Classification of finite $p$-groups by the size of their Schur multipliers

TL;DR

This paper characterizes the structure of finite p-groups that reach the maximum possible order of their Schur multiplier, based on their minimal generating set and derived subgroup size, and shows these groups are capable.

Contribution

It determines the structure of all p-groups attaining the upper bound of their Schur multiplier's order and proves their capability.

Findings

Identified the structure of p-groups reaching the Schur multiplier bound.

Proved all such groups are capable.

Provided a classification for these extremal p-groups.

Abstract

Let $d(G)$ be the minimum number of elements required to generated a group $G.$ For a group $G $ of order $p^n$ with derived subgroup of order $ p^k $ and $d(G) = d,$ we knew the order of the Schur multiplier of $G$ is bounded by $ p^{\frac{1}{2}(d-1)(n-k+2)+1}. $ In the current paper, we find the structure of all $p$-groups that attains the mentioned bound. Moreover, we show that all of them are capable.

Equations

$$|\mathcal{M}(G)|\le p^{\frac{1}{2}(n-k-1)(n+k-2)+1}.$$

$$|\mathcal{M}(G)|\le p^{\frac{1}{2}(d-1)(n+k-2)+1}.$$

$$1_{G\wedge G}$$

$$\displaystyle=\overset{p^{m-1}-1}{\prod_{i=1}}(b\wedge[b,a])^{i}\overset{p^{m-1}-1}{\prod_{i=0}}(b\wedge a)=\big{(}\overset{p^{m-1}-1}{\prod_{i=1}}(b\wedge[b,a])^{i}\big{)}(b\wedge a)^{p^{m-1}}.$$

$$\stackrel{p^{m-1}-1}{\prod _{i=1}}(b\wedge [b,a]{)}^i$$

$$=b\wedge [b,a{]}^{\frac{1}{2}{p}^{m-1}(p^{m-1}-1)}$$

$$b^{p^{m-1}}\wedge a=(b\wedge a{)}^{p^{m-1}}=1_{G\wedge G}.$$

$$a^{p^{m-1}}\wedge b$$

$$=\stackrel{p^{m-1}-1}{\prod _{i=0}}(a\wedge [a^i,b]b)$$

$$=\stackrel{p^{m-1}-1}{\prod _{i=0}}(a\wedge b)=(a\wedge b{)}^{p^{m-1}},$$

$$\Psi_{2}:(G/Z(G))^{(ab)}\otimes(G/Z(G))^{(ab)}\otimes(G/Z(G))^{(ab)}\rightarrow\big{(}G^{\prime}/\gamma_{3}(G)\big{)}\otimes G/G^{\prime}$$

$$([x,y]{\gamma }_3(G)\otimes z{G}^{\prime })([z,x]{\gamma }_3(G)\otimes y{G}^{\prime })([y,z]{\gamma }_3(G)\otimes x{G}^{\prime })$$

$${\Psi }_3:$$

$$\to {\gamma }_3(G)/{\gamma }_4(G)\otimes (G/Z(G){)}^{(ab)}\text{given by}$$

$$\displaystyle\big{(}xG^{\prime}Z(G)\big{)}\otimes\big{(}yG^{\prime}Z(G)\big{)}\otimes\big{(}zG^{\prime}Z(G)\big{)}\otimes\big{(}wG^{\prime}Z(G)\big{)}$$

$$\displaystyle\mapsto\big{(}[[x,y],z]\gamma_{4}(G)\otimes wG^{\prime}Z(G)\big{)}\big{(}[w,[x,y]]\gamma_{4}(G)\otimes zG^{\prime}Z(G)\big{)}$$

$$\displaystyle\big{(}[[z,w],x]\gamma_{4}(G)\otimes yG^{\prime}Z(G)\big{)}\big{(}[y,[z,w]]\gamma_{4}(G)\otimes xG^{\prime}Z(G)\big{)}$$

$$|G\wedge G||\mathrm{Im}{\Psi }_2||\mathrm{Im}{\Psi }_3|$$

$$=|\mathcal{M}(G/G^{\prime })|\prod _{i=2}^c|{\gamma }_i(G)/{\gamma }_{i+1}(G)\otimes G/G^{\prime }|\le |\mathcal{M}(G/G^{\prime })|p^{kd},$$

$${\alpha }_i:{\gamma }_i(G)\wedge G\to {\gamma }_{i-1}(G)\wedge G$$

$$x\wedge z\mapsto x\wedge z$$

$$|G\wedge G|\prod _{i=2}^c|\ker {\alpha }_i|=|G/G^{\prime }\wedge G/G^{\prime }|\prod _{i=2}^c|{\gamma }_i(G)/{\gamma }_{i+1}(G)\otimes G/G^{\prime }|.$$

$$|G\wedge G|\prod _{i=2}^c|\ker {\alpha }_i|=|\mathcal{M}(G)||G^{\prime }|\prod _{i=2}^c|\ker {\alpha }_i|$$

$$=|\mathcal{M}(G/G^{\prime })|\prod _{i=2}^c|{\gamma }_i(G)/{\gamma }_{i+1}(G)\otimes G/G^{\prime }|\le |\mathcal{M}(G/G^{\prime })|p^{kd}.$$

$${\delta }_i:{\gamma }_i(G)/{\gamma }_{i+1}(G)\otimes G/G^{\prime }\to {\gamma }_i(G)/{\gamma }_{i+1}(G)\otimes G/Z(G)G^{\prime }$$

$$|\mathcal{M}(G)||G^{\prime }||\mathrm{Im}{\Psi }_2||\mathrm{Im}{\Psi }_3|$$

$$=|\mathcal{M}(G/G^{\prime })|\prod _{i=2}^c|{\gamma }_i(G)/{\gamma }_{i+1}(G)\otimes G/G^{\prime }|/|\ker {\delta }_i|$$

$$=|\mathcal{M}(G)||G^{\prime }|\prod _{i=2}^c|\ker {\alpha }_i|/|\ker {\delta }_i|$$

$$|G\wedge G||\mathrm{Im}{\Psi }_2||\mathrm{Im}{\Psi }_3|$$

$$=|\mathcal{M}(G)||G^{\prime }|\prod _{i=2}^c|\ker {\alpha }_i|$$

$$=|\mathcal{M}(G/G^{\prime })|\prod _{i=2}^c|{\gamma }_i(G)/{\gamma }_{i+1}(G)\otimes G/G^{\prime }|$$

$$\le |\mathcal{M}(G/G^{\prime })|p^{kd}.$$

$$\eta :G^{\prime }\otimes G/G^{\prime }\to \mathcal{M}(G)=(R\cap F^{\prime })/[R,F]$$

$$x\otimes (z{G}^{\prime })\mapsto [\tilde{x},\tilde{z}][R,F]$$

$$⟨([x,y]\otimes z{G}^{\prime })([z,x]\otimes y{G}^{\prime })([y,z]\otimes x{G}^{\prime }),w^{p^s}\otimes w{G}^{\prime }\mid x,y,z,w\in G⟩\subseteq \ker \eta .$$

$$1\to \ker \eta \to G^{\prime }\otimes G/G^{\prime }\stackrel{\eta }{\to }\mathcal{M}(G)\to \mathcal{M}(G/G^{\prime })\to G^{\prime }\to 1$$

$$\eta :G^{\prime }\otimes G/G^{\prime }\to \mathcal{M}(G)=(R\cap F^{\prime })/[R,F]$$

$$x\otimes z{G}^{\prime }\mapsto [\tilde{x},\tilde{z}][R,F]$$

Full text

Mathematics Subject Classification 2010. Primary 20D15; Secondary 20E34. 20F18

1. Motivation and Preliminaries

The Schur multiplier, $\mathcal{M}(G),$ of a group $G$ first appeared in $1904$ in the work of Schur on projective representations of groups.

The Schur multiplier was studied by several authors and proved to be an important tool in the classification of $p$-groups. For a group $G$ of order $p^{n},$ by a result of Green in [10], we have $|\mathcal{M}(G)|\leq p^{\frac{1}{2}n(n-1)-t(G)}$ with $t(G)\geq 0.$ Several authors characterized the structure of $p$-groups by using $t(G).$ The reader can find the structure of $p$-groups when $t(G)\in\{0,\ldots,6\}$ (see [1, 7, 11, 17, 22, 23]). Later, the first author [16] improved Green’s bound and showed for any non-abelian group $G$ of order $p^{n}$ with $|G^{\prime}|=p^{k},$ we have

[TABLE]

He also characterized all of $p$-groups that attain the upper bound when $k=1.$ Recently, Rai [18] improved this bound. He showed for a $p$-group $G$ of order $p^{n}$ with $|G^{\prime}|=p^{k}$ and $d(G)=d,$ we have

[TABLE]

In the present paper, we are going to find the structure of all $p$-groups that attain the bound (1.2), and then we show that all of them are capable.

The concept of the non-abelian tensor square $G\otimes G$ of a group $G$ is a special case of the non-abelian tensor product of two arbitrary groups that was introduced by Brown and Loday [5]. It is easy to check that $\kappa:G\otimes G\rightarrow G^{\prime}$ given by $g\otimes g^{\prime}\rightarrow[g,g^{\prime}]$ for all $g,g^{\prime}\in G$ is an epimorphism. Let $J_{2}(G)$ be the kernel of $\kappa,$ and let $\bigtriangledown(G)$ be a subgroup of $G\otimes G$ generated by the set $\{g\otimes g\mid g\in G\}.$ Clearly, $\bigtriangledown(G)$ is a central subgroup of $G\otimes G.$ The non-abelian exterior square $G\wedge G$ is the quotient group $\dfrac{G\otimes G}{\bigtriangledown(G)}.$ The element $(g\otimes g^{\prime})\bigtriangledown(G)$ in $G\wedge G$ is denoted by $g\wedge g^{\prime}$ for all $g,g^{\prime}\in G.$ The map $\kappa$ induces the epimorphism $\kappa^{\prime}:G\wedge G\rightarrow G^{\prime}$ given by $g\wedge g^{\prime}\rightarrow[g,g^{\prime}]$ for all $g,g^{\prime}\in G.$ The kernel of the map $\kappa^{\prime}$ is isomorphic to the Schur multiplier of $G$ (for more information, see [5]).

Recall that a group $G$ is called capable provided that $G\cong H/Z(H)$ for a group $H.$ Beyl et al. [3] gave a criterion for detecting capable groups. They showed that a group $G$ is capable if and only if the epicenter of $G,$ $Z^{*}(G),$ is trivial. Ellis [9] showed $Z^{\wedge}(G)=Z^{*}(G),$ where $Z^{\wedge}(G)$ is the exterior center of $G,$ i.e. the set of all elements $g$ of $G$ for which $g\wedge h=1_{G\wedge G}$ for all $h\in G$ (see for instance [9] to find more information in these topics).

The following technical result characterizes the structure of all minimal non-abelian $p$-groups.

Lemma 1.1**.**

[2, Exercise 8a.]** and [21] Let $G$ be a minimal non-abelian $p$-group. Then $|G^{\prime}|=p$ and $G$ is isomorphic to one of the following groups:

• $(a).$

$G\cong\langle a,b\mid a^{p^{m}}=1,b^{p^{n}}=1,[a,b]=a^{p^{m-1}},[a,b,a]=[a,b,b]=1\rangle$* for all $m,n$ such that $m\geq 2,n\geq 1.$ Moreover, $|G|=p^{m+n}$ and $Z(G)=\langle a^{p}\rangle\times\langle b^{p}\rangle$.*

• $(b).$

$G\cong\langle a,b\mid a^{p^{m}}=b^{p^{n}}=[a,b]^{p}=1,[a,b,a]=[a,b,b]=1\rangle$* is of order $p^{m+n+1}$ and if $p=2,$ then $m+n>2.$ Moreover, $Z(G)=\langle a^{p}\rangle\times\langle b^{p}\rangle\times\langle[a,b]\rangle.$*

• $(c).$

$G\cong Q_{8}.$**

Let $\mathbb{Z}_{n}^{(t)}$ denote the direct sum of $t$ copies of $\mathbb{Z}_{n},$ in which $\mathbb{Z}_{n}$ is the cyclic group of order $n.$

Proposition 1.2**.**

Let $G$ be a minimal non-abelian $p$-group as in Lemma 1.1$(a),$ where $n\geq 1,$ and $m\geq 2$ if $p>2,$ $m\geq 3$ if $p=2.$ If $G/G^{\prime}$ is homocyclic, then $n=m-1,$ $G$ is non-capable, and $Z^{\wedge}(G)=G^{\prime}.$

Proof.

Since $G/G^{\prime}$ is homocyclic and $G/G^{\prime}\cong\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m-1}},$ $n=m-1.$ We claim that $Z^{\wedge}(G)=G^{\prime}.$ [3, Corollary 7.4] implies $G/G^{\prime}$ is capable and so $Z^{\wedge}(G)\subseteq G^{\prime}\cong\mathbb{Z}_{p},$ by [3, Corollary 2.2]. It is sufficient to show that $G^{\prime}=\langle a^{p^{m-1}}\rangle\subseteq Z^{\wedge}(G).$ Since $b^{{}^{p^{m-1}}}=1,$ we have

[TABLE]

Clearly,

[TABLE]

Since $[a,b]^{p}=1,$ we get $[b,a]^{\frac{1}{2}p^{m-1}(p^{m-1}-1)}=1,$ and so $\displaystyle\overset{p^{m-1}-1}{\prod_{i=1}}(b\wedge[b,a])^{i}=1.$ Thus

[TABLE]

We will show that $a^{p^{m-1}}\wedge b=(a\wedge b)^{p^{m-1}}.$ Since

[TABLE]

we have $a^{p^{m-1}}\wedge b=(a\wedge b)^{p^{m-1}}=(b\wedge a)^{-p^{m-1}}=(b^{p^{m-1}}\wedge a)^{-1}=1_{G\wedge G}.$ Moreover, $a\wedge[a,b]=a\wedge a^{p^{m-1}}=1_{G\wedge G}.$ Therefore $1\neq a^{p^{m-1}}\in Z^{\wedge}(G)$ and so $Z^{\wedge}(G)=G^{\prime}.$ Hence $G$ is non-capable. ∎

Let $d(G)$ denote the minimum number of elements required to generate a group $G.$

Lemma 1.3**.**

Let $G$ be a capable non-abelian $p$-group of order $p^{n}$ such that $|G^{\prime}|=p,$ $d(G)=d,$ and $e(G/G^{\prime})>p.$ Then $|G/Z(G)|=p^{2}$ and $G=NZ(G),$ where $N$ is a minimal non-abelian $p$-group.

Proof.

[12, Theorem C] implies that $|G/Z(G)|=p^{2}.$ Using [2, Lemma 4.2], we have $G=NZ(G),$ where $N$ is minimal non-abelian. ∎

Let $\gamma_{i}(G)$ be denoted the $i$-th term of the lower central series of a group $G.$ We need the following result in the proof of Theorem 1.5.

Proposition 1.4**.**

[8, Proposition 1]** and [6, 11] Let $G$ be a finite non-abelian $p$-group of class $c.$

• $(i)$

The map

[TABLE]

*given by *$xG^{\prime}Z(G)\otimes yG^{\prime}Z(G)\otimes zG^{\prime}Z(G)\mapsto$

[TABLE]

is a homomorphism. If any two elements of the set $\{x,y,z\}$ are linearly dependent, then \Psi_{2}(xG^{\prime}Z(G)\otimes yG^{\prime}Z(G)\otimes zG^{\prime}Z(G))=1_{\big{(}G^{\prime}/\gamma_{3}(G)\big{)}\otimes G/G^{\prime}}.

• $(ii)$

The map

[TABLE]

is a homomorphism.

Theorem 1.5**.**

Let $G$ be a non-abelian group of order $p^{n}$ of class $c$ with $|G^{\prime}|=p^{k}$ and $d=d(G).$ Then

[TABLE]

where

[TABLE]

[TABLE]

is a natural homomorphism for all $i$ such that $i\geq 2.$

Proof.

Similar to the proof of [6, Proposition 5], we have

[TABLE]

Since $|G\wedge G|=|\mathcal{M}(G)||G^{\prime}|$ and using the proof of [18, Theorem 1.2], we have

[TABLE]

We claim that $|\mathrm{Im}\Psi_{2}||\mathrm{Im}\Psi_{3}|\leq\prod_{i=2}^{c}|\ker\alpha_{i}|.$ Clearly, the map

[TABLE]

given by $x\gamma_{i+1}(G)\otimes yG^{\prime}\mapsto x\gamma_{i+1}(G)\otimes yG^{\prime}Z(G)$ is a natural epimorphism for all $i\geq 2.$ By [20, Proposition 2.1], we have

[TABLE]

and so $|\mathrm{Im}\Psi_{2}||\mathrm{Im}\Psi_{3}|\leq\prod_{i=2}^{c}|\ker\alpha_{i}|/|\ker\delta_{i}|\leq\prod_{i=2}^{c}|\ker\alpha_{i}|.$ Therefore

[TABLE]

The proof is completed. ∎

Let $1\rightarrow R\rightarrow F\xrightarrow{\pi}G\rightarrow 1$ be a free presentation for a group $G$ and the exponent of a group $X$ is denoted by $e(X).$

The next result is extracted from the work of Blackburn and Evens in [4, Remark, Section 3].

Theorem 1.6**.**

Let $G$ be a non-abelian $p$-group of class two and $e(G/G^{\prime})=p^{s}$. Then $1\rightarrow\ker\eta\rightarrow G^{\prime}\otimes G/G^{\prime}\xrightarrow{\eta}\mathcal{M}(G)\rightarrow\mathcal{M}(G/G^{\prime})\rightarrow G^{\prime}\rightarrow 1$ is exact, in which

[TABLE]

[TABLE]

such that $\pi(\tilde{x}R)=x$ and $\pi(\tilde{z}R)=z.$ Moreover,

[TABLE]

Proof.

By [14, Corollary 3.2.4], we get

[TABLE]

is exact, in which

[TABLE]

[TABLE]

such that $\pi(\tilde{x}R)=x$ and $\pi(\tilde{z}R)=z.$ Similar to the proof of [4, Theorem 3.1], we have $\langle([x,y]\otimes zG^{\prime})([z,x]\otimes yG^{\prime})([y,z]\otimes xG^{\prime}),w^{p^{s}}\otimes wG^{\prime}\mid x,y,z,w\in G\rangle\subseteq\ker\eta,$ as required. ∎

Lemma 1.7**.**

Let $G$ be a group of class two such that $d(G/Z(G))=d$ is finite. Then $d(G^{\prime})\leq\frac{1}{2}d(d-1).$

Proof.

We can choose a generating set $\{x_{1}Z(G),\ldots,x_{d}Z(G)\}$ for $G/Z(G)$ such that $[x_{i},x_{j}]$ is non-trivial for $i\neq j.$ It is clear to see that $\{[x_{i},x_{j}]\mid 1\leq i<j\leq d\}$ generates $G^{\prime},$ as required. ∎

Lemma 1.8**.**

Let $G$ be a group.

• $(i)$

If $G\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p^{m}}=1,[a,b,a]=[a,b,b]=1,m\geq 2\rangle$ for $p\neq 2,$ then $\mathcal{M}(G)\cong\mathbb{Z}_{p^{m}}\times\mathbb{Z}_{p^{m}}.$

• $(ii)$

If $G\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1,1\leq k<m\rangle,$ then $\mathcal{M}(G)\cong\mathbb{Z}_{p^{m-k}}\times\mathbb{Z}_{p^{k}}\times\mathbb{Z}_{p^{k}}.$

Proof.

It is clearly obtained by [15, Theorems 49 and 50]. ∎

Theorem 1.9**.**

[19, Theorem 1.1]** Let $G$ be a non-abelian group of order $p^{n}$ of class two and $|G^{\prime}|=p^{k}.$ Then $|\mathcal{M}(G)|=p^{\frac{1}{2}(n-k-1)(n+k-2)+1}$ if and only if $G$ is isomorphic to one of the following groups:

For $p\neq 2,$ $G_{1}\cong E_{1}\times\mathbb{Z}_{p}^{(n-3)},$ where $E_{1}$ is the extra-special $p$-group of order $p^{3}$ and exponent $p.$

For $p\neq 2,$ $G_{2}\cong\mathbb{Z}_{p}^{(4)}\rtimes\mathbb{Z}_{p}.$

For $p\neq 2,$

[TABLE]

[TABLE]

2. main results

As proven in [18, Theorem 1.1], the Schur multiplier of a non-abelian group $G$ of order $p^{n}$ with $|G^{\prime}|=p^{k}$ and $d(G)=d$ is bounded by $p^{\frac{1}{2}(d-1)(n+k-2)+1}.$ Let $p$ is an odd prime number. The main result of this paper is devoted to characterizing the structure of all finite $p$-groups that attain the mentioned upper bound. Moreover, we show that all $p$-groups that attain the bound are capable. Throughout the paper, we say that $|\mathcal{M}(G)|$ attains the bound provided that $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}.$

Main Theorem**.**

Let $G$ be a non-abelian group of order $p^{n}$ with $p\neq 2,$ $|G^{\prime}|=p^{k},$ and $d(G)=d.$ Then $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}$ if and only if $G$ is isomorphic to one of the following groups:

• $(i)$

$H_{1}\cong E_{1}\times\mathbb{Z}_{p}^{(n-3)},$ where $E_{1}$ is the extra-special $p$-group of order $p^{3}$ and exponent $p.$

• $(ii)$

$H_{2}\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p}=1,[a,b,a]=[a,b,b]=1\rangle,$ where $m>1.$

• $(iii)$

$H_{3}\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p^{m}}=1,[a,b,a]=[a,b,b]=1\rangle,$ where $m\geq 2.$

• $(iv)$

$H_{4}\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1,k\geq 2,m\geq 2\rangle.$

• $(v)$

$H_{5}\cong\mathbb{Z}_{p}^{(4)}\rtimes\mathbb{Z}_{p}.$

• $(vi)$

$H_{6}\cong\langle x_{1},x_{2},x_{3}\mid[x_{1},x_{2}]^{p}=[x_{2},x_{3}]^{p}=[x_{3},x_{1}]^{p}=x_{i}^{p}=1,$

[TABLE]

• $(vii)$

$H_{7}\cong\langle x_{1},y_{1},x_{2},y_{2},x_{3},y_{3},z\mid[x_{1},x_{2}]=y_{3},[x_{2},x_{3}]=y_{1},[x_{3},x_{1}]=y_{2},$

[TABLE]

Let $cl(X)$ be used to denote nilpotency class of a group $X$. We begin with the following lemma for the future convenience.

Lemma 2.1**.**

Let $G$ be a non-abelian group of order $p^{n}$ such that $|G^{\prime}|=p^{k},$ $d(G)=d,$ and $G/G^{\prime}\cong\mathbb{Z}_{p^{\alpha_{1}}}\oplus\ldots\oplus\mathbb{Z}_{p^{\alpha_{d}}}$, where $\alpha_{1}\geq\ldots\geq\alpha_{d}.$ If $|\mathcal{M}(G)|$ attains the bound, then

• $(i)$

$G/G^{\prime}$* is homocyclic.*

• $(ii)$

$\text{Im}~{}\Psi_{2}\cong\mathbb{Z}_{p}^{(d-2)},$* $\text{Im}~{}\Psi_{3}=1,$ and $|\text{Im}~{}\Psi_{2}|=|\ker\alpha_{2}|.$*

Proof.

• $(i)$

By using [18, Corollary 1.3], we have

[TABLE]

Thus $\alpha_{1}=\alpha_{d}$ and so $G/G^{\prime}$ is homocyclic.

• $(ii)$

Let $cl(G)=c.$ By part $(i),$ $G/G^{\prime}$ is homocyclic and so $n=d\alpha_{1}+k.$ Theorem 1.5 and [14, Corollary 2.2.12] imply

[TABLE]

Thus

[TABLE]

Consider $G=\langle x_{1},x_{2},\ldots,x_{d}\rangle$ and $1\neq[x_{1},x_{2}]\in G^{\prime}\setminus\gamma_{3}(G)$. We claim that $\text{Im}\Psi_{2}\cong\mathbb{Z}_{p}^{(d-2)}.$ Similar to the proof of [8, Theorem 2], the set $A=\{\Psi_{2}(x_{1}G^{\prime}\otimes x_{2}G^{\prime}\otimes x_{j}G^{\prime})\mid 3\leq j\leq d\}$ consists of $d-2$ linearly independent elements of order at least $p$ in the abelian $p$-group $G/G^{\prime}\otimes G^{\prime}/\gamma_{3}(G).$ Hence $p^{d-2}\leq|\langle A\rangle|\leq|\text{Im}\Psi_{2}|$. By using (2.1) and the proof of Theorem 1.5,

[TABLE]

Hence $|\langle A\rangle|=|\text{Im}\Psi_{2}|=|\ker\alpha_{2}|=p^{d-2}$ and $|\text{Im}\Psi_{3}|=1$ and so $\langle A\rangle=\text{Im}\Psi_{2}\cong\mathbb{Z}_{p}^{(d-2)}.$

∎

Proposition 2.2**.**

Let $G$ be a non-abelian group of order $p^{n}$ with $|G^{\prime}|=p^{k},$ $d(G)=d$ and $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}.$ If $k\geq 2$ and $K$ is a central subgroup of order $p$ contained in $Z(G)\cap G^{\prime},$ then $|\mathcal{M}(G/K)|$ also attains the bound, that is

[TABLE]

Proof.

Let $K\subseteq Z(G)\cap G^{\prime}$ and $|K|=p.$ Lemma 2.1$(i)$ implies $G/G^{\prime}\cong\mathbb{Z}^{(d)}_{p^{m}}$ for some $m\geq 1,$ and so $G/G^{\prime}\otimes K\cong\mathbb{Z}^{(d)}_{p}.$ Using [13, Theorem 4.1], we have

[TABLE]

Thus

[TABLE]

Therefore $|\mathcal{M}(G/K)|=p^{\frac{1}{2}(d-1)(n+k-4)+1},$ as required. ∎

The following lemma shows that if the order of the Schur multiplier of a non-abelian $p$-group $G$ attains the bound, then $G$ is capable.

Lemma 2.3**.**

Let $G$ be a non-abelian $p$-group of order $p^{n}$ with $|G^{\prime}|=p^{k}$ and $d(G)=d.$ If $|\mathcal{M}(G)|$ attains the bound, then $G$ is capable.

Proof.

Lemma 2.1$(i)$ implies $G/G^{\prime}$ is homocyclic. Therefore $G/G^{\prime}$ is capable, by [3, Corollary 7.4]. Hence $Z^{*}(G)\subseteq G^{\prime},$ by [3, Corollary 2.2]. Assume to the contrary that $G$ is non-capable. Then there is a normal subgroup $K$ of order $p$ in $Z^{*}(G).$ By using [3, Theorem 4.2] and [14, Theorem 2.5.6$(i)$], we have $|\mathcal{M}(G)|=|\mathcal{M}(G/K)|p^{-1}.$ Proposition 2.2 implies that $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-4)},$ which is a contradiction, since by our assumption, we have $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}.$ Thus $Z^{*}(G)=1$ and so the result follows from [3, Corollary 2.3]. ∎

Proposition 2.4**.**

Let $G$ be a non-abelian finite $p$-group of order $p^{n}$ such that $|G^{\prime}|=p$ and $d(G)=d.$ If $e(G/G^{\prime})>p,$ then $|\mathcal{M}(G)|$ attains the bound if and only if $G\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p}=1,[a,b,a]=[a,b,b]=1\rangle,$ where $m\geq 1.$

Proof.

Assume that $|\mathcal{M}(G)|$ attains the bound. By Lemmas 1.3 and 2.3, we have $|G/Z(G)|=p^{2}$ and $G=NZ(G),$ where $N$ is a minimal non-abelian $p$-group. Since $G/G^{\prime}$ is homocyclic, we get $G/G^{\prime}\cong\mathbb{Z}_{p^{\alpha_{1}}}^{(d)}$ and $G/Z(G)\cong\mathbb{Z}_{p}\oplus\mathbb{Z}_{p}.$ Since $G=NZ(G),$ we have $G^{\prime}=N^{\prime}=\langle c\rangle$ such that $N=\langle a,b\rangle$ and $c=[a,b].$ We claim that $G=N.$

Assume that $G\neq N.$ Choose an element $y\in Z(G)\setminus G^{\prime}.$ Thus $c\otimes yG^{\prime}$ is a non-trivial element in $G^{\prime}\otimes G/G^{\prime}.$ By using Theorem 1.6, we can see that $c\otimes yG^{\prime}$ is non-trivial in $\ker\eta$ and so

[TABLE]

Thus $|\mathcal{M}(G)|\leq p^{\frac{1}{2}d(d-1)\alpha_{1}+d-2}.$ Now, since $n=\alpha_{1}d+1,$ by our assumption we have $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+1-2)+1}=p^{\frac{1}{2}d(d-1)\alpha_{1}+1}.$ It is a contradiction. Now $G=NZ(G)=N.$ So, $G$ is a capable minimal non-abelian $p$-group. By Lemma 1.1, Proposition 1.2, and [3, Corollary 8.2], we get

[TABLE]

where $m>1.$ The converse holds by Lemma 1.8. ∎

Proposition 2.5**.**

Let $G$ be a non-abelian group of order $p^{n}$ of class $t$ such that $|G^{\prime}|=p^{k},$ $d(G)=d$ and $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}$ for all $k$ such that $k\geq 2.$ If $K$ is a non-trivial central subgroup of order $p^{m}$ contained in $Z(G)\cap G^{\prime}\neq G^{\prime},$ then $|\mathcal{M}(G/K)|$ also attains the bound, that is

[TABLE]

Proof.

Let $K$ be a non-trivial central subgroup of order $p^{m}$ contained in $Z(G)\cap G^{\prime}.$ We have $|G/K|=p^{n-m}$ and $|(G/K)^{\prime}|=p^{k-m}.$ We prove the result by using induction on $m.$ If $m=1,$ then the result holds by Proposition 2.2. Now let $m\geq 2.$ Consider a normal subgroup $K_{1}$ in $K$ of order $p^{m-1}$ and using the induction hypothesis, we have

[TABLE]

Since $K/K_{1}\subseteq Z(G/K_{1})\cap(G^{\prime}/K_{1})$ and $|K/K_{1}|=p,$ Proposition 2.2 implies that

[TABLE]

This completes the proof. ∎

The proof of the following corollary is similar to that of [20, Theorem 1.2].

Corollary 2.6**.**

Let $G$ be a non-abelian group of order $p^{n}$ of class $t\geq 3$ such that $|G^{\prime}|=p^{k},$ $d(G)=d$ and $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}$ for $k\geq 2.$ Then $|\mathcal{M}(G/\gamma_{i}(G))|$ also attains the bound for all $i$ such that $3\leq i\leq t.$

Proof.

We prove the result by using induction on $j=t-i+3$ for all $i$ such that $3\leq i\leq t.$ If $j=3,$ then since $k\geq 2,$ by using Proposition 2.5, $|\mathcal{M}(G/\gamma_{t}(G))|$ attains the bound. Using the induction hypothesis, $|\mathcal{M}(G/\gamma_{i}(G))|$ attains the bound.

Since $\gamma_{i-1}(G)/\gamma_{i}(G)\subseteq Z(G/\gamma_{i}(G))\cap G^{\prime}/\gamma_{i}(G),$ Proposition 2.5 implies that

[TABLE]

attains the bound, as required. ∎

Proposition 2.7**.**

Let $G$ be a non-abelian group of order $p^{n}$ such that $G/G^{\prime}\cong\mathbb{Z}_{p^{m}}^{(2)}$ and $G^{\prime}\cong\mathbb{Z}_{p^{k}}$ with $k\geq 2.$ Then $|\mathcal{M}(G)|$ attains the bound if and only if

[TABLE]

where $m\geq 2$ and $p\neq 2$ or

[TABLE]

Proof.

Let $|\mathcal{M}(G)|$ attain the bound. Now by [15, Theorem 1], we get

[TABLE]

where $\alpha+\beta+\gamma=n,$ $1\leq\gamma\leq\beta\leq\alpha,$ and $0\leq\rho,\sigma\leq\gamma.$ Since $G/G^{\prime}\cong\mathbb{Z}_{p^{m}}^{(2)},$ we have $\alpha=\beta.$ By Lemma 2.3 and [15, Theorems 63 and 67$(i)$], $\rho=\sigma=\gamma.$ Using [15, Theorem 1], we conclude that

[TABLE]

where $m\geq 2$ and $p\neq 2$ or

[TABLE]

The converse holds by Lemma 1.8. ∎

Now we are ready to obtain the structure of a $p$-group $G$ of class two such that $|\mathcal{M}(G)|$ attains the bound.

Lemma 2.8**.**

Let $G$ be a non-abelian group of order $p^{n}$ of class two such that $|G^{\prime}|\geq p^{2}$ and $d(G)=d.$ If $|\mathcal{M}(G)|$ attains the bound, then $2\leq d\leq 3.$

Proof.

If $e(G/G^{\prime})=p,$ then $d=n-k.$ By using Theorem 1.9, $2\leq d\leq 3.$ Let $e(G/G^{\prime})>p.$ If $k=1,$ then Proposition 2.4 implies $d=2.$ Now, assume that $k\geq 2$ and $K\subsetneqq G^{\prime}$ such that $|G^{\prime}/K|=p.$ By Proposition 2.5, $|\mathcal{M}(G/K)|$ attains the bound so $d=d(G/K)=2,$ by Proposition 2.4. Hence, $2\leq d\leq 3.$ ∎

Lemma 2.9**.**

There exists no non-abelian group $G$ of order $p^{n}$ of class two such that $d(G)=3,$ $e(G^{\prime})\geq p^{2},$ $e(G/G^{\prime})\geq p^{2},$ and $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}.$

Proof.

Assume to the country that there is a such group $G.$ Clearly, $e(G/Z(G))=e(G^{\prime})=p^{k}.$ Using Lemma 1.7, $d(G^{\prime})\leq 3.$ Now let $G^{\prime}\cong\mathbb{Z}_{p^{k}}.$ Consider the factor group $G/G^{\prime p}.$ Obviously, $(G/G^{\prime p})^{\prime}\cong\mathbb{Z}_{p}$ and $d(G/G^{\prime p})=3.$ Using Proposition 2.5, $|\mathcal{M}(G/G^{\prime p})|$ also attains the bound, so Proposition 2.4 implies $d(G/G^{\prime p})=2.$ It is a contradiction. By a similar way, if $2\leq d(G^{\prime})\leq 3,$ then we get a contradiction. ∎

Theorem 2.10**.**

Let $G$ be a non-abelian group of order $p^{n}$ of class two such that $|G^{\prime}|=p^{k}$ and $d(G)=d.$ Then $|\mathcal{M}(G)|$ attains the bound if and only if $G$ is isomorphic to one of the following groups:

• $(i)$

For $p\neq 2,$ $H_{1}\cong E_{1}\times\mathbb{Z}_{p}^{(n-3)},$ where $E_{1}$ is the extra-special $p$-group of order $p^{3}$ and exponent $p.$

• $(ii)$

$H_{2}\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p}=1,[a,b,a]=[a,b,b]=1\rangle,$* where $m>1.$*

• $(iii)$

For $p\neq 2,$ $H_{3}\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p^{m}}=1,[a,b,a]=[a,b,b]=1\rangle,$ where $m\geq 2.$

• $(iv)$

$H_{4}\cong\langle a,b\mid a^{p^{m}}=b^{p^{m}}=[a,b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1,k\geq 2,m\geq 2\rangle.$**

• $(v)$

For $p\neq 2,$ $H_{5}\cong\mathbb{Z}_{p}^{(4)}\rtimes\mathbb{Z}_{p}.$

• $(vi)$

*For $p\neq 2,$ *$H_{6}\cong\langle x_{1},x_{2},x_{3}\mid[x_{1},x_{2}]^{p}=[x_{2},x_{3}]^{p}=[x_{3},x_{1}]^{p}=x_{i}^{p}=1,$

[TABLE]

Proof.

Suppose that $|\mathcal{M}(G)|$ attains the bound. First assume that $|G^{\prime}|=p.$ By Lemma 2.1$(i),$ $G/G^{\prime}$ is homocyclic. Theorem 1.9$(i)$ and Proposition 2.4 imply $G\cong E_{1}\times\mathbb{Z}_{p}^{(n-3)}\cong H_{1}$ or

[TABLE]

where $m>1.$

Now suppose that $|G^{\prime}|\geq p^{2}.$ By using Lemma 2.8, $2\leq d\leq 3.$ Let $d=2.$ By Proposition 2.7, we have

[TABLE]

where $m\geq 2$ and $p\neq 2$ or

[TABLE]

Thus $G\cong H_{3}$ or $G\cong H_{4}.$

Let now $d=3.$ By Lemma 2.9, $G/G^{\prime}$ is of exponent $p$ and hence $G^{\prime}=\phi(G).$ Thus $d(G)=n-k=3.$ Therefore

[TABLE]

Using Theorem 1.9, $G\cong H_{5}$ or $G\cong H_{6}.$ The converse follows from Propositions 2.4, 2.7 and Theorem 1.9. The proof is completed. ∎

Proposition 2.11**.**

There exists no non-abelian $2$-generator group $G$ of order $p^{n}$ of class $t\geq 3$ such that $|G^{\prime}|\geq p^{2}$ and $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}$ with $p\neq 2.$

Proof.

Assume to the contrary that there is a such group $G.$ Let $G=\langle x,y\rangle,$ where $x,y\in G\setminus\phi(G).$ Without loss of generality, we may assume that $[x,y,x]\neq 1.$ So, $\Psi_{3}(xG^{\prime}\otimes yG^{\prime}\otimes xG^{\prime}\otimes yG^{\prime})=([[x,y],x]\otimes yG^{\prime})^{2}([y,[x,y]]\otimes xG^{\prime})^{2}\neq 1.$ On the other hand, by Lemma 2.1$(ii),$ $|\mathrm{Im}\Psi_{3}|=1.$ Hence we have a contradiction. ∎

We are ready to obtain the structures of $G$ when $|\mathcal{M}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}.$

Proof of the Main Theorem**.**

Suppose that $G$ is nilpotent of class two and $|\mathcal{M}(G)|$ attains the bound. Then $G$ is isomorphic to one of the groups $H_{1},H_{2},H_{3},H_{4},H_{5}$ or $H_{6},$ by Theorem 2.10. Now, let $G$ be nilpotent of class at least $3.$ Using Corollary 2.6, $|\mathcal{M}(G/\gamma_{3}(G))|$ attains the bound. Theorem 2.10 implies that $G/\gamma_{3}(G)\cong H_{i}$ for some $i$ in $1\leq i\leq 6$ and so $2\leq d(G/\gamma_{3}(G))\leq 3.$ Since $d(G)=d(G/\gamma_{3}(G)),$ we have $2\leq d(G)\leq 3.$ Thus by Proposition 2.11, $d(G)=3$ and so $d(G/\gamma_{3}(G))=3.$ By Theorem 2.10, $G/\gamma_{3}(G)\cong H_{5}$ or $G/\gamma_{3}(G)\cong H_{6}.$ Hence $(G/\gamma_{3}(G))^{ab}$ is elementary abelian. Thus $d(G)=n-k.$ By using [20, Theorem 1.2]$(iv),$ $G\cong H_{7}.$ The converse holds by Theorem 2.10 and [20, Theorem 1.2].**