International Journal of Group Theory Vol. XX No. X (201X), pp XX-XX.

# On the Order of the Schur Multiplier of a Pair of Finite p-Groups II

Fahimeh Mohammadzadeh, Azam Hokmabadi∗ and Behrooz Mashayekhy

###### Abstract.

Let $G$ be a finite $p$-group and $N$ be a normal subgroup of $G$, with
$|N|=p^{n}$ and $|G/N|=p^{m}$. A result of Ellis (1998) shows that the order
of the Schur multiplier of such a pair $(G,N)$ of finite $p$-groups is bounded
by $p^{\frac{1}{2}n(2m+n-1)}$ and hence it is equal to
$p^{\frac{1}{2}n(2m+n-1)-t}$, for some non-negative integer $t$. Recently the
authors characterized the structure of $(G,N)$ when $N$ has a complement in
$G$ and $t\leq 3$. This paper is devoted to classify the structure of $(G,N)$
when $N$ has a normal complement in $G$ and $t=4,5$.

MSC(2010): Primary: 20E34; Secondary: 20D15.
Keywords: pair of groups, Schur multiplier, finite $p$-groups.
$*$Corresponding author
$\copyright$ 2011 University of Isfahan

Communicated by

##  1\. Introduction

By a pair of groups $(G,N)$ we mean a group $G$ with a normal subgroup $N$. In
1998 Ellis [2] defined the Schur multiplier of a pair $(G,N)$ to be the
abelian group $M(G,N)$ appearing in a natural exact sequence

$\displaystyle H_{3}(G)$ $\displaystyle\rightarrow$ $\displaystyle
H_{3}(G/N)\rightarrow M(G,N)\rightarrow M(G)\rightarrow M(G/N)$
$\displaystyle\rightarrow$ $\displaystyle
N/[N,G]\rightarrow(G)^{ab}\rightarrow(G/N)^{ab}\rightarrow 0,$

in which $H_{3}(G)$ is the third homology of $G$ with integer coefficients. In
1956, Green [4] showed that if $G$ is a group of order $p^{n}$ , then its
Schur multiplier is of order at most $p^{\frac{n(n-1)}{2}}$, and hence equals
to $p^{\frac{n(n-1)}{2}-t}$, for some non-negative integer $t$. Berkovich [1],
Zhou [11], Ellis [3] and Niroomand [7, 8] determined the structure of $G$ for
$t=0,1,2,3,4,5$ by different methods.

In 1998, Ellis [2] gave an upper bound for the order of the Schur multiplier
of a pair of finite $p$-groups. He proved that if $G$ is a finite $p$-group
with normal subgroup $N$ of order $p^{n}$ and quotient $G/N$ of order $p^{m}$,
then the Schur multiplier of $(G,N)$ is bounded by $p^{\frac{1}{2}n(2m+n-1)}$
and hence equals to $p^{\frac{1}{2}n(2m+n-1)-t}$, for some nonnegative integer
$t$.

Let $(G,N)$ be a pair of groups and $K$ be the complement of $N$ in $G$. In
2004, Sallemkar, Moghaddam and Saeedi [10] characterized the structure of such
a pair $(G,N)$ when $t=0,1$ with some conditions. Recently the authors [5]
determined the structure of the pair $(G,N)$, for $t=0,1$, without any
condition and also gave the structure of $(G,N)$ for $t=2,3$, when $K$ is
normal. In this paper we are going to determine the structure of $(G,N)$ for
$t=4,5$ when $K$ is a normal subgroup of $G$.

In this paper $D$ and $Q$ denote the dihedral and quaternion group of order 8
and, $E_{1}$ and $E_{2}$ denote the extra special $p$-groups of order $p^{3}$
of odd exponent $p$ and $p^{2}$, respectively. $E_{4}$ denotes the unique
central product of a cyclic group of order $p^{2}$ and a non-abelian group of
order $p^{3}.$ Also ${\bf{Z}}_{n}^{(m)}$ denote the direct product of $m$
copies of ${\bf{Z}}_{n}$.

The following result is essential to prove the main theorems.

######  Theorem 1.1.

[2]Let $(G,N)$ be a pair of groups and $K$ be the complement of $N$ in $G$.
Then

$M(G)\cong M(G,N)\times M(K).$

In 1907 Schur [6] gave an structure for the Schur multiplier of a direct
product of finite groups. He showed that

$M(G_{1}\times G_{2})=M(G_{1})\times M(G_{2})\times(G_{1}^{ab}\otimes
G_{2}^{ab}).$

As a consequence of this fact we have the following important result.

######  Corollary 1.2.

Let $(G,N)$ be a pair of groups and $K$ be the complement of $N$ in $G$. Then

$|M(G,N)|=|M(N)||N^{ab}\otimes K^{ab}|.$

The following theorems gave the structure of a finite $p$-group, in terms of
the order of its Schur multiplier.

######  Theorem 1.3.

[3] Let $G$ be a group of prime-power order $p^{n}$. Suppose that
$|M(G)|=p^{\frac{1}{2}n(n-1)-t}$. Then
$i)$ $t=0$ if and only if $G$ is elementary abelian;
$ii)$ $t=1$ if and only if $G\cong{\bf{Z}}_{p^{2}}$ or $G\cong E_{1}$;
$iii)$ $t=2$ if and only if $G\cong{\bf{Z}}_{p}\times{\bf{Z}}_{p^{2}}$,
$G\cong D$ or $G\cong{\bf{Z}}_{p}\times E_{1}$;
$iv)$ $t=3$ if and only if $G\cong{\bf{Z}}_{p^{3}}$,
$G\cong{\bf{Z}}_{p}\times{\bf{Z}}_{p}\times{\bf{Z}}_{p^{2}}$, $G\cong Q$,
$G\cong E_{2}$, $G\cong D\times{\bf{Z}}_{2}$ or $G\cong
E_{1}\times{\bf{Z}}_{p}\times{\bf{Z}}_{p}$.

######  Theorem 1.4.

[9] Let $G$ be an abelian group of order $p^{n}$. Suppose that
$|M(G)|=p^{\frac{1}{2}n(n-1)-4}$. Then $G$ is isomorphic to
${\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p^{2}}$ or
${\bf{Z}}_{p^{2}}\times{{\bf{Z}}_{p}}^{(3)}$.

The following result can be easily obtained by using a method similar to the
proof of the above theorem in [9].

######  Theorem 1.5.

Let $G$ be an abelian group of order $p^{n}$. Suppose that
$|M(G)|=p^{\frac{1}{2}n(n-1)-5}$. Then $G$ is isomorphic to
${\bf{Z}}_{p^{3}}\times{\bf{Z}}_{p}$ or
${\bf{Z}}_{p^{2}}\times{{\bf{Z}}_{p}}^{(4)}$.

######  Theorem 1.6.

[7] Let $G$ be a non-abelian group of order $p^{n}$. Suppose that
$|M(G)|=p^{\frac{1}{2}n(n-1)-4}$. Then $G$ is isomorphic to one of the
following groups:
For $p=2$,
1) $D\times{\bf{Z}}_{p}^{(2)};$
2) $Q\times{\bf{Z}}_{2};$
3) $\langle a,b|a^{4}=b^{4}=1,[a,b,a]=[a,b,b]=1,[a,b]=a^{2}b^{2}\rangle;$
4) $\langle a,b,c|a^{2}=b^{2}=c^{2}=1,abc=bca=cab\rangle;$
For $p\neq 2$
5) $E_{4}$;
6) $E_{1}\times{\bf{Z}}_{p}^{(3)}$;
7) ${\bf{Z}}_{p}^{(4)}>\\!\\!\\!\lhd_{\theta}{\bf{Z}}_{p}$;
8) $E_{2}\times{\bf{Z}}_{p}$;
9) $\langle a,b|a^{p^{2}}=b^{p}=1,[a,b,a]=[a,b,b]=1\rangle$;
10) $\langle a,b|a^{9}=b^{3}=1,[a,b,a]=1,[a,b,b]=a^{6},[a,b,b,b]=1\rangle;$
11) $\langle a,b|a^{p}=b^{p}=1,[a,b,a]=[a,b,b,a]=[a,b,b,b]=1\rangle(p\neq 3);$

######  Theorem 1.7.

[8] Let $G$ be a non-abelian group of order $p^{n}$. Suppose that
$|M(G)|=p^{\frac{1}{2}n(n-1)-5}$. Then $G$ is isomorphic to one of the
following groups:
1) $D\times{\bf{Z}}_{2}^{(3)}$;
2) $E_{1}\times{\bf{Z}}_{p}^{(4)}$;
3) $E_{2}\times{\bf{Z}}_{p}^{(2)}$;
4) $E_{4}\times{\bf{Z}}_{p};$
5) extra special $p$-group of order $p^{5}$;
6) $\langle a,b|a^{p^{2}}=b^{p^{2}}=1,[a,b,a]=[a,b,b]=1,[a,b]=a^{p}\rangle$;
7) $\langle a,b|a^{p^{2}}=b^{p}=1,[a,b,a]=[a,b,b]=a^{p},[a,b,b,b,]=1\rangle$;
8) $\langle a,b|a^{p^{2}}=b^{p}=1,[a,b,a]=[a,b,b,b,]=1,[a,b,b]=a^{np}\rangle$
where $n$ is a fixed quadratic non-residue of $p$ and $p\neq 3$;
9) $\langle
a,b|a^{p^{2}}=1,b^{3}=a^{3},,[a,b,a]=[a,b,b,b,]=1,[a,b,b]=a^{6}\rangle$;
10) $\langle a,b|a^{p}=1,b^{p}=[a,b,b],[a,b,a]=[a,b,b,b,]=[a,b,b,a]=1\rangle$;
11) $D_{16}$;
12) $\langle a,b|a^{4}=b^{4}=1,a^{-1}ba=b^{-1}\rangle$;
13) $Q\times{\bf{Z}}_{2}^{(2)}$;
14) $(D\times{\bf{Z}}_{2})>\\!\\!\\!\lhd{\bf{Z}}_{2}$;
15) $(Q\times{\bf{Z}}_{2})>\\!\\!\\!\lhd{\bf{Z}}_{2}$;
16) ${\bf{Z}}_{2}\times\langle a,b,c|a^{2}=b^{2}=c^{2}=1,abc=bca=cab\rangle;$

##  2\. Main Results

In this section, let $(G,N)$ be a pair of groups such that $G\cong N\times K$,
with $|N|=p^{n}$ and $|K|=p^{m}$. It is obtained that
$|M(G,N)|=p^{\frac{1}{2}n(2m+n-1)-t}$, for some nonnegative integer $t$.
Recently, all these pairs of finite $p$-groups are listed in [5] by the
authors, when $t=0,1,2,3$. The aim of this paper is characterizing the
structure of such pairs of finite $p$-groups, when $t=4,5$.

######  Theorem 2.1.

By the above assumption, $t=4$ if and only if $G$ is isomorphic to one of the
following groups:
1) $G\cong N\times K$ where $N\cong{\bf{Z}}_{p}$ and $K$ is any group with
$d(K)=m-4$;
2) $G\cong N\times K$ where $N\cong{\bf{Z}}_{p}\times{\bf{Z}}_{p}$ and $K$ is
any group with $d(K)=m-2$;
3) $G\cong N\times K$ where $N\cong{{\bf{Z}}_{p}}^{(4)}$ and $K$ is any group
with $d(K)=m-1$;
4) $G=N\cong D\times{\bf{Z}}_{2}^{(2)};$
5) $G=N\cong Q\times{\bf{Z}}_{2};$
6) $G=N\cong\langle
a,b|a^{4}=b^{4}=1,[a,b,a]=[a,b,b]=1,[a,b]=a^{2}b^{2}\rangle;$
7) $G=N\cong\langle a,b,c|a^{2}=b^{2}=c^{2}=1,abc=bca=cab\rangle;$
8) $G=N\cong E_{4}$;
9) $G=N\cong E_{1}\times{\bf{Z}}_{p}^{(3)}$;
10) $G=N\cong{\bf{Z}}_{p}^{(4)}>\\!\\!\\!\lhd_{\theta}{\bf{Z}}_{p}$;
11) $G=N\cong E_{2}\times{\bf{Z}}_{p}$;
12) $G=N\cong\langle a,b|a^{p^{2}}=b^{p}=1,[a,b,a]=[a,b,b]=1\rangle$;
13) $G=N\cong\langle
a,b|a^{9}=b^{3}=1,[a,b,a]=1,[a,b,b]=a^{6},[a,b,b,b]=1\rangle;$
14) $G=N\cong\langle
a,b|a^{p}=b^{p}=1,[a,b,a]=[a,b,b,a]=1,[a,b,b,b]=1\rangle(p\neq 3);$
15) $G=N\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p^{2}};$
16) $G=N\cong{\bf{Z}}_{p^{2}}\times{{\bf{Z}}_{p}}^{(3)};$
17) $G\cong N\times K$ where $K={\bf{Z}}_{p}$ and
$N\cong{{\bf{Z}}_{p}}^{(2)}\times{\bf{Z}}_{p^{2}};$
18) $G\cong N\times K$ where $K={\bf{Z}}_{p}$ and $N\cong Q;$
19) $G\cong N\times K$ where $K={\bf{Z}}_{p}$ and $N\cong E_{2};$
20) $G\cong N\times K$ where $K={\bf{Z}}_{p}$ and $N\cong
D\times{\bf{Z}}_{2};$
21) $G\cong N\times K$ where $K={\bf{Z}}_{p}$ and $N\cong
E_{1}\times{{\bf{Z}}_{p}}^{(2)}$;
22) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}$ and
$N\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}$;
23) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}$ and $N\cong D$;
24) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}$ and $N\cong
E_{1}\times{\bf{Z}}_{p}$;
25) $G\cong N\times K$ where $K={\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}$ and
$N\cong{\bf{Z}}_{p^{2}}$;
26) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(3)}$ and $N\cong E_{1}$;
27) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(3)}$ and
$N\cong{\bf{Z}}_{p^{2}}$.

###### Proof.

The necessity of theorem follows from the fact that $G=N\times K$ and
Corollary 1.2. For sufficiency first suppose that $N$ is an elementary abelian
$p$-group. Then $nm-4=nd(K)$. Since $|N\otimes K^{ab}|=p^{nm-4}$ by Corollary
1.2 and also it is of order $p^{nd(K)}$. So $n(m-d(K))=4$, which implies that
$n=1,2$ or 4. Therefore $N\cong{\bf{Z}}_{p}$ and $K$ is any group with
$d(K)=m-4$ or $N\cong{{\bf{Z}}_{p}}^{(2)}$ and $K$ is any group with
$d(K)=m-2$, or $N\cong{{\bf{Z}}_{p}}^{(4)}$ and $K$ is any group with
$d(K)=m-1$.

Now suppose that $N$ is not an elementary abelian $p$-group . Then using
Corollary 1.2, we have $|N^{ab}\otimes K^{ab}|>p^{nm-4}$ and so $md(N)>nm-4$
which implies that $m(n-d(N))<4.$ Therefore $m=0,1,2,3$ .

If $m=0$, then $K=1$ and $N$ is one of the groups which are listed in Theorems
1.4 and 1.6.

If $m=1$, then $K={\bf{Z}}_{p}$ and $d(N)=n-1,n-2$ or $n-3$, respectively. It
follows that $|N^{ab}\otimes K|=p^{n-1},p^{n-2}$ or $p^{n-3}$. So Corollary
1.2 implies that $|M(N)|=p^{\frac{n^{2}-n}{2}-3},p^{\frac{n^{2}-n}{2}-2}$, or
$p^{\frac{n^{2}-n}{2}-1}$. In the first case $N$ is
${\bf{Z}}_{p}\times{\bf{Z}}_{p}\times{\bf{Z}}_{p^{2}},Q,E_{2},,D\times{\bf{Z}}_{2}$
or $E_{1}\times{\bf{Z}}_{p}\times{\bf{Z}}_{p}$ and other cases are impossible,
by Theorem 1.3.

If $m=2$, then $d(N)=n-1$ and $K={\bf{Z}}_{p}\times{\bf{Z}}_{p}$ or
$K={\bf{Z}}_{p^{2}}$. In the first case $|N^{ab}\otimes K|=p^{2(n-1)}$ and so
$|M(N)|=p^{\frac{n^{2}-n}{2}-2}$. Therefore $N$ is
${\bf{Z}}_{p}\times{\bf{Z}}_{p^{2}},D$ or ${\bf{Z}}_{p}\times E_{1}$. In the
second case $N^{ab}\cong{{\bf{Z}}_{p}}^{(n-1)}$ or
$N^{ab}\cong{\bf{Z}}_{p^{2}}\times{{\bf{Z}}_{p}}^{(n-2)}$. If $N^{ab}$ is an
elementary abelian $p$-group, then $|N^{ab}\otimes K|=p^{(n-1)}$ and so
$|M(N)|=p^{\frac{n^{2}+n-6}{2}}$ which is impossible and if
$N^{ab}\cong{\bf{Z}}_{p^{2}}\times{{\bf{Z}}_{p}}^{(n-2)}$, then
$|M(N)|=p^{\frac{n^{2}+n-8}{2}}$ which is impossible too.

If $m=3$, then $d(N)=n-1$ and $K$ is an abelian $p$-group of order $p^{3}$ or
an extra special $p$-group of order $p^{3}$. In the first case we have three
possibility for $K$. The first possibility is $K\cong{\bf{Z}}_{p}^{{(3)}},$.
Then similar to the previous part, one can see that
$|M(N)|=p^{\frac{n^{2}-n}{2}-1}$ and so $N\cong E_{1}$ or ${\bf{Z}}_{p^{2}}$.
The second possibility is $K\cong{\bf{Z}}_{p^{3}}$. This implies that $n=1$
which is a contradiction. The third possibility is
$K\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}.$ which implies that $n=2$ and
$N\cong{\bf{Z}}_{p^{2}}$.
In the second case, if $K$ is an extra special $p$-group of order $p^{3}$,
then $N^{ab}\cong{\bf{Z}}_{p}^{(n-1)}$ or
$N^{ab}\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}^{(n-2)}$. This implies that
$|N^{ab}\otimes K|=|N^{ab}\otimes{\bf{Z}}_{p}^{(2)}|=p^{2n-2}$ and so $n=1$
which is a contradiction. Hence the proof is complete. ∎

######  Theorem 2.2.

By the previous assumption, $t=5$ if and only if $G$ is isomorphic to one of
the following groups:
1) $G\cong N\times K$ where $N\cong{\bf{Z}}_{p}$ and $K$ is any group with
$d(K)=m-5$;
2) $G\cong N\times K$ where $N\cong{{\bf{Z}}_{p}}^{(5)}$ and $K$ is any group
with $d(K)=m-1$;
3) $G=N\cong D\times{\bf{Z}}_{2}^{(3)}$;
4) $G=N\cong E_{1}\times{\bf{Z}}_{p}^{(4)}$;
5) $G=N\cong E_{2}\times{\bf{Z}}_{p}^{(2)}$;
6) $G=N\cong E_{4}\times{\bf{Z}}_{p};$
7) $G=N\cong$ extra special $p$-group of order $p^{5}$;
8) $G=N\cong\langle
a,b|a^{p^{2}}=b^{p^{2}}=1,[a,b,a]=[a,b,b]=1,[a,b]=a^{p}\rangle$;
9) $G=N\cong\langle
a,b|a^{p^{2}}=b^{p}=1,[a,b,a]=[a,b,b]=a^{p},[a,b,b,b,]=1\rangle$;
10) $G=N\cong\langle
a,b|a^{p^{2}}=b^{p}=1,[a,b,a]=[a,b,b,b,]=1,[a,b,b]=a^{np}\rangle$ where $n$ is
a fixed quadratic non-residue of $p$ and $p\neq 3$ ;
11) $G=N\cong\langle
a,b|a^{p^{2}}=1,b^{3}=a^{3},[a,b,a]=[a,b,b,b,]=1,[a,b,b]=a^{6}\rangle$;
12) $G=N\cong\langle
a,b|a^{p}=1,b^{p}=[a,b,b],[a,b,a]=[a,b,b,b,]=[a,b,b,a]=1\rangle$;
13) $G=N\cong D_{16}$;
14) $G=N\cong\langle a,b|a^{4}=b^{4}=1,a^{-1}ba=b^{-1}\rangle$;
15) $G=N\cong Q\times{\bf{Z}}_{2}^{(2)}$;
16) $G=N\cong(D\times{\bf{Z}}_{2})>\\!\\!\\!\lhd{\bf{Z}}_{2}$;
17) $G=N\cong(Q\times{\bf{Z}}_{2})>\\!\\!\\!\lhd{\bf{Z}}_{2}$;
18) $G=N\cong{\bf{Z}}_{2}\times<a,b,c|a^{2}=b^{2}=c^{2}=1,abc=bca=cab>;$
19) $G=N\cong{\bf{Z}}_{p^{3}}\times{\bf{Z}}_{p}$;
20) $G=N\cong{\bf{Z}}_{p^{2}}\times{{\bf{Z}}_{p}}^{(4)};$
21) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}$ and $N\cong
D\times{{\bf{Z}}_{2}}^{(2)}$;
22) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}$ and $N\cong
Q\times{{\bf{Z}}_{2}}$;
23) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}$ and
$N\cong<a,b,c|a^{2}=b^{2}=c^{2}=1,abc=bca=cab>$;
24) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}$ and $N\cong E_{4}$;
25) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}$ and $N\cong
E_{1}\times{{\bf{Z}}_{p}}^{(3)}$;
26) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}$ and
$N\cong{\bf{Z}}_{p}^{(4)}>\\!\\!\\!\lhd{\bf{Z}}_{p}$;
27) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}$ and $N\cong
E_{2}\times{\bf{Z}}_{p}$;
28) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}$ and $N\cong
E_{2}\times{{\bf{Z}}_{p}}$;
29) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}$ and $N\cong
E_{1}\times{{\bf{Z}}_{p}}^{(2)}$;
30) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}$ and
$N\cong{{\bf{Z}}_{p}}^{(2)}\times{{\bf{Z}}_{p^{2}}}$;
31) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}$ and $N\cong Q$;
32) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}$ and $N\cong E_{2}$;
33) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}$ and $N\cong
D\times{{\bf{Z}}_{2}};$
34) $G\cong N\times K$ where $K={{\bf{Z}}_{p^{2}}}$ and $N\cong E_{1}$;
35) $G\cong N\times K$ where $K={{\bf{Z}}_{p^{2}}}$ and
$N\cong{{\bf{Z}}_{p}}\times{{\bf{Z}}_{p^{2}}}$;
36) $G\cong N\times K$ where $K={{\bf{Z}}_{p^{3}}}$ and
$N\cong{{\bf{Z}}_{p^{2}}}$;
37) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(3)}$ and $N\cong D$;
38) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(3)}$ and $N\cong
E_{1}\times{{\bf{Z}}_{p}}$;
39) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(3)}$ and
$N\cong{{\bf{Z}}_{p}}\times{{\bf{Z}}_{p^{2}}}$;
40) $G\cong N\times K$ where $K$ is an extra special $P$-group and
$N\cong{{\bf{Z}}_{p^{2}}}$;
41) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(4)}$ and $N\cong E_{1}$,
42) $G\cong N\times K$ where $K={{\bf{Z}}_{p}}^{(2)}\times{{\bf{Z}}_{p}}$ and
$N\cong{{\bf{Z}}_{p^{2}}}.$

###### Proof.

The proof of this theorem is similar to the proof of the previous theorem so
we left the details to the reader. Necessity is straight forward. For
sufficiency first suppose that $N$ is an elementary abelian $p$-group . Then
$n(m-d(K))=5$, so $n=1$ or 5. If $n=1$, then $N\cong{\bf{Z}}_{p}$ and $K$ is
any group with $d(K)=m-5$ and $n=5$ implies that $N\cong{\bf{Z}}_{p}^{(5)}$
and $K$ is any group with $d(K)=m-1.$

Suppose that $N$ is not an elementary abelian $p$-group. Then we have
$|N^{ab}\otimes K^{ab}|>p^{nm-5}$, by Corollary 1.2. It follows that
$md(N)>nm-5$ and thus $m(n-d(N))<5$, which implies that $m=0,1,2,3$ or $4.$ If
$m=0$, then $K=1$ and $N$ is one of the groups that are listed in Theorems 1.5
and 1.7.

If $m=1$, then $K={\bf{Z}}_{p}$ and $d(N)=n-i$, for $1\leq i\leq 4$. Therefore
by Corollary 1.2, $|M(N)|=p^{\frac{n(n-1)}{2}-(5-i)}$ for $1\leq i\leq 4$,
respectively. It follows that
$N\cong{\bf{Z}}_{p^{2}}\times{{\bf{Z}}_{p}}^{(3)}$,
$D\times{\bf{Z}}_{2}^{(2)}$, $Q\times{\bf{Z}}_{2}$, $E_{4}$,
$E_{1}\times{\bf{Z}}_{p}^{(3)}$, $E_{2}\times{\bf{Z}}_{p}$,
$N\cong{\bf{Z}}_{p^{3}}$ or $\langle
a,b,c|a^{2}=b^{2}=c^{2}=1,abc=bca=cab\rangle$, by Theorems 1.6, 1.4 and 1.3.

If $m=2$, then $K={\bf{Z}}_{p^{2}}$ or $K={\bf{Z}}_{p}\times{\bf{Z}}_{p}$ and
$d(N)=n-1$ or $d(N)=n-2$. First suppose that
$K={\bf{Z}}_{p}\times{\bf{Z}}_{p}$ . If $d(N)=n-1$, then
$|M(N)|=p^{\frac{n^{2}-n}{2}-3}$. It follows that
$N\cong{\bf{Z}}_{p}\times{\bf{Z}}_{p}\times{\bf{Z}}_{p^{2}}$, $N\cong
E_{1}\times{\bf{Z}}_{p}\times{\bf{Z}}_{p}$, $N\cong Q$, $N\cong E_{2}$ or
$N\cong D\times{\bf{Z}}_{2}$. If $d(N)=n-2$, then $|N\otimes K|=p^{2(n-2)}$.
This implies that $|M(N)|=p^{\frac{n^{2}+n-2}{2}}$ and so $n<1$ which is
impossible.
Now suppose that $K={\bf{Z}}_{p^{2}}$. If $d(N)=n-1$ and
$N^{ab}\cong{{\bf{Z}}_{p}}^{(n-1)}$, then $|M(N)|=p^{\frac{n^{2}+n-8}{2}}$
which implies that $n=3$ and $|M(N)|=p^{2}$. Therefore $N\cong E_{1}$. If
$d(N)=n-1$ and $N^{ab}\cong{\bf{Z}}_{p}^{2}\times{{\bf{Z}}_{p}}^{(n-2)}$, then
$n=3$ or $n=4$. For $n=3$ there is not any structure for $N$ and $n=4$ implies
that $N\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}$.
If $d(N)=n-2$, then $N^{ab}\cong{\bf{Z}}_{p}^{(n-2)}$ or
$N^{ab}\cong{\bf{Z}}_{p}^{3}\times{\bf{Z}}_{p}^{(n-3)}$ or
$N^{ab}\cong{\bf{Z}}_{p}^{2}\times{\bf{Z}}_{p}^{2}\times{\bf{Z}}_{p}^{(n-4)}$.
Therefore similar to the previous case one can see that $n<5$ which is
impossible.

If $m=3$, then $d(N)=n-1$ and $K$ is an abelian $p$-group of order $p^{3}$ or
is an extra special $p$-group of order $p^{3}$. In the first case we have
three possibility for $K$. The first possibility is $K\cong{\bf{Z}}_{p^{3}}$.
Now if $N^{ab}\cong{\bf{Z}}_{p}^{(n-1)}$, then $n=1$ which is impossible and
if $N^{ab}\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}^{(n-2)}$, then
$N\cong{\bf{Z}}_{p^{2}}$. The second possibility is
$K\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}.$ In this case there is no structure
for $N$. The third possibility is $K\cong{\bf{Z}}_{p}^{(3)}$. Thus
$|M(N)|=p^{\frac{n^{2}-n}{2}-2}$ and so $N\cong D$, $N\cong
E_{1}\times{\bf{Z}}_{p}$ or $N\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}$, by
Theorem 1.3.
Now suppose that $K$ is an extra special $p$-group of order $p^{3}$. Then
$N^{ab}\cong{\bf{Z}}_{p}^{(n-1)}$ or
$N^{ab}\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}^{(n-2)}$. If $N^{ab}$ is an
elementary abelian, then there is no structure for $N$. Otherwise
$N={\bf{Z}}_{p^{2}}$.

If $m=4$, then $d(N)=n-1$. So $N^{ab}\cong{\bf{Z}}_{p}^{(n-1)}$ or
$N^{ab}\cong{\bf{Z}}_{p}^{2}\times{\bf{Z}}_{p}^{(n-2)}$. In the first case
$|N^{ab}\otimes K^{ab}|=|{\bf{Z}}_{p}^{(n-1)}\otimes K^{ab}|\leq
p^{d(K)(n-1)}$. Now suppose that $d(K)<4$. Then we have $|N^{ab}\otimes
K^{ab}|\leq p^{3(n-1)}$. Therefore $|M(N)|\geq p^{-3(n-1)+(8n+n^{2}-n-10)/2}$,
by Corollary 1.2. So $n<2$ which is impossible. If $d(K)=4$, then we have
$K\cong{\bf{Z}}_{p}^{(4)}$. Hence $|N^{ab}\otimes K^{ab}|=p^{4(n-1)}$ which
implies that $|M(N)|=p^{(n^{2}-n)/2-1}$. So $N\cong E_{1}$, by Theorem 1.3.
In the second case, suppose that $K$ is not abelian. Then $|N^{ab}\otimes
K^{ab}|=|{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}^{(n-2)}\otimes
K^{ab}|=|{\bf{Z}}_{p^{2}}\otimes K^{ab}||{\bf{Z}}_{p}^{(n-2)}\otimes
K^{ab}|\leq p^{3}p^{d(K)(n-2)}\leq p^{3(n-1)}$ which implies that $n<2$ and it
is impossible.
If $K$ is abelian, then $K\cong{\bf{Z}}_{p^{2}}\times{\bf{Z}}_{p}^{(2)}$. thus
$|N^{ab}\otimes K|={p}^{3n-2}$. It follows that $N\cong{\bf{Z}}_{p^{2}}$. This
completes the proof. ∎

## References

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Fahimeh Mohammadzadeh
Department of Mathematics, Payame Noor University 19395-4697 Tehran, Iran
Email<EMAIL_ADDRESS>

Azam Hokmabadi
Department of Mathematics, Payame Noor University 19395-4697 Tehran, Iran
Email<EMAIL_ADDRESS>

Behrooz Mashayekhy
Department of Mathematics, Ferdowsi University of Mashhad, P.O.Box 1159-91775,
Mashhad, Iran
Email<EMAIL_ADDRESS>