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

TL;DR

This paper classifies the structure of pairs of finite p-groups with a normal subgroup and specific bounds on the Schur multiplier, extending previous work to cases where the parameter t equals 4 or 5.

Contribution

It extends the classification of pairs of finite p-groups with a normal subgroup to cases where t=4 or 5, building on earlier results for t ≤ 3.

Findings

Classified the structure of (G,N) when t=4

Classified the structure of (G,N) when t=5

Extended the understanding of Schur multiplier bounds in p-groups

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 $t=4,5$.

Equations

$$H_3(G)$$

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

$$M(G_1\times G_2)=M(G_1)\times M(G_2)\times (G_1^{ab}\otimes G_2^{ab}).$$

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

Full text

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

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

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. ∎