Classification of Finite p-Groups by the Size of Their Schur Multipliers

Abstract

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

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 $|{\mathcal {M}}(G)|\le p^{\frac{1}{2}n(n-1)-t(G)}$ with $t(G)\ge 0.$ 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 author in [16] improved Green’s bound and showed for any non-abelian group G of order $p^n$ with $|G'|=p^k,$ we have

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

(1.1)

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

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

(1.2)

In this 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 $\kappa : G\otimes G \rightarrow G'$ given by $g \otimes g'\rightarrow [g, g' ]$ for all $g,g'\in G$ 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 $(g\otimes g') \bigtriangledown (G)$ in $G\wedge G$ is denoted by $g\wedge g'$ for all $g,g'\in G.$ The map $\kappa$ induces the epimorphism $\kappa ': G\wedge G \rightarrow G'$ given by $g \wedge g'\rightarrow [g, g' ]$ for all $g,g'\in G.$ The kernel of the map $\kappa '$ 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 on 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, $|G'|= p$ 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 $m\ge 2, n\ge 1.$ 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 $n \ge 1,$ and $m \ge 2$ if $p>2,$ $m\ge 3$ if $p=2.$ If $G/G'$ is homocyclic, then $n=m-1,$ G is non-capable, and $Z^{\wedge }(G)=G'.$

Proof

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

$$\begin{aligned} 1_{G\wedge G}&=b^{^{p^{m-1}}}\wedge a=\overset{p^{m-1}-1}{\prod _{i=0}} (^{b^i}(b\wedge a))= \overset{p^{m-1}-1}{\prod _{i=0}} (b\wedge [b^i,a]a)\\ {}&=\overset{p^{m-1}-1}{\prod _{i=1}} (b\wedge [b, a])^i\overset{p^{m-1}-1}{\prod _{i=0}} (b\wedge a)=\left( \overset{p^{m-1}-1}{\prod _{i=1}} (b\wedge [b,a])^i \right) (b\wedge a)^{p^{m-1}}. \end{aligned}$$

Clearly,

$$\begin{aligned} \overset{p^{m-1}-1}{\prod _{i=1}} (b\wedge [b,a])^i&=(b\wedge [b, a])^{\frac{1}{2}p^{m-1}(p^{m-1}-1)} \\&=b\wedge [b, a]^{\frac{1}{2}p^{m-1}(p^{m-1}-1)} \end{aligned}$$

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

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

By a similar way, we will show that $a^{p^{m-1}}\wedge b=(a\wedge b)^{p^{m-1}}.$ We have

$$\begin{aligned} a^{^{p^{m-1}}}\wedge b&=\overset{p^{m-1}-1}{\prod _{i=0}} (^{a^i}(a\wedge b))\\ {}&= \overset{p^{m-1}-1}{\prod _{i=0}} (a\wedge [a^i,b]b)\\ {}&=\overset{p^{m-1}-1}{\prod _{i=1}} (a\wedge [a, b])^i\overset{p^{m-1}-1}{\prod _{i=0}} (a\wedge b)\\ {}&=\left( \overset{p^{m-1}-1}{\prod _{i=1}} (a\wedge a^{p^{m-1}})^i \right) (a\wedge b)^{p^{m-1}}. \end{aligned}$$

Since $[a,b]=a^{p^{m-1}},$ we get $a\wedge [ a, b]= a\wedge a^{p^{m-1}}=1_{G\wedge G}.$ Thus,

$$\begin{aligned} a^{^{p^{m-1}}}\wedge b= (a\wedge b)^{p^{m-1}}. \end{aligned}$$

We conclude that

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

Therefore, $1 \ne a^{p^{m-1}}\in Z^{\wedge }(G)$ and so $Z^{\wedge }(G)=G'.$ Hence, G is non-capable.

$\square$

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 $|G'|=p,$ $d(G)=d,$ and $e(G/G')>p.$ 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. $\square$

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

  $$\begin{aligned} \Psi _2: (G/Z(G))^{(ab)}\otimes (G/Z(G))^{(ab)}\otimes (G/Z(G))^{(ab)} \rightarrow \big (G'/\gamma _3(G)\big )\otimes G/G' \end{aligned}$$

  given by $xG'Z(G)\otimes yG'Z(G) \otimes zG'Z(G)\mapsto$

  $$\begin{aligned} ([x,y]\gamma _3(G)\otimes zG' ) ([z,x]\gamma _3(G)\otimes yG')([y,z]\gamma _3(G)\otimes xG') \end{aligned}$$

  is a homomorphism. If any two elements of the set $\{ x,y,z\}$ are linearly dependent, then $\Psi _2(xG'Z(G) \otimes yG'Z(G) \otimes zG'Z(G))=1_{\big ( G'/\gamma _3(G)\big ) \otimes G/G'}.$

• (ii) The map

  $$\begin{aligned} \Psi _3:&(G/Z(G))^{(ab)}\otimes (G/Z(G))^{(ab)}\otimes (G/Z(G))^{(ab)}\otimes (G/Z(G))^{(ab)}\\ {}&\rightarrow \gamma _3(G)/\gamma _4(G)\otimes (G/Z(G))^{(ab)}~\text {given by}~\\ {}&(x G'Z(G) )\otimes (yG'Z(G) ) \otimes (z G'Z(G))\otimes (w G'Z(G) )\\&\mapsto ([[x,y],z]\gamma _4(G)\otimes wG'Z(G) ) ([w,[x,y]]\gamma _4(G)\otimes zG'Z(G))\\ {}&([[z,w],x]\gamma _4(G)\otimes yG'Z(G))([y,[z,w]]\gamma _4(G)\otimes xG'Z(G)) \end{aligned}$$

  is a homomorphism.

Theorem 1.5

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

$$\begin{aligned} | G\wedge G|| \mathrm {Im}\Psi _2|| \mathrm {Im} \Psi _3|&\le | G\wedge G|\prod _{i=2}^c | \ker \alpha _i|=| {\mathcal {M}}(G)| |G'|\prod _{i=2}^c | \ker \alpha _i|\\ {}&=| {\mathcal {M}}(G/G')|\prod _{i=2}^c|\gamma _i(G)/\gamma _{i+1}(G)\otimes G/G'|\le | {\mathcal {M}}(G/G')| p^{kd},\ \end{aligned}$$

where

$$\begin{aligned}&\alpha _{i}: \gamma _i(G)\wedge G \rightarrow \gamma _{i-1}(G)\wedge G\\&x\wedge z \mapsto x \wedge z \end{aligned}$$

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

Proof

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

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

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

$$\begin{aligned}&| G\wedge G|\prod _{i=2}^c | \ker \alpha _i|=| {\mathcal {M}}(G)| |G'|\prod _{i=2}^c | \ker \alpha _i|\\ {}&=| {\mathcal {M}}(G/G')|\prod _{i=2}^c|\gamma _i(G)/\gamma _{i+1}(G)\otimes G/G'|\le | {\mathcal {M}}(G/G')| p^{kd}. \end{aligned}$$

We claim that $| \mathrm {Im}\Psi _2| | \mathrm {Im} \Psi _3|\le \prod _{i=2}^c | \ker \alpha _i|.$ Clearly, the map

$$\begin{aligned} \delta _i: \gamma _i(G)/\gamma _{i+1}(G)\otimes G/G' \rightarrow \gamma _i(G)/\gamma _{i+1}(G)\otimes G/Z(G)G' \end{aligned}$$

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

$$\begin{aligned} | {\mathcal {M}}(G)| |G'|| \mathrm {Im}\Psi _2|| \mathrm {Im} \Psi _3|&\le | {\mathcal {M}}(G/G')|\prod _{i=2}^c|\gamma _i(G)/\gamma _{i+1}(G)\otimes G/G'Z(G)|\\ {}&=| {\mathcal {M}}(G/G')|\prod _{i=2}^c|\gamma _i(G)/\gamma _{i+1}(G)\otimes G/G'|/|\ker \delta _i |\\ {}&=| {\mathcal {M}}(G)| |G'|\prod _{i=2}^c | \ker \alpha _i |/|\ker \delta _i | \end{aligned}$$

and so $| \mathrm {Im}\Psi _2|| \mathrm {Im} \Psi _3|\le \prod _{i=2}^c | \ker \alpha _i |/|\ker \delta _i | \le \prod _{i=2}^c | \ker \alpha _i |.$ Therefore,

$$\begin{aligned} | G\wedge G|| \mathrm {Im}\Psi _2|| \mathrm {Im} \Psi _3|&\le | G\wedge G|\prod _{i=2}^c | \ker \alpha _i|\\ {}&=| {\mathcal {M}}(G)| |G'|\prod _{i=2}^c | \ker \alpha _i|\\ {}&=| {\mathcal {M}}(G/G')|\prod _{i=2}^c|\gamma _i(G)/\gamma _{i+1}(G)\otimes G/G' |\\ {}&\le | {\mathcal {M}}(G/G')| p^{kd}. \end{aligned}$$

The proof is completed. $\square$

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, Sect. 3]. The proof of Theorem 1.6 is similar to proofs of [14,  Corollary 3.2.4] and [4,  Theorem 3.1].

Theorem 1.6

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

$$\begin{aligned}&\eta : G'\otimes G/G'\rightarrow {\mathcal {M}}(G)=(R\cap F')/[R,F]\\&x\otimes (zG')\mapsto [{\tilde{x}},{\tilde{z}}][R,F] \end{aligned}$$

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

$$\begin{aligned} \langle ([x,y]\otimes zG') ([z,x]\otimes y G')([y,z]\otimes xG'),w^{p^s}\otimes w G'\mid x,y,z,w\in G\rangle \subseteq \ker \eta . \end{aligned}$$

Proof

By [14,  Corollary 3.2.4], we get

$$\begin{aligned} 1\rightarrow \ker \eta \rightarrow G'\otimes G/G' \xrightarrow {\eta } {\mathcal {M}}(G)\rightarrow {\mathcal {M}}(G/G')\rightarrow G' \rightarrow 1 \end{aligned}$$

is exact, in which

$$\begin{aligned}&\eta : G'\otimes G/G'\rightarrow {\mathcal {M}}(G)=(R\cap F')/[R,F]\\&x\otimes zG' \mapsto [{\tilde{x}},{\tilde{z}}][R,F] \end{aligned}$$

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

Lemma 1.7

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

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 $i\ne j.$ It is clear to see that $\{[x_i,x_j]\mid 1\le i<j\le d\}$ generates $G',$ as required. $\square$

Lemma 1.8

Let G be a group.

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

• (ii) If $G\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1,1\le k< m \rangle ,$ 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]. $\square$

It is well known that a p-group G is termed extra-special if its center, derived subgroup and Frattini subgroup all coincide, and moreover, each of these subgroups is of prime order. Also, $E_1$ is denoted the extra-special p-group of order $p^3$ and exponent p.

Theorem 1.9

[19,  Theorem 1.1] Let G be a non-abelian group of order $p^n$ of class two and $|G'|=p^k.$ 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:

• (1) For $p\ne 2,$ $G_1\cong E_1\times {\mathbb {Z}}_{p}^{(n-3)}.$

• (2) For $p\ne 2,$ $G_2\cong {\mathbb {Z}}_{p}^{(4)}\rtimes {\mathbb {Z}}_{p}.$

• (3) For $p\ne 2,$

$$\begin{aligned}&G_3\cong \langle x_1,x_2,x_3\mid [x_1,x_2]^p=[x_2,x_3]^p=[x_3,x_1]^p=x_i^p=1,\\&[x_1,x_2,x_i]=[x_3,x_1,x_i]=[x_2,x_3,x_i]=1, 1\le i\le 3\rangle . \end{aligned}$$

2 Main Results

As proven in [18,  Theorem 1.1], the Schur multiplier of a non-abelian group G of order $p^n$ with $|G'|=p^k$ 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 $p\ne 2,$ $|G'|=p^k,$ 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) $H_1\cong E_1\times {\mathbb {Z}}_{p}^{(n-3)}.$

• (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 $m\ge 2.$

• (iv) $H_4\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1,k\ge 2,m\ge 2 \rangle .$

• (v) $H_5\cong {\mathbb {Z}}_{p}^{(4)} \rtimes {\mathbb {Z}}_{p}.$

• (vi) $H_6\cong \langle x_1,x_2,x_3\mid [x_1,x_2]^p=[x_2,x_3]^p=[x_3,x_1]^p=x_i^p=1,$

  $$\begin{aligned}=[x_3,x_1,x_i]=[x_2,x_3,x_i]=1, 1\le i\le 3\rangle .\end{aligned}$$

• (vii) $H_7\cong \langle x_1,y_1,x_2,y_2,x_3,y_3,z\mid [x_1,x_2]=y_3,[x_2,x_3]=y_1,[x_3,x_1]=y_2,$

  $$\begin{aligned}=z,[x_i,y_j]=1,x_i^3=y_i^3=z^3=1, (i=1,2,3,~j=2,3)\rangle . \end{aligned}$$

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 $|G'|=p^k,$ $d(G)=d,$ and $G/G'\cong {\mathbb {Z}}_{p^{\alpha _1}}\oplus \ldots \oplus {\mathbb {Z}}_{p^{\alpha _d}}$, where $\alpha _1 \ge \ldots \ge \alpha _d.$ If $|{\mathcal {M}}(G)|$ attains the bound, then

• (i) $G/G'$ 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

  $$\begin{aligned}&|{\mathcal {M}}(G)|= p^{\frac{1}{2}(d-1)(n+k-2)+1}\le p^{\frac{1}{2}(d-1)(n+k-2-\alpha _1-\alpha _d)+1}. \end{aligned}$$

  Thus, $\alpha _1=\alpha _d$ and so $G/G'$ is homocyclic.

• (ii) Let $cl(G)=c.$ By part (i),  $G/G'$ is homocyclic and so $n=d\alpha _1+k.$ Theorem 1.5 and [14,  Corollary 2.2.12] imply

  $$\begin{aligned} p^{\frac{1}{2}(d-1)(n+k-2)+1}p^{k}\prod _{i=2}^c | \ker \alpha _i |&= p^{\frac{1}{2}(d-1)(d\alpha _1+2k-2)+1}p^{k}\prod _{i=2}^c | \ker \alpha _i |\\ {}&=| {\mathcal {M}}(G)||G'|\prod _{i=2}^c | \ker \alpha _i |\\ {}&= | {\mathcal {M}}(G/G')|\prod _{i=2}^c| \gamma _i(G)/\gamma _{i+1}(G)\otimes G/G'|\\ {}&\le p^{\frac{1}{2}d (d-1)\alpha _1}p^{kd}. \end{aligned}$$

  Thus,

  $$\begin{aligned} \prod _{i=2}^c | \ker \alpha _i | \le p^{d-2}. \end{aligned}$$

  (2.1)

  Consider $G= \langle x_1,x_2,\ldots ,x_d\rangle$ and $1\ne [x_1,x_2]\in G'{\setminus } \gamma _3(G)$. We claim that $\text {Im} \Psi _2\cong {\mathbb {Z}}_p^{(d-2)}.$ Similar to the proof of [8,  Theorem 2], the set $A=\{\Psi _2(x_1 G'\otimes x_2 G'\otimes x_j G')\mid 3\le j\le d\}$ consists of $d-2$ linearly independent elements of order at least p in the abelian p-group $G/G'\otimes G'/\gamma _3(G).$ Hence, $p^{d-2}\le |\langle A\rangle |\le |\text {Im} \Psi _2|$. By using (2.1) and the proof of Theorem 1.5,

  $$\begin{aligned} p^{d-2}&\le |\langle A\rangle |\le |\text {Im} \Psi _2|\\ {}&\le |\text {Im} \Psi _2||\text {Im} \Psi _3|\le | \ker \alpha _2| \\ {}&\le \prod _{i=2}^c | \ker \alpha _i | \\ {}&\le p^{d-2}. \end{aligned}$$

  Hence, $|\langle A\rangle |=|\text {Im} \Psi _2|=| \ker \alpha _2|=p^{d-2}$ and $|\text {Im} \Psi _3|=1$ and so $\langle A\rangle =\text {Im} \Psi _2\cong {\mathbb {Z}}_p^{(d-2)}.$

$\square$

Proposition 2.2

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

$$\begin{aligned} |{\mathcal {M}}(G/K)| =p^{\frac{1}{2}(d-1)(n + k - 4) + 1}. \end{aligned}$$

Proof

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

$$\begin{aligned} | {\mathcal {M}}(G)||G'\cap K|&\le | {\mathcal {M}}(G/K) || {\mathcal {M}}(K)| |G/G' \otimes K|\\&= | {\mathcal {M}}(G/K)||G/G' \otimes K|. \end{aligned}$$

Thus,

$$\begin{aligned} | {\mathcal {M}}(G)|&\le | {\mathcal {M}}(G/K) ||G/G' \otimes K|p^{-1}\\ {}&\le p^{\frac{1}{2}(d- 1)(n + k - 4) + 1 + d - 1}\\&=p^{\frac{1}{2} (d - 1)(n + k - 2) + 1}=| {\mathcal {M}}(G)|. \end{aligned}$$

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

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 $|G'|=p^k$ and $d(G)=d.$ If $|{\mathcal {M}}(G)|$ attains the bound, then G is capable.

Proof

Lemma 2.1(i) implies $G/G'$ is homocyclic. Therefore, $G/G'$ is capable, by [3,  Corollary 7.4]. Hence, $Z^*(G)\subseteq G',$ 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]. $\square$

Proposition 2.4

Let G be a non-abelian finite p-group of order $p^n$ such that $|G'|=p$ and $d(G)=d.$ If $e(G/G')>p,$ 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 $m \ge 1.$

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 $G/G'$ is homocyclic, we get $G/G'\cong {\mathbb {Z}}_{p^{\alpha _1}}^{(d)}$ and $G/Z(G)\cong {\mathbb {Z}}_{p}\oplus {\mathbb {Z}}_{p}.$ Since $G= NZ(G),$ we have $G'=N'=\langle c\rangle$ such that $N=\langle a,b\rangle$ and $c=[a,b].$ We claim that $G=N.$ Assume that $G\ne N.$ Choose an element $y\in Z(G){\setminus } G'.$ Thus $c\otimes yG'$ is a non-trivial element in $G'\otimes G/G'.$ By using Theorem 1.6, we can see that $c\otimes yG'$ is non-trivial in $\ker \eta$ and so

$$\begin{aligned} | {\mathcal {M}}(G)|p^{2}&\le | {\mathcal {M}}(G)||G'||\ker \eta | \\ {}&=|G/G'\wedge G/G'|| G'\otimes G/G'|\le p^{\frac{1}{2} d(d-1)\alpha _1}p^{d}. \end{aligned}$$

Thus, $| {\mathcal {M}}(G)|\le p^{\frac{1}{2} d(d-1)\alpha _1+d-2}.$ 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

$$\begin{aligned} G\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^p=1,[a,b,a]=[a,b,b]=1 \rangle , \end{aligned}$$

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

Proposition 2.5

Let G be a non-abelian group of order $p^n$ of class t such that $|G'|=p^k,$ $d(G)=d$ and $|{\mathcal {M}}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}$ for all k such that $k\ge 2.$ If K is a non-trivial central subgroup of order $p^m$ contained in $Z(G)\cap G'\ne G',$ then $|{\mathcal {M}}(G/K)|$ also attains the bound, that is

$$\begin{aligned} |{\mathcal {M}}(G/K)| =p^{\frac{1}{2}(d-1)(n + k - 2(m+1)) + 1}. \end{aligned}$$

Proof

Let K be a non-trivial central subgroup of order $p^m$ contained in $Z(G)\cap G'.$ We have $| G/K|=p^{n-m}$ and $|(G/K)'|=p^{k-m}.$ We prove the result by using induction on m. If $m=1,$ then the result holds by Proposition 2.2. Now let $m\ge 2.$ Consider a normal subgroup $K_1$ in K of order $p^{m-1}$ and using the induction hypothesis, we have

$$\begin{aligned}|{\mathcal {M}}(G/K_1)| =p^{\frac{1}{2}(d - 1)(n + k - 2m) + 1}.\end{aligned}$$

Since $K/K_1\subseteq Z(G/K_1)\cap (G'/K_1)$ and $|K/K_1| =p,$ Proposition 2.2 implies that

$$\begin{aligned}|{\mathcal {M}}(G/K)|=| {\mathcal {M}}(\dfrac{G/K_1}{K/K_1})| =p^{\frac{1}{2}(d - 1)(n + k - 2(m+1)) + 1}.\end{aligned}$$

This completes the proof. $\square$

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 $t\ge 3$ such that $|G'|=p^k,$ $d(G)=d$ and $|{\mathcal {M}}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}$ for $k\ge 2.$ Then, $|{\mathcal {M}}( G/\gamma _i(G))|$ also attains the bound for all i such that $3 \le i\le t.$

Proof

We prove the result by using induction on $j=t-i+3$ for all i such that $3 \le i\le t.$ If $j=3,$ then since $k\ge 2,$ 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 $\gamma _{i-1}(G)/\gamma _i(G)\subseteq Z(G/\gamma _i(G))\cap G'/\gamma _i(G),$ Proposition 2.5 implies that

$$\begin{aligned} |{\mathcal {M}}( G/ \gamma _{i-1}(G))|=\left| {\mathcal {M}}(\dfrac{G/\gamma _i(G)}{\gamma _{i-1}(G)/\gamma _i(G)} )\right| \end{aligned}$$

attains the bound, as required. $\square$

Proposition 2.7

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

$$\begin{aligned} G\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{m}}=1,[a,b,a]=[a,b,b]=1 \rangle , \end{aligned}$$

where $m\ge 2$ and $p\ne 2$ or

$$\begin{aligned} G\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1,k\ge 2,m\ge 2 \rangle . \end{aligned}$$

Proof

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

$$\begin{aligned} G&\cong G(\alpha ,\beta ,\gamma ;\rho ,\sigma )\\ {}&=\langle a, b\mid [a,b]^{p^{\gamma }}=[a,b,a]=[a,b,b]=1,a^{p^{\alpha }}=[a,b]^{p^{\rho }},b^{p^{\beta }}=[a,b]^{p^{\sigma }} \rangle , \end{aligned}$$

where $\alpha +\beta +\gamma =n,$ $1\le \gamma \le \beta \le \alpha ,$ and $0\le \rho ,\sigma \le \gamma .$ Since $G/G'\cong {\mathbb {Z}}_{p^{m}}^{(2)},$ 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

$$\begin{aligned} G\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{m}}=1,[a,b,a]=[a,b,b]=1 \rangle , \end{aligned}$$

where $m\ge 2$ and $p\ne 2$ or

$$\begin{aligned} G\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1,k\ge 2,m\ge 2 \rangle .\end{aligned}$$

The converse holds by Lemma 1.8. $\square$

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 $|G'|\ge p^2$ and $d(G)=d.$ If $|{\mathcal {M}}(G)|$ attains the bound, then $2\le d\le 3.$

Proof

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

Lemma 2.9

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

Proof

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

Theorem 2.10

Let G be a non-abelian group of order $p^n$ of class two such that $|G'|=p^k$ 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 $p\ne 2,$ $H_1\cong E_1\times {\mathbb {Z}}_{p}^{(n-3)},$ 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 $p\ne 2,$ $H_3\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{m}}=1,[a,b,a]=[a,b,b]=1 \rangle ,$ where $m\ge 2.$

• (iv) $H_4\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1, k\ge 2,m\ge 2 \rangle .$

• (v) For $p\ne 2,$ $H_5\cong {\mathbb {Z}}_{p}^{(4)} \rtimes {\mathbb {Z}}_{p}.$

• (vi) For $p\ne 2,$ $H_6\cong \langle x_1,x_2,x_3\mid [x_1,x_2]^p=[x_2,x_3]^p=[x_3,x_1]^p=x_i^p=1,$

  $$\begin{aligned}=[x_3,x_1,x_i]=[x_2,x_3,x_i]=1, 1\le i\le 3\rangle .\ \end{aligned}$$

Proof

Suppose that $|{\mathcal {M}}(G)|$ attains the bound. First assume that $|G'|=p.$ By Lemma 2.1(i),  $G/G'$ 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

$$\begin{aligned}G\cong H_2\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^p=1,[a,b,a]=[a,b,b]=1 \rangle , \end{aligned}$$

where $m > 1.$ Now suppose that $|G'|\ge p^2.$ By using Lemma 2.8, $2\le d\le 3.$ Let $d=2.$ By Proposition 2.7, we have

$$\begin{aligned}G\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{m}}=1,[a,b,a]=[a,b,b]=1 \rangle ,\end{aligned}$$

where $m\ge 2$ and $p\ne 2$ or

$$\begin{aligned} G\cong \langle a, b\mid a^{p^m}= b^{p^m}=[ a, b]^{p^{k}}=1,[a,b,a]=[a,b,b]=1,2\le k\le m \rangle . \end{aligned}$$

Thus, $G\cong H_3$ or $G\cong H_4.$ Let now $d=3.$ By Lemma 2.9, $G/G'$ is of exponent p and hence $G'=\phi (G).$ Thus, $d(G)=n-k=3.$ Therefore,

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

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. $\square$

Proposition 2.11

There exists no non-abelian 2-generator group G of order $p^n$ of class $t\ge 3$ such that $| G'|\ge p^2$ and $|{\mathcal {M}}(G)|=p^{\frac{1}{2}(d-1)(n+k-2)+1}$ with $p\ne 2.$

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 $[x,y,x]\ne 1.$ So, $\Psi _3(x G'\otimes y G'\otimes x G' \otimes y G')=([[x,y],x]\otimes yG')^2([y,[x,y]]\otimes xG')^2 \ne 1.$ On the other hand, by Lemma 2.1(ii),  $|\mathrm {Im} \Psi _3|= 1.$ Hence, we have a contradiction. $\square$

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 $1\le i\le 6$ and so $2\le d(G/\gamma _3(G)) \le 3.$ Since $d(G)=d(G/\gamma _3(G)),$ we have $2\le d(G)\le 3.$ 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].