# Conjugacy classes of maximal cyclic subgroups of metacyclic $p$-groups

M. Bianchi, R.D. Camina, & Mark L. Lewis

( )

###### Abstract

In this paper, we set $\eta(G)$ to be the number of conjugacy classes of
maximal cyclic subgroups of a finite group $G$. We compute $\eta(G)$ for all
metacyclic $p$-groups. We show that if $G$ is a metacyclic $p$-group of order
$p^{n}$ that is not dihedral, generalized quaternion, or semi-dihedral, then
$\eta(G)\geq n-2$, and we determine when equality holds.
Keywords: group covering, metacyclic group
2020 Mathematics Subject Classification: 20D15

##  1 Introduction

Unless otherwise stated, all groups in this paper are finite, and we will
follow standard notation from [6]. As in [3] and [4], we set $\eta(G)$ to be
the number of conjugacy classes of maximal cyclic subgroups of a group $G$.
For $p=2$, we have that $\eta(G)=3$ when $G$ is a dihedral $2$-group, a
generalized quaternion $2$-group, or a semi-dihedral group. In [1], the second
and third authors along with Yiftach Barnea and Mikhail Ershov have shown that
for every prime $p\geq 5$ there are infinitely many $p$-groups with $\eta=p+2$
and for $p=3$ there are infinitely many $3$-groups with $\eta=9$. This answers
negatively Question 5.0.9 from [9] which asked whether $\eta(G)$ grows with
the order of $G$ when $G$ is a $p$-group and $p$ is odd.

On the other hand, it is rare for this to occur. Indeed, the only $2$-groups
(in fact the only $p$-groups) that have $\eta=3$ are the Klein $4$-group, the
dihedral groups, the generalized quaternion groups, and the semi-dihedral
groups. To see this, we know that $\eta(G)\geq\eta(G/G^{\prime})$ (see [3]),
and for $p$-groups $\eta(G/G^{\prime})\geq p+1$ when $G/G^{\prime}$ is not
cyclic (see [4]). Thus, $\eta=3$ can only occur when $p=2$. Also, in [4], we
show that $\eta(G/G^{\prime})=3$ if and only if $G/G^{\prime}\cong C_{2}\times
C_{2}$. It is well known that if $G$ is a $2$-group of order at least $8$ and
$|G:G^{\prime}|=4$, then $G$ is either dihedral, generalized quaternion, or
semi-dihedral. (See Problem 6B.8 of [6].)

Now, dihedral groups, generalized quaternion groups, and semi-dihedral groups
are examples of metacyclic groups. I.e., groups $G$ with a normal subgroup $N$
so that $N$ and $G/N$ are both cyclic groups. This motivated us to investigate
the invariant $\eta$ for all metacyclic $p$-groups. Indeed this project began
before the results of [1] were known and we were originally curious as to
whether we would find another family of metacyclic $p$-groups with fixed
$\eta$. However, we prove the following:

###### Theorem 1.1

Let $G$ be a metacyclic $p$-group of order $p^{n}$ that is not a dihedral
group, generalized quaternion group, or semi-dihedral group. Then $\eta(G)\geq
n-2$.

In fact, we compute $\eta(G)$ for every metacyclic $p$-group $G$. Thus, we
list the metacyclic $p$-groups where equality occurs in Theorem 1.1. King in
[7] gave a description of all metacyclic $p$-groups. We will give this
description of these groups in Section 3. In particular, King divided the
metacyclic $p$-groups into two families of groups which he called positive
type and negative type. The negative type groups only occur when $p=2$, so if
$p$ is an odd prime, then all of the metacyclic $p$-groups are of positive
type. We have the following result for the metacyclic groups of positive type.

###### Theorem 1.2

Let $G$ be a metacyclic group of positive type. Then
$\eta(G)=\eta(G/G^{\prime})$.

We note that Rogério in [8] has a formula to compute $\eta(A)$ for an abelian
group $A$. His formula involves the Euler $\phi$-function and a second number
theoretic function. When $G$ is a metacyclic abelian $p$-group, we prove in
[4] a formula for $\eta(G)$ that is only in terms of the sizes of the direct
factors of $G$. Notice in Theorem 1.2 that $G/G^{\prime}$ will be a metacyclic
abelian $p$-group, and so, our formula will compute $\eta(G/G^{\prime})$ and
hence, $\eta(G)$.

When $G$ is a metacyclic $p$-group of negative type, it is not usually the
case that $\eta(G)$ and $\eta(G/G^{\prime})$ are equal. However, we will find
that there usually is a proper quotient whose value of $\eta$ equals
$\eta(G)$. We will also see for most metacyclic groups of negative type that
the formula for $\eta$ is dependent on the formula for $\eta$ that we found
for the metacyclic abelian $p$-groups.

The authors would like to thank Emanuele Pacifici for a number of helpful
conversations while working on this paper.

##  2 Preliminaries

In our preprint [3], we prove two results that we need in this paper. The
first is a criteria for determining when the quotient of a $p$-group $G$ has
the same value for $\eta$ as $\eta(G)$. Given a prime $p$, we set
$G^{\\{p\\}}=\\{g^{p}\mid g\in G\\}$. I.e., $G^{\\{p\\}}$ is the set of $p$-th
powers in $G$.

###### Theorem 2.1

Let $N$ be a normal subgroup of the $p$-group $G$. Then
$\eta(G/N)\leq\eta(G).$ Furthermore, $\eta(G/N)=\eta(G)$ if and only if
$N\subseteq G^{\\{p\\}}$ and for all $x\in G\setminus G^{\\{p\\}}$ every
element of $xN$ is conjugate to a generator of $\langle x\rangle$. In
particular, if $\eta(G/N)=\eta(G)$, then $G^{\\{p\\}}$ is a union of
$N$-cosets and $G^{\\{p\\}}N=G^{\\{p\\}}$.

This second Proposition relates $\eta(G)$ to the number of $G$-orbits of
maximal cyclic subgroups of a normal subgroup.

###### Proposition 2.2

Let $N$ be a normal subgroup of a group $G$ and let $\eta^{*}(N)$ be the
number of $G$-orbits on the $N$-conjugacy classes of maximal cyclic subgroups
of $N$. Then $\eta(G)\geq\eta^{*}(N)$. In particular,

  1. (i)

if $N$ is central in $G$, then $\eta(G)\geq\eta(N)$.

  2. (ii)

if $|G:N|=k$, then $\eta(G)\geq\eta(N)/k$.

Let $p$ be a prime, and let $a$ and $b$ be positive integers. We take $k={\rm
max}(a,b)$ and $l={\rm min}(a,b)$. We set
$g_{p}(a,b)=p^{(l-1)}((k-l)(p-1)+p+1)$. In [4], we prove the following lemma.

###### Lemma 2.3

If $p$ is a prime and $a$ and $b$ are positive integers so that
$G=C_{p^{a}}\times C_{p^{b}}$, then $g_{p}(a,b)=\eta(G)$.

We close this section with an easy lemma that computes $g_{2}$ for small
values and gives a lower bound for larger values. We remark that when $p=2$,
this function is much easier to work with.

###### Lemma 2.4

Suppose $k\geq l$. Then the following hold:

  1. 1.

If $l=1$, then $g_{2}(k,1)=k+2$.

  2. 2.

If $l=2$, then $g_{2}(k,2)=2(k+1)$.

  3. 3.

If $l=3$, then $g_{2}(k,3)=4k$.

  4. 4.

If $l\geq 4$, then $g_{2}(k,l)\geq 4k+2l$.

Proof. We have $g_{2}(a,b)=g_{2}(k,l)=2^{l-1}(k-l+3)$. Conclusions (1), (2),
and (3) are immediate. We focus on (4). Begin with $g_{2}(4,4)=24$; so the
result holds for $g_{2}(4,4)$. Next, $g_{2}(l,l)-6l=3\cdot 2^{l-1}-6l$ is
clearly increasing when $l\geq 3$. Thus, we have $g_{2}(l,l)\geq 4l+2l$ when
$l\geq 3$. Let $k=l+m$ for $m\geq 0$. Then
$g_{2}(k,l)=g_{2}(l+m,m)=2^{l-1}(m+3)$ and $4k+2l=4(l+m)+2l=6l+4m$. Fixing
$l\geq 4$, we note that $2^{l-1}(m+3)-6l-4m$ will be an increasing function in
$m$. We conclude that $g_{2}(k,l)\geq 4k+2l$ for $l\geq 4$. $\Box$

##  3 Metacyclic $p$-Groups

For the rest of the paper, we will focus on metacyclic $p$-groups. A finite
metacyclic $p$-group can be described as follows. This description is taken
from [7],

$G_{p}(\alpha,\beta,\epsilon,\delta,\pm)=\langle x,y\mid
x^{p^{\alpha}}=1,y^{p^{\beta}}=x^{p^{\alpha-\epsilon}},x^{y}=x^{r}\rangle$

where $r=p^{\alpha-\delta}+1$ (positive type) or $r=p^{\alpha-\delta}-1$
(negative type). The integers $\alpha$, $\beta$, $\delta$, $\epsilon$ satisfy
$\alpha,\beta>0$ and $\delta,\epsilon$ nonnegative, furthermore
$\delta\leq{\rm min}\\{\alpha-1,\beta\\}$ and $\delta+\epsilon\leq\alpha$.
When $G$ has negative type, only $\epsilon=0$ or $1$ occur. For $p$ odd

$G\cong G_{p}(\alpha,\beta,\epsilon,\delta,+).$

In other words, the negative type only occurs when $p=2$; when $p$ is odd,
only the positive type occurs. Metacyclic $2$-groups can be of either positive
type or negative type. We note that dihedral, semi-dihedral and generalized
quaternion groups are all of negative type.

If $p=2$, then in addition $\alpha-\delta>1$ and

$G\cong G_{2}(\alpha,\beta,\epsilon,\delta,+)\;\;{\rm or}\;\;G\cong
G_{2}(\alpha,\beta,\epsilon,\delta,-).$

Note, the above presentation does not guarantee nonisomorphic groups for
different parameters (see [2]). However, the parameters do determine some
structural information about $G$. For example, $|G|=p^{\alpha+\beta}$ and
$G^{\prime}=\langle x^{p^{\alpha-\delta}}\rangle$ if $G$ is of positive type
and $G^{\prime}=\langle x^{2}\rangle$ if $G$ is of negative type. All elements
of $G$ can be written as $y^{b}x^{a}$ for some integers $a$ and $b$. Also if
$G$ is of positive type then $Z(G)=\langle
x^{p^{\delta}},y^{p^{\delta}}\rangle$ and $|Z(G)|=p^{\alpha+\beta-2\delta}$,
if $G$ is of negative type $Z(G)=\langle
x^{2^{\alpha-1}},y^{2^{\max\\{1,\delta\\}}}\rangle$, [2, Prop. 2.5]. Note that
if $G$ is of positive type and $\delta=0$, then $G$ will be abelian.

As we mentioned above, the dihedral groups, the generalized quaternion groups,
and the semi-dihedral groups are the only $p$-groups $G$ that satisfy
$\eta(G)=3$. These are also precisely the $2$-groups of maximal class. We have
also mentioned that they are metacyclic. In terms of our notation, the
dihedral groups are $G_{2}(\alpha,1,0,0,-)$, the generalized quaternion groups
are $G_{2}(\alpha,1,1,0,-)$, and the semi-dihedral groups are
$G_{2}(\alpha,1,0,1,-)$.

For Lemmas 3.1 and 3.3, we are writing
$G_{p}(\alpha,\beta,\epsilon,\delta,\pm)$ as
$G_{p}(\alpha,\beta,\epsilon,\delta,\gamma)$ where we take $\gamma=+$ when $G$
is of positive type and $\gamma=-$ when $G$ is of negative type. We consider
quotients of $G$. Note that this lemma would not be well defined if $\delta=0$
and would not say anything if $\delta=1$.

###### Lemma 3.1

Suppose $G$ is $G_{p}(\alpha,\beta,\epsilon,\delta,\gamma)$ with $\delta\geq
2$. Then $N=\langle x^{p^{\alpha-\delta+1}}\rangle$ is a normal subgroup of
$G$ and $G/N$ is isomorphic to

$G_{p}(\alpha-\delta+1,\beta,(\epsilon-\delta+1)^{*},1,\gamma)$

where $(\epsilon-\delta+1)^{*}=\epsilon-\delta+1$ when $\epsilon\geq\delta-1$
and $(\epsilon-\delta+1)^{*}=0$ when $\epsilon<\delta-1$.

Proof. Set $Z=\langle x^{p^{\alpha-1}}\rangle\leq Z(G)$. We first prove that
$G/Z$ is isomorphic to $G_{p}(\alpha-1,\beta,\epsilon-1,\delta-1,\gamma)$ when
$\epsilon\geq 1$ and $G_{p}(\alpha-1,\beta,0,\delta-1,\gamma)$ when
$\epsilon=0$. We know that $G/Z=\langle xZ,yZ\rangle$ where $xZ$ has order
$p^{\alpha-1}$. Observe that
$(yZ)^{p^{\beta}}=y^{p^{\beta}}Z=x^{p^{\alpha-\epsilon}}Z$. When $\epsilon\geq
1$, we have

$x^{p^{\alpha-\epsilon}}Z=x^{p^{(\alpha-1)-(\epsilon-1)}}Z$

and when $\epsilon=0$, we have

$x^{p^{\alpha-\epsilon}}Z=x^{p^{\alpha}}Z=Z.$

Also,

$(xZ)^{yZ}=x^{y}Z=x^{p^{\alpha-\delta}+\gamma}Z=x^{p^{(\alpha-1)-(\delta-1)}+\gamma}Z.$

Hence, $G/Z$ satisfies the hypotheses for
$G_{p}(\alpha-1,\beta,\epsilon-1,\delta-1,\gamma)$ when $\epsilon\geq 1$ and
$G_{p}(\alpha-1,\beta,0,\delta-1,\gamma)$ when $\epsilon=0$.

We know that $X=\langle x\rangle$ is a cyclic, normal subgroup of $G$. Observe
that $N$ is contained in $X$ and so is characteristic. This implies that $N$
is normal in $G$. Observe that $Z\leq N$ and we have shown that $G/Z\cong
G_{p}(\alpha-1,\beta,\epsilon-1,\delta-1,\gamma)$ or
$G_{p}(\alpha-1,\beta,0,\delta-1,\gamma)$. If $\delta=2$, then $N=Z$, and we
have the desired result. Otherwise, we have $\delta\geq 3$. Using induction,
we have $G/N\cong(G/Z)/(N/Z)$ is isomorphic to either

$G_{p}((\alpha-1)-(\delta-1)+1,\beta,(\epsilon-1)-(\delta-1)+1,1,\gamma)\cong
G_{p}(\ \alpha-\delta+1,\beta,\epsilon-\delta+1,1,\gamma)$

or

$G_{p}((\alpha-1)-(\delta-1)+1,\beta,0,1,\gamma)\cong
G_{p}(\alpha-\delta+1,\beta,0,1,\gamma).~{}~{}\Box$

We consider the metacyclic groups of positive type and use Theorem 2.1 and
Lemma 2.3. Thus, we first analyze $G/G^{\prime}$.

###### Lemma 3.2

Suppose $G=G_{p}(\alpha,\beta,\epsilon,\delta,+)$.

  1. (i)

If $\delta\geq\epsilon$ or $\delta<\epsilon$ and $\alpha\geq\beta+\epsilon$,
then $G/G^{\prime}=C_{p^{\alpha-\delta}}\times C_{p^{\beta}}$.

  2. (ii)

If $\delta<\epsilon$ and $\alpha<\beta+\epsilon$, then
$G/G^{\prime}=C_{p^{\alpha-\epsilon}}\times C_{p^{\beta+\epsilon-\delta}}$.

Proof. Now $G^{\prime}=\langle x^{p^{\alpha-\delta}}\rangle$, so
$|G^{\prime}|=p^{\delta}$. Also $|G|=p^{\alpha+\beta}$, so
$|G:G^{\prime}|=p^{\alpha+\beta-\delta}.$

If $\delta\geq\epsilon$, then $\langle y\rangle\cap G^{\prime}=\langle
x^{p^{\alpha-\epsilon}}\rangle=\langle x\rangle\cap\langle y\rangle$. We see
that $xG^{\prime}$ has order $p^{\alpha-\delta}$, and $yG^{\prime}$ has order
$p^{\beta}$ and $G/G^{\prime}=\langle xG^{\prime}\rangle\times\langle
yG^{\prime}\rangle$ yielding the desired result.

Now suppose $\delta<\epsilon$. In this case, we see that $G^{\prime}<\langle
x^{p^{\alpha-\epsilon}}\rangle=\langle x\rangle\cap\langle y\rangle$. We see
that $xG^{\prime}$ has order $p^{\alpha-\delta}$ and $yG^{\prime}$ has order
$p^{\beta+\epsilon-\delta}$. Since $G^{\prime}<\langle x\rangle\cap\langle
y\rangle$, we do not have that $G/G^{\prime}$ is a direct product of $\langle
xG^{\prime}\rangle$ and $\langle yG^{\prime}\rangle$. We see that
$G/G^{\prime}$ is abelian and generated by $xG^{\prime}$ and $yG^{\prime}$, so
every element of $G/G^{\prime}$ has order $\leq{\rm
max}\\{p^{\alpha-\delta},p^{\beta+\epsilon-\delta}\\}$. If
$\alpha\geq\beta+\epsilon$, then $\alpha-\delta\geq\beta+\epsilon-\delta$. In
this case, $xG^{\prime}$ has the largest order of any element in
$G/G^{\prime}$, and so we get $G/G^{\prime}=C_{p^{\alpha-\delta}}\times
C_{p^{\beta}}$ since $|G/G^{\prime}|=p^{\alpha+\beta-\delta}$. On the other
hand, if $\alpha<\beta+\epsilon$, then $\alpha-\delta<\beta+\epsilon-\delta$.
In this case, $yG^{\prime}$ has the largest order of any element in
$G/G^{\prime}$ and we get $G/G^{\prime}=C_{p^{\alpha-\epsilon}}\times
C_{p^{\beta+\epsilon-\delta}}$. $\Box$

Given an element $g\in G$, we write ${\rm cl}(g)$ to denote the conjugacy
class of $g$ in $G$.

###### Lemma 3.3

Let $G=G_{p}(\alpha,\beta,\epsilon,\delta,\gamma)$. If $g=y^{pl+a}x^{m}$ for
integers $l$, $m$, and $a$ so that $a\in\\{1,\dots,p-1\\}$, then ${\rm
cl}(g)=gG^{\prime}$.

Proof. We first claim that $G=\langle x,g\rangle$. We know that $G=\langle
x,y\rangle$. Obviously, $\langle x,g\rangle\leq G$. Observe that
$y^{pl+a}=gx^{-m}\in\langle x,g\rangle$. Since the order of $y$ is a power of
$p$, this implies that $y\in\langle x,g\rangle$. We conclude that $G=\langle
x,y\rangle\leq\langle x,g\rangle\leq G$. This proves the claim.

Because $\langle x\rangle$ is normal in $G$, we obtain $G=\langle
x\rangle\langle g\rangle$. Observe that $\langle g\rangle\leq C_{G}(g)$. By
Dedekind’s lemma (see Lemma X.3 on page 328 of [6]), it follows that
$C_{G}(g)=(C_{G}(g)\cap\langle x\rangle)\langle g\rangle=C_{\langle
x\rangle}(g)\langle g\rangle$. Since $x$ centralizes $x^{m}$, we have

$C_{\langle x\rangle}(g)=C_{\langle x\rangle}(y^{pl+a}x^{m})=C_{\langle
x\rangle}(y^{pl+a})=C_{\langle x\rangle}(y)=\langle x^{p^{t}}\rangle,$

where $t=\delta$ if $\gamma=+$ and $t=\alpha-1$ when $\gamma=-$. We see that
$C_{G}(g)=\langle g,x^{p^{t}}\rangle$. We deduce that

$|G:C_{G}(g)|=|\langle x\rangle:\langle x^{p^{t}}\rangle|=p^{t}=|G^{\prime}|.$

Since ${\rm cl}(g)\subseteq gG^{\prime}$, we conclude that ${\rm
cl}(g)=gG^{\prime}$. $\Box$

Given a group $G$ and a prime $p$, we define $G^{p}=\langle
G^{\\{p\\}}\rangle$. I.e., $G^{p}$ is the subgroup generated by $G^{\\{p\\}}$.
In a similar fashion, we define $G^{4}=\langle g^{4}\mid g\in G\rangle$.
Following the literature, we say that a finite $p$-group $G$ is powerful if
(i) $G^{\prime}\leq G^{p}$ when $p$ is odd and (ii) $G^{\prime}\leq G^{4}$
when $p=2$. If $G$ is a powerful $p$-group, then it is known that
$G^{p}=G^{\\{p\\}}$, i.e. the set of $p$-powers of elements of $G$ is equal to
the subgroup the $p$-powers generate. (See Section 2 of [5] and in particular
Propostion 2.6 of that citation.)

We claim that metacyclic $p$-groups of positive type are powerful. Let $G$ be
$G_{p}(\alpha,\beta,\epsilon,\delta,+)$, then $G^{\prime}=\langle
x^{p^{\alpha-\delta}}\rangle.$ As $\alpha-\delta\geq 1$ it follows immediately
that $G$ is powerful when $p$ is odd. For $p=2$, we note that
$\alpha-\delta\geq 2$ so again we have that $G$ is powerful.

When $G$ is of positive type, we extend Lemma 3.3.

###### Lemma 3.4

Let $G=G_{p}(\alpha,\beta,\epsilon,\delta,+)$ and $g\in G\setminus
G^{\\{p\\}}$. Then ${\rm cl}(g)=gG^{\prime}$.

Proof. Let $g\in G$ then $g=y^{n}x^{m}$ for some integers $n$ and $m$. As $G$
is powerful, it follows that if $g\in G\setminus G^{\\{p\\}}$, then $g\not\in
G^{p}$, and thus, one of $n$ and $m$ is not divisible by $p$. When $n$ is not
divisible by $p$, we obtain the conclusion by Lemma 3.3.

We now suppose that $g=y^{n}x^{m}$ where $m$ is not divisible by $p$. We want
to prove that ${\rm cl}(g)=gG^{\prime}$. We know that ${\rm cl}(g)\subseteq
gG^{\prime}$. It suffices to prove that $|{\rm
cl}(g)|\geq|gG^{\prime}|=|G^{\prime}|=p^{\delta}$. On the other hand, we know
that $y$ acts as an automorphism of order $p^{\delta}$ on $\langle x\rangle$,
so $x$ has $p^{\delta}$ distinct images under powers of $y$. Thus, if $1\leq
a,b\leq p^{\delta}$, then $x^{y^{a}}=x^{y^{b}}$ if and only if $a=b$. Since
$m$ is coprime to $p$, we see that $(x^{y^{a}})^{m}=(x^{y^{b}})^{m}$ if and
only if $a=b$. Hence, we have that $g^{y^{a}}=g^{y^{b}}$ if and only if
$(y^{n}x^{m})^{y^{a}}=(y^{n}x^{m})^{y^{b}}$ and this occurs if and only if
$a=b$. We deduce that $g$ has at least $p^{\delta}$ distinct conjugates under
$\langle y\rangle$ and so $|{\rm cl}(g)|\geq p^{\delta}$ as desired. This
proves the lemma. $\Box$

We now prove that if $G$ is metacyclic of positive type, then
$\eta(G)=\eta(G/G^{\prime})$. Combining this fact with Lemmas 2.3 and 3.2, we
are able to compute $\eta(G)$ for all primes $p$.

###### Corollary 3.5

Suppose $G$ is $G_{p}(\alpha,\beta,\epsilon,\delta,+)$. Then
$\eta(G)=\eta(G/G^{\prime})$.

Proof. As $G$ is powerful, by Theorem 2.1, we need to show that for all $g\in
G\setminus G^{\\{p\\}}$ every element of $gG^{\prime}$ is conjugate to a
generator of $\langle g\rangle$, this follows from Lemma 3.4. $\Box$

For the record, we explicitly record the value of $\eta(G)$ when $G$ is a
metacyclic group of positive type.

###### Corollary 3.6

Suppose $G$ is $G_{p}(\alpha,\beta,\epsilon,\delta,+)$.

  1. (i)

If $\delta\geq\epsilon$ or $\delta<\epsilon$ and $\alpha\geq\beta+\epsilon$,
then $\eta(G)=g_{p}(\alpha-\delta,\beta)$.

    1. (a)

If $\beta\leq\alpha-\delta$, then
$\eta(G)=p^{\beta-1}((\alpha-\delta-\beta)(p-1)+p+1)$.

    2. (b)

If $\beta>\alpha-\delta$, then
$\eta(G)=p^{\alpha-\delta-1}((\beta-\alpha+\delta)(p-1)+p+1)$.

  2. (ii)

If $\delta<\epsilon$ and $\alpha<\beta+\epsilon$, then
$\eta(G)=g_{p}(\alpha-\epsilon,\beta+\epsilon-\delta)=p^{\alpha-\epsilon-1}((\beta-\alpha+2\epsilon-\delta)(p-1)+p+1)$.

Proof. Using Corollary 3.5, we have $\eta(G)=\eta(G/G^{\prime})$. If
$\delta\geq\epsilon$ or $\delta<\epsilon$ and $\alpha\geq\beta+\epsilon$, then
in view of Lemma 3.2, we see that $G/G^{\prime}=C_{p^{\alpha-\delta}}\times
C_{p^{\beta}}$ and $\eta(G)=g_{p}(\alpha-\delta,\beta)$. The remainder of (i)
follows from the definition of $g_{p}$. Suppose $\delta<\epsilon$ and
$\alpha<\beta+\epsilon$. Applying Lemma 3.2, we see that
$G/G^{\prime}=C_{p^{\alpha-\epsilon}}\times C_{p^{\beta+\epsilon-\delta}}$.
Observe that $\alpha<\beta+\epsilon$ yields
$\alpha-\epsilon<\beta<\beta+\epsilon-\delta$ as we are assuming
$\delta<\epsilon$. In light of the definition of $g_{p}$, we obtain conclusion
(ii). $\Box$

When $G$ is metacyclic of positive type, we show that
$\eta(G)\geq\alpha+\beta$.

###### Corollary 3.7

If $G$ is $G_{p}(\alpha,\beta,\epsilon,\delta,+)$, then
$\eta(G)\geq\alpha+\beta$.

Proof. We consider separately the cases given in Corollary 3.6. We use the
fact that $2^{\beta-1}\geq\beta$ for $\beta$ a positive integer. First,
(i)(a), where $\alpha-\delta\geq\beta$,

$\displaystyle\eta(G)$ $\displaystyle=$ $\displaystyle
p^{\beta-1}((\alpha-\delta-\beta)(p-1)+p+1)$ $\displaystyle\geq$
$\displaystyle 2^{\beta-1}(\alpha-\delta-\beta+3)$ $\displaystyle\geq$
$\displaystyle\alpha-\delta-\beta+3\beta$ $\displaystyle\geq$
$\displaystyle\alpha+\beta$

since $\beta\geq\delta$.

Now, case (i)(b), so $\alpha-\delta<\beta$. First assume $\alpha-\delta>1$,
then

$\displaystyle\eta(G)$ $\displaystyle=$ $\displaystyle
p^{\alpha-\delta-1}((\beta-\alpha+\delta)(p-1)+p+1)$ $\displaystyle\geq$
$\displaystyle 2^{\alpha-\delta-1}(\beta-\alpha+\delta+3)$ $\displaystyle\geq$
$\displaystyle 2(\beta-\alpha+\delta)+3(\alpha-\delta)$ $\displaystyle=$
$\displaystyle 2\beta+(\alpha-\delta)$ $\displaystyle\geq$
$\displaystyle\beta+\alpha+(\beta-\delta)$ $\displaystyle\geq$
$\displaystyle\beta+\alpha$

since $\beta\geq\delta$. If $\alpha-\delta=1$ then $p\geq 3$, also note
$\alpha=1+\delta\leq 1+\beta$. So, we have

$\eta(G)\geq 2(\beta-\alpha+\delta)+4=2\beta+2>\beta+\alpha.$

Case (ii) follows similarly to (i)(a), we have
$\alpha-\epsilon\leq\beta+\epsilon-\delta$,

$\displaystyle\eta(G)$ $\displaystyle=$ $\displaystyle
p^{\alpha-\epsilon-1}((\beta-\alpha+2\epsilon-\delta)(p-1)+p+1)$
$\displaystyle\geq$ $\displaystyle
2^{\alpha-\epsilon-1}(\beta-\alpha+2\epsilon-\delta+3)$ $\displaystyle\geq$
$\displaystyle\beta-\alpha+2\epsilon-\delta+3(\alpha-\epsilon)$
$\displaystyle=$ $\displaystyle\beta+\alpha+(\alpha-\epsilon-\delta)$
$\displaystyle\geq$ $\displaystyle\beta+\alpha$

since $\alpha\geq\delta+\epsilon$.$\Box$

##  4 Metacyclic Groups of Negative Type

The goal of this section is to compute $\eta$ when $G$ is a metacyclic group
of negative type. We begin by looking at quotients of $G$. We begin with a
preliminary lemma that is useful in understanding the quotients.

Using the notation of Section 3 and applying Theorem 2.1, we have that if
$G=G_{2}(\alpha,\beta,\epsilon,\delta,-)$ with $\delta\geq 1$ and $N=\langle
x^{2^{\alpha-\delta+1}}\rangle$, then $\eta(G)\geq\eta(G/N)$. We now show that
in fact this is an equality. We remind the reader that $\alpha-\delta\geq 2$
when $p=2$.

We now prove the promised equality between $\eta(G)$ and $\eta(G/N)$.

###### Theorem 4.1

Let $G=G_{2}(\alpha,\beta,\epsilon,\delta,-)$ where $\delta\geq 1$. Then
$\eta(G)=\eta(G/N)$ where $N=\langle x^{2^{\alpha-\delta+1}}\rangle$.

Proof. Note that $N$ does not make sense if $\delta=0$; so that it is why we
assume $\delta\geq 1$. Also, if $\delta=1$, then $N=1$; so the conclusion is
trivial in this case. Hence, we will assume $\delta\geq 2$.

We first prove that $\eta(G)=\eta(G/Z)$ where $Z=\langle
x^{2^{\alpha-1}}\rangle$. Recall from Theorem 2.1 that to prove
$\eta(G)=\eta(G/Z)$, we need to prove that $Z\subseteq G^{\\{2\\}}$ and if
$g\in G\setminus G^{\\{2\\}}$, then every element of $gZ$ is conjugate to a
generator of $\langle g\rangle$. Observe that $Z\subseteq G^{\\{2\\}}$. Since
$x^{2^{\alpha-1}}$ is the only nonidentity element of $Z$, it suffices to
prove that if $g\not\in G^{\\{2\\}}$, then $\langle g\rangle$ and $\langle
gx^{2^{\alpha-1}}\rangle$ are conjugate. We know from [2] that
$G^{\prime}=\langle x^{2}\rangle$.

We prove the claim by working by induction on $\delta$. We begin with the case
that $\delta=2$. We know that $x^{y}=x^{2^{\alpha-2}-1}$. It follows that

$(x^{2})^{y}=(x^{y})^{2}=(x^{2^{\alpha-2}-1})^{2}=x^{2^{\alpha-1}-2}=(x^{-2})x^{2^{\alpha-1}}.$

Observe that this yields that $(x^{-2})^{y}=x^{2}x^{2^{\alpha-1}}$. Using this
fact and the observation that $x^{2^{\alpha-1}}$ is central, we then have

$(x^{2})^{y^{2}}=(x^{-2}x^{2^{\alpha-1}})^{y}=x^{2}x^{2^{\alpha-1}}x^{2^{\alpha-1}}=x^{2}.$

It follows that $x^{2}$ and $y^{2}$ commute. Let $A=\langle
x^{2},y^{2}\rangle$, and observe that $G^{\prime}\leq A$, so $A$ is a normal,
abelian subgroup of $G$.

We know that every element of $G$ has the form $y^{k}x^{m}$ where $0\leq k\leq
2^{\beta}-1$ and $0\leq m\leq 2^{\alpha}-1$ are integers. Notice that if $4$
divides both $k$ and $m$, then $g\in A^{\\{2\\}}\subseteq G^{\\{2\\}}$. Also,
$x^{2},y^{2}\in G^{\\{2\\}}$.

If $g=y^{2l+1}x^{m}$ for integers $l$ and $m$, then we can appeal to Lemma 3.3
to see that $g$ is conjugate to $gx^{2^{\alpha-1}}$ and so, $\langle g\rangle$
and $\langle gx^{2^{\alpha-1}}\rangle$ are conjugate, as desired.

Since $x^{y}=x^{2^{\alpha-2}-1}$, we have
$x^{y^{2}}=x^{2^{2\alpha-4}-2^{\alpha-1}+1}$. Since $\delta\geq 2$, we know
that $\alpha\geq 4$ (this is using the fact that $\alpha-\delta\geq 2$), so
$2\alpha-4\geq\alpha$. Hence, we have $x^{y^{2}}=x^{-2^{\alpha-1}+1}$. In
addition, $x^{2^{\alpha-1}}$ has order $2$, so
$x^{-2^{\alpha-1}}=x^{2^{\alpha-1}}$. Thus, we have shown
$x^{y^{2}}=x^{2^{\alpha-1}+1}$.

Suppose now that $g=y^{2l}x^{2h+1}$ for integers $l$ and $h$. From above, we
have

$g^{y^{2}}=(y^{2l}x^{2h+1})^{y^{2}}=y^{2l}(x^{y^{2}})^{2h+1}=y^{2l}(x^{2^{\alpha-1}+1})^{2h+1}=y^{2l}x^{2h+1}x^{2^{\alpha-1}}=gx^{2^{\alpha-1}}.$

We deduce that $\langle g\rangle$ and $\langle gx^{2^{\alpha-1}}\rangle$ are
conjugate, as desired.

We have shown that $x^{y^{2}}=xx^{2^{\alpha-1}}$. This implies that
$x^{-1}y^{-2}x=x^{2^{\alpha-1}}y^{-2}$. Inverting, we obtain
$(y^{2})^{x}=y^{2}x^{{2^{\alpha}-1}}$. Now, suppose that $g=y^{2l}x^{2h}$. We
can assume from above that either $l$ is odd or $h$ is odd. Assume first that
$l$ is odd. We have

$g^{x}=(y^{2l}x^{2h})^{x}=((y^{2})^{x})^{l}x^{2h}=(y^{2}x^{2^{\alpha-1}})^{l}x^{2h}=y^{2l}x^{2h}x^{2^{\alpha-1}}=gx^{2^{\alpha}-1}.$

We obtain $\langle g\rangle$ and $\langle gx^{2^{\alpha-1}}\rangle$ are
conjugate, as desired.

We are left with the case that $g=y^{4l}x^{2(2h+1)}$ for integers $h$ and $l$.
We claim that $g\in G^{\\{2\\}}$. Notice that there is an integer $k$ so that
$\langle g\rangle=\langle y^{4k}x^{2}\rangle$ and that $g\in G^{\\{2\\}}$ if
and only if $y^{4k}x^{2}\in G^{\\{2\\}}$. We show that $y^{4k}x^{2}\in
G^{\\{2\\}}$. We have $x^{y^{2}}=xx^{2^{\alpha-1}}$. It follows that
$xy^{2}=y^{2}xx^{2^{\alpha-1}}$ and

$(y^{2k}x)^{2}=y^{2k}xy^{2k}x=y^{2k}y^{2k}xx^{2^{\alpha-1}k}x=y^{4k}x^{2}x^{2^{\alpha-1}k}.$

When $k$ is even, we see that $(y^{2k}x)^{2}=y^{4k}x^{2}$. Now assume that $k$
is odd. We have

$\displaystyle(y^{2k}xx^{2^{\alpha-2}})^{2}$ $\displaystyle=$ $\displaystyle
y^{2k}xx^{2^{\alpha-2}}y^{2k}xx^{2^{\alpha-2}}=y^{2k}y^{2k}xx^{2^{\alpha-1}k}xx^{2^{\alpha-2}2}$
$\displaystyle=$ $\displaystyle y^{4k}x^{2}x^{2^{\alpha-1}(k+1)}=y^{4k}x^{2}.$

Note that we are using the fact that $x^{2^{\alpha-2}}$ commutes with both $x$
and $y^{2}$ here. Thus, this yields $g\in G^{\\{2\\}}$. We conclude for all
elements $g\in G\setminus G^{\\{2\\}}$ that $g$ and $gx^{2^{\alpha-1}}$ are
conjugate and we have proved that $\eta(G)=\eta(G/Z)$ when $\delta=2$.

We now assume that $\delta>2$. Let $M=\langle x^{2},y\rangle$. Since
$x^{y}=x^{2^{\alpha-\delta}-1}$, we see that
$(x^{2})^{y}=(x^{2^{\alpha-\delta}-1})^{2}=(x^{2})^{2^{(\alpha-1)-(\delta-1)}-1}$.
Also,
$y^{2^{\beta}}=x^{2^{\alpha-\epsilon}}=(x^{2})^{2^{(\alpha-1)-\epsilon}}$.
Observe that $(x^{2})^{2^{(\alpha-1)-1}}=x^{2^{\alpha-1}}$. We conclude that
$M=G_{2}(\alpha-1,\beta,\epsilon,\delta-1,-)$. Let $g\in G\setminus
G^{\\{2\\}}$. If $g\in M$, then $g\in M\setminus M^{\\{2\\}}$. By induction,
we have that $g$ is conjugate to $g(x^{2})^{2^{(\alpha-1)-1}}$, and so, $g$
and $gx^{2^{\alpha-1}}$ are conjugate. Thus, we may assume that $g\not\in M$.
This implies that $g=y^{l}x^{2m+1}$ for integers $l$ and $m$. We know that $y$
induces an automorphism of $\langle x\rangle$ of order $2^{\delta}$. It
follows that $y^{2^{\delta-1}}$ induces an automorphism of $\langle x\rangle$
of order $2$. Since $\delta\geq 3$, we know that this automorphism is a
square. It is not difficult to see that $x\mapsto xx^{2^{\alpha-1}}$ is the
unique automorphism of $\langle x\rangle$ that has order $2$ and is a square.
Hence, we see that $x^{y^{2^{\delta-1}}}=xx^{2^{\alpha-1}}$. We conclude that
$g^{y^{2^{\delta-1}}}=(y^{l}x^{2m+1})^{y^{2^{\delta-1}}}=y^{l}(x^{y^{2^{\delta-1}}})^{2m+1}=y^{l}(xx^{2^{\alpha-1}})^{2m+1}=y^{l}x^{2m+1}x^{2^{\alpha-1}}=gx^{2^{\alpha-1}}$.
This completes the proof of the claim that $\eta(G)=\eta(G/Z)$.

We now work to prove $\eta(G)=\eta(G/N)$. We work by induction on $\delta$. If
$\delta=2$, then $N=Z$, and the above claim yields the result. We assume that
$\delta\geq 3$. We have that $\eta(G)=\eta(G/Z)$. By induction,
$\eta(G/Z)=\eta((G/Z)/(N/Z))$, and the First Isomorphism Theorem implies that
$G/N\cong(G/Z)/(N/Z)$, so $\eta(G/N)=\eta((G/Z)/(N/Z))$, and we have the
desired equality. $\Box$

In light of Theorem 4.1 and Lemma 3.1, we see that if we can compute $\eta$
for $G_{2}(\alpha,\beta,\epsilon,\delta,-)$ when $\delta=0,1$, then we can
compute $\eta$ for all metacyclic $2$-groups of negative type. There are a
number of cases to consider when $\delta=0$ or $1$, and then using these
cases, we will compute $\eta$ when $\delta\geq 2$. Recall that the dihedral
$2$-groups are the groups of the form $G_{2}(\alpha,1,0,0,-)$, the generalized
quaternion $2$-groups are of the form $G_{2}(\alpha,1,1,0,-)$, and the semi-
dihedral groups are of the form $G_{2}(\alpha,1,0,1,-)$. Also, it is known
that $G_{2}(\alpha,\beta,1,0,-)$ and $G_{2}(\alpha,\beta,1,1,-)$ are
isomorphic for all $\alpha\geq 3$ and $\beta\geq 2$. Since $\delta\leq\beta$,
it follows that dihedral, generalized quaternion, and semi-dihedral are the
only groups of negative type where $\beta=1$.

Thus, we need to analyze the negative metacyclic $2$-groups of type

$G_{2}(\alpha,\beta,\epsilon,\delta,-)$

with $\beta\geq 2$. We recall a few facts about the classification of such
groups. In particular, for negative type $\epsilon$ is either $0$ or $1$ only.
Also the parameters satisfy: $\alpha\geq\delta+2$ and $\beta\geq\delta$ when
$\epsilon=0$ and $\beta\geq\delta+1$ when $\epsilon=1$.

When $\delta=0$ or $1$, there is a particular abelian normal subgroup $M$ of
$G$. For this subgroup $M$, we determine which maximal cyclic subgroups of $M$
are maximal in $G$ and how many maximal cyclic subgroups of $G$ lie outside of
$M$. This yields the following result. Recall that $\eta^{*}(M)$ is the number
of $G$-orbits on the $M$-conjugacy classes of maximal cyclic subgroups of $M$.

###### Proposition 4.2

Suppose $G$ is $G_{2}(\alpha,\beta,\epsilon,\delta,-)$ where $\delta=0$ or $1$
and $\beta\geq 2$. Let $M=\langle x,y^{2}\rangle$. Then $M$ is a normal
abelian subgroup of $G$ and the following holds:

  1. (i)

If $\delta=0$, then $\eta(G)=\eta^{*}(M)+1$ and every maximal cyclic subgroup
of $M$ is maximal cyclic in $G$ except $\langle y^{2}\rangle$.

  2. (ii)

If $\delta=1$, then $\eta(G)=\eta^{*}(M)$ and every maximal cyclic subgroup of
$M$ is maximal cyclic in $G$ except $\langle y^{2}\rangle$ and $\langle
y^{2}x^{2^{\alpha-1}}\rangle$.

Proof. As $M$ is a subgroup of index $2$ in $G$ it follows that $M$ is normal
in $G$. Let $Y=\langle y^{2}\rangle$. Observe that $y^{2}$ centralizes
$\langle x\rangle$ and is obviously central in $\langle y\rangle$; so
$Y=\langle y^{2}\rangle$ is central in $G$. Now, $M$ is central-by-cyclic, so
$M$ is abelian.

We now prove that there are exactly two conjugacy classes of maximal cyclic
subgroups of $G$ outside of $M$. Since $\langle x\rangle$ is normal in $G$ and
$G=\langle x\rangle\langle y\rangle=\langle x\rangle\langle xy\rangle$, we see
that $C_{G}(\langle y\rangle)=C_{\langle x\rangle}(\langle y\rangle)\langle
y\rangle=\langle x^{2^{\alpha-1}}\rangle\langle y\rangle$ and $C_{G}(\langle
xy\rangle)=\langle x^{2^{\alpha-1}}\rangle\langle xy\rangle$. It follows that
both $\langle y\rangle$ and $\langle xy\rangle$ lie in conjugacy classes of
size $|\langle x\rangle:\langle x^{2^{\alpha-1}}\rangle|=2^{\alpha-1}$. It is
not difficult to see now that every cyclic subgroup of $G$ outside of $M$ is
conjugate to either $\langle y\rangle$ or $\langle xy\rangle$.

$(i)$ For $\delta=0$ we show that every maximal cyclic subgroup of $M$ is a
maximal cyclic subgroup of $G$ except $\langle y^{2}\rangle$ which lies in
exactly $2$ different conjugacy classes of maximal cyclic subgroups of $G$,
namely $\langle y\rangle$ and $\langle xy\rangle$.

Observe that $yY$ acts on $M/Y$ inverting every element. Thus, $M/Y$ is a
cyclic subgroup of index $2$ in $G/Y$. We have $(yY)^{2}=Y$, so $G/Y$ is a
dihedral group. It follows that if $g\in G\setminus M$, then $(gY)^{2}=Y$ and
so, $g^{2}\in Y$. Hence, $Y$ is the only maximal cyclic subgroup of $M$ that
is not maximal cyclic in $G$. Notice that $Y\leq\langle y\rangle$. Also, we
know that $\langle yY\rangle$ and $\langle xyY\rangle$ are in different
conjugacy classes of subgroups of $G/Y$, so $\langle y\rangle$ and $\langle
xy\rangle$ are in different conjugacy classes of $G$. Since $x^{y}=x^{-1}$, so
$xy=yx^{-1}$. It follows that $(yx)^{2}=yxyx=y(yx^{-1})x=y^{2}$.

$(ii)$ For $\delta=1$ we show that the only maximal cyclic subgroups of $M$
that are not maximal in $G$ are $\langle y^{2}\rangle$ and $\langle
y^{2}x^{2^{\alpha-1}}\rangle$. Again there are exactly $2$ different conjugacy
classes of maximal cyclic subgroups outside of $M$ given by $\langle y\rangle$
and $\langle xy\rangle$. Note that $\langle y\rangle$ contains $\langle
y^{2}\rangle$ and $\langle xy\rangle$ contains $\langle
y^{2}x^{2^{\alpha-1}}\rangle$.

Note that $M/Y$ is cyclic in $G/Y$ of order $2^{\alpha}$. Also, $(yY)^{2}=Y$
and $(xY)^{yY}=x^{2^{\alpha-1}-1}Y=(xY)^{2^{\alpha-1}-1}$. It follows that
$G/Y$ is isomorphic to a semi-dihedral group. Let $Z=\langle
x^{2^{\alpha-1}},Y\rangle$, and observe that $Z/Y=Z(G/Y)$. Notice that if
$g\in G\setminus M$, then $(gY)^{2}\in Z/Y$. This implies that $g^{2}\in Z$.
Observe that $\langle y^{2}\rangle$ and $\langle y^{2}x^{2^{\alpha-1}}\rangle$
are central (and hence normal) in $G$. It follows that the square of any
conjugate of $y$ will be $y^{2}$. Since $\delta=1$, we have
$x^{y}=x^{2^{\alpha-1}-1}$, so $xy=yx^{2^{\alpha-1}-1}$. We have
$(yx)^{2}=yxyx=y(yx^{2^{\alpha-1}-1})x=y^{2}x^{2^{\alpha-1}}$. This implies
that the square of any conjugate of $xy$ will be $y^{2}x^{2^{\alpha-1}}$.
Hence, any other subgroup of $M$ that is maximal cyclic in $M$ will be maximal
cyclic in $G$. $\Box$

We now work to compute $\eta$ for the groups with negative type and $\delta$
equal to $0$ or $1$. We will first handle the case when $\epsilon=0$ and
$\beta=2$. For the following lemma recall that $\alpha\geq\delta+2$ when
$p=2$, so when $\delta=1$ we must have $\alpha\geq 3$.

###### Lemma 4.3

Suppose $G$ is $G_{2}(\alpha,2,0,\delta,-)$. Then

  1. (i)

$\eta(G)=\alpha+3$ if $\delta=0$ and

  2. (ii)

$\eta(G)=\alpha+2$ if $\delta=1$.

Proof. Following Proposition 4.2, we take $M=\langle x,y^{2}\rangle$; so $M$
is abelian. We have $M\cong C_{2^{\alpha}}\times C_{2}$ and $\eta(M)=\alpha+2$
by Lemma 2.3. We claim that all subgroups of $M$ are normal in $G$. To see
this, note that if $K$ is a subgroup of $M$ then (1) $K$ is a subgroup of
$\langle x\rangle$, (2) $K=\langle x^{a},y^{2}\rangle$ for some integer $1\leq
a\leq 2^{\alpha}-1$ or (3) $K=\langle x^{a}y^{2}\rangle$ for some integer
$1\leq a\leq 2^{\alpha}-1$. When $\delta=0$, we know that $x^{y}=x^{-1}$, so
$(x^{a})^{y}=(x^{a})^{-1}$ and $(x^{a}y^{2})^{y}=(x^{a}y^{2})^{-1}$ for every
integer $a$. When $\delta=1$, we have $(x^{a})^{y}=x^{a(2^{\alpha-1}-1)}$. The
observation is that $\langle x^{a}\rangle=\langle
x^{a(2^{\alpha-1}-1)}\rangle$, $\langle x^{a},y^{2}\rangle=\langle
x^{a(2^{\alpha-1}-1)},y^{2}\rangle$, and $\langle x^{a}y^{2}\rangle=\langle
x^{a(2^{\alpha-1}-1)}y^{2}\rangle$. This proves the claim. Therefore
$\eta^{*}(M)=\eta(M)$ and the result follows from Proposition 4.2. $\Box$.

We continue with the case where $\epsilon=0$. We now consider the case that
$\beta\geq 3$. Recall that $g_{p}(a,b)=p^{(l-1)}((k-l)(p-1)+p+1)$ where $p$ a
prime, $a$ and $b$ are positive integers, and we take $k={\rm max}(a,b)$ and
$l={\rm min}(a,b)$. Recall also that $g_{p}(a,b)=\eta(C_{p^{a}}\times
C_{p^{b}})$. The following can be viewed as an improvement on Proposition
2.2(ii).

###### Theorem 4.4

Suppose $G$ is $G_{2}(\alpha,\beta,0,\delta,-)$ with $\beta\geq 3$. As
previously let $M=\langle x,y^{2}\rangle$. Then the following hold:

  1. 1.

If $\delta=1$, then $\eta(G)=\eta(M)/2+2=g_{2}(\alpha,\beta-1)/2+2$.

  2. 2.

If $\delta=0$, then $\eta(G)=\eta(M)/2+3=g_{2}(\alpha,\beta-1)/2+3$.

Proof. Note that we are assuming $\delta$ is $0$ or $1$. As in Proposition
4.2, we let $M=\langle x,y^{2}\rangle$; so it follows that $M$ is abelian. In
particular, since we are assuming that $\epsilon=0$, we have $M\cong\langle
x\rangle\times\langle y^{2}\rangle=C_{2^{\alpha}}\times C_{2^{\beta-1}}$.
Using Lemma 2.3, we obtain $\eta(M)=g_{2}(\alpha,\beta-1)$. Let $k$ be the
maximum of $\alpha$ and $\beta-1$ and let $l$ be the minimum of $\alpha$ and
$\beta-1$; so that $\eta(M)=g_{2}(\alpha,\beta-1)=2^{l-1}(k-l+3)$. We now work
to prove that $\eta^{*}(M)=g_{2}(\alpha,\beta-1)/2+2$. Once this is done, then
we will have the conclusion via Proposition 4.2.

It is not difficult to see that $\langle x\rangle$, $\langle y^{2}\rangle$,
and $\langle y^{2}x^{2^{\alpha-1}}\rangle$ are maximal cyclic subgroups of $M$
that are normal in $G$. We claim that $\langle y^{2(2^{\beta-2})}x\rangle$ is
a maximal cyclic subgroup of $M$ that is normal in $G$. It is easy to see that
it is maximal cyclic. When $\delta=0$, we see that $(\langle
y^{2(2^{\beta-2})}x\rangle)^{y}=\langle
y^{2(2^{\beta-2})}x^{-1}\rangle=\langle(y^{2(2^{\beta-2})}x)^{-1}\rangle$, and
when $\delta=1$, we have $(\langle y^{2(2^{\beta-2})}x\rangle)^{y}=\langle
y^{2(2^{\beta-2})}x^{2^{\alpha}-1}\rangle=\langle(y^{2(2^{\beta-2})}x)^{2^{\alpha}-1}\rangle$.
This proves that it is normal in $G$.

We will prove that all the other maximal cyclic subgroups of $M$ will be in
conjugacy classes of size $2$ in $G$. Thus,
$\eta^{*}(M)=(\eta(M)-4)/2+4=\eta(M)/2-2+4=g_{2}(\alpha,\beta-1)/2+2$.

Let $C$ be a maximal cyclic subgroup of $M$. It is not difficult to see that
$C$ will be generated by an element of the form $y^{2l}x$ or one of the form
$y^{2}x^{l}$. When $\delta=0$, we have that $(y^{2l}x)^{y}=y^{2l}x^{-1}$ and
$(y^{2}x^{l})^{y}=y^{2}x^{-l}$. For $C$ to be normal, we need this conjugate
to be in $C$. When the generator is $y^{2l}x$, we need
$y^{2l}x^{-1}=(y^{2l}x)^{k}=y^{2lk}x^{k}$ for some integer $k$. This implies
that $y^{2l-2lk}=x^{k+1}$. Since $\epsilon=0$, we have that
$y^{2l-2lk}=x^{k+1}=1$. We see that we must have $2^{\alpha}$ dividing $k+1$
and $2^{\beta}$ must divide $2l(1-k)$. Thus, there is an integer $r$ so that
$k+1=2^{\alpha}r$, and thus, $k=2^{\alpha}r-1$. We obtain that $2^{\beta-1}$
must divide $l(1-(2^{\alpha}r-1))=l(2-2^{\alpha}r)=2l(1-2^{\alpha-1}r)$. Since
we know that $\alpha\geq 2$, this implies that $2^{\beta-2}$ must divide $l$.
It follows that $\langle x\rangle$ and $\langle y^{2^{\beta-1}}x\rangle$ are
the only two maximal cyclic subgroups of $M$ that are normal in $G$ that are
generated by an element of the form $y^{2l}x$ when $\delta=0$.

When the generator is $y^{2}x^{l}$, we need
$y^{2}x^{-l}=(y^{2}x^{l})^{k}=y^{2k}x^{lk}$ for some integer $k$. This implies
that $y^{2-2k}=x^{lk+l}=1$. This implies that $2^{\beta}$ divides $2(1-k)$ and
so, $2^{\beta-1}$ divides $1-k$. Hence, there is an integer $r$ so that
$1-k=r2^{\beta-1}$, and hence, $k=1-r2^{\beta-1}$. We see that $2^{\alpha}$
divides $l(1+k)=l(1+(1-r2^{\beta-1}))=l(2-r2^{\beta-1})=2l(1-r2^{\beta-2})$.
Since $\beta\geq 3$, we deduce that $2^{\alpha-1}$ must divide $l$. It follows
that $\langle y^{2}\rangle$ and $\langle y^{2}x^{2^{\alpha-1}}\rangle$ are the
only maximal cyclic subgroups of $M$ that are normal in $G$ that are generated
by an element of the form $y^{2}x^{l}$ when $\delta=0$. This proves the result
when $\delta=0$.

Now we suppose that $\delta=1$. Recall that $\alpha\geq\delta+2$, so
$\alpha\geq 3$. We have that $(y^{2l}x)^{y}=y^{2l}x^{2^{\alpha-1}-1}$ and
$(y^{2}x^{l})^{y}=y^{2}x^{l(2^{\alpha-1}-1)}$. For $C$ to be normal, we need
this conjugate to be in $C$. Suppose the generator is $y^{2l}x$. We need
$y^{2l}x^{2^{\alpha-1}-1}=(y^{2l}x)^{k}=y^{2lk}x^{k}$ for some integer $k$.
This implies that $y^{2l-2lk}=x^{k-2^{\alpha-1}+1}=1$. We deduce that
$2^{\alpha}$ must divide $k-2^{\alpha-1}+1$, and so, there is an integer $r$
so that $k-2^{\alpha-1}+1=2^{\alpha}r$. We obtain
$k=2^{\alpha}r+2^{\alpha-1}-1$. We have that $2^{\beta}$ divides
$2l(1-k)=2l(1-2^{\alpha}r-2^{\alpha-1}+1)$. It follows that $2^{\beta-2}$
divides $l(1-2^{\alpha-1}r-2^{\alpha-2})$. Since $\alpha\geq 3$, we see that
$2^{\beta-2}$ divides $l$. We conclude that $\langle x\rangle$ and $\langle
y^{2^{\beta-1}}x\rangle$ are the only two maximal cyclic subgroups of $M$ that
are normal in $G$ that are generated by an element of the form $y^{2l}x$ when
$\delta=1$.

When the generator is $y^{2}x^{l}$, we need
$y^{2}x^{l(2^{\alpha-1}-1)}=(y^{2}x^{l})^{k}=y^{2k}x^{lk}$ for some integer
$k$. We see that $y^{2-2k}=x^{lk-l(2^{\alpha-1}-1)}=1$. It follows that
$2^{\beta}$ divides $2(1-k)$, and so, $2^{\beta-1}$ divides $1-k$. There is an
integer $r$ so that $1-k=2^{\beta-1}r$ which yields $k=1-2^{\beta-1}r$. We now
determine that $2^{\alpha}$ divides
$l(k-2^{\alpha-1}+1)=l(1-2^{\beta-1}r-2^{\alpha-1}+1)=2l(1-2^{\beta-2}r-2^{\alpha-2})$.
Since $\alpha\geq 3$ and $\beta\geq 3$, we have that $2^{\alpha-1}$ divides
$l$. We conclude that $\langle y^{2}\rangle$ and $\langle
y^{2}x^{2^{\alpha-1}}\rangle$ are the only maximal cyclic subgroups of $M$
that are normal in $G$ that are generated by an element of the form
$y^{2}x^{l}$ when $\delta=1$. This proves the result when $\delta=1$. $\Box$

In this next corollary, recall that $\delta\leq\beta$, so when $\beta=2$, we
must have $\delta=2$. We are able to use Theorem 4.4 to compute $\eta$ for
groups of negative type where $\delta\geq 2$.

###### Corollary 4.5

Suppose $G$ is $G_{2}(\alpha,\beta,\epsilon,\delta,-)$ with $\delta\geq 2$,
then

  1. 1.

$\eta(G)=\alpha-\delta+3=\alpha+1$ if $\beta=2$.

  2. 2.

$\eta(G)=g_{2}(\alpha-\delta+1,\beta-1)/2+2$ if $\beta\geq 3$.

Proof. By Theorem 4.1, we have that $\eta(G)=\eta(G/N)$ where $N=\langle
x^{2^{\alpha-\delta+1}}\rangle$. Applying Lemma 3.1, we see that $G/N\cong
G_{2}(\alpha-\delta+1,\beta,0,1,-)$. Using Lemma 4.3, we see that
$\eta(G/N)=\alpha-\delta+1+2=\alpha+3-\delta$ when $\beta=2$. Since
$2\leq\delta\leq\beta=2$, we see that $\delta=2$, and so, $\eta(G)=\alpha+1$.
When $\beta\geq 3$, we apply Theorem 4.4 to see that
$\eta(G)=\eta(G/N)=g_{2}(\alpha-\delta+1,\beta-1)/2+2$. $\Box$

We now compute $\eta$ for groups of negative type with $\delta=0$ and
$\epsilon=1$. We first handle the case where $\beta=2$.

###### Lemma 4.6

Suppose $G$ is $G_{2}(\alpha,2,1,0,-)$ then $\eta(G)=\alpha+2$.

Proof. Define $M=\langle x,y^{2}\rangle$. By Proposition 4.2, we know that $M$
is a normal abelian subgroup of $G$. First note that
$(x^{2^{\alpha-2}}y^{2})^{2}=x^{2^{\alpha-1}}y^{4}=x^{2^{\alpha-1}}x^{2^{\alpha-1}}=x^{2^{\alpha}}=1.$
Thus, $M=\langle x\rangle\times\langle x^{2^{\alpha-2}}y^{2}\rangle\cong
C_{2^{\alpha}}\times C_{2}$ and $\eta(M)=\alpha+2$. Consideration of the
maximal cyclic subgroups of $M$ shows that all are normal except
$\langle(1,x^{2^{\alpha-2}}y^{2})\rangle$ and
$\langle(x^{2^{\alpha-1}},x^{2^{\alpha-2}}y^{2})\rangle$ which are conjugate
in $G$ via $y$. To see that these two subgroups are conjugate, observe that
$M$ has three subgroups of order $2$ and that $\langle
x^{2^{\alpha-1}}\rangle=\langle y^{2^{\beta}}\rangle$ is central in $G$ and
that $Z(G)$ is cyclic. Either $y$ normalizes both of the other two subgroups
of order $2$ or it permutes them. However, if $y$ were to normalize them, they
would be normal in $G$ and since they have order $2$, that would imply that
they would be central in $G$. This however would contradict the fact that the
center of $G$ is cyclic. Thus $\eta^{*}(M)=\alpha+1$. The result follows from
Proposition 4.2. $\Box$

We continue with the groups of negative type where $\delta=0$ and
$\epsilon=1$. We next consider $\beta\geq 3$ and $\alpha=2$.

###### Lemma 4.7

Suppose $G$ is $G_{2}(2,\beta,1,0,-)$ with $\beta\geq 3$. Then
$\eta(G)=\beta+2$.

Proof. Define $M=\langle x,y^{2}\rangle$. By Proposition 4.2, we know that $M$
is a normal abelian subgroup of $G$. Note that
$(xy^{2^{\beta-1}})^{2}=x^{2}y^{2^{\beta}}=x^{2}x^{2}=x^{4}=1$. So $M=\langle
xy^{2^{\beta-1}}\rangle\times\langle y^{2}\rangle\cong C_{2}\times
C_{2^{\beta}}$ and $\eta(M)=\beta+2$. Consideration of the maximal cyclic
subgroups of $M$ shows that all are normal except for
$\langle(xy^{2^{\beta-1}},1)\rangle$ and
$\langle(xy^{2^{\beta-1}},y^{2^{\beta}})\rangle$ which are conjugate in $G$
via $y$. The proof that these two subgroups are conjugate is similar to the
proof of Lemma 4.6. In particular, $Z(G)$ is cyclic, $M$ has three subgroups
of order $2$, and if $y$ normalized these two subgroups, then it would
centralize them and contradict the fact that $Z(G)$ is cyclic. Thus
$\eta^{*}(M)=\beta+1$. The result follows from Proposition 4.2. $\Box$

We conclude by computing $\eta$ when $\delta=0$, $\epsilon=1$, $\alpha\geq 3$,
and $\beta\geq 3$. Note this also covers the cases $\delta=1$, $\epsilon=1$
and $\alpha,\beta\geq 3$.

###### Theorem 4.8

Suppose $G$ is $G_{2}(\alpha,\beta,1,0,-)$ with $\alpha\geq 3$ and $\beta\geq
3$. Let $M=\langle x,y^{2}\rangle$.

  1. 1.

If $\alpha\geq\beta$, then $\eta(G)=\eta(M)/2+3=g_{2}(\alpha,\beta-1)/2+3$.

  2. 2.

If $\alpha<\beta$, then $\eta(G)=\eta(M)/2+3=g_{2}(\alpha-1,\beta)/2+3$.

Proof. As in Proposition 4.2, we let $M=\langle x,y^{2}\rangle$; so it follows
that $M$ is abelian. We know that $|M|=2^{\alpha+\beta-1}$, that $x$ has order
$2^{\alpha}$ and $y^{2}$ has order $2^{\beta}$. Suppose $\alpha\geq\beta$,
then $M\cong C_{2^{\alpha}}\times C_{2^{\beta-1}}$, and so
$\eta(M)=g_{2}(\alpha,\beta-1)$. Let $w=y^{2}x^{2^{\alpha-\beta}}$. Observe
that
$w^{2^{\beta-2}}=(y^{2}x^{2^{\alpha-\beta}})^{2^{\beta-2}}=y^{2^{\beta-1}}x^{2^{\alpha-2}}\not\in\langle
x\rangle$ and
$w^{2^{\beta-1}}=(y^{2}x^{2^{\alpha-\beta}})^{2^{\beta-1}}=y^{2^{\beta}}x^{2^{\alpha-1}}=x^{2^{\alpha-1}}x^{2^{\alpha-1}}=1$.
It follows that $M=\langle x\rangle\times\langle w\rangle$.

If $\beta\geq\alpha+1$, then $M\cong C_{2^{\alpha-1}}\times C_{2^{\beta}}$,
and so $\eta(M)=g_{2}(\alpha-1,\beta)$. Let $u=y^{2^{\beta-\alpha+1}}x$. We
compute
$u^{2^{\alpha-2}}=(y^{2^{\beta-\alpha+1}}x)^{2^{\alpha-2}}=y^{2^{\beta-1}}x^{2^{\alpha-2}}\not\in\langle
y\rangle$ and
$u^{2^{\alpha-1}}=(y^{2^{\beta-\alpha+1}}x)^{2^{\alpha-1}}=y^{2^{\beta}}x^{2^{\alpha-1}}=x^{2^{\alpha-1}}x^{2^{\alpha-1}}=1$.
We deduce that $M=\langle u\rangle\times\langle y\rangle$.

In both cases, we will show that $\eta^{*}(M)=\eta(M)/2+2$, and we obtain the
conclusion by applying Proposition 4.2. Notice that a maximal cyclic subgroup
of $M$ will be generated either by an element of the form $y^{2l}x$ for some
integer $l$ or by an element of the form $y^{2}x^{l}$ for some integer $l$.
Observe that $\langle x\rangle$ and $\langle y^{2}\rangle$ are maximal cyclic
subgroups of $M$ that are normal in $G$.

We next show that $\langle y^{2^{\beta-1}}x\rangle$ and $\langle
y^{2}x^{2^{\alpha-2}}\rangle$ are normal subgroups in $G$. Since $M$ is
abelian and has index $2$ in $G$, it suffices to show that $y$ normalizes
these subgroups. We compute
$(y^{2^{\beta-1}}x)^{y}=y^{2^{\beta-1}}x^{-1}=(y^{2^{\beta-1}}x)^{-1}$. Since
$y$ conjugates the generator of $\langle y^{2^{\beta-1}}x\rangle$ to its
inverse, this implies that $\langle y^{2^{\beta-1}}x\rangle$ is normal in $G$.

We now turn to $\langle y^{2}x^{2^{\alpha-2}}\rangle$. We begin with the
observation that $(y^{2}x^{2^{\alpha-2}})^{4}=y^{8}$. Since $\beta\geq 3$, we
see that $x^{2^{\alpha-1}}=y^{2^{\beta}}\in\langle
y^{2}x^{2^{\alpha-2}}\rangle$. Conjugating yields
$(y^{2}x^{2^{\alpha-2}})^{y}=y^{2}x^{-2^{\alpha-2}}$. Note that
$x^{-2^{\alpha-2}}=x^{2^{\alpha-2}}x^{2^{\alpha-1}}$. We have
$(y^{2}x^{2^{\alpha-2}})^{y}=y^{2}x^{2^{\alpha-2}}x^{2^{\alpha-1}}$. Since
both $y^{2}x^{2^{\alpha-2}}$ and $x^{2^{\alpha-1}}$ lie in $\langle
y^{2}x^{2^{\alpha-2}}\rangle$, we conclude that $(y^{2}x^{2^{\alpha-2}})^{y}$
lies in $\langle y^{2}x^{2^{\alpha-2}}\rangle$. We deduce that $\langle
y^{2}x^{2^{\alpha-2}}\rangle$ is normal in $G$.

We prove that the remaining maximal cyclic subgroups of $M$ lie in orbits of
size $2$. We have noted that a maximal cyclic subgroup $C$ of $M$ will have a
generator of the form $y^{2l}x$ or of the form $y^{2}x^{l}$ for some integer
$l$. If $C$ has a generator of the form $y^{2l}x$, then for $C$ to be normal
we need $(y^{2l}x)^{y}=y^{2l}x^{-1}\in C$. This implies that
$y^{2l}x^{-1}=(y^{2l}x)^{k}$ for some integer $k$. We have
$y^{2l-2lk}=x^{k+1}=u\in\langle x\rangle\cap\langle y^{2}\rangle=\langle
x^{2^{\alpha-1}}\rangle$. Hence, $u$ is either $1$ or $x^{2^{\alpha-1}}$. If
$u=1$, then $2^{\alpha}$ divides $k+1$ and $2^{\beta+1}$ divides $2l(1-k)$. We
see that there is an integer $r$ so that $k+1=2^{\alpha}r$, and hence,
$k=2^{\alpha}r-1$. This implies that $2^{\beta+1}$ divides
$2l(1-k)=2l(1-2^{\alpha}r+1)=4l(1-2^{\alpha-1}r)$. Since $\alpha\geq 2$, this
yields $2^{\beta-1}$ divides $l$. When $u=x^{2^{\alpha-1}}$, we obtain that
$k+1\equiv 2^{\alpha-1}~{}({\rm mod}~{}2^{\alpha})$. Hence, there is an
integer $r$ so that $k+1=2^{\alpha-1}+r2^{\alpha}$, and so,
$k=2^{\alpha-1}+r2^{\alpha}-1$. We see that $2l(1-k)\equiv 2^{\beta}~{}({\rm
mod}~{}2^{\beta+1})$. This implies that $2^{\beta+1}$ divides
$2l(1-k)-2^{\beta}=2l(1-2^{\alpha-1}-r2^{\alpha}+1)-2^{\beta}=4l(1-2^{\alpha-2}-r2^{\alpha-1})-2^{\beta}$.
We deduce that $2^{\beta-2}$ divides $l$. We conclude that $\langle x\rangle$
and $\langle y^{2^{\beta-1}}x\rangle$ are the only maximal cyclic subgroups of
$M$ having the form $\langle y^{2l}x\rangle$ that are normal in $G$.

We now suppose that $C$ has a generator of the form $y^{2}x^{l}$. We need
$(y^{2}x^{l})^{y}=y^{2}x^{-l}\in C$. Hence, we have that
$y^{2}x^{-l}=(y^{2}x^{l})^{k}=y^{2k}x^{lk}$ for some integer $k$. We have
$y^{2-2k}=x^{lk+l}=u$. As in the previous paragraph, we see that $u$ is either
$1$ or $x^{2^{\alpha-1}}$. If $u=1$, then we have that $2^{\beta+1}$ divides
$2(1-k)$, and so, there is an integer $r$ so that $1-k=2^{\beta}r$. We
determine that $2^{\alpha}$ divides
$l(k+1)=l(1-2^{\beta}r+1)=2l(1-2^{\beta-1}r)$. It follows that $2^{\alpha-1}$
divides $l$. Now, suppose that $u=x^{2^{\alpha-1}}$. We must have that
$2(1-k)\equiv 2^{\beta}~{}({\rm mod}~{}2^{\beta+1})$ and $l(k+1)\equiv
2^{\alpha-1}~{}({\rm mod}~{}2^{\alpha})$. Hence, there is an integer $r$ so
that $2(1-k)=2^{\beta}+2^{\beta+1}r$. This implies that
$k=1-2^{\beta-1}-2^{\beta}r$. We then obtain that $2^{\alpha}$ divides
$l(k+1)-2^{\alpha-1}=l(1-2^{\beta-1}-2^{\beta}r+1)-2^{\alpha-1}=2(l(1-2^{\beta-2}-2^{\beta-1}r)-2^{\alpha-2})$.
This implies that $2^{\alpha-1}$ divides
$l(1-2^{\beta-2}-2^{\beta-1}r)-2^{\alpha-2}$. Hence, there is an integer $s$
so that $l(1-2^{\beta-2}-2^{\beta-1}r)-2^{\alpha-2}=2^{\alpha-1}s$. This leads
to
$l(1-2^{\beta-2}-2^{\beta-1}r)=2^{\alpha-1}s+2^{\alpha-2}=2^{\alpha-2}(2s+1)$.
This yields $2^{\alpha-2}$ divides $l$. Observe that
$x^{2^{\alpha-1}}=y^{2^{\beta}}$, and so, $\langle
y^{2}x^{2^{\alpha-1}}\rangle=\langle y^{2}\rangle$. We deduce that $\langle
y^{2}\rangle$ and $\langle y^{2}x^{2^{\alpha-2}}\rangle$ are the only maximal
cyclic subgroups of $M$ having the form $\langle y^{2}x^{l}\rangle$ that are
normal in $G$,

We now see that the number of $G$-orbits of maximal cyclic subgroups of $M$ is
$(\eta(M)-4)/2+4=\eta(M)/2-2+4=\eta(M)+2$, which completes the proof of the
result. $\Box$

We close by proving that when $G$ is metacyclic of minus type that is not
dihedral, generalized quaternion, or semi-dihedral, then
$\eta(G)\geq\alpha+\beta-2$ and we determine when equality occurs. We first
handle when $\delta$ equals $0$ or $1$. In this case, we have
$\eta(G)\geq\alpha+\beta$.

###### Proposition 4.9

Suppose $G=G_{2}(\alpha,\beta,\epsilon,\delta,-)$ with $\delta=0$ or 1 and
$\beta\geq 2$. Then $\eta(G)\geq\alpha+\beta.$

Proof. (i) Suppose $\epsilon=0$. Denote $l=\min(\alpha,\beta-1)$ and
$k=\max(\alpha,\beta-1)$. First, consider $l\geq 3$. Then $\beta\geq 4$ and by
Theorem 4.4 and Lemma 2.4

$\eta(G)\geq g_{2}(\alpha,\beta-1)/2+2\geq 2k+2\geq\alpha+\beta.$

Next, assume $l=2$. So $\beta\geq 3$ and by Theorem 4.4 and Lemma 2.4

$\eta(G)\geq g_{2}(\alpha,\beta-1)/2+2=k+3\geq\alpha+\beta.$

Finally, set $l=1$. As $\alpha\geq 2$, we have $\beta=2$. The result follows
from Lemma 4.3.

(ii) Now suppose $\epsilon=1$. Assume $\alpha\geq\beta$, then
$l=\min(\alpha,\beta-1)=\beta-1$ and $k=\max(\alpha,\beta-1)=\alpha$. If
$l\geq 3$, then $\beta\geq 4$ and $\alpha\geq 4$, so we can assume $\delta=0$.
Applying Theorem 4.8 and Lemma 2.4 yields

$\eta(G)=g_{2}(\alpha,\beta-1)/2+3\geq 2k+3\geq\alpha+\beta.$

If $l=2$, then $\beta=3$, and we again appeal to Theorem 4.8 to obtain

$\eta(G)=g_{2}(\alpha,\beta-1)/2+3=g_{2}(k,2)/2+3=k+4\geq\alpha+\beta.$

If $l=1$, then $\beta=2$. If $\alpha=2$ then $\delta=0$ and if $\alpha\geq 3$
we can assume $\delta=0$. Thus we apply Lemma 4.6.

Finally, suppose $\epsilon=1$ and $\alpha<\beta$. We set
$l=\min(\alpha-1,\beta)=\alpha-1$ and $k=\max(\alpha-1,\beta)=\beta$. When
$l\geq 3$, we apply Theorem 4.8 and Lemma 2.4 to get

$\eta(G)=g_{2}(\alpha-1,\beta)/2+3\geq 2k+3\geq\alpha+\beta.$

If $l=2$, then $\alpha=3$ and $\beta>3$. Apply Theorem 4.8 with Lemma 2.4 to
give

$\eta(G)=g_{2}(\beta,2)/2+3=\beta+4\geq\alpha+\beta.$

If $l=1$, then $\alpha=2$ and $\delta=0$, the result follows from Lemma 4.7.
$\Box$

We now have the case where $\delta\geq 2$.

###### Proposition 4.10

Suppose $G=G_{2}(\alpha,\beta,\epsilon,\delta,-)$ with $\delta\geq 2$. Then
$\eta(G)\geq\alpha+\beta-2$. Equality holds if and only if $\beta=\delta$ and
either (i) $\beta=3$ or (ii) $\beta\geq 4$ and $\alpha-\beta=2$.

Proof. Set $l=\min(\alpha-\delta+1,\beta-1)$ and
$k=\max(\alpha-\delta+1,\beta-1)$. We consider various cases according to the
value of $l$.

First, suppose $l\geq 4$. Then by Corollary 4.5 and Lemma 2.4

$\displaystyle\eta(G)$ $\displaystyle=$ $\displaystyle
g_{2}(\alpha-\delta+1,\beta-1)/2+2$ $\displaystyle=$ $\displaystyle
g_{2}(k,l)/2+2\geq 2k+l+2$ $\displaystyle=$ $\displaystyle(k+l)+k+2$
$\displaystyle\geq$ $\displaystyle\alpha-\delta+\beta+\beta-1+2$
$\displaystyle\geq$ $\displaystyle\alpha+\beta+1$

since $\delta\leq\beta$.

Now consider $l=3$. We use Corollary 4.5 and Lemma 2.4 to find an exact value
for $\eta(G)$.

$\eta(G)=g_{2}(\alpha-\delta+1,\beta-1)/2+2=g_{2}(k,3)/2+2=2k+2.$

If $\alpha-\delta+1>\beta-1=3$, then $\delta\leq 4$ and

$\eta(G)=2(\alpha-\delta+1)+2=\alpha+(\alpha-\delta+2)+(-\delta+2)>\alpha+\beta-2.$

On the other hand, when $\beta-1\geq\alpha-\delta+1=3$, we obtain $\beta\geq
4$ and $\alpha-\delta=2$, so $\alpha-2\leq\beta$ and

$\eta(G)=2(\beta-1)+2=2\beta\geq\beta+\alpha-2$

with equality if and only if $\beta=\delta$.

Next suppose $l=2$. Since $\alpha-\delta+1\geq 2+1=3$, we must have $\beta=3$.
Applying Corollary 4.5 and Lemma 2.4,

$\displaystyle\eta(G)$ $\displaystyle=$ $\displaystyle
g_{2}(\alpha-\delta+1,\beta-1)/2+2=g_{2}(k,2)/2+2$ $\displaystyle=$
$\displaystyle k+3=\alpha-\delta+4$ $\displaystyle\geq$
$\displaystyle\alpha+1=\alpha+\beta-2$

with equality if and only if $\delta=3=\beta.$

Lastly consider $l=1$. In this case $\beta=2$ and the result follows from
Corollary 4.5. $\Box$

## References

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Mariagrazia Bianchi: Dipartimento di Matematica F. Enriques, Università degli
Studi di Milano, via Saldini 50, 20133 Milano, Italy.
<EMAIL_ADDRESS>
Rachel Camina: Fitzwilliam College, Cambridge, CB3 0DG, UK.
<EMAIL_ADDRESS>
Mark L. Lewis: Department of Mathematical Sciences, Kent State University,
Kent, Ohio, 44242 USA.
<EMAIL_ADDRESS>