt_{1}\equiv 1\pmod{2^{m}}$. Hence
$x_{i}^{2j_{1}}=z_{1}^{2l_{1}j_{1}}z_{2}$ and
$y_{i}^{2t_{1}}=z_{1}^{2s_{1}t_{1}}z_{2}$. Further,
$(x_{i}^{j_{1}}y_{i}^{-t_{1}}z_{1}^{s_{1}t_{1}-l_{1}j_{1}}z_{2}^{2^{m-2}})^{2}=1$
when $m\geq 2$. It follows that

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}},x_{i}^{j_{1}}y_{i}^{-t_{1}}z_{1}^{s_{1}t_{1}-l_{1}j_{1}}z_{2}^{2^{m-2}},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}.$

When $m=1$, we have

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}},y_{i}z_{1}^{-s_{1}},z_{1}\rangle\cong
Q_{8}\times\mathbb{Z}_{2^{n}}.$

Suppose that $(j,2)=1$ and $2|t$. Let $t=2t_{2}$. Since the orders of
$x_{i}z_{1}^{-l_{1}}$ and $y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{2}}$ are $2^{m+1}$
and $2$, we have

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}},y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{2}},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}.$

Similarly, we may obtain the same result for the case $2|j$ and $(t,2)=1$.

Suppose that $2|j$ and $2|t$. Let $j=2j_{2}$ and $t=2t_{3}$. We may obtain

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}}z_{2}^{-j_{2}},y_{i}z_{1}^{-s_{1}}z_{2}^{-t_{3}},z_{2},z_{1}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}.$

(v) Let

$G_{i}=\langle x_{i},y_{i},c\ |\
x_{i}^{2^{n+1}}=y_{i}^{2}=c^{2}=1,[x_{i},y_{i}]=c,[x_{i},c]=[y_{i},c]=1\rangle\cong
M_{2}(n+1,1,1).$

Let $x_{i}^{2}=z_{1}^{l}z_{2}^{j}$, where $1\leq l\leq 2^{n},1\leq j\leq
2^{m}$. If both $l$ and $j$ can be divisible by $2$, then the order of
$z_{1}^{l}z_{2}^{j}$ is less than $2^{n}$, which is impossible. If $l$ can be
divisible by $2$, then $(j,2)=1$ and
$x_{i}^{2^{n}}=z_{1}^{2^{n-1}l}z_{2}^{2^{n-1}j}=z_{2}^{2^{n-1}j}$, which is a
contradiction. Hence $(l,2)=1$. From this, we have

$G_{i}\mathbin{\mathsf{Y}}N=\langle x_{i},y_{i},z_{2}\rangle\cong
M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}.$

(vi) Let

$G_{i}=\langle x_{i},y_{i},c\ |\
x_{i}^{2^{w}}=y_{i}^{2}=c^{2}=1,[x_{i},y_{i}]=c,[x_{i},c]=[y_{i},c]=1\rangle\cong
M_{2}(w,1,1),$

where $1\leq w<n+1$. Let $x_{i}^{2}=z_{1}^{l}z_{2}^{j}$, where $1\leq l\leq
2^{n},1\leq j\leq 2^{m}$. Since
$1=x_{i}^{2^{w}}=z_{1}^{2^{w-1}l}z_{2}^{2^{w-1}j}$, we have $l$ can be
divisible by $2$. Let $l=2l_{1}$.

If $(j,2)=1$, then $G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z_{1}^{-l_{1}},y_{i},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}.$

Suppose $j$ can be divisible by $2$ and let $j=2j_{1}$. Thus

$G_{i}\mathbin{\mathsf{Y}}N=\langle
x_{i}z^{-l_{1}}z_{2}^{-j_{1}},y_{i},z_{2},z_{1}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}.$

According to Lemmas 3.1 and 3.5, for arbitrary factor $G_{i}$ of the central
product of $E$, $G_{i}\mathbin{\mathsf{Y}}N$ has the following isomorphism
classes: $M_{2}(n+1,m+1)$, $M_{2}(n+1,1)\times\mathbb{Z}_{2^{m}}$,
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$,
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$ and
$Q_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$. From
this, the central product of arbitrary two factors $G_{1}$ and $G_{2}$ may
only be considered that among each other of $M_{2}(n+1,m+1)$, $M_{2}(n+1,1)$,
$M_{2}(m+1,1,1)$, $D_{8}$ and $Q_{8}$. Note that
$Q_{8}\mathbin{\mathsf{Y}}Q_{8}\cong D_{8}\mathbin{\mathsf{Y}}D_{8}$. Hence we
only consider other cases.

######  Lemma 3.9.

Suppose $n\geq m\geq 1$ and $c\in\langle z_{1}\rangle$. Let $G_{1}$ and
$G_{2}$ are arbitrary two factors which are isomorphic to
$M_{2}(n+1,m+1),M_{2}(n+1,1)$ or $M_{2}(m+1,1,1)$.

(i) If $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(n+1,m+1)$, then

$G_{1}\mathbin{\mathsf{Y}}G_{2}\cong\left\\{\begin{aligned}
&M_{2}(2,1,1)\mathbin{\mathsf{Y}}D_{8}\ \mbox{or}\
M_{2}(2,1,1)\mathbin{\mathsf{Y}}Q_{8},\ \ n=1.\\\
&M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}M_{2}(n+1,1),\ \ n>1.\end{aligned}\right.$

(ii) If $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(n+1,m+1)$ and
$M_{2}(n+1,1)$, respectively, then $G_{1}\mathbin{\mathsf{Y}}G_{2}\cong
M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}D_{8}$.

(iii) If $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(n+1,m+1)$ and
$M_{2}(m+1,1,1)$, respectively, then $G_{1}\mathbin{\mathsf{Y}}G_{2}\cong
M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}D_{8}$.

(iv) If both $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(m+1,1,1)$, then
$G_{1}\mathbin{\mathsf{Y}}G_{2}\cong M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}D_{8}$
or $M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1)$.

(v) If both $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(n+1,1)$, then
$G_{1}\mathbin{\mathsf{Y}}G_{2}\cong M_{2}(n+1,1)\mathbin{\mathsf{Y}}D_{8}$ or
$M_{2}(n+1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1,1)$.

######  Proof 3.10.

(i) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{n+1}}=y_{i}^{2^{m+1}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{n}}\rangle\cong
M_{2}(n+1,m+1),$

where $i=1,2$. Since $\zeta G_{1}=\zeta G_{2}=N$, we may let
$x_{2}^{2}=x_{1}^{2l}y_{1}^{2j}$ and $y_{2}^{2}=x_{1}^{2s}y_{1}^{2t}$.

First, we suppose that $n>m$. Obviously, $(l,2)=1$ and $s$ is divisible by
$2$. Let $s=2s_{1}$. If $t$ is divisible by $2$, then $1\neq
y_{2}^{2^{m}}=x_{1}^{2^{m}s}y_{1}^{2^{m}t}=x_{1}^{2^{m}s}$, which is
impossible. Hence we obtain that $(t,2)=1$. If $(j,2)=1$, then
$x_{1}^{l}y_{1}^{j}x_{2}^{-2^{n-1}-1}$ is of order $2$. From this, we have

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
y_{1},x_{1}^{l}y_{1}^{j}x_{2}^{-2^{n-1}-1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}^{-1}x_{1}^{2s_{1}}y_{1}^{t}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}M_{2}(n+1,1).$

If $j$ is divisible by $2$, then we have

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
y_{1},x_{1}^{l}y_{1}^{j}x_{2}^{-1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}^{-1}x_{1}^{2s_{1}}y_{1}^{t}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}M_{2}(n+1,1).$

We next suppose that $n=m$. Obviously, $l$ and $j$ can not simultaneously be
divisible by $2$, nor can $s$ and $t$. If $j$ and $2$ are coprime, then
$c=x_{2}^{2^{m}}=x_{1}^{2^{m}l}y_{1}^{2^{m}j}=c^{l}y_{1}^{2^{m}}$, which is a
contradiction. From this, we have $j$ is divisible by $2$ and $(l,2)=1$. Let
$j=2j_{1}$. Similarly, $t$ are coprime to $2$.

Suppose that $s$ is coprime to $2$. If $m\geq 2$, then

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
y_{2},x_{1}^{l}y_{1}^{2j_{1}}x_{2}^{-1}\rangle\mathbin{\mathsf{Y}}\langle
x_{1},x_{1}^{s-2^{m-1}}y_{1}^{t}y_{2}^{-1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1).$

If $m=1$, then

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
y_{2},x_{1}^{l}y_{1}^{2j_{1}}x_{2}^{-1}\rangle\mathbin{\mathsf{Y}}\langle
x_{1},y_{1}^{t}y_{2}^{-1}\rangle\cong M_{2}(2,1,1)\mathbin{\mathsf{Y}}Q_{8}.$

If $s$ is divisible by $2$, then we similarly have
$G_{1}\mathbin{\mathsf{Y}}G_{2}\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1)$.

(ii) Let

$G_{1}=\langle x_{1},y_{1}\ |\
x_{1}^{2^{n+1}}=y_{1}^{2^{m+1}}=1,x_{1}^{y_{1}}=x_{1}^{1+2^{n}}\rangle\cong
M_{2}(n+1,m+1)$

and

$G_{2}=\langle x_{2},y_{2}\ |\
x_{2}^{2^{n+1}}=y_{2}^{2}=1,x_{2}^{y_{2}}=x_{2}^{1+2^{n}}\rangle\cong
M_{2}(n+1,1).$

Since $\zeta G_{1}=N\geq\zeta G_{2}=\langle x_{2}^{2}\rangle$, we may let
$x_{2}^{2}=x_{1}^{2l}y_{1}^{2j}$. Further, we have $(l,2)=1$ since
$c=x_{2}^{2^{n}}=x_{1}^{2^{n}l}y_{1}^{2^{n}j}=c^{l}y_{1}^{2^{n}j}$.

Suppose that $n>m$. If $(j,2)=1$, then we have

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
x_{1}y_{2},y_{1}y_{2}\rangle\mathbin{\mathsf{Y}}\langle
x_{2}^{-2^{n-1}-1}x_{1}^{l}y_{1}^{j},y_{2}\rangle\cong
M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}D_{8}.$

If $j$ is divisible by $2$, then we have

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
x_{1},y_{1}y_{2}\rangle\mathbin{\mathsf{Y}}\langle
x_{2}^{-1}x_{1}^{l}y_{1}^{j},y_{2}\rangle\cong
M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}D_{8}.$

If $n=m$, then $j$ is divisible by $2$. Similarly, the result is true.

(iii) Let

$G_{1}=\langle x_{1},y_{1}\ |\
x_{1}^{2^{n+1}}=y_{1}^{2^{m+1}}=1,x_{1}^{y_{1}}=x_{1}^{1+2^{n}}\rangle\cong
M_{2}(n+1,m+1)$

and

$G_{2}=\langle x_{2},y_{2},c\ |\
x_{2}^{2^{m+1}}=y_{2}^{2}=c^{2}=1,[x_{2},y_{2}]=c,[x_{2},c]=[y_{2},c]=1\rangle\cong
M_{2}(m+1,1,1).$

Since $\zeta G_{1}=N\geq\zeta G_{2}=\langle x_{2}^{2},c\rangle$, we may let
$x_{2}^{2}=x_{1}^{2l}y_{1}^{2j}$. If $j$ can be divisible by $2$, then
$x_{2}^{2^{m}}=x_{1}^{2^{m}l}y_{1}^{2^{m}j}=x_{1}^{2^{m}l}=c$, which is
impossible. From this, we have $(j,2)=1$.

First we suppose that $n>m\geq 1$. Obviously $l$ is divisible by $2$. Let
$l=2l_{1}$. It follows that

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
x_{1}y_{2},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2}^{-1}x_{1}^{2l_{1}}y_{1}^{j},y_{2}\rangle\cong
M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}D_{8}.$

We next suppose that $n=m$. For the case when $2|l$, the result is similarly
obtained. If $(l,2)=1$, then we have

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
x_{1}y_{2},y_{1}y_{2}\rangle\mathbin{\mathsf{Y}}\langle
x_{1}^{l}y_{1}^{j}x_{2}^{-1},y_{2}\rangle\cong
M_{2}(m+1,m+1)\mathbin{\mathsf{Y}}D_{8}.$

(iv) Let

$G_{i}=\langle x_{i},y_{i},c\ |\
x_{i}^{2^{m+1}}=y_{i}^{2}=c^{2}=1,[x_{i},y_{i}]=c,[x_{i},c]=[y_{i},c]=1\rangle\cong
M_{2}(m+1,1,1),$

where $i=1,2$. Let $x_{1}^{2}=z_{1}^{l}z_{2}^{j}$. If $j$ is divisible by $2$,
then $x_{1}^{2^{m}}=z_{1}^{2^{m-1}l}z_{2}^{2^{m-1}j}=z_{1}^{2^{m-1}l}$, which
is impossible. Hence we have $(j,2)=1$. Note that $\langle
x_{1}^{2}\rangle\times\langle z_{1}\rangle=N=\langle
x_{2}^{2}\rangle\times\langle z_{1}\rangle$. Let
$x_{2}^{2}=x_{1}^{2s}z_{1}^{t}$. Obviously, $(s,2)=1$.

If $t$ is divisible by $2$, then we may let $t=2t_{1}$ and replace $x_{1}$ by
$x_{1}^{s}z_{1}^{t_{1}}$. At this time, $x_{1}^{2}=x_{2}^{2}$. It follows that

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
x_{1},y_{1}y_{2}\rangle\mathbin{\mathsf{Y}}\langle
x_{2}x_{1}^{-1},y_{2}\rangle\cong M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}D_{8}.$

If (t,2)=1, then $n=m$ and

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
x_{1},y_{1}y_{2}\rangle\mathbin{\mathsf{Y}}\langle
x_{2}x_{1}^{-s},y_{2}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1).$

(v) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{n+1}}=y_{i}^{2}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{n}}\rangle\cong
M_{2}(n+1,1),$

where $i=1,2$. Note that $\langle x_{1}^{2}\rangle\times\langle
z_{2}\rangle=N=\langle x_{2}^{2}\rangle\times\langle z_{2}\rangle$. Let
$x_{2}^{2}=x_{1}^{2s}z_{2}^{t}$. Obviously, $(s,2)=1$.

If $t$ is divisible by $2$, then we may let $t=2t_{1}$ and replace $x_{1}$ by
$x_{1}^{s}z_{2}^{t_{1}}$. At this time, $x_{1}^{2}=x_{2}^{2}$. It follows that

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
x_{1},y_{1}y_{2}\rangle\mathbin{\mathsf{Y}}\langle
x_{2}x_{1}^{-1},y_{2}\rangle\cong M_{2}(n+1,1)\mathbin{\mathsf{Y}}D_{8}.$

If (t,2)=1, then

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
x_{1},y_{1}y_{2}\rangle\mathbin{\mathsf{Y}}\langle
x_{2}x_{1}^{-s},y_{2}\rangle\cong
M_{2}(n+1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1,1).$

######  Theorem 3.11.

Suppose that $n\geq m\geq 1$ and $c\in\langle z_{1}\rangle$. Then the
isomorphism classes of $G$ are as follows.

(i)
$M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(ii)
$M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(iii)
$M_{2}(n+1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(iv)
$M_{2}(n+1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-3}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 3$;

(v)
$M_{2}(n+1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\times\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(vi)
$M_{2}(n+1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\times\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(vii)
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(viii)
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(ix)
$\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(x)
$Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$.

######  Proof 3.12.

First, we rewrite $G$ by
$(G_{1}\mathbin{\mathsf{Y}}N)\mathbin{\mathsf{Y}}(G_{2}\mathbin{\mathsf{Y}}N)\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}(G_{k}\mathbin{\mathsf{Y}}N)\mathbin{\mathsf{Y}}A_{1}$.
According to Lemmas 3.1 and 3.5, we have the isomorphism classes of
$G_{i}\mathbin{\mathsf{Y}}N^{\prime}s$ are $M_{2}(n+1,m+1)$,
$M_{2}(n+1,1)\times\mathbb{Z}_{2^{m}}$,
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$,
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$ and
$Q_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$.

First, we suppose that $n>m$. If $\mathrm{Exp}(E)$ is equal to $2^{n+1}$, then
there exists some $G_{i}$ such that $\mathrm{Exp}(G_{i})=2^{n+1}$. Without
loss of generality, suppose $\mathrm{Exp}(G_{1})=2^{n+1}$. Thus we have
$G_{1}\mathbin{\mathsf{Y}}N\cong M_{2}(n+1,m+1)$ or
$M_{2}(n+1,1)\times\mathbb{Z}_{2^{m}}$ by Lemma 3.5. If the number $\lambda$
of $(G_{i}\mathbin{\mathsf{Y}}N)^{\prime}s$ is odd, which are isomorphic to
$M_{2}(n+1,m+1)$, then we may obtain the isomorphism classes (i) and (ii) of
$G$ by Lemma 3.9. If $\lambda$ is a nonzero even number, then isomorphism
classes of $G$ are (iii) and (iv). Suppose $\lambda=0$. If there exists some
$G_{i}\mathbin{\mathsf{Y}}N$ such that $G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$, then
$M_{2}(n+1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1,1)$ will produce
$M_{2}(n+1,m+1)$, which implies $\lambda\neq 0$. At the time, we may obtain
the isomorphism classes (v) and (vi) of $G$.

Suppose that $\mathrm{Exp}(E)$ is less than $2^{n+1}$. At this case, the
isomorphism classes of $(G_{i}\mathbin{\mathsf{Y}}N)^{\prime}s$ are
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$,
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$ and
$Q_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$. If
there exists some $G_{i}\mathbin{\mathsf{Y}}N$ such that
$G_{i}\mathbin{\mathsf{Y}}N\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$, then we may obtain
isomorphism classes (vii) and (viii) of $G$ according to (iv) of Lemma 3.9
avoiding repeating the above types. Otherwise, (ix) and (x) may be obtained.

We next suppose that $n=m$. For convenience, we will distinguish the case
$m>1$ from $m=1$. First, suppose that $m>1$. If $\mathrm{Exp}(E)$ is equal to
$2^{m+1}$, then we may suppose $\mathrm{Exp}(G_{1})=2^{m+1}$. Thus we have
$G_{1}\mathbin{\mathsf{Y}}N\cong M_{2}(m+1,m+1)$,
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}$ or
$M_{2}(m+1,1)\times\mathbb{Z}_{2^{m}}$ by Lemma 3.5. If the number $\mu$ of
$(G_{i}\mathbin{\mathsf{Y}}N)^{\prime}s$ is odd, which are isomorphic to
$M_{2}(m+1,m+1)$, then we may obtain the isomorphism classes (i) and (ii) of
$G$ by Lemma 3.9. If $\mu$ is a nonzero even number, then isomorphism classes
of $G$ are (iii) and (iv). Suppose that $\mu=0$. Obviously, $M_{2}(m+1,1,1)$
and $M_{2}(m+1,1)$ can not appear simultaneously. When $M_{2}(m+1,1)$ comes
into being, we may obtain the isomorphism classes (v) and (vi) of $G$. When
$M_{2}(m+1,1,1)$ appears, (vii) and (viii) may be obtained. Suppose that
$\mathrm{Exp}(E)$ is less than $2^{m+1}$. At this case, (ix) and (x) may be
obtained.

If $m=1$, then we have $G_{i}\mathbin{\mathsf{Y}}N\cong M_{2}(2,2)$,
$M_{2}(2,1,1)$, $D_{8}\times\mathbb{Z}_{2}$ or $Q_{8}\times\mathbb{Z}_{2}$ by
Lemma 3.5 for arbitrary $i$. At this time, the following arbitrary two are the
same classes: (iii) and (vii), (iv) and (viii), (v) and (ix), (vi) and (x).
Similarly, we may obtain six isomorphism classes of $G$, that is, (i), (ii),
(iii), (iv),(v) and (vi).

According to Lemma 3.7, we only consider the central products among
$M_{2}(m+1,n+1),M_{2}(m+1,1),M_{2}(n+1,1,1),D_{8}$ and $Q_{8}$. Hence we have
the following lemma.

######  Lemma 3.13.

Suppose $n>m\geq 1$ and $c\in\langle z_{2}\rangle$. Let $G_{1}$ and $G_{2}$
are arbitrary two factors which are isomorphic to
$M_{2}(m+1,n+1),M_{2}(m+1,1)$ or $M_{2}(n+1,1,1)$.

(i) If $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(m+1,n+1)$, then

$G_{1}\mathbin{\mathsf{Y}}G_{2}\cong\left\\{\begin{aligned}
&M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}D_{8}\ or\
M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}Q_{8},\ \ m=1.\\\
&M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1)\,\ \ m>1.\end{aligned}\right.$

(ii) If $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(m+1,n+1)$ and
$M_{2}(m+1,1)$, respectively, then $G_{1}\mathbin{\mathsf{Y}}G_{2}\cong
M_{2}(m+1,n+1)\mathbin{\mathsf{Y}}D_{8}$.

(iii) If $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(m+1,n+1)$ and
$M_{2}(n+1,1,1)$, respectively, then $G_{1}\mathbin{\mathsf{Y}}G_{2}\cong
M_{2}(m+1,n+1)\mathbin{\mathsf{Y}}D_{8}$.

(iv) If both $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(n+1,1,1)$, then
$G_{1}\mathbin{\mathsf{Y}}G_{2}\cong M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}D_{8}$
or $M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1)$.

(v) If both $G_{1}$ and $G_{2}$ are isomorphic to $M_{2}(m+1,1)$, then
$G_{1}\mathbin{\mathsf{Y}}G_{2}\cong M_{2}(m+1,1)\mathbin{\mathsf{Y}}D_{8}$.

######  Proof 3.14.

(i) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{m+1}}=y_{i}^{2^{n+1}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{m}}\rangle\cong
M_{2}(m+1,n+1),$

where $i=1,2$. Since $\zeta G_{1}=\zeta G_{2}=N$, we may let
$x_{2}^{2}=x_{1}^{2l}y_{1}^{2j}$ and $y_{2}^{2}=x_{1}^{2s}y_{1}^{2t}$.
Obviously, $j$ is divisible by $2$ and let $j=2j_{1}$. If $l$ is divisible by
$2$, then $c=x_{2}^{2^{m}}=x_{1}^{2^{m}l}y_{1}^{2^{m}j}=y_{1}^{2^{m}j}$, which
is impossible. Hence we obtain that $(l,2)=1$. If $t$ is divisible by $2$,
then $1\neq y_{2}^{2^{n}}=x_{1}^{2^{n}s}y_{1}^{2^{n}t}=1$ since $n>m\geq 1$, a
contradiction. Hence $t$ is coprime to $2$.

If $m=1$, then $\langle x_{1},y_{2}y_{1}^{-t}\rangle$ is isomorphic to $D_{8}$
(when $2|s$) or $Q_{8}$ (when $(s,2)=1$). It follows that

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
y_{2},x_{1}^{l}y_{1}^{2j_{1}}x_{2}^{-1}\rangle\mathbin{\mathsf{Y}}\langle
x_{1},y_{2}y_{1}^{-t}\rangle\cong M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}D_{8}\ or\
M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}Q_{8}.$

Suppose that $m>1$. We may take $\varepsilon:=2^{m-1}$ (when $(s,2)=1$) and
$0$ (when $2|s$). Hence $y_{2}y_{1}^{-t}x_{1}^{\varepsilon-s}$ is of order $2$
and

$G_{1}\mathbin{\mathsf{Y}}G_{2}=\langle
y_{2},x_{1}^{l}y_{1}^{2j_{1}}x_{2}^{-1}\rangle\mathbin{\mathsf{Y}}\langle
x_{1},y_{2}y_{1}^{-t}x_{1}^{\varepsilon-s}\rangle\cong
M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1).$

(ii)-(v) may be obtained similar to Lemma 3.9.

######  Theorem 3.15.

Suppose that $n>m\geq 1$ and $c\in\langle z_{2}\rangle$. Then the isomorphism
classes of $G$ are as follows.

(i)
$M_{2}(m+1,n+1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(ii)
$M_{2}(m+1,n+1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(iii)
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(iv)
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-3}\mathbin{\mathsf{Y}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 3$;

(v)
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(vi)
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(vii)
$M_{2}(m+1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\times\mathbb{Z}_{2^{n}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(viii)
$M_{2}(m+1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\times\mathbb{Z}_{2^{n}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(ix)
$\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(x)
$Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$.

######  Proof 3.16.

According to Lemma 3.1, we have that

$G=E\mathbin{\mathsf{Y}}A_{1}=G_{1}\mathbin{\mathsf{Y}}G_{2}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}G_{k}\mathbin{\mathsf{Y}}A_{1}=(G_{1}\mathbin{\mathsf{Y}}N)\mathbin{\mathsf{Y}}(G_{2}\mathbin{\mathsf{Y}}N)\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}(G_{k}\mathbin{\mathsf{Y}}N)\mathbin{\mathsf{Y}}A_{1}.$

According to Lemmas 3.1 and 3.7, the isomorphism classes of
$G_{i}\mathbin{\mathsf{Y}}N^{\prime}s$ are $M_{2}(m+1,n+1)$,
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}$,
$M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}$,
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}$ and
$Q_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}$.

If $\mathrm{Exp}(E)$ is equal to $2^{n+1}$, then there exists some $G_{i}$
such that $\mathrm{Exp}(G_{i})=2^{n+1}$. Without loss of generality, suppose
$\mathrm{Exp}(G_{1})=2^{n+1}$. Thus we have $G_{1}\mathbin{\mathsf{Y}}N\cong
M_{2}(m+1,n+1)$ or $M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}$ by
Lemma 3.7. If the number $\mu$ of $G_{i}\mathbin{\mathsf{Y}}N^{\prime}s$ is
odd, which are isomorphic to $M_{2}(m+1,n+1)$, then we may obtain the
isomorphism classes (i) and (ii) of $G$ by Lemma 3.13. If $\lambda$ is a
nonzero even number, then isomorphism classes of $G$ are (iii) and (iv).
Suppose $\mu=0$. If there exists some $G_{i}\mathbin{\mathsf{Y}}N$ such that
$G_{i}\mathbin{\mathsf{Y}}N\cong M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}$, then
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1)$ will produce
$M_{2}(m+1,n+1)$, which implies $\mu\neq 0$. Obviously, At the time, if two
factors $G_{i}$ and $G_{j}$ are isomorphic to $M_{2}(n+1,1,1)$, then
$G_{1}\mathbin{\mathsf{Y}}G_{2}$ is only isomorphic to
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}D_{8}$ in (iv) of Lemma 3.13. It follows
the isomorphism classes (v) and (vi) of $G$.

Suppose that $\mathrm{Exp}(E)$ is less than $2^{n+1}$. At this case, the
isomorphism classes of $(G_{i}\mathbin{\mathsf{Y}}N)^{\prime}s$ are
$M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}$,
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}$ and
$Q_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n}}$. If
there exists some $G_{i}\mathbin{\mathsf{Y}}N$ such that
$G_{i}\mathbin{\mathsf{Y}}N\cong M_{2}(m+1,1)\times\mathbb{Z}_{2^{n}}$, then
we may obtain isomorphism classes (vii) and (viii) of $G$. Otherwise, (ix) and
(x) may be obtained.

###  3.2  The isomorphism types $A_{2}$ and $A_{3}$ of $\zeta G$

In the section, we first consider

$\zeta G=\langle z_{1}\rangle\times\langle
z_{2}\rangle\times\cdots\times\langle
z_{r}\rangle\cong\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2},N=\langle
z_{1}^{2}\rangle\times\langle z_{2}\rangle,$

Without loss of generality, we may always let $c\in\langle z_{1}^{2}\rangle$
or $c\in\langle z_{2}\rangle$. $E$ is the central product of
$G_{1},G_{2},\ldots,G_{k}$, whose isomorphism classes are $Q_{8}$,$M_{2}(u,v)$
and $M_{2}(w,1,1)$, where $u,v,w\leq n+1$ in (2) of Lemma 3.1. First we
determine the types of $G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle$.

######  Lemma 3.17.

Suppose $n\geq m\geq 1$ and $c\in\langle z_{1}\rangle$.

(i) If $G_{i}\cong M_{2}(n+1,v)$, then $1\leq v\leq m+1$ and
$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}$ or
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}$.

(ii) If $n=m$ and $G_{i}\cong M_{2}(u,m+1)$, where $2\leq u<m+1$, then
$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m+1}}$. If $n>m$, then
$G_{i}$ is not isomorphic to $M_{2}(u,n+1)$, where $2\leq u\leq m+1$.

(iii) If $G_{i}\cong M_{2}(u,v)$, where $2\leq u<n+1$ and $1\leq v<n+1$, then
$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}$ or
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}$.

(iv) If $G_{i}\cong M_{2}(w,1,1)$, then $1\leq w<n+1$ (If $n>m$) or $1\leq
w\leq m+1$ (If $n=m$). Further, $G_{i}\mathbin{\mathsf{Y}}\langle
z_{1},z_{2}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}$ or
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}$.

######  Proof 3.18.

(i) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{n+1}}=y_{i}^{2^{v}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{n}}\rangle\cong
M_{2}(n+1,v).$

Since $\zeta G_{i}=\langle x_{i}^{2},y_{i}^{2}\rangle\leq N$, we have
$2^{n+v-1}\leq|N|=2^{n+m}$ and $v\leq m+1$. Let
$x_{i}^{2}=z_{1}^{2l}z_{2}^{j}$ and $y_{i}^{2}=z_{1}^{2s}z_{2}^{t}$, where
$1\leq l,s\leq 2^{n},1\leq j,t\leq 2^{m}$.

First, we suppose that $v=m+1$. Obviously, $\zeta G_{i}=N$ by comparing their
orders. If $t$ is divisible by $2$, then $1\neq
y_{i}^{2^{m}}=z_{1}^{2^{m}s}z_{2}^{2^{m-1}t}=z_{1}^{2^{m}s}$, which is
impossible. Hence $t$ and $2$ are coprime.

We will next consider two cases $(j,2)=1$ and $2|j$. When $j$ and $2$ are
coprime, we may suppose that $x_{i}^{2}=z_{1}^{2l}z_{2}$ and
$y_{i}^{2}=z_{1}^{2s}z_{2}$ without loss of generality. At this time, we have

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
y_{i}z_{1}^{-s},x_{i}y_{i}^{-1}z_{1}^{s-l+2^{n-1}}\rangle\mathbin{\mathsf{Y}}\langle
z_{1}\rangle\cong M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}.$

When $j$ is divisible by $2$, let $j=2j_{1}$, then we have

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
y_{i}z_{1}^{-s},x_{i}z_{1}^{-l}z_{2}^{-j_{1}}\rangle\mathbin{\mathsf{Y}}\langle
z_{1}\rangle\cong M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}.$

We next suppose $1\leq v<m+1$. Since
$1=y_{i}^{2^{v}}=z_{1}^{2^{v}s}z_{2}^{2^{v-1}t}$, we have $t$ is divisible by
$2^{m-v+1}$. Let $t=2t_{1}$. If $j$ and $2$ are coprime, then

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},y_{i}z_{1}^{-s}z_{2}^{-t_{1}},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}.$

Suppose that $j$ is divisible by $2$ and let $j=2j_{1}$. Hence

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l}z_{2}^{-j_{1}},y_{i}z_{1}^{-s}z_{2}^{-t_{1}}\rangle\mathbin{\mathsf{Y}}\langle
z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}.$

(ii) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{u}}=y_{i}^{2^{n+1}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{u-1}}\rangle\cong
M_{2}(u,n+1),$

where $2\leq u\leq m+1$. Obviously, $\zeta G_{i}=\langle
x_{i}^{2},y_{i}^{2}\rangle\leq\langle z_{1}^{2},z_{2}\rangle$. At this time,
let $x_{i}^{2}=z_{1}^{2l}z_{2}^{j}$ and $y_{i}^{2}=z_{1}^{2s}z_{2}^{t}$.

If $n=m$, then we only consider the case $2\leq u<m+1$ in order to avoid
repeating (i). Since $1=x_{i}^{2^{u}}=z_{1}^{2^{u}l}z_{2}^{2^{u-1}j}$, both
$l$ and $j$ are divisible by $2^{m+1-u}$. Let $j=2j_{1}$. Since $1\neq
y_{i}^{2^{m}}=z_{1}^{2^{m}s}z_{2}^{2^{m-1}t}=c^{s}z_{2}^{2^{m-1}t}$, we have
$t$ and $2$ are coprime. Hence

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
y_{i}z_{1}^{-s},x_{i}z_{1}^{-l}z_{2}^{-j_{1}},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m+1}}.$

Suppose that $n>m$. Note that $1\neq
y_{i}^{2^{n}}=z_{1}^{2^{n}s}z_{2}^{2^{n-1}t}=z_{1}^{2^{n}s}=c^{s}$, a
contradiction.

(iii) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{u}}=y_{i}^{2^{v}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{u-1}}\rangle\cong
M_{2}(u,v),$

where $u,v\leq n$. Let $x_{i}^{2}=z_{1}^{2l}z_{2}^{j}$ and
$y_{i}^{2}=z_{1}^{2s}z_{2}^{t}$, where $1\leq l,s\leq 2^{n},1\leq j,t\leq
2^{m}$. Obviously, both $l$ and $s$ are divisible by $2$. Let $l=2l_{1}$ and
$s=2s_{1}$.

If $(j,2)=1=(t,2)$, then we may let $x_{i}^{2}=z_{1}^{2l}z_{2}$ and
$y_{i}^{2}=z_{1}^{2s}z_{2}$ without loss of generality. At this time, we have

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},x_{i}y_{i}^{-1}z_{1}^{s-l+2^{n-1}},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}.$

If $(j,2)=1$ and $2|t$. Let $t=2t_{1}$, then we have

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},y_{i}z_{1}^{-s}z_{2}^{-t_{1}},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}.$

Similarly, we may obtain the same result for the case $2|j$ and $(t,2)=1$.

Suppose that $2|j$ and $2|t$. Let $j=2j_{2}$ and $t=2t_{2}$. We may obtain

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l}z_{2}^{-j_{2}},y_{i}z_{1}^{-s}z_{2}^{-t_{2}},z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}.$

(iv) Let

$G_{i}=\langle x_{i},y_{i},c\ |\
x_{i}^{2^{w}}=y_{i}^{2}=c^{2}=1,[x_{i},y_{i}]=c,[x_{i},c]=[y_{i},c]=1\rangle\cong
M_{2}(w,1,1).$

Since $\langle x_{i}^{2},c\rangle\leq\langle z_{1}^{2},z_{2}\rangle$, we may
let $x_{i}^{2}=z_{1}^{2l}z_{2}^{j}$, where $1\leq l\leq 2^{n},1\leq j\leq
2^{m}$. Obviously, we have $1\leq w\leq n+1$. If $w=n+1$ and $n>m$, then
$1\neq x_{i}^{2^{n}}=z_{1}^{2^{n}l}z_{2}^{2^{n-1}j}=z_{1}^{2^{n}l}=c^{l}$,
which is impossible. Hence $w$ satisfies the following conditions: $1\leq
w\leq n+1$ (If $n=m$) or $1\leq w<n+1$ (If $n>m$).

If $j$ and $2$ are coprime, then we have

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},y_{i},z_{1}\rangle\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}.$

Suppose that $j$ is divisible by $2$ and let $j=2j_{1}$. It follows that

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l}z_{2}^{-j_{1}},y_{i},z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}.$

By Lemmas 3.1 and 3.17, the factor $G_{i}$ of the central product of $E$ may
only be considered as three types: $M_{2}(m+1,1,1),D_{8},Q_{8}$. Note that
$G_{1}\mathbin{\mathsf{Y}}G_{2}$ is only isomorphic to
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}D_{8}$ if both $G_{1}$ and $G_{2}$ are
isomorphic to $M_{2}(m+1,1,1)$ according to (iv) of Lemma 3.9. From this, we
may obtain the following theorem similar to Theorem 3.11.

######  Theorem 3.19.

Suppose that $n\geq m\geq 1$ and $c\in\langle z_{1}\rangle$. Then the
isomorphism classes of $G$ are as follows.

(i)
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(ii)
$M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(iii)
$\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(iv)
$Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$.

######  Lemma 3.20.

Suppose $n\geq m\geq 1$ and $c\in\langle z_{2}\rangle$.

(i) If $G_{i}\cong M_{2}(u,n+1)$, then $2\leq u\leq m+1$ and

$G_{i}\mathbin{\mathsf{Y}}\langle
z_{1},z_{2}\rangle\cong\left\\{\begin{aligned}
&D_{8}\times\mathbb{Z}_{2^{n+1}}\ or\ Q_{8}\times\mathbb{Z}_{2^{n+1}},\
m=1.\\\ &M_{2}(m+1,1)\times\mathbb{Z}_{2^{n+1}}\ or\
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n+1}},\
m>1.\end{aligned}\right.$

(ii) If $n=m$ and $G_{i}\cong M_{2}(m+1,v)$, where $1\leq v<m+1$, then
$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{m+1}}$. If $n>m$, then $G_{i}$ is not
isomorphic to $M_{2}(n+1,v)$, where $1\leq v\leq m+1$.

(iii) If $G_{i}\cong M_{2}(u,v)$, where $2\leq u<n+1$ and $1\leq v<n+1$, then
we may obtain the same results of $(i)$.

(iv) If $G_{i}\cong M_{2}(w,1,1)$, where $1\leq w\leq n+1$, then
$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n+1}}$ or
$D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n+1}}$.

######  Proof 3.21.

(i) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{u}}=y_{i}^{2^{n+1}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{u-1}}\rangle\cong
M_{2}(u,n+1).$

Since $\zeta G_{i}=\langle x_{i}^{2},y_{i}^{2}\rangle\leq N$, we have
$2^{n+u-1}\leq|N|=2^{n+m}$ and $2\leq u\leq m+1$. Let
$x_{i}^{2}=z_{1}^{2l}z_{2}^{j}$ and $y_{i}^{2}=z_{1}^{2s}z_{2}^{t}$, where
$1\leq l,s\leq 2^{n},1\leq j,t\leq 2^{m}$.

First, we suppose that $u=m+1$. Obviously, $\zeta G_{i}=N$ by comparing their
orders. If $j$ is divisible by $2$, then
$c=x_{i}^{2^{m}}=z_{1}^{2^{m}l}z_{2}^{2^{m-1}j}=z_{1}^{2^{m}l}$, which is
impossible. Hence $j$ and $2$ are coprime.

We will next consider two cases $(t,2)=1$ and $2|t$. When $t$ is divisible by
$2$, let $t=2t_{1}$, then

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},y_{i}z_{1}^{-s}z_{2}^{-t_{1}},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n+1}}.$

When $t$ and $2$ are coprime, we may suppose that $x_{i}^{2}=z_{1}^{2l}z_{2}$
and $y_{i}^{2}=z_{1}^{2s}z_{2}$ without loss of generality. If $m\geq 2$, then
we have

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},x_{i}y_{i}^{-1}z_{1}^{s-l}z_{2}^{2^{m-2}},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n+1}}.$

If $m=1$, then $u=2$ and

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i},y_{i}z_{1}^{-s},z_{1}\rangle\cong Q_{8}\times\mathbb{Z}_{2^{n+1}}.$

We next suppose $2\leq u<m+1$. Since
$1=x_{i}^{2^{u}}=z_{1}^{2^{u}l}z_{2}^{2^{u-1}j}$, we have $j$ is divisible by
$2^{m-u+1}$. Let $j=2j_{1}$. If $t$ and $2$ are coprime, then

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
y_{i}z_{1}^{-s},x_{i}z_{1}^{-l}z_{2}^{-j_{1}},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n+1}}.$

Suppose that $t$ is divisible by $2$ and let $t=2t_{2}$. Hence

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l}z_{2}^{-j_{1}},y_{i}z_{1}^{-s}z_{2}^{-t_{2}}\rangle\mathbin{\mathsf{Y}}\langle
z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n+1}}.$

(ii) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{n+1}}=y_{i}^{2^{v}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{n}}\rangle\cong
M_{2}(n+1,v),$

where $1\leq v\leq m+1$. Obviously, $\zeta G_{i}=\langle
x_{i}^{2},y_{i}^{2}\rangle\leq N=\langle z_{1}^{2},z_{2}\rangle$. At this
time, let $x_{i}^{2}=z_{1}^{2l}z_{2}^{j}$ and $y_{i}^{2}=z_{1}^{2s}z_{2}^{t}$.

If $n=m$, then we only consider the case $1\leq v<m+1$. Since
$c=x_{i}^{2^{m}}=z_{1}^{2^{m}l}z_{2}^{2^{m-1}j}=z_{1}^{2^{m}l}c^{j}$, $j$ and
$2$ are coprime. Since $1=y_{i}^{2^{v}}=z_{1}^{2^{v}s}z_{2}^{2^{v-1}t}$, $t$
are divisible by $2^{m-v+1}$, that is, $2|t$. Let $t=2t_{1}$. Hence

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},y_{i}z_{1}^{-s}z_{2}^{-t_{1}},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{m+1}}.$

Suppose that $n>m$. Note that
$c=x_{i}^{2^{n}}=z_{1}^{2^{n}l}z_{2}^{2^{n-1}j}=z_{1}^{2^{n}l}$, a
contradiction.

(iii) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{u}}=y_{i}^{2^{v}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{u-1}}\rangle\cong
M_{2}(u,v),$

where $u,v\leq n$. Let $x_{i}^{2}=z_{1}^{2l}z_{2}^{j}$ and
$y_{i}^{2}=z_{1}^{2s}z_{2}^{t}$, where $1\leq l,s\leq 2^{n},1\leq j,t\leq
2^{m}$.

If $(j,2)=1$ and $2|t$. Let $t=2t_{1}$, then we have

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},y_{i}z_{1}^{-s}z_{2}^{-t_{1}},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n+1}}.$

Similarly, we may obtain the same result for the case $2|j$ and $(t,2)=1$.

Suppose that $2|j$ and $2|t$. Let $j=2j_{2}$ and $t=2t_{2}$. Then

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l}z_{2}^{-j_{2}},y_{i}z_{1}^{-s}z_{2}^{-t_{2}},z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n+1}}.$

If $(j,2)=1=(t,2)$, then we may let $x_{i}^{2}=z_{1}^{2l}z_{2}$ and
$y_{i}^{2}=z_{1}^{2s}z_{2}$ without loss of generality. At this time, we will
distinguish the case $m=1$ from the case $m>1$. When $m=1$, we have

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},y_{i}z_{1}^{-s},z_{1}\rangle\cong
Q_{8}\times\mathbb{Z}_{2^{n+1}}.$

When $m>1$, then

$G_{i}\ast\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},x_{i}y_{i}^{-1}z_{1}^{s-l}z_{2}^{2^{m-2}},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n+1}}.$

(iv) Let

$G_{i}=\langle x_{i},y_{i},c\ |\
x_{i}^{2^{w}}=y_{i}^{2}=c^{2}=1,[x_{i},y_{i}]=c,[x_{i},c]=[y_{i},c]=1\rangle\cong
M_{2}(w,1,1).$

Note that $\langle x_{i}^{2},c\rangle\leq\langle z_{1}^{2},z_{2}\rangle$. Thus
we may let $x_{i}^{2}=z_{1}^{2l}z_{2}^{j}$, where $1\leq l\leq 2^{n},1\leq
j\leq 2^{m}$. If $j$ and $2$ are coprime, then

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l},y_{i},z_{1}\rangle\cong
M_{2}(m+1,1)\times\mathbb{Z}_{2^{n+1}}.$

Suppose that $j$ is divisible by $2$ and let $j=2j_{1}$. Then

$G_{i}\ast\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l}z_{2}^{-j_{1}},y_{i},z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n+1}}.$

By Lemmas 3.1 and 3.20, the factor $G_{i}$ of the central product of $E$ may
only be considered as three types: $M_{2}(m+1,1),D_{8},Q_{8}$. Note that
$G_{1}\mathbin{\mathsf{Y}}G_{2}$ is only isomorphic to
$M_{2}(m+1,1)\mathbin{\mathsf{Y}}D_{8}$ if both $G_{1}$ and $G_{2}$ are
isomorphic to $M_{2}(m+1,1)$. From this, we may obtain the following theorem
similar to Theorem 3.15.

######  Theorem 3.22.

Suppose that $n\geq m\geq 1$ and $c\in\langle z_{2}\rangle$. Then the
isomorphism classes of $G$ are as follows.

(i)
$M_{2}(m+1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\times\mathbb{Z}_{2^{n+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(ii)
$M_{2}(m+1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\times\mathbb{Z}_{2^{n+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(iii)
$\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(iv)
$Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$.

Suppose that $N=\langle z_{1}\rangle\times\langle z_{2}^{2}\rangle.$ Similar
to the case $A_{2}$, we have the following the results about $A_{3}$.

######  Corollary 3.23.

Suppose that $n\geq m\geq 1$ and $c\in\langle z_{1}\rangle$. Then the
isomorphism classes of $G$ are as follows.

(i)
$M_{2}(n+1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\times\mathbb{Z}_{2^{m+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(ii)
$M_{2}(n+1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\times\mathbb{Z}_{2^{m+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(iii)
$\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(iv)
$Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$.

######  Corollary 3.24.

Suppose that $n\geq m\geq 1$ and $c\in\langle z_{2}\rangle$. Then the
isomorphism classes of $G$ are as follows.

(i)
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(ii)
$M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-2}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m+1}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 2$;

(iii)
$\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2^{n}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(iv)
$Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2^{n}}\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$.

###  3.3  The isomorphism type $A_{4}$ of $\zeta G$

In the section,

$\zeta G=\langle z_{1}\rangle\times\langle
z_{2}\rangle\times\cdots\times\langle
z_{r}\rangle\cong\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2},N=\langle
z_{1}^{2}\rangle\times\langle z_{2}^{2}\rangle.$

Without loss of generality, we may always let $c\in\langle z_{1}^{2}\rangle$
or $c\in\langle z_{2}^{2}\rangle$. First we determine the types of
$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle$, where $i=1,2,\ldots,k$.

######  Lemma 3.25.

Suppose $n\geq m\geq 1$ and $c\in\langle z_{1}\rangle$ or $\langle
z_{2}\rangle$.

(i) If $G_{i}\cong M_{2}(u,v)$, where $2\leq u\leq n+1$ and $1\leq v\leq n+1$,
then $G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}(\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m+1}})$

(ii) If $G_{i}\cong M_{2}(w,1,1)$, where $1\leq w\leq n+1$, then
$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}(\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m+1}})$.

######  Proof 3.26.

(i) Let

$G_{i}=\langle x_{i},y_{i}\ |\
x_{i}^{2^{u}}=y_{i}^{2^{v}}=1,x_{i}^{y_{i}}=x_{i}^{1+2^{u-1}}\rangle\cong
M_{2}(u,v).$

Note that $\zeta G_{i}=\langle x_{i}^{2},y_{i}^{2}\rangle\leq N$. Thus we may
let $x_{i}^{2}=z_{1}^{2l}z_{2}^{2j}$ and $y_{i}^{2}=z_{1}^{2s}z_{2}^{2t}$,
where $1\leq l,s\leq 2^{n},1\leq j,t\leq 2^{m}$. Hence

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l}z_{2}^{-j},y_{i}z_{1}^{-s}z_{2}^{-t},z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}(\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m+1}}).$

(ii) Let

$G_{i}=\langle x_{i},y_{i},c\ |\
x_{i}^{2^{w}}=y_{i}^{2}=c^{2}=1,[x_{i},y_{i}]=c,[x_{i},c]=[y_{i},c]=1\rangle\cong
M_{2}(w,1,1).$

where $1\leq w\leq n+1$. Let $x_{i}^{2}=z_{1}^{2l}z_{2}^{2j}$, where $1\leq
l\leq 2^{n},1\leq j\leq 2^{m}$. Thus

$G_{i}\mathbin{\mathsf{Y}}\langle z_{1},z_{2}\rangle=\langle
x_{i}z_{1}^{-l}z_{2}^{-j},y_{i},z_{1},z_{2}\rangle\cong
D_{8}\mathbin{\mathsf{Y}}(\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m+1}}).$

By Lemmas 3.1 and 3.25, it is easy to obtain the following theorem since
$Q_{8}\mathbin{\mathsf{Y}}Q_{8}\cong Q_{8}\mathbin{\mathsf{Y}}D_{8}$.

######  Theorem 3.27.

Suppose that $n\geq m\geq 1$ and $c\in\langle z_{1}\rangle$ or $\langle
z_{2}\rangle$. Then the isomorphism classes of $G$ are as follows.

(i)
$\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k}\mathbin{\mathsf{Y}}(\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m+1}})\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$;

(ii)
$Q_{8}\mathbin{\mathsf{Y}}\underbrace{D_{8}\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}D_{8}}_{k-1}\mathbin{\mathsf{Y}}(\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2^{m+1}})\times\underbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}_{r-2}$,
$k\geq 1$.

##  4  The unitary subgroup

Let $G$ be a nonabelian $2$-group given by a central extension of the form

$1\longrightarrow N\longrightarrow
G\longrightarrow\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}\longrightarrow
1$

and $G^{\prime}=\langle c\rangle\cong\mathbb{Z}_{2}$,
$N\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}$, $n\geq m\geq 1$.

In section 3, we see there is an elementary abelian $2$-group in direct
product factors of $G$. According to Lemmas 2.6 and 3.3, we only consider the
case $\zeta G=\langle z_{1}\rangle\times\langle z_{2}\rangle\geq N$ for
computing the order of $U_{*}(FG)$.

Let $\bar{G}=G/G^{\prime}$ and let $\Psi:FG\rightarrow F\overline{G}$ be the
natural homomorphism. Consider the sets

$N_{\Psi}^{*}:=\\{x\in V(FG)\ |\ \Psi(x)\in V_{*}(F\overline{G})\\},$
$\mathrm{Ker}\Psi^{+}:=\\{1+x\ |\ x\in\mathrm{Ker}\Psi\\}.$

It is easy to verify that $N_{\Psi}^{*}$ forms a subgroup of $V(FG)$ and
$\mathrm{Ker}\Psi^{+}$ forms a normal subgroup of $V(FG)$. Let
$S_{G^{\prime}}:=\\{xx^{*}\ |\ x\in N_{\Psi}^{*}\\}$. We have that
$S_{G^{\prime}}\subseteq\mathrm{Ker}\Psi^{+}\leq N_{\Psi}^{*}$.

######  Lemma 4.1 ([2]).

(i) $\mathrm{Supp}(xx^{*})\cap\Omega_{1}(G)=\\{1\\}$ for every $x\in V(FG)$.

(ii) If $1+g\widehat{G^{\prime}}\in S_{G^{\prime}}$ for some $g\in G$, then
$g^{2}=c$.

(iii) If $h^{2}$ is not included in $G^{\prime}$ for any $h\in G$, then
$1+\alpha(h+h^{-1})\widehat{G^{\prime}}\in S_{G}^{\prime}$ for any $\alpha\in
F$.

We denote by $\Theta(G)$ the order of the group $\langle
1+\sum\limits_{g\in\Omega_{c}(G)}\alpha_{g}g\widehat{G^{\prime}}\in
S_{G^{\prime}},\alpha_{g}\in F\rangle$. From this, we may obtain the following
lemma.

######  Lemma 4.2.

$S_{G^{\prime}}$ is a subgroup of $N_{\Psi}^{*}$. And

$|V_{*}(FG)|=\frac{1}{\Theta(G)}|F|^{\frac{|G|+|\Omega_{1}(G)|+|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|.$

In particular, if $1+\alpha g\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any
$g\in\Omega_{c}(G)$ and $\alpha\in F$, then

$|V_{*}(FG)|=|F|^{\frac{|G|+|\Omega_{1}(G)|-|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|.$

######  Proof 4.3.

Note that the ideal $\mathrm{Ker}\Psi$ of $FG$ is generated by $1+c$. By Lemma
4.1, we may see the elements of $S_{G^{\prime}}$ are the following form

$1+\sum_{g\in\Omega_{c}(G)}\alpha_{g}g\widehat{G^{\prime}}+\sum_{h^{2}\notin
G^{\prime}}\beta_{h}(h+h^{-1})\widehat{G^{\prime}}.$

For any $1+\alpha x\widehat{G^{\prime}}$,
$\sum\limits_{i=1}^{|G|}\alpha_{i}g_{i}\in V(FG)$, where $\alpha,\alpha_{i}\in
F$, $x,g_{i}\in G$, $i=1,2,\ldots,|G|$, we have

$\displaystyle\left(1+\alpha
x\widehat{G^{\prime}}\right)\left(\sum_{i=1}^{|G|}\alpha_{i}g_{i}\right)$
$\displaystyle=\sum_{i=1}^{|G|}\alpha_{i}g_{i}+\sum_{i=1}^{|G|}\alpha\alpha_{i}xg_{i}\widehat{G^{\prime}}=\sum_{i=1}^{|G|}\alpha_{i}g_{i}+\sum_{i=1}^{|G|}\alpha\alpha_{i}g_{i}x[x,g_{i}]\widehat{G^{\prime}}$
$\displaystyle=\sum_{i=1}^{|G|}\alpha_{i}g_{i}+\sum_{i=1}^{|G|}\alpha\alpha_{i}g_{i}x\widehat{G^{\prime}}=\left(\sum_{i=1}^{|G|}\alpha_{i}g_{i}\right)\left(1+\alpha
x\widehat{G^{\prime}}\right).$

Thus $S_{G^{\prime}}$ is included in the center of $V(FG)$ and
$S_{G^{\prime}}$ is a subgroup of $N_{\Psi}^{*}$. It is clear to see that
$\langle 1+\sum_{h^{2}\notin
G^{\prime}}\beta_{h}(h+h^{-1})\widehat{G^{\prime}},\beta_{h}\in F\rangle$ has
order $|F|^{\frac{|G|-|\Omega_{1}(G)|-|\Omega_{c}(G)|}{4}}$. Hence
$|S_{G^{\prime}}|=\Theta(G)|F|^{\frac{|G|-|\Omega_{1}(G)|-|\Omega_{c}(G)|}{4}}$
according to the hypothesis.

Consider the mapping $\Phi:N_{\Psi}^{*}\rightarrow S_{G^{\prime}}$. Obviously,
$\Phi$ is an epimorphism. Note that
$\Psi(x)^{-1}=\Psi(x^{-1})=\Psi(x^{*})=\Psi(x)^{*}$ if $x\in V_{*}(FG)$. Thus

$\mathrm{Ker}(\Phi)=\\{x\in N_{\Psi}^{*}\ |\ \Phi(x)=1\\}=\\{x\in V(FG)\ |\
x^{*}=x^{-1}\\}=V_{*}(FG).$

From this, we have $N_{\Psi}^{*}/V_{*}(FG)\cong S_{G^{\prime}}$. Consider the
restriction $\Psi|_{N_{\Psi}^{*}}:N_{\Psi}^{*}\rightarrow
V_{*}(F\overline{G})$. Obviously $\Psi|_{N_{\Psi}^{*}}$ is surjection and
$\mathrm{Ker}(\Psi|_{N_{\Psi}^{*}})=\mathrm{Ker}\Psi^{+}$. It follows that

$|V_{*}(FG)|=\frac{|N_{\Psi}^{*}|}{|S_{G^{\prime}}|}=\frac{|\mathrm{Ker}\Psi^{+}|\cdot|V_{*}(F\overline{G})|}{|S_{G^{\prime}}|}.$

Since

$|\mathrm{Ker}\Psi^{+}|=|\mathrm{Ker}\Psi|=\frac{|FG|}{|F\overline{G}|}=F^{|G|/2},$

we obtain

$|V_{*}(FG)|=\frac{|F|^{|G|/2}\cdot|V_{*}(F\overline{G})|}{|S_{G^{\prime}}|}=\frac{1}{\Theta(G)}|F|^{\frac{|G|+|\Omega_{1}(G)|+|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|.$

In particular, if $1+\alpha g\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any
$g\in\Omega_{c}(G)$ and $\alpha\in F$, then
$\Theta(G)=|F|^{\frac{|\Omega_{c}(G)|}{2}}$ and

$|V_{*}(FG)|=|F|^{\frac{|G|+|\Omega_{1}(G)|-|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|.$

######  Lemma 4.4.

Suppose that $h_{1},h_{2},h\in G$ and $\alpha\in F$. Then

(i) If $h_{1}^{4}=h_{2}^{4}=1$, $h_{1}^{2}=h_{2}^{2}=c$ and $[h_{1},h_{2}]=1$,
then $1+\alpha(h_{1}+h_{2})\widehat{G^{\prime}}\in S_{G^{\prime}}$.

(ii) If $h_{1}^{4}=h_{2}^{2}=1$, $h_{1}^{2}=c$ and $[h_{1},h_{2}]=1$, then
$1+\alpha
h_{1}\widehat{G^{\prime}}+\alpha^{2}h_{1}h_{2}\widehat{G^{\prime}}\in
S_{G^{\prime}}$.

(iii) If $h_{1}^{2}=h_{2}^{2}=1$ and $[h_{1},h_{2}]=c$, then $1+\alpha
h_{1}h_{2}\widehat{G^{\prime}}\in S_{G^{\prime}}$.

(iv) If $h_{1}^{4}=h_{2}^{2}=1$ and $[h_{1},h_{2}]=h_{1}^{2}=c$, then
$1+\alpha h_{1}\widehat{G^{\prime}}\in S_{G^{\prime}}$.

(v) If $h^{8}=1$ and $h^{4}=c$, then $1+\alpha h^{2}\widehat{G^{\prime}}\in
S_{G^{\prime}}$.

######  Proof 4.5.

(i) follows since $(1+\alpha h_{1}+\alpha h_{2})(1+\alpha h_{1}+\alpha
h_{2})^{*}=1+\alpha(h_{1}+h_{2})\widehat{G^{\prime}}$.

(ii) follows since $(1+\alpha h_{1}+\alpha h_{2})(1+\alpha h_{1}+\alpha
h_{2})^{*}=1+\alpha
h_{1}\widehat{G^{\prime}}+\alpha^{2}h_{1}h_{2}\widehat{G^{\prime}}$.

(iii) follows since $(1+\alpha h_{1}h_{2}+\alpha h_{2})(1+\alpha
h_{1}h_{2}+\alpha h_{2})^{*}=1+\alpha h_{1}h_{2}\widehat{G^{\prime}}$.

(iv) follows since $(1+\alpha h_{1}+\alpha h_{2})(1+\alpha h_{1}+\alpha
h_{2})^{*}=1+\alpha h_{1}\widehat{G^{\prime}}$.

It is easy to verify $(\alpha+\alpha h^{2}+h)(\alpha+\alpha
h^{2}+h)^{*}=1+\alpha^{2}h^{2}\widehat{G^{\prime}}$. Note that the mapping:
$\phi:F\rightarrow F,\alpha\mapsto\alpha^{2}$ is an automorphism. From this,
for any $\alpha\in F$, we have $1+\alpha h^{2}\widehat{G^{\prime}}\in
S_{G^{\prime}}$.

For convenience, we describe the relations of $D_{8}$ or $Q_{8}$ are as
follows in the following part,

$\langle x,y\ |\ x^{2}=y^{2}=1,[x,y]=c\rangle\cong D_{8},\langle x,y\ |\
x^{4}=1,x^{2}=y^{2}=[x,y]=c\rangle\cong Q_{8}.$

######  Lemma 4.6.

(i) Let $H_{1}=\langle a,b\ |\
a^{4}=b^{4}=1,[a,b]=a^{2}=c\rangle\cong\mathbb{Z}_{4}\rtimes\mathbb{Z}_{4}$.
Then $\Theta(H_{1})$ is equal to
$\frac{1}{2}|F|^{\frac{|\Omega_{c}(H_{1})|}{2}}$.

(ii) Let $H_{2}=\langle x,y\rangle\cong Q_{8}$. Then $\Theta(H_{2})$ is equal
to $\frac{1}{4}|F|^{\frac{|\Omega_{c}(H_{2})|}{2}}$.

######  Proof 4.7.

By Lemmas 2.4, 2.5 and 2.6, we know
$|V_{*}(FH_{i})|=4|F|^{\frac{|H_{i}|+|\Omega_{1}(H_{i})|}{2}-1}$. Suppose that
$\overline{H_{i}}=H_{i}/H^{\prime}_{i}$, where $i=1,2$, then

$|V_{*}(FH_{i})|=\frac{1}{\Theta(H_{i})}|F|^{\frac{|H_{i}|+|\Omega_{1}(H_{i})|+|\Omega_{c}(H_{i})|}{4}}\cdot|V_{*}(F\overline{H_{i}})|.$

according to Lemma 4.2.

For the group $H_{1}$, $|\Omega_{1}(H_{1})|=4=|\Omega_{c}(H_{1})|$ and
$\overline{H_{1}}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{4}$. Since

$\frac{1}{\Theta(H_{1})}|F|^{\frac{16+4+4}{4}}\cdot(2|F|^{5})=4|F|^{\frac{16+4}{2}-1},$

we have
$\Theta(H_{1})=\frac{|F|^{2}}{2}=\frac{1}{2}|F|^{\frac{|\Omega_{c}(H_{1})|}{2}}$.

For the group $H_{2}$, $|\Omega_{1}(H_{2})|=2$, $|\Omega_{c}(H_{2})|=6$ and
$\overline{H_{1}}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Since

$\frac{1}{\Theta(H_{2})}|F|^{\frac{8+2+6}{4}}\cdot(|F|^{3})=4|F|^{\frac{8+2}{2}-1},$

we have
$\Theta(H_{2})=\frac{|F|^{3}}{4}=\frac{1}{4}|F|^{\frac{|\Omega_{c}(H_{2})|}{2}}$.

######  Lemma 4.8.

Suppose that $H=\langle x,y\ |\ x^{4}=y^{2^{v}}=1,[x,y]=x^{2}=c\rangle\cong
M_{2}(2,v)$, where $v\geq 3$. Then
$\Theta(H)=\frac{1}{2}|F|^{\frac{|\Omega_{c}(H)|}{2}}$.

######  Proof 4.9.

It is easy to verify that $\Omega_{c}(H)=\\{xy^{i}c^{j}\ |\
i=0,2^{v-1},j=0,1\\}$.

Let $X:=\langle 1+\alpha x\widehat{H^{\prime}}\in S_{H^{\prime}}\ |\ \alpha\in
F\rangle$. First, we assert that $X$ has order $\frac{1}{2}|F|$. Since
$(1+\beta x^{2}+\beta x)(1+\beta x^{2}+\beta
x)^{*}=1+(\beta+\beta^{2})x\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any
$\beta\in F$, the the order of $X$ is at less $\frac{1}{2}|F|$. Every element
of $FH$ can be written in the following form $a=a_{1}+a_{2}x+a_{3}y+a_{4}xy$,
where $a_{i}=a_{i1}+a_{i2}x^{2}$, $a_{ij}\in F\langle y^{2}\rangle$.

$\displaystyle aa^{*}$
$\displaystyle=a_{1}a_{1}^{*}+a_{2}a_{2}^{*}+a_{3}a_{3}^{*}+a_{4}a_{4}^{*}+(a_{1}^{*}a_{2}+a_{3}^{*}a_{4})x+(a_{1}a_{2}^{*}+a_{3}a_{4}^{*})x^{3}$
$\displaystyle+a_{1}a_{3}^{*}y^{-1}+a_{1}^{*}a_{3}y+a_{1}a_{4}^{*}y^{-1}x^{-1}+a_{1}^{*}a_{4}xy$
$\displaystyle+a_{2}a_{3}^{*}xy^{-1}+a_{2}^{*}a_{3}yx^{-1}+a_{2}a_{4}^{*}x^{2}y^{-1}+a_{2}^{*}a_{4}x^{2}y$
$\displaystyle=a_{1}a_{1}^{*}+a_{2}a_{2}^{*}+a_{3}a_{3}^{*}+a_{4}a_{4}^{*}+(a_{1}^{*}a_{2}+a_{3}^{*}a_{4})x+(a_{1}a_{2}^{*}+a_{3}a_{4}^{*})x^{3}$
$\displaystyle+b_{1}y^{-1}+b_{2}x^{2}y^{-1}+b_{3}y+b_{4}x^{2}y+b_{5}xy^{-1}+b_{6}x^{3}y^{-1}+b_{7}xy+b_{8}x^{3}y,$

where

$\displaystyle
b_{1}=a_{11}a_{31}^{*}+a_{12}a_{32}^{*}+a_{21}a_{42}^{*}+a_{22}a_{41}^{*}=b_{3}^{*}$
$\displaystyle
b_{2}=a_{11}a_{32}^{*}+a_{12}a_{31}^{*}+a_{21}a_{41}^{*}+a_{22}a_{42}^{*}=b_{4}^{*}$
$\displaystyle
b_{5}=a_{11}a_{41}^{*}+a_{12}a_{42}^{*}+a_{21}a_{31}^{*}+a_{22}a_{32}^{*}=b_{7}^{*}$
$\displaystyle
b_{6}=a_{11}a_{42}^{*}+a_{12}a_{41}^{*}+a_{21}a_{32}^{*}+a_{22}a_{31}^{*}=b_{8}^{*}$

Consider the augmentation mapping of $FH$ to $F$, which is denoted by $\chi$.
Set $w_{i}=\chi(a_{i})$ and $w_{ij}=\chi(a_{ij})$. If $aa^{*}\in\in
S_{H^{\prime}}$ and $1+\alpha x\widehat{H^{\prime}}$ in $\langle
1+\sum\limits_{g\in\Omega_{c}(H)}\alpha_{g}g\widehat{H^{\prime}}\in
S_{H^{\prime}},\alpha_{g}\in F\rangle$ only appears in the expression of
$aa^{*}$, then we have

$\displaystyle\chi(a_{1}a_{1}^{*}+a_{2}a_{2}^{*}+a_{3}a_{3}^{*}+a_{4}a_{4}^{*})=w_{1}^{2}+w_{2}^{2}+w_{3}^{2}+w_{4}^{2}=1$
$\displaystyle\chi(a_{1}^{*}a_{2}+a_{3}^{*}a_{4})=w_{1}w_{2}+w_{3}w_{4}=\alpha$
$\displaystyle
w_{11}w_{31}+w_{12}w_{32}+w_{21}w_{42}+w_{22}w_{41}=w_{11}w_{32}+w_{12}w_{31}+w_{21}w_{41}+w_{22}w_{42}$
$\displaystyle
w_{11}w_{41}+w_{12}w_{42}+w_{21}w_{31}+w_{22}w_{32}=w_{11}w_{42}+w_{12}w_{41}+w_{21}w_{32}+w_{22}w_{31}.$

From this,

$w_{1}+w_{2}+w_{3}+w_{4}=1,w_{1}w_{2}+w_{3}w_{4}=\alpha,w_{1}w_{3}+w_{2}w_{4}=0,w_{1}w_{4}+w_{2}w_{3}=0.$

According to Lemma 10 of [2], the number of $\alpha$ is $\frac{1}{2}|F|$.
Hence the assertion is true.

According to (i) of Lemma 4.4, we have
$1+\alpha(x+xy^{2^{v-1}})\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any
$\alpha\in F$. From this, the order $\Theta(H)$ of $\langle
1+\sum\limits_{g\in\Omega_{c}(H)}\alpha_{g}g\widehat{H^{\prime}}\in
S_{H^{\prime}},\alpha_{g}\in F\rangle$ is $\frac{1}{2}|F|^{2}$, that is,
$\Theta(H)=\frac{1}{2}|F|^{\frac{|\Omega_{c}(H)|}{2}}$.

######  Lemma 4.10.

Suppose that $G=H\mathbin{\mathsf{Y}}K$, $H\bigcap K=\langle c\rangle$ and
$H\cong M_{2}(u,v)$, where $u\geq 2$ and $v\geq 1$. If $(u,v)=(2,1)$ or $u>2$
, then $1+\alpha g\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any
$g\in\Omega_{c}(G)$, $\alpha\in F$.

######  Proof 4.11.

Let

$H=\langle x,y\ |\ x^{2^{u}}=y^{2^{v}}=1,x^{y}=x^{1+2^{u-1}}\rangle\cong
M_{2}(u,v).$

For any $g\in\Omega_{c}(G)$, suppose that $g=x^{i}y^{j}g_{1}$, where $0\leq
i<2^{u-1}$, $0\leq j<2^{v}$ and $g_{1}\in K$. Further we have
$x^{2i+2^{u-1}ij}y^{2j}g_{1}^{2}=c$.

First, we consider $u=2$ and $v=1$. At this time, $H\cong D_{8}$. By taking
$h_{1}=x$ and $h_{2}=y$ in (iv) of Lemma 4.4, we have $1+\alpha
x\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any $\alpha\in F$.

If $g_{1}\in\Omega_{c}(G)$, then $x^{2i+2ij}=1$. Hence
$(i,j)=(0,0),(0,1),(1,1)$.

$1+\alpha(x+g_{1})\widehat{G^{\prime}}\in S_{G^{\prime}}$ follows by taking
$h_{1}=x$ and $h_{2}=g_{1}$ in (i) of Lemma 4.4;

$1+\alpha g_{1}\widehat{G^{\prime}}+\alpha^{2}yg_{1}\widehat{G^{\prime}}\in
S_{G^{\prime}}$ follows by taking $h_{1}=g_{1}$ and $h_{2}=y$ in (ii) of Lemma
4.4;

$1+\alpha xyg_{1}\widehat{G^{\prime}}\in S_{G^{\prime}}$ follows by taking
$h_{1}=xg_{1}$ and $h_{2}=y$ in (iii) of Lemma 4.4.

According to the above results, we have $1+\alpha g\widehat{G^{\prime}}\in
S_{G^{\prime}}$ for three cases about $(i,j)$.

If $g_{1}\in\Omega_{1}(G)$, then $x^{2i+2ij}=c$. Hence $i=1$ and $j=0$.
Further, by taking $h_{1}=x$ and $h_{2}=xg_{1}$ in (i) of Lemma 4.4, we have
$1+\alpha xg_{1}\widehat{G^{\prime}}\in S_{G^{\prime}}$ since $1+\alpha
x\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any $\alpha\in F$.

Suppose that $u$ is more than $2$. We have $1+\alpha
x^{2^{u-2}}\widehat{G^{\prime}}\in S_{G^{\prime}}$ by taking $h=x^{2^{u-3}}$
in (v) of Lemma 4.4.

If $g_{1}\in\Omega_{c}(G)$, then $x^{2i+2^{u-1}ij}y^{2j}=1$. Hence $j=0$ or
$2^{v-1}$. From this, we have $i=0$. Otherwise, $i=2^{u-2}$ and
$x^{2^{u-1}(1+2^{u-2}j)}=1$, which is impossible. By taking
$h_{1}=x^{2^{u-2}}$ and $h_{2}=y^{j}g_{1}$ in (i) of Lemma 4.4, we have
$1+\alpha y^{j}g_{1}\widehat{G^{\prime}}\in S_{G^{\prime}}$ since $1+\alpha
x^{2^{u-2}}\widehat{G^{\prime}}\in S_{G^{\prime}}$.

If $g_{1}\in\Omega_{1}(G)$, then $x^{2i+2^{u-1}ij}y^{2j}=c$. Hence $j=0$ or
$2^{v-1}$. From this, we have $i=2^{u-2}$. For any $\alpha\in F$, by taking
$h_{1}=x^{2^{u-2}}$ and $h_{2}=y^{j}g_{1}$ in (ii) of Lemma 4.4, we have
$1+\alpha x^{2^{u-2}}y^{j}g_{1}\widehat{G^{\prime}}\in S_{G^{\prime}}$.

######  Lemma 4.12.

Suppose that $G=H\mathbin{\mathsf{Y}}K$, $H\bigcap K=\langle c\rangle$ and
$H\cong\mathbb{Z}_{2^{u}}$, where $u\geq 3$. Then $1+\alpha
g\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any $g\in\Omega_{c}(G)$ and
$\alpha\in F$.

######  Proof 4.13.

Let $H=\langle z\rangle$. By the hypothesis, $z^{2^{u-1}}=c$. $1+\alpha
z^{2^{u-2}}\widehat{G^{\prime}}\in S_{G^{\prime}}$ follows from (v) of Lemma
4.4. For any $g\in\Omega_{c}(G)$, obviously $[z,g]=1$.
$1+\alpha(z^{2^{u-2}}+g)\widehat{G^{\prime}}\in S_{G^{\prime}}$ follows from
(i) of Lemma 4.4, which implies $1+\alpha g\widehat{G^{\prime}}\in
S_{G^{\prime}}$.

######  Lemma 4.14.

$\Omega_{1}(D_{8}^{\mathbin{\mathsf{Y}}k})=\Omega_{c}(Q_{8}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-1)})=2\gamma_{1}(k)$
and
$\Omega_{c}(D_{8}^{\mathbin{\mathsf{Y}}k})=\Omega_{1}(Q_{8}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-1)})=2\gamma_{2}(k)$.

######  Proof 4.15.

Let

$G_{l}=\langle x_{1},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}\rangle\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}\langle
x_{k},y_{k}\rangle\cong
H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-1)},$

where $\langle x_{i},y_{i}\rangle\cong D_{8},i=2,\ldots,k,$ and $H_{1}\cong
D_{8}$, $H_{2}\cong Q_{8}$ , $l=1,2$. For any $g\in G$, we may suppose that
$g=x_{1}^{i_{1}}y_{1}^{j_{1}}\cdots x_{k}^{i_{k}}y_{k}^{j_{k}}c^{s},$ where
$i_{t},j_{t},s=0$ or $1$, $t=1,\ldots,k$.

For the group $G_{1}$, we denote by $r_{1}$ the number of set
$\\{(i_{t},j_{t})\ |\ i_{t}=j_{t}=1,t\in\\{1,\ldots,k\\}\\}$. If $g^{2}=1$,
then $r_{1}$ is even. At this time, we have
$\Omega_{1}(G_{1})=2\gamma_{1}(k)$. If $g^{2}=c$, then $r_{1}$ is odd. At this
time, we have $\Omega_{c}(G_{1})=2\gamma_{2}(k)$.

For the group $G_{2}$, we denote by $r_{2}$ the number of set
$\\{(i_{t},j_{t})\ |\ i_{t}=j_{t}=1,t\in\\{2,\ldots,k\\}\\}$. We first suppose
$g^{2}=1$. If $r_{2}$ is even, then $i_{1}=j_{1}=0$; If $r_{2}$ is odd, then
$i_{1}+j_{1}\neq 0$. At this time, we have

$\Omega_{1}(G_{2})=2\gamma_{1}(k-1)+6\gamma_{2}(k-1)=2^{2k}-2^{k}=2\gamma_{2}(k).$

We next suppose $g^{2}=c$. If $r_{2}$ is even, then $i_{1}+j_{1}\neq 0$; If
$r_{2}$ is odd, then $i_{1}=j_{1}=0$. At this time, we have

$\Omega_{c}(G_{2})=6\gamma_{1}(k-1)+2\gamma_{2}(k-1)=2^{2k}+2^{k}=2\gamma_{1}(k).$

######  Theorem 4.16.

Suppose that $G\cong
M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-1)}$, where
$n\geq m\geq 1$, $k\geq 1$. Then

(i) $|\Omega_{1}(G)|=|\Omega_{c}(G)|=2^{2k}$.

(ii) $|V_{*}(FG)|=\left\\{\begin{aligned}
&2|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1},\ \mathrm{if}\ n=1\ \mathrm{and}\
k\geq 2.\\\ &4|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

######  Proof 4.17.

(i) Let $G=\langle x_{1},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}\rangle\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}\langle
x_{k},y_{k}\rangle$, where $\langle x_{i},y_{i}\rangle\cong
D_{8},i=2,\ldots,k,$ and

$\langle x_{1},y_{1}\ |\
x_{1}^{2^{n+1}}=y_{1}^{2^{m+1}}=1,x_{1}^{y_{1}}=x_{1}^{1+2^{n}}\rangle\cong
M_{2}(n+1,m+1).$

For any $g\in G$, let $g=g_{1}g_{2}$, where
$g_{1}=x_{1}^{i_{1}}y_{1}^{j_{1}}$, $0\leq i_{1}<2^{n}$, $0\leq
j_{1}<2^{m+1}$, and $g_{2}\in\langle x_{2},y_{2},\ldots,x_{k},y_{k}\rangle$.

If $g_{1}^{2}=1$, then $x_{1}^{2i_{1}}y_{1}^{2j_{1}}c^{i_{1}j_{1}}=1$. From
this, we have $j_{1}=0$ or $2^{m}$ and $i_{1}=0$. If $g_{1}^{2}=c$, then
$x_{1}^{2i_{1}}y_{1}^{2j_{1}}c^{i_{1}j_{1}-1}=1$. From this, we have $j_{1}=0$
or $2^{m}$ and $i_{1}=2^{n-1}$. By Lemma 4.14, we have

$|\Omega_{1}(G)|=4\gamma_{1}(k-1)+4\gamma_{2}(k-1)=2^{2k}=|\Omega_{c}(G)|.$

(ii) We first consider the case $k=1$. At this time, $G=\langle
x_{1},y_{1}\rangle.$ If $n=1$, then $m=1$ and
$|V_{*}(FG)|=4|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1}$ from Lemma 2.5.

Suppose that $n\geq 2$. By Lemma 4.10, we may obtain $1+\alpha
g\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any $g\in\Omega_{c}(G)$ and
$\alpha\in F$. By Lemma 4.2, we have

$|V_{*}(FG)|=|F|^{\frac{|G|+|\Omega_{1}(G)|-|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|=|F|^{2^{n+m}}\cdot|V_{*}(F\overline{G})|.$

Note that $\overline{G}\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m+1}}$.
According to Lemma 2.3, we have

$|V_{*}(F\overline{G})|=|\overline{G}^{2}[2]|\cdot|F|^{\frac{|\overline{G}|+|\Omega_{1}(\overline{G})|}{2}-1}=4|F|^{2^{n+m}+1}.$

Hence $|V_{*}(FG)|=4|F|^{2^{n+m+1}+1}$. Also since
$\frac{|G|+|\Omega_{1}(G)|}{2}-1=\frac{2^{n+m+2}+4}{2}-1$ by (i),
$|V_{*}(FG)|=4|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1}$.

We next suppose that $k\geq 2$. At this time, $D_{8}$ appears in the central
product of $G$. By Lemma 4.10, $1+\alpha g\widehat{G^{\prime}}\in
S_{G^{\prime}}$ for any $g\in\Omega_{c}(G),\alpha\in F$. Hence

$|V_{*}(FG)|=|F|^{\frac{|G|+|\Omega_{1}(G)|-|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|=|F|^{2^{n+m+2k-2}}|V_{*}(F\overline{G})|$

by Lemma 4.2. Note that
$\overline{G}\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m+1}}\times(\mathbb{Z}_{2}\times\mathbb{Z}_{2})^{(k-1)}$.
Hence $|V_{*}(F\overline{G})|=2|F|^{2^{n+m+2k-2}+2^{2k-1}-1}$ (If $n=1$) or
$4|F|^{2^{n+m+2k-2}+2^{2k-1}-1}$ (If $n\geq 2$) by Lemma 2.3. It follows that

$|V_{*}(FG)|=\left\\{\begin{aligned} &2|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1},\
\mathrm{if}\ n=1\ \mathrm{and}\ k\geq 2.\\\
&4|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1},\ \mathrm{if}\ n\geq 2\ \mathrm{and}\
k\geq 2.\end{aligned}\right.$

######  Theorem 4.18.

Suppose that $G\cong
M_{2}(n+1,m+1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-2)}$,
where $n\geq m\geq 1$, $k\geq 2$. Then

(i) $|\Omega_{1}(G)|=|\Omega_{c}(G)|=2^{2k}$.

(ii) $|V_{*}(FG)|=\left\\{\begin{aligned}
&2|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1},\ \mathrm{if}\ n=1\ \mathrm{and}\
k\geq 3.\\\ &4|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

######  Proof 4.19.

Let $G=\langle x_{1},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}\rangle\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}\langle
x_{k},y_{k}\rangle$, where $k\geq 2$, $\langle x_{2},y_{2}\rangle\cong Q_{8}$,
$\langle x_{i},y_{i}\rangle\cong D_{8},i=3,\ldots,k,$ and

$\langle x_{1},y_{1}\ |\
x_{1}^{2^{n+1}}=y_{1}^{2^{m+1}}=1,x_{1}^{y_{1}}=x_{1}^{1+2^{n}}\rangle\cong
M_{2}(n+1,m+1).$

(i) follows similar to (i) of Theorem 4.16.

(ii) We will use the same notations in (i) and suppose that $g^{2}=c$. We
first consider the case $k=2$. At this time, the forms of $g$ are as follows:

$x_{1}^{2^{n-1}}y_{1}^{j_{1}}c^{s},y_{1}^{j_{1}}x_{2}c^{s},y_{1}^{j_{1}}y_{2}c^{s},y_{1}^{j_{1}}x_{2}y_{2}c^{s}.$

If $n=1$, then $\langle
x_{1},y_{1}\rangle\cong\mathbb{Z}_{4}\rtimes\mathbb{Z}_{4}$. By taking
$h_{1}:=x_{1}$ and $h_{2}:=x_{1}y_{1}^{2}$ in (i) of Lemma 4.4, we have
$1+\alpha(x_{1}+x_{1}y_{1}^{2})\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any
$\alpha\in F$. For any $\beta\in F$, it is easy to verify

$(1+\beta x_{1}+\beta c)(1+\beta x_{1}+\beta
c)^{*}=1+(\beta+\beta^{2})x_{1}\widehat{G^{\prime}}\in S_{G^{\prime}}.$

Hence
$1+(\beta+\beta^{2})x_{1}\widehat{G^{\prime}}+\alpha(x_{1}+x_{1}y_{1}^{2})\widehat{G^{\prime}}\in
S_{G^{\prime}}$ for any $\alpha,\beta\in F$. We know that the number of
$\gamma$ in $F$ which satisfies $1+\gamma x_{1}\widehat{G^{\prime}}\in
S_{G^{\prime}}$ is $|F|/2$ by (i) of Lemma 4.6. Suppose that $g$ is any other
type except for $x_{1}$. At this time, $g^{4}=x_{1}^{4}=1$,
$g^{2}=x_{1}^{2}=c$ and $[x_{1},g]=1$. By (i) of Lemma 4.4, we have
$1+\alpha(x_{1}+g)\widehat{G^{\prime}}\in S_{G^{\prime}}$ for any $\alpha\in
F$. In a word, $\langle
1+\sum_{g\in\Omega_{c}(G)}\alpha_{g}g\widehat{G^{\prime}}\in
S_{G^{\prime}},\alpha_{g}\in F\rangle$ has order
$\frac{1}{2}|F|^{\frac{|\Omega_{c}(G)|}{2}}$. By Lemma 4.2,

$|V_{*}(FG)|=\frac{2}{|F|^{\frac{|\Omega_{c}(G)|}{2}}}|F|^{\frac{|G|+|\Omega_{1}(G)|+|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|=2|F|^{\frac{|G|+|\Omega_{1}(G)|-|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|.$

Note that
$\overline{G}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{4}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.
According to Lemma 2.3, we have

$|V_{*}(F\overline{G})|=|\overline{G}^{2}[2]|\cdot|F|^{\frac{|\overline{G}|+|\Omega_{1}(\overline{G})|}{2}-1}=2|F|^{23}.$

Hence $|V_{*}(FG)|=4|F|^{39}$. Also since
$\frac{|G|+|\Omega_{1}(G)|}{2}-1=\frac{2^{6}+2^{4}}{2}-1=39$ by (i),
$|V_{*}(FG)|=4|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1}$.

If $n\geq 2$, then $1+\alpha g\widehat{C}\in S_{G^{\prime}}$ for any
$g\in\Omega_{c}(G),\alpha\in F$ by Lemma 4.10. From this, by Lemma 4.2,

$|V_{*}(FG)|=|F|^{\frac{|G|+|\Omega_{1}(G)|-|\Omega_{c}(G)|}{4}}\cdot|V_{*}(F\overline{G})|=|F|^{2^{n+m+2}}|V_{*}(F\overline{G})|.$

Note that
$\overline{G}\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.
Thus $|V_{*}(F\overline{G})|=4|F|^{2^{n+m+2}+7}$. It follows that
$|V_{*}(FG)|=4|F|^{2^{n+m+3}+7}=4|F|^{\frac{|G|+|\Omega_{1}(G)|}{2}-1}$.

We next suppose that $k\geq 3$. At this time, $D_{8}$ appears in the central
product of $G$. By Lemma 4.10, $1+\alpha g\widehat{G^{\prime}}\in
S_{G^{\prime}}$ for any $g\in\Omega_{c}(G),\alpha\in F$. Similar to the proof
of Theorem 4.16, we may obtain (ii).

######  Theorem 4.20.

Suppose that $G_{l}\cong
M_{2}(m+1,n+1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-2)}$,
where $n>m\geq 1$, $H_{1}\cong D_{8}$ and $H_{2}\cong Q_{8}$, $l=1,2$ and
$k\geq 2$. Then

(i) $|\Omega_{1}(G_{l})|=|\Omega_{c}(G_{l})|=2^{2k}$.

(ii) $|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ l=1=m,k\geq 2\
\mathrm{or}\ m=1,l=2,k\geq 3.\\\
&4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
otherwise.\end{aligned}\right.$

######  Proof 4.21.

(i) Let $G_{l}=\langle x_{1},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}\rangle\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}\langle
x_{k},y_{k}\rangle$, where $\langle x_{2},y_{2}\rangle=H_{l}$, $\langle
x_{i},y_{i}\rangle\cong D_{8},i=3,\ldots,k,$ and

$\langle x_{1},y_{1}\ |\
x_{1}^{2^{m+1}}=y_{1}^{2^{n+1}}=1,x_{1}^{y_{1}}=x_{1}^{1+2^{m}}\rangle\cong
M_{2}(m+1,n+1).$

For any $g\in G$, let $g=g_{1}g_{2}$, where
$g_{1}=x_{1}^{i_{1}}y_{1}^{j_{1}}$, $0\leq i_{1}<2^{m}$, $0\leq
j_{1}<2^{n+1}$, and $g_{2}\in\langle x_{2},y_{2},\ldots,x_{k},y_{k}\rangle$.

(i) follows similar to (i) of Theorem 4.16.

(ii) First, we suppose that $m=1$. If $G_{1}\cong M_{2}(2,n+1)$. By Lemma 4.8,
we have $\Theta(G_{1})=\frac{1}{2}|F|^{\frac{\Omega_{c}(G_{1})}{2}}$. Hence

$|V_{*}(FG_{1})|=2|F|^{\frac{|G_{1}|}{4}}\cdot|V_{*}(F\overline{G}_{1})|=2|F|^{2^{1+n}}\cdot|V_{*}(F\overline{G}_{1})|$

by Lemma 4.2. Also since
$\overline{G}_{1}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2^{n+1}}$, we have

$|V_{*}(FG_{1})|=2|F|^{2^{1+n}}\cdot
2|F|^{2^{1+n}+1}=4|F|^{2^{2+n}+1}=4|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1}.$

If $G_{2}\cong M_{2}(2,n+1)\mathbin{\mathsf{Y}}Q_{8}$, then we have
$\Theta(G_{2})=\frac{1}{2}|F|^{\frac{\Omega_{c}(G_{2})}{2}}$ similar to
Theorem 4.18. Note that
$\overline{G}_{2}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2^{n+1}}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.
Hence we have

$|V_{*}(FG_{2})|=2|F|^{2^{3+n}}\cdot|V_{*}(F\overline{G}_{2})|=4|F|^{2^{4+n}+7}=4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1}.$

For other cases, by Lemma 4.10, we may obtain $1+\alpha
g\widehat{G_{l}^{\prime}}\in S_{G_{l}^{\prime}}$ for any
$g\in\Omega_{c}(G_{l})$ and $\alpha\in F$. Hence
$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|}{4}}\cdot|V_{*}(F\overline{G}_{l})|$ by
Lemma 4.2. Note that
$\overline{G}_{l}\cong\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2^{n+1}}\times(\mathbb{Z}_{2}\times\mathbb{Z}_{2})^{(k-1)}$.
Hence

$|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&2|F|^{2^{n+m+2k-1}+2^{2k-1}-1}=2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ m=1.\\\
&4|F|^{2^{n+m+2k-1}+2^{2k-1}-1}=4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ m\geq 2.\end{aligned}\right.$

######  Theorem 4.22.

Suppose that $G_{l}\cong
M_{2}(n+1,1)\mathbin{\mathsf{Y}}M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-3)}$
, where $n\geq m\geq 1$, $k\geq 3$ and $l=1,2$, $H_{1}\cong D_{8}$ and
$H_{2}\cong Q_{8}$.

(i) If $n=1$, then $\Omega_{1}(G_{1})=2^{2k}+2^{k+1}=\Omega_{c}(G_{2})$ and
$\Omega_{c}(G_{1})=2^{2k}-2^{k+1}=\Omega_{1}(G_{2})$. If $n\geq 2$, then
$\Omega_{1}(G_{l})=2^{2k}=\Omega_{c}(G_{l})$.

(ii) $|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n=1.\\\
&4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n\geq
2.\end{aligned}\right.$

######  Proof 4.23.

(i) Let $G_{l}=\langle x_{1},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}\rangle\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}\langle
x_{k},y_{k}\rangle$, where $\langle x_{i},y_{i}\rangle\cong
D_{8},i=4,\ldots,k$, $\langle x_{3},y_{3}\rangle=H_{l}$ and

$\langle x_{1},y_{1}\ |\
x_{1}^{2^{n+1}}=y_{1}^{2}=1,x_{1}^{y_{1}}=x_{1}^{1+2^{n}}\rangle\cong
M_{2}(n+1,1),$ $\langle x_{2},y_{2}\ |\
x_{2}^{2^{m+1}}=y_{2}^{2}=1,[x_{2},y_{2}]=c\rangle\cong M_{2}(m+1,1,1).$

For any $g\in G$, we may let $g=g_{1}g_{2}$, where
$g_{1}=x_{1}^{i_{1}}y_{1}^{j_{1}}x_{2}^{i_{2}}y_{2}^{j_{2}}$, $0\leq
i_{1}<2^{n}$, $0\leq i_{2}<2^{m+1}$, $j_{1},j_{2}=0$ or $1$ and
$g_{2}\in\langle x_{3},y_{3},\ldots,x_{k},y_{k}\rangle$.

If $g_{1}^{2}=1$, then
$x_{1}^{2i_{1}}x_{2}^{2i_{2}}c^{i_{1}j_{1}+i_{2}j_{2}}=1$. Hence $i_{2}=0$ or
$2^{m}$, and $x_{1}^{2i_{1}+2^{n}i_{1}j_{1}}=1$. From this, we have $i_{1}=0$
if $j_{1}=0$, $i_{1}=0$ or $2^{n-1}$ if $n=1$ and $j_{1}=1$, $i_{1}=0$ if
$n\geq 2$ and $j_{1}=1$. If $g_{1}^{2}=c$, then $i_{2}=0$ or $2^{m}$, and
$x_{1}^{2i_{1}+2^{n}(i_{1}j_{1}-1)}=1$. From this, we have $i_{1}=2^{n-1}$ if
$j_{1}=0$, $i_{1}$ is non-value if $n=1$ and $j_{1}=1$, $i_{1}=2^{n-1}$ if
$n\geq 2$ and $j_{1}=1$. By Lemma 4.14, we have the following results:

(1) If $n=1$, then we have
$\Omega_{1}(G_{1})=24\gamma_{1}(k-2)+8\gamma_{2}(k-2)=2^{2k}+2^{k+1}=\Omega_{c}(G_{2})$
and
$\Omega_{c}(G_{1})=8\gamma_{1}(k-2)+24\gamma_{2}(k-2)=2^{2k}-2^{k+1}=\Omega_{1}(G_{2})$.

(2) If $n\geq 2$, then we have
$\Omega_{1}(G_{l})=16\gamma_{1}(k-2)+16\gamma_{2}(k-2)=2^{2k}=\Omega_{c}(G_{l})$.

(ii) According to Lemma 4.10, we have $1+\alpha g\widehat{G_{l}^{\prime}}\in
S_{G_{l}^{\prime}}$ for any $g\in\Omega_{c}(G_{l})$ and $\alpha\in F$. Hence
$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|-|\Omega_{c}(G_{l})|}{4}}\cdot|V_{*}(F\overline{G}_{l})|$
by Lemma 4.2. Note that
$\overline{G}_{l}\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2}^{(2k-2)}$.
From this, it follows that

$|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&2|F|^{2^{n+m+2k-1}+2^{2k-1}+\epsilon_{l}2^{k}-1}=2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ n=1.\\\
&4|F|^{2^{n+m+2k-1}+2^{2k-1}-1}=4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ n\geq 2.\end{aligned}\right.$

where $\epsilon_{1}=1$ and $\epsilon_{2}=-1$.

According to Theorem 4.22, it is easy to obtain the following result.

######  Corollary 4.24.

Suppose that $G_{l}\cong
M_{2}(m+1,1)\mathbin{\mathsf{Y}}M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-3)}$
, where $n>m\geq 1$, $k\geq 3$ and $l=1,2$, $H_{1}\cong D_{8}$ and $H_{2}\cong
Q_{8}$. Then $|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ m=1.\\\
&4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ m\geq
2.\end{aligned}\right.$

######  Theorem 4.25.

Suppose that $G_{l}\cong
M_{2}(n+1,1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-2)}\times\mathbb{Z}_{2^{m}}$
, where $n\geq m\geq 1$, $k\geq 2$ and $l=1,2$, $H_{1}\cong D_{8}$ and
$H_{2}\cong Q_{8}$. Then

(i) If $n=1$, $|\Omega_{1}(G_{1})|=2^{2k+1}+2^{k+1}=|\Omega_{c}(G_{2})|$ and
$|\Omega_{c}(G_{1})|=2^{2k+1}-2^{k+1}=|\Omega_{1}(G_{2})|$; If $n\geq 2$,
$|\Omega_{1}(G_{l})|=2^{2k+1}=|\Omega_{c}(G_{l})|$, where $l=1,2$.

(ii) $|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n=m=1.\\\
&2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n>m=1.\\\
&4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n\geq m\geq
2.\end{aligned}\right.$

######  Proof 4.26.

(i) Let $G_{l}=\langle x_{1},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}\rangle\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}\langle
x_{k},y_{k}\rangle\times\langle z\rangle$, where $\langle
z\rangle\cong\mathbb{Z}_{2^{m}}$ and

$\langle x_{1},y_{1}\ |\
x_{1}^{2^{n+1}}=y_{1}^{2}=1,x_{1}^{y_{1}}=x_{1}^{1+2^{n}}\rangle\cong
M_{2}(n+1,1),$ $\langle x_{2},y_{2}\rangle=H_{l},l=1,2,\ \mathrm{and}\ \langle
x_{i},y_{i}\rangle\cong D_{8},i=3,\ldots,k.$

For any $g\in G$, let $g=g_{1}g_{2}$, where
$g_{1}=x_{1}^{i_{1}}y_{1}^{j_{1}}z^{s}$, $0\leq i_{1}<2^{n}$, $0\leq j_{1}<2$,
$0\leq s<2^{m}$ and $g_{2}\in\langle x_{2},y_{2},\ldots,x_{k},y_{k}\rangle$.

If $g_{1}^{2}=1$, that is, $x_{1}^{2i_{1}}c^{i_{1}j_{1}}z^{2s}=1$, then $s=0$
or $2^{m-1}$. Furthermore, we have $(i_{1},j_{1})=(0,0),(0,1),(1,1)$ when
$n=1$; $(i_{1},j_{1})=(0,0),(0,1)$ when $n\geq 2$.

If $g_{1}^{2}=c$, that is, $x_{1}^{2i_{1}}c^{i_{1}j_{1}}z^{2s}=c$, then $s=0$
or $2^{m-1}$. Furthermore, we have $(i_{1},j_{1})=(1,0)$ when $n=1$;
$(i_{1},j_{1})=(2^{n-1},0),(2^{n-1},1)$ when $n\geq 2$. By Lemma 4.14, we have
the following results:

(1) If $n=1$, then we have
$\Omega_{1}(G_{1})=12\gamma_{1}(k-1)+4\gamma_{2}(k-1)=2^{2k+1}+2^{k+1}=\Omega_{c}(G_{2})$
and
$\Omega_{c}(G_{1})=4\gamma_{1}(k-1)+12\gamma_{2}(k-1)=2^{2k+1}-2^{k+1}=\Omega_{1}(G_{2})$.

(2) If $n\geq 2$, then we have
$\Omega_{1}(G_{l})=8\gamma_{1}(k-1)+8\gamma_{2}(k-1)=2^{2k+1}=\Omega_{c}(G_{l})$.

(ii) According to Lemma 4.10, we have $1+\alpha g\widehat{G_{l}^{\prime}}\in
S_{G_{l}^{\prime}}$ for any $g\in G_{l}$ and $\alpha\in F$. From this, by
Lemma 4.2, we obtain

$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|-|\Omega_{c}(G_{l})|}{4}}\cdot|V_{*}(F\overline{G}_{l})|=\left\\{\begin{aligned}
&|F|^{2^{n+m+2k-2}+2^{k}\varepsilon_{l}}\cdot|V_{*}(F\overline{G}_{l})|,\
\mathrm{if}\ n=1.\\\ &|F|^{2^{n+m+2k-2}}\cdot|V_{*}(F\overline{G}_{l})|,\
\mathrm{if}\ n\geq 2.\end{aligned}\right.$

where $\varepsilon_{1}=1$ and $\varepsilon_{2}=-1$. Note that
$\overline{G}\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2}^{(2k-1)}$.
According to Lemma 2.3, we have

$|V_{*}(F\overline{G})=\left\\{\begin{aligned} &|F|^{2^{n+m+2k-2}+2^{2k}-1},\
\mathrm{if}\ n=m=1.\\\ &2|F|^{2^{n+m+2k-2}+2^{2k}-1},\ \mathrm{if}\ n>m=1.\\\
&4|F|^{2^{n+m+2k-2}+2^{2k}-1},\ \mathrm{if}\ n\geq m\geq
2.\end{aligned}\right.$

From this, we have

$|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&|F|^{2^{n+m+2k-1}+2^{2k}+2^{k}\varepsilon_{l}-1}=|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ n=m=1.\\\
&2|F|^{2^{n+m+2k-1}+2^{2k}-1}=2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ n>m=1.\\\
&4|F|^{2^{n+m+2k-1}+2^{2k}-1}=4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ n\geq m\geq 2\end{aligned}\right.$.

Obviously, by Lemma 2.6, we do not consider an elementary abelian $2$-group as
a direct product term of $G$. According to Theorem 4.25, the unitary subgroups
of (i) and (ii) of Corollary 3.23 are as follows.

######  Corollary 4.27.

Suppose that $G_{l}\cong
M_{2}(n+1,1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-2)}\times\mathbb{Z}_{2^{m+1}}$
, where $n\geq m\geq 1$, $k\geq 2$ and $l=1,2$, $H_{1}\cong D_{8}$ and
$H_{2}\cong Q_{8}$. Then

$|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n=m=1.\\\
&4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n\geq
2.\end{aligned}\right.$

######  Theorem 4.28.

Suppose that $G_{l}\cong
M_{2}(m+1,1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-2)}\times\mathbb{Z}_{2^{n}}$
, where $n>m\geq 1$, $k\geq 2$ and $l=1,2$, $H_{1}\cong D_{8}$ and $H_{2}\cong
Q_{8}$. Then

(i) If $m=1$, $|\Omega_{1}(G_{1})|=2^{2k+1}+2^{k+1}=|\Omega_{c}(G_{2})|$ and
$|\Omega_{c}(G_{1})|=2^{2k+1}-2^{k+1}=|\Omega_{1}(G_{2})|$; If $m\geq 2$,
$|\Omega_{1}(G_{l})|=2^{2k+1}=|\Omega_{c}(G_{l})|$.

(ii) $|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n>m=1.\\\
&4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\ \mathrm{if}\ n>m\geq
2.\end{aligned}\right.$

######  Proof 4.29.

(i) follows from (i) of Theorem 4.25.

(ii) According to Lemma 4.10, we have $1+\alpha g\widehat{G_{l}^{\prime}}\in
S_{G_{l}^{\prime}}$ for any $g\in G_{l}$ and $\alpha\in F$. From this, by
Lemma 4.2, we obtain

$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|-|\Omega_{c}(G_{l})|}{4}}\cdot|V_{*}(F\overline{G}_{l})|=\left\\{\begin{aligned}
&|F|^{2^{n+m+2k-2}+2^{k}\varepsilon_{l}}\cdot|V_{*}(F\overline{G}_{l})|,\
\mathrm{if}\ m=1.\\\ &|F|^{2^{n+m+2k-2}}\cdot|V_{*}(F\overline{G}_{l})|,\
\mathrm{if}\ m\geq 2.\end{aligned}\right.$

where $\varepsilon_{1}=1$ and $\varepsilon_{2}=-1$. Note that
$\overline{G}_{l}\cong\mathbb{Z}_{2^{n}}\times\mathbb{Z}_{2^{m}}\times\mathbb{Z}_{2}^{(2k-1)}$.
According to Lemma 2.3, we have

$|V_{*}(F\overline{G}_{l})=\left\\{\begin{aligned}
&2|F|^{2^{n+m+2k-2}+2^{2k}-1},\ \mathrm{if}\ m=1.\\\
&4|F|^{2^{n+m+2k-2}+2^{2k}-1},\ \mathrm{if}\ m\geq 2.\end{aligned}\right.$

From this, we have

$|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&2|F|^{2^{n+m+2k-1}+2^{2k}+2^{k}\varepsilon_{l}-1}=2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ m=1.\\\
&4|F|^{2^{n+m+2k-1}+2^{2k}-1}=4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ m\geq 2.\end{aligned}\right.$

Obviously, by Lemma 2.6, we do not consider an elementary abelian $2$-group as
a direct product term of $G$. Furthermore, the unitary subgroups of (i) and
(ii) of Theorem 3.22 are the special cases of Theorem 4.28.

######  Theorem 4.30.

Suppose that $G_{l}\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-2)}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n}}$
, where $n\geq m\geq 1$, $k\geq 2$ and $l=1,2$, $H_{1}\cong D_{8}$ and
$H_{2}\cong Q_{8}$. Then

(i) If $n=1$, $\Omega_{1}(G_{1})=2^{2k}+2^{k+1}=\Omega_{c}(G_{2})$ and
$\Omega_{c}(G_{1})=2^{2k}-2^{k+1}=\Omega_{1}(G_{2})$; If $n\geq 2$,
$\Omega_{1}(G_{l})=2^{2k+1}=\Omega_{c}(G_{l})$.

(ii) $|V_{*}(FG_{1})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\ \mathrm{if}\ n=1\
\mathrm{or}\ n=2\ \mathrm{and}\ k\geq 2.\\\
&4|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

$|V_{*}(FG_{2})|=\left\\{\begin{aligned}
&8|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\ \mathrm{if}\ n=1\
\mathrm{and}\ k=2.\\\ &2|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{if}\ 1\leq n\leq 2\ \mathrm{and}\ k\geq 3.\\\
&4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

######  Proof 4.31.

Let $G_{l}=\langle x_{1},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}\rangle\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}\langle
x_{k},y_{k}\rangle\mathbin{\mathsf{Y}}\langle z\rangle$, where $k\geq 2$,
$\langle x_{2},y_{2}\rangle=H_{l}$, $\langle x_{i},y_{i}\rangle\cong
D_{8},i=3,\ldots,k$, $\langle z\rangle\cong\mathbb{Z}_{2^{n}}$ and

$\langle x_{1},y_{1}\ |\
x_{1}^{2^{m+1}}=y_{1}^{2}=c^{2}=1,[x_{1},y_{1}]=c\rangle\cong M_{2}(m+1,1,1).$

(i) For any $g\in G$, we may let $g=g_{1}g_{2}$, where
$g_{1}=x_{1}^{i_{1}}y_{1}^{j_{1}}z^{s}$, $0\leq i_{1}<2^{m+1}$, $0\leq
j_{1}<2$, $0\leq s<2^{n-1}$ and $g_{2}\in\langle
x_{2},y_{2},\ldots,x_{k},y_{k}\rangle$. If $g_{1}^{2}=1$, then
$x_{1}^{2i_{1}}y_{1}^{2j_{1}}z^{2s+2^{n-1}i_{1}j_{1}}=1$. Hence $i_{1}=0$ or
$2^{m}$, $j_{1}=0$ or $1$, and $s=0$. If $g_{1}^{2}=c$, then $n\geq 2$ and
$x_{1}^{2i_{1}}y_{1}^{2j_{1}}z^{2s+2^{n-1}(i_{1}j_{1}-1)}=1$. Hence $i_{1}=0$
or $2^{m}$, $j_{1}=0$ or $1$, and $s=2^{n-2}$.

(1) If $n=1$, then we have
$\Omega_{1}(G_{1})=8\gamma_{1}(k-1)=2^{2k}+2^{k+1}=\Omega_{c}(G_{2})$ and
$\Omega_{c}(G_{1})=8\gamma_{2}(k-1)=2^{2k}-2^{k+1}=\Omega_{1}(G_{2})$.

(2) If $n\geq 2$, then we have
$\Omega_{1}(G_{l})=8\gamma_{1}(k-1)+8\gamma_{2}(k-1)=2^{2k+1}=\Omega_{c}(G_{l})$.

(ii) First, we consider the case $n=1$. If $G_{1}\cong M_{2}(2,1,1)$, then
$\Omega_{c}(G_{1})=\emptyset$. Hence $\Theta(G_{1})=1$ in Lemma 4.2.
Furthermore,

$|V_{*}(FG_{1})|=|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|+|\Omega_{c}(G_{1})|}{4}}\cdot|V_{*}(F\overline{G}_{1})|.$

Note that $\overline{G}_{1}\cong\mathbb{Z}_{2^{2}}\times\mathbb{Z}_{2}$. It
follows that
$|V_{*}(FG_{1})|=2|F|^{11}=2|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1}$.

Suppose that $G_{2}\cong M_{2}(2,1,1)\mathbin{\mathsf{Y}}Q_{8}$. Note that the
group $\langle
1+\sum\limits_{g\in\Omega_{c}(G_{2})}\alpha_{g}g\widehat{G_{2}^{\prime}}\in
S_{G_{2}^{\prime}},\alpha_{g}\in F\rangle$ is generated by the following
elements

$1+(\alpha^{2}+\alpha)x_{2}\widehat{G_{2}^{\prime}},1+(\alpha^{2}+\alpha)y_{2}\widehat{G_{2}^{\prime}},1+\alpha(x_{2}+y_{2}+x_{2}y_{2})\widehat{G_{2}^{\prime}},$
$1+\alpha(x_{2}+x_{2}g)\widehat{G_{2}^{\prime}},1+\alpha(y_{2}+y_{2}g)\widehat{G_{2}^{\prime}},1+\alpha(x_{2}y_{2}+x_{2}y_{2}g)\widehat{G_{2}^{\prime}},$

where $\alpha\in F,g\in\\{x_{1}^{2},y_{1},y_{1}x_{1}^{2}\\}$, according to (i)
of Lemma 4.4 and (ii) of Lemma 4.6. Hence

$\Theta(G_{2})=\frac{1}{4}|F|^{12}=\frac{1}{4}|F|^{\frac{\Omega_{c}(G_{2})}{2}}.$

By Lemma 4.2,

$|V_{*}(FG_{2})|=\frac{4}{|F|^{\frac{|\Omega_{c}(G_{2})|}{2}}}|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|+|\Omega_{c}(G_{2})|}{4}}\cdot|V_{*}(F\overline{G}_{2})|=4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|-|\Omega_{c}(G_{2})|}{4}}\cdot|V_{*}(F\overline{G}_{2})|.$

Note that
$\overline{G}_{2}\cong\mathbb{Z}_{2^{2}}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.
According to Lemma 2.3, we have

$|V_{*}(F\overline{G}_{2})|=|\overline{G}_{2}^{2}[2]|\cdot|F|^{\frac{|\overline{G}_{2}|+|\Omega_{1}(\overline{G}_{2})|}{2}-1}=2|F|^{23}.$

Hence
$|V_{*}(FG_{2})|=8|F|^{35}=8|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1}$.

For the other cases when $n=1$, $D_{8}$ will appear in the central product of
$G_{l}$, where $l=1,2$. At this time, we have $1+\alpha
g\widehat{G_{l}^{\prime}}\in S_{G_{l}^{\prime}}$ by Lemma 4.10. From this, by
Lemma 4.2,

$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|-|\Omega_{c}(G_{l})|}{4}}\cdot|V_{*}(F\overline{G}_{l})|.$

Note that
$\overline{G}_{l}\cong\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2}^{(2k-1)}$.
Hence

$|V_{*}(FG_{1})|=2|F|^{2^{m+2k}+2^{2k-1}+2^{k}-1}=2|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1}.$
$|V_{*}(FG_{2})|=2|F|^{2^{m+2k}+2^{2k-1}-2^{k}-1}=2|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1}.$

Suppose that $n=2$ and $G_{1}\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}\mathbb{Z}_{4}$ or $G_{2}\cong
M_{2}(m+1,1,1)\mathbin{\mathsf{Y}}Q_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{4}$.

We have $|V_{*}(F\langle z\rangle)|=2|F|^{2}$ by Lemma 2.3. From this,
$\Theta\langle z\rangle=\frac{1}{2}|F|$ according to Lemma 4.2. Note that
$1+(\beta^{2}+\beta)z\widehat{G_{l}^{\prime}}\in S_{G_{l}^{\prime}}$ since
$(1+\beta z+\beta c)(1+\beta z+\beta c)^{*}\in S_{G_{l}^{\prime}}$ for any
$\beta\in F$, where $l=1,2$, which implies $\langle 1+\alpha
z\widehat{G_{l}^{\prime}}\in S_{G_{l}^{\prime}},\alpha\in F\rangle$ has exact
$\frac{1}{2}|F|$ elements. For any $g\in\Omega_{c}(G_{l})$, we have
$1+\alpha(z+g)\widehat{G_{l}^{\prime}}\in S_{G_{l}^{\prime}}$ for any
$\alpha\in F$ by (i) of Lemma 4.4. It follows that
$\Theta(G_{l})=\frac{1}{2}|F|^{\frac{\Omega_{c}(G_{l})}{2}}$, where $l=1,2$.
By Lemma 4.2,

$|V_{*}(FG_{l})|=2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|-|\Omega_{c}(G_{l})|}{4}}\cdot|V_{*}(F\overline{G}_{l})|.$

Note that
$\overline{G}_{1}\cong\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2}^{(2)}$ and
$\overline{G}_{2}\cong\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2}^{(4)}$. Hence

$|V_{*}(FG_{1})|=4|F|^{2^{m+3}+3}=4|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1}.$
$|V_{*}(FG_{2})|=4|F|^{2^{m+5}+15}=4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1}.$

For the other cases when $n=2$, $D_{8}$ will appear in the central product of
$G_{l}$, where $l=1,2$. At this time, we have $1+\alpha
g\widehat{G_{l}^{\prime}}\in S_{G_{l}^{\prime}}$ for any $\alpha\in F$ and
$g\in\Omega_{c}(G_{l})$ by Lemma 4.10. From this, by Lemma 4.2 and (i),

$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|}{4}}\cdot|V_{*}(F\overline{G}_{l})|.$

Note that
$\overline{G}_{l}\cong\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2}^{(2k)}$. Hence

$|V_{*}(FG_{l})|=2|F|^{2^{m+2k+1}+2^{2k}-1}=2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1}.$

If $n\geq 3$, then $1+\alpha g\widehat{G_{l}^{\prime}}\in S_{G_{l}^{\prime}}$
for any $\alpha\in F$ and $g\in\Omega_{c}(G_{l})$ by Lemma 4.12. From this, by
Lemma 4.2,

$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|}{4}}\cdot|V_{*}(F\overline{G}_{l})|.$

Note that
$\overline{G}_{l}\cong\mathbb{Z}_{2^{m+1}}\times\mathbb{Z}_{2}^{(2k-1)}\times\mathbb{Z}_{2^{n-1}}$.
It follows that

$|V_{*}(FG_{l})|=4|F|^{2^{n+m+2k-1}+2^{2k}-1}=4|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1}.$

Obviously, by Lemma 2.6, the unitary subgroups of (i) and (ii) of Theorem 3.19
are the special cases of Theorem 4.30. Furthermore, it is easy to obtain the
following results similar to Theorem 4.30.

######  Corollary 4.32.

Suppose that $G_{l}\cong
M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-2)}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m}}$
, where $n>m\geq 1$, $k\geq 2$ and $l=1,2$, $H_{1}\cong D_{8}$ and $H_{2}\cong
Q_{8}$. Then

$|V_{*}(FG_{1})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\ \mathrm{if}\ m=1\
\mathrm{or}\ m=2\ \mathrm{and}\ k\geq 2.\\\
&4|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

$|V_{*}(FG_{2})|=\left\\{\begin{aligned}
&8|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\ \mathrm{if}\ m=1\
\mathrm{and}\ k=2.\\\ &2|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{if}\ 1\leq m\leq 2\ \mathrm{and}\ k\geq 3.\\\
&4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

######  Corollary 4.33.

Suppose that $G_{l}\cong
M_{2}(n+1,1,1)\mathbin{\mathsf{Y}}H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-2)}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{m+1}}$
, where $n\geq m\geq 1$, $k\geq 2$ and $l=1,2$, $H_{1}\cong D_{8}$ and
$H_{2}\cong Q_{8}$. Then

$|V_{*}(FG_{1})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\ \mathrm{if}\ m=1\
\mathrm{and}\ k\geq 2.\\\ &4|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

$|V_{*}(FG_{2})|=\left\\{\begin{aligned}
&2|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\ \mathrm{if}\ m=1\
\mathrm{and}\ k\geq 3.\\\ &4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

######  Theorem 4.34.

Suppose that $G_{l}\cong
H_{l}\mathbin{\mathsf{Y}}D_{8}^{\mathbin{\mathsf{Y}}(k-1)}\mathbin{\mathsf{Y}}\mathbb{Z}_{2^{n_{1}}}\times\mathbb{Z}_{2^{m_{1}}}$
, where $n_{1},m_{1},k\geq 1$ and $l=1,2$, $H_{1}\cong D_{8}$, $H_{2}\cong
Q_{8}$. Then

(i) If $n_{1}=1$, $\Omega_{1}(G_{1})=2^{2k+1}+2^{k+1}=\Omega_{c}(G_{2})$ and
$\Omega_{c}(G_{1})=2^{2k+1}-2^{k+1}=\Omega_{1}(G_{2})$; If $n_{1}\geq 2$,
$\Omega_{1}(G_{l})=2^{2k+2}=\Omega_{c}(G_{l})$, where $l=1,2$.

(ii) $|V_{*}(FG_{1})|=\left\\{\begin{aligned}
&|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\ \mathrm{if}\ 1\leq n_{1}\leq
2,\ \ m_{1}=1.\\\ &4|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\
\mathrm{if}\ n_{1}\geq 3,\ m_{1}\geq 2.\\\
&2|F|^{\frac{|G_{1}|+|\Omega_{1}(G_{1})|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

$|V_{*}(FG_{2})|=\left\\{\begin{aligned}
&8|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\ \mathrm{if}\ n_{1}=k=1,\
m_{1}\geq 2.\\\ &4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\begin{aligned} &\ \mathrm{if}\ n_{1}=m_{1}=k=1\\\ &\mathrm{or}\
n_{1}=2,m_{1}\geq 2,\ k=1\\\ &\mathrm{or}\ n_{1}\geq 3,\ m_{1}\geq
2.\end{aligned}\\\ &|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{if}\ 1\leq n_{1}\leq 2,m_{1}=1,\ k\geq 2.\\\
&2|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{otherwise}.\end{aligned}\right.$

######  Proof 4.35.

Let $G_{l}=\langle x_{1},y_{1}\rangle\mathbin{\mathsf{Y}}\langle
x_{2},y_{2}\rangle\mathbin{\mathsf{Y}}\cdots\mathbin{\mathsf{Y}}\langle
x_{k},y_{k}\rangle\mathbin{\mathsf{Y}}\langle z_{1}\rangle\times\langle
z_{2}\rangle$, where $k\geq 1$, $\langle x_{1},y_{1}\rangle=H_{l}$, $\langle
x_{i},y_{i}\rangle\cong D_{8},i=2,\ldots,k$, $\langle
z_{1}\rangle\cong\mathbb{Z}_{2^{n_{1}}}$ and $\langle
z_{2}\rangle\cong\mathbb{Z}_{2^{m_{1}}}$.

(i) For any $g\in G$, we may let $g=g_{1}g_{2}$, where
$g_{1}=z_{1}^{i}z_{2}^{j}$, $0\leq i<2^{n_{1}-1}$, $0\leq j<2^{m_{1}}$, and
$g_{2}\in\langle x_{1},y_{1},\ldots,x_{k},y_{k}\rangle$. If $g_{1}^{2}=1$,
then $z_{1}^{2i}z_{2}^{2j}=1$. Hence $i=0$, $j=0$ or $2^{m_{1}-1}$. If
$g_{1}^{2}=c$, then $n_{1}\geq 2$ and
$z_{1}^{2i}z_{2}^{2j}=z_{1}^{2^{n_{1}-1}}$. Hence $i=2^{n_{1}-2}$, $j=0$ or
$2^{m_{1}-1}$.

(1) If $n_{1}=1$, then we have
$\Omega_{1}(G_{1})=4\gamma_{1}(k)=2^{2k+1}+2^{k+1}=\Omega_{c}(G_{2})$ and
$\Omega_{c}(G_{1})=4\gamma_{2}(k)=2^{2k+1}-2^{k+1}=\Omega_{1}(G_{2})$.

(2) If $n_{1}\geq 2$, then we have
$\Omega_{1}(G_{l})=4\gamma_{1}(k)+4\gamma_{2}(k)=2^{2k+2}=\Omega_{c}(G_{l})$.

(ii) Suppose $n_{1}=1$ and $G_{2}=Q_{8}\times\langle z_{2}\rangle$. Note that
the group $\langle
1+\sum\limits_{g\in\Omega_{c}(G_{2})}\alpha_{g}g\widehat{G_{2}^{\prime}}\in
S_{G_{2}^{\prime}},\alpha_{g}\in F\rangle$ is generated by the following
elements

$1+(\alpha^{2}+\alpha)x_{1}\widehat{G_{2}^{\prime}},1+(\alpha^{2}+\alpha)y_{1}\widehat{G_{2}^{\prime}},1+\alpha(x_{1}+y_{1}+x_{1}y_{1})\widehat{G_{2}^{\prime}},$
$1+\alpha(x_{1}+x_{1}z_{2}^{2^{m_{1}-1}})\widehat{G_{2}^{\prime}},1+\alpha(y_{1}+y_{1}z_{2}^{2^{m_{1}-1}})\widehat{G_{2}^{\prime}},1+\alpha(x_{1}y_{1}+x_{1}y_{1}z_{2}^{2^{m_{1}-1}})\widehat{G_{2}^{\prime}},$

where $\alpha\in F$, according to (i) of Lemma 4.4 and (ii) of Lemma 4.6.
Hence
$\Theta(G_{2})=\frac{1}{4}|F|^{6}=\frac{1}{4}|F|^{\frac{\Omega_{c}(G_{2})}{2}}.$
At this time, by Lemma 4.2,
$|V_{*}(FG_{2})|=4|F|^{\frac{|G_{2}|-2^{3}}{4}}\cdot|V_{*}(F\overline{G}_{2})|.$
Note that
$\overline{G}_{2}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2^{m_{1}}}$.
Hence

$|V_{*}(FG_{2})|=\left\\{\begin{aligned}
&4|F|^{2^{m_{1}+2}+1}=4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{if}\ m_{1}=1,\\\
&8|F|^{2^{m_{1}+2}+1}=8|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{if}\ m_{1}\geq 2.\end{aligned}\right.$

For the other cases when $n_{1}=1$, $D_{8}$ will appear in the central product
of $G_{l}$. At this time, we have $1+\alpha g\widehat{G_{l}^{\prime}}\in
S_{G_{l}^{\prime}}$ for any $\alpha\in F$ and $g\in\Omega_{c}(G_{l})$ by Lemma
4.10. From this, by Lemma 4.2,

$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|-|\Omega_{c}(G_{l})|}{4}}\cdot|V_{*}(F\overline{G}_{l})|.$

Note that
$\overline{G}_{l}\cong\mathbb{Z}_{2}^{(2k)}\times\mathbb{Z}_{2^{m_{1}}}$.
Hence

$|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&|F|^{2^{2k+m_{1}}+2^{2k}+2^{k}\epsilon_{l}-1}=|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ m_{1}=1,\\\
&2|F|^{2^{2k+m_{1}}+2^{2k}+2^{k}\epsilon_{l}-1}=2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ m_{1}\geq 2,\end{aligned}\right.$

where $\epsilon_{1}=1$ and $\epsilon_{2}=-1$.

Suppose that $n_{1}=2$ and $G_{2}\cong
Q_{8}\mathbin{\mathsf{Y}}\mathbb{Z}_{4}\times\mathbb{Z}_{2^{m_{1}}}$. Similar
to (ii) of Lemma 4.30, we
$\Theta(G_{2})=\frac{1}{2}|F|^{\frac{\Omega_{c}(G_{2})}{2}}$. By Lemma 4.2,
$|V_{*}(FG_{2})|=2|F|^{\frac{|G_{2}|}{4}}\cdot|V_{*}(F\overline{G}_{2})|.$
Note that
$\overline{G}_{2}\cong\mathbb{Z}_{2}^{(3)}\times\mathbb{Z}_{2^{m_{1}}}$. Hence

$|V_{*}(FG_{2})|=\left\\{\begin{aligned}
&2|F|^{2^{m_{1}+3}+7}=2|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{if}\ m_{1}=1,\\\
&4|F|^{2^{m_{1}+3}+7}=4|F|^{\frac{|G_{2}|+|\Omega_{1}(G_{2})|}{2}-1},\
\mathrm{if}\ m_{1}\geq 2.\end{aligned}\right.$

For the other cases when $n_{1}=2$, we have $1+\alpha
g\widehat{G_{l}^{\prime}}\in S_{G_{l}^{\prime}}$ for any $\alpha\in F$ and
$g\in\Omega_{c}(G_{l})$ by Lemma 4.10. From this, by Lemma 4.2,
$|V_{*}(FG_{l})|=|F|^{\frac{|G_{l}|}{4}}\cdot|V_{*}(F\overline{G}_{l})|.$ Note
that
$\overline{G}_{l}\cong\mathbb{Z}_{2}^{(2k)}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2^{m_{1}}}$.
Hence

$|V_{*}(FG_{l})|=\left\\{\begin{aligned}
&|F|^{2^{2k+m_{1}+1}+2^{2k+1}-1}=|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ m_{1}=1,\\\
&2|F|^{2^{2k+m_{1}+1}+2^{2k+1}-1}=2|F|^{\frac{|G_{l}|+|\Omega_{1}(G_{l})|}{2}-1},\
\mathrm{if}\ m_{1}\geq 2.\end{aligned}\right.$

Suppose that $n_{1}\geq 3$, we have $1+\alpha g\widehat{G_{l}^{\prime}}\in
S_{G_{l}^{\prime}}$ for any $\alpha\in F$ and $g\in\Omega_{c}(G_{l})$ by Lemma
4.10. From this, we have

$|V_{*}(FG_{l})|=\left\\{\begin{aligned}