Schur multipliers of special p-groups of rank 2

TL;DR

This paper explicitly determines the Schur multiplier for special p-groups of rank 2, which are groups with specific elementary abelian p-group properties related to their center and commutator subgroup.

Contribution

It provides an explicit calculation of the Schur multiplier for a specific class of special p-groups of rank 2, advancing understanding of their structure.

Findings

Explicit formulas for Schur multipliers of rank 2 special p-groups

Enhanced understanding of the structure of these groups

Potential applications in group cohomology and classification

Abstract

A group G is called special p-group of rank k if the commutator subgroup [G,G] and centre Z(G) are equal, which is elementary abelian p-group of rank k and G/[G,G] is also elementary abelian p-group. In this article we determine the Schur multiplier of special p-groups of rank 2 explicitly.

Equations

$$E{S}_{p^2}(p^{2m+1})\times {\mathbb{Z}}_p^{(d-2m)}(m\ge 2),E{S}_{p^2}(p^3)\times {\mathbb{Z}}_p^{(d-2)}\text{or}{Q}_8\times {\mathbb{Z}}_2^{(d-2)}.$$

$$(v_1,v_2)=[g_1,g_2]$$

$$v_1\otimes (v_2,v_3)+v_2\otimes (v_3,v_1)+v_3\otimes (v_1,v_2)$$

$$f(g{G}^{\prime })=g^p$$

$$X:=X_1+X_2.$$

$$\sigma(v_{1}\wedge v_{2})=\big{(}v_{1}\otimes f(v_{2})+(_{2}^{p})v_{2}\otimes(v_{1},v_{2})\big{)}+X.$$

$$1\to N\to M^{\ast }\stackrel{\xi }{\to }V\wedge V\to 1$$

$$\sigma \xi (\alpha )={\alpha }^p\text{for}\alpha \in M^{\ast }.$$

$$\rho (v_1\wedge v_2)=(v_1,v_2)$$

$$\{{\bar{x}}_1\otimes [x_2,x_3]+{\bar{x}}_2\otimes [x_3,x_1]+{\bar{x}}_3\otimes [x_1.x_2]\mid x_1,x_2,x_3\in S\},$$

$$(G/G^{\prime })\otimes Z\stackrel{{\lambda }_Z}{\to }M(G)\stackrel{\mu }{\to }M(G/Z)\to G^{\prime }\cap Z\to 1.$$

$$\{v_i\otimes f(v_i),v_l\otimes f(v_i),(v_i\otimes f(v_j)+v_j\otimes f(v_i))\mid 1\le i<j\le r,(r+1)\le l\le d\}$$

$$\frac{|M(G)|}{|\mathrm{Im}{\lambda }_Z|}=\frac{|M(G/Z)|}{|G^{\prime }\cap Z|}.$$

$$|M(G)|=p^{\frac{1}{2}d(d-1)-2}.\frac{p^{2d}}{|X|}$$

$$\{v_{i}\otimes f(v_{i}),v_{j}\otimes f(v_{i}),\;\big{(}v_{1}\otimes f(v_{2})+v_{2}\otimes f(v_{1})\big{)}\mid i=1,2\;\text{and}\;3\leq j\leq d\}.$$

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

$$|M(G)|=p^{\frac{1}{2}d(d-1)-2}.$$

$$|M(G)|=p^{\frac{1}{2}d(d-1)-1}.$$

$$G/Z$$

$$M^{\prime }:=\{v_i\otimes f(v_1),1\le i\le d\}.$$

$$|M(G)|=\frac{|M(G/G^p)|}{p}$$

$$N^{\prime }:=\{v_i\otimes (v_1,v_2)+v_1\otimes (v_2,v_i)+v_2\otimes (v_i,v_1)\mid 3\le i\le d\}$$

$$p^{2d}\ge |X|\ge p^{2d-2}.$$

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

$$|M(G/G^p)|=p^{\frac{1}{2}d(d-1)-1},$$

$$G/G^p\cong E{S}_p(p^{2m+1})\times {\mathbb{Z}}_p^{(d-2m)},m\ge 2,$$

$$|M(G/G^p)|=p^{\frac{1}{2}d(d-1)}.$$

$$|M(G/G^p)|=p^{\frac{1}{2}d(d-1)+1},$$

$$G/G^p\cong E{S}_p(p^3)\times {\mathbb{Z}}_p^{(d-2)},$$

$$A:=\{x_i\otimes [x_1,x_2]+x_1\otimes [x_2,x_i]+x_2\otimes [x_i,x_1]\mid 3\le i\le d\}$$

Full text

1. Introduction

Let $G$ be a finite $p$-group and $d(G)$ denotes the cardinality of minimal generating set of $G$. The commutator subgroup and center of $G$ are denoted by $G^{\prime}$ and $\operatorname{Z}(G)$ respectively. By $ES_{p^{k}}(p^{n})$, we denote the extraspecial $p$-group of order $p^{n}$ of exponent $p^{k}$, $k=1,2$ and by $\mathbb{Z}_{p}^{(k)}$, we denote the elementary abelian $p$-group of rank $k$ for $k\geq 1$. For a prime $p$, $G^{p}$ denotes the group generated by the set $\{g^{p}\mid g\in G\}$. A finite $p$-group $G$ is called special $p$-group of rank $k$ if $G^{\prime}=\operatorname{Z}(G)$ is an elementary abelian $p$-group of order $p^{k}$ and $G/G^{\prime}$ is elementary abelian. A group $G$ is called capable group if there exists a group $H$ such that $G\cong H/\operatorname{Z}(H)$. The epicenter of a group $G$ is denoted by $Z^{*}(G)$, which is the smallest central subgroup $K$ of $G$ such that $G/K$ is capable.

The Schur multiplier of a group $G$, denoted by $\operatorname{M}(G)$, is the second integral homology group $\operatorname{H}_{2}(G,\mathbb{Z})$ which was introduced by Schur in 1904 in his fundamental work on projective representation of groups. There has been a great importance in understanding the Schur multipliers of finite $p$-groups in recent past. Here we are interested to compute Schur multiplier of special $p$-groups of rank $2$. The special $p$-groups of minimum rank are the extraspecial $p$-groups and their Schur multiplier was studied in [3]. The Schur multiplier of special $p$-groups having maximum rank was studied in [12]. In this article we determine the Schur multiplier of special $p$-groups of rank $2$ and that answered the question which was asked by Berkovich and Janko in [1, Problem 2027].

Recall that, there are two extraspecial $p$-groups of order $8$ upto isomorphism, both are of exponent $4$, one is quaternian group $Q_{8}$ which has trivial Schur multiplier and another is dihedral group $D_{8}$ which has Schur multiplier of order $2$ (see [8, Theorem 3.3.6]).

We state our main results now. The following result describes the Schur multiplier of $G$ when $G^{p}=G^{\prime}\cong\mathbb{Z}_{p}\times\mathbb{Z}_{p}$.

Theorem 1.1**.**

Let $G$ be a special $p$-group of rank $2$ with $d=d(G)$ and $G^{p}=G^{\prime}$. Then the following assertions hold:

1. (a)

Either $Z^{*}(G)=\operatorname{Z}(G)$ or $G$ is capable. 2. (b)

$\operatorname{M}(G)$* is elementary abelian of order $p^{\frac{1}{2}d(d-1)-2}$ if and only if $Z^{*}(G)=\operatorname{Z}(G)$.* 3. (c)

$\operatorname{M}(G)$* is of order $p^{\frac{1}{2}d(d-1)-1}$ of exponent at most $p^{2}$ if and only if $G$ is capable.* 4. (d)

For every central subgroup $Z$ of order $p$, $G/Z$ is isomorphic to

[TABLE]

By [5, Proposition 3] and Theorem 1.1 we have the immediate corollary.

Corollary 1.2**.**

If $G$ is a special $p$-group of rank $2$ of order $p^{n}\;(n\geq 8)$ with $G^{p}=G^{\prime}$, then $\operatorname{M}(G)$ is elementary abelian of order $p^{\frac{1}{2}(n-2)(n-3)-2}$.

The following result describes the Schur multiplier of $G$ when $G^{p}$ is cyclic of prime order.

Theorem 1.3**.**

If $G$ is a special $p$-group of rank $2$ with $d=d(G)$ and $G^{p}\cong\mathbb{Z}_{p}$, then the following assertions hold:

1. (a)

$G$* is not capable and either $Z^{*}(G)=\operatorname{Z}(G)$ or $Z^{*}(G)=G^{p}$.* 2. (b)

$\operatorname{M}(G)$* is elementary abelian.* 3. (c)

The following are equivalent:

1. (i)

$\operatorname{M}(G)$* is of order $p^{\frac{1}{2}d(d-1)-2}$.* 2. (ii)

$Z^{*}(G)=\operatorname{Z}(G)$. 3. (iii)

$G/G^{p}\cong ES_{p}(p^{2m+1})\times\mathbb{Z}_{p}^{(d-2m)},m\geq 2$. 4. (d)

The following are equivalent:

1. (i)

$\operatorname{M}(G)$* is of order $p^{\frac{1}{2}d(d-1)}$.* 2. (ii)

$Z^{*}(G)=G^{p}$. 3. (iii)

$G/G^{p}\cong ES_{p}(p^{3})\times\mathbb{Z}_{p}^{(d-2)}$.

Now we are left with only one case when $G^{p}$ is trivial and in this case the Schur multiplier of $G$ is studied in the following result.

Theorem 1.4**.**

If $G$ is a special $p$-group of rank $2$ with $d=d(G)$ and $G^{p}=1$, $p$ odd, then the following assertions hold:

1. (a)

$\operatorname{M}(G)$* is elementary abelian.* 2. (b)

$p^{\frac{1}{2}d(d-1)-2}\leq|\operatorname{M}(G)|\leq p^{\frac{1}{2}d(d-1)+3}$. 3. (c)

$G$* is capable if and only if $G$ is isomorphic to one of the following groups:*

1. (i)

$\Phi_{4}(1^{5})=\langle\alpha,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\mid[\alpha_{i},\alpha]=\beta_{i},\alpha^{p}=\alpha_{i}^{p}=\beta_{i}^{p}=1\;(i=1,2)\rangle$. 2. (ii)

$\Phi_{12}(1^{6})=ES_{p}(p^{3})\times ES_{p}(p^{3})$, 3. (iii)

$\Phi_{13}(1^{6})=\langle\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\beta_{1},\beta_{2}\mid[\alpha_{i},\alpha_{i+1}]=\beta_{i},[\alpha_{2},\alpha_{4}]=\beta_{2},\alpha_{i}^{p}=\alpha_{3}^{p}=\alpha_{4}^{p}=\beta_{i}^{p}=1\;(i=1,2)\rangle$, 4. (iv)

$\Phi_{15}(1^{6})=\langle\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\beta_{1},\beta_{2}\mid[\alpha_{i},\alpha_{i+1}]=\beta_{i},[\alpha_{3},\alpha_{4}]=\beta_{1},[\alpha_{2},\alpha_{4}]=\beta_{2}^{g},\alpha_{i}^{p}=\alpha_{3}^{p}=\alpha_{4}^{p}=\beta_{i}^{p}=1\;(i=1,2)\rangle$, where $g$ is non-quadratic residue modulo $p$. 5. (v)

$T=\langle x_{1},\cdots,x_{5},c_{1},c_{2}\mid[x_{2},x_{1}]=[x_{5},x_{3}]=c_{1},[x_{3},x_{1}]=[x_{5},x_{4}]=c_{2},x_{i}^{p}=c_{j}^{p}=1,\;1\leq i\leq 5,1\leq j\leq 2\rangle$. 4. (d)

$|\operatorname{M}(G)|=p^{\frac{1}{2}d(d-1)+3}$* if and only if $G$ is isomorphic to $\Phi_{4}(1^{5})$.* 5. (e)

$|\operatorname{M}(G)|=p^{\frac{1}{2}d(d-1)+2}$* if and only if $G$ is isomorphic to $\Phi_{12}(1^{6}),\Phi_{13}(1^{6}),\;\text{or}\;\Phi_{15}(1^{6})$.* 6. (f)

$|\operatorname{M}(G)|=p^{\frac{1}{2}d(d-1)-1}$* if and only if $G$ is isomorphic to $T$.* 7. (g)

$\operatorname{M}(G)$* is of order $p^{\frac{1}{2}d(d-1)-2}$ if and only if $Z^{*}(G)=\operatorname{Z}(G)$. In this case, $G/Z\cong ES_{p}(p^{2m+1})\times\mathbb{Z}_{p}^{(d-2m)},m\geq 2$, for every central subgroup $Z$ of order $p$.* 8. (h)

$\operatorname{M}(G)$* is of order $p^{\frac{1}{2}d(d-1)}$ if and only if $Z^{*}(G)\cong\mathbb{Z}_{p}$. In this case, $G/Z^{*}(G)\cong ES_{p}(p^{3})\times\mathbb{Z}_{p}^{(d-2)}$.*

The following result is for $p=2$.

Theorem 1.5**.**

Let $G$ be a special $2$-group of rank $2$. Then $G^{2}=G^{\prime}$ holds.

2. Preliminaries

For a finite group $G$ of class $2$ with $G/G^{\prime}$ elementary abelian, the following construction is given in [3]. We consider $G/G^{\prime}$ and $G^{\prime}$ as vector spaces over $\mathbb{F}_{p}$, which we denote by $V,W$ respectively. The bilinear map $(-,-):V\times V\to W$ is defined by

[TABLE]

for $v_{1},v_{2}\in V$ such that $v_{i}=g_{i}G^{\prime},i\in\{1,2\}$ for some $g_{1},g_{2}\in G$. Let $X_{1}$ be the subspace of $V\otimes W$ spanned by all

[TABLE]

for $v_{1},v_{2},v_{3}\in V$. Consider a map $f:V\rightarrow W$ given by

[TABLE]

for $g\in G$. Let $X_{2}$ be the subspace spanned by all $v\otimes f(v)$, $v\in V$, and take

[TABLE]

Now consider a homomorpism $\sigma:V\wedge V\rightarrow(V\otimes W)/X$ given by

[TABLE]

Then there exists an abelian group $M^{*}$ with a subgroup $N$ isomorphic to $(V\otimes W)/X$, such that

[TABLE]

is exact and

[TABLE]

Now we consider a homomorphism $\rho:V\wedge V\rightarrow W$ given by

[TABLE]

for all $v_{1},v_{2}\in V.$ Notice that $\rho$ is an epimorphism. We let $M$ be the subgroup of $M^{*}$ containing $N$ such that $M/N\cong\operatorname{Ker}\rho$. We use this notation throughout the paper without further reference.

With the above setting, we have

Theorem 2.1**.**

$($[3, Theorem 3.1]$)$* $\operatorname{M}(G)\cong M$.*

Note: It is easy to observe that $X_{1}$ is generated by the set

[TABLE]

where $S$ is a set of generators of $G$ and $\bar{x}$ is the image of $x$ in $G/G^{\prime}$.

Suppose $G$ has a free presentation $F/R$. Let $Z=S/R$ be a central subgroup of $G$. Then the map from $(F/F^{\prime}R)\times(S/R)$ to $(F^{\prime}\cap R)/[F,R]$ defined by $(xF^{\prime}R,sR)\mapsto[x,s][F,R]$ is a well-defined bilinear map and induces a homomorphism $\lambda_{Z}:(G/G^{\prime})\otimes Z\to\operatorname{M}(G)$, called the Ganea map.

Theorem 2.2** ([4]).**

Let $Z$ be a central subgroup of a finite group $G$. Then the following sequence is exact

[TABLE]

Theorem 2.3** ([2]).**

Let $Z$ be a central subgroup of a finite group $G$. Then $Z\subseteq Z^{*}(G)$ if and only if $(G/G^{\prime})\otimes Z=\operatorname{Ker}\lambda_{Z}$.

By [8, Corollary 3.2.4], we have $X=\operatorname{Ker}\lambda_{\operatorname{Z}(G)}$. Hence by Theorem 2.3, we have the following result:

Lemma 2.4**.**

Let $Z$ be a central subgroup of a group $G$ of nilpotency class $2$. Then $Z\subseteq Z^{*}(G)$ if and only if $(G/G^{\prime})\otimes Z$ is contained in $X$.

Let $v_{1},v_{2},\ldots,v_{d}$ be the generators of $V$ such that $\{f(v_{1}),f(v_{2}),\ldots,f(v_{r})\}$ is a basis of $G^{p}$. Then the set

[TABLE]

forms a basis of $X_{2}$, from which the following result follows.

Proposition 2.5** (Proposition 3.3 of [12]).**

Let $G$ be a special $p$-group with $d=d(G)$ and $G^{p}$ of order $r$. Then $|X_{2}|=p^{rd-\frac{1}{2}r(r-1)}$.

By Theorem 2.2, we have

[TABLE]

As $X=\operatorname{Ker}\lambda_{\operatorname{Z}(G)}$, so $|\operatorname{Im}\lambda_{\operatorname{Z}(G)}|=\frac{p^{2d}}{|X|}$. Hence, by (2.1), taking $Z=\operatorname{Z}(G)$, we have

[TABLE]

Now we recall the following results which will be used in the proof of the main results.

Theorem 2.6** (Main Theorem of [9]).**

Let $G$ be a $p$-group of order $p^{n}$. Then $|\operatorname{M}(G)|=p^{\frac{1}{2}(n-1)(n-2)+1}$ if and only if $G\cong ES_{p}(p^{3})\times\mathbb{Z}_{p}^{(n-3)}$.

Theorem 2.7** (Theorem 21 of [10]).**

Let $G$ be a $p$-group of order $p^{n}$. Then $|\operatorname{M}(G)|=p^{\frac{1}{2}(n-1)(n-2)}$ if and only if $G$ is isomorphic to one of the following groups.

1. (i)

$G\cong\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}^{(n-2)}$, 2. (ii)

$G\cong D_{8}\times\mathbb{Z}_{2}^{(n-3)}$, 3. (iii)

$G\cong\mathbb{Z}_{p}^{(4)}\rtimes\mathbb{Z}_{p}\;(p\neq 2)$.

Theorem 2.8** (Theorem 11 of [11]).**

Let $G$ be a group of order $p^{n}$ such that $G/G^{\prime}$ is elementary abelian of order $p^{n-1}$. Then $|\operatorname{M}(G)|=p^{\frac{1}{2}(n-1)(n-2)-1}$ if and only if $G$ is isomorphic to one of the following groups.

1. (i)

$ES_{p^{2}}(p^{3})\times\mathbb{Z}_{p}^{(n-3)}$, 2. (ii)

$Q_{8}\times\mathbb{Z}_{2}^{(n-3)}$, 3. (iii)

$ES(p^{2m+1})\times\mathbb{Z}_{p}^{(n-2m-1)}\;(m\geq 2)$.

3. Proofs

In this section we prove our main results.

Proof of Theorem 1.1: Consider $G^{p}=G^{\prime}\cong\mathbb{Z}_{p}\times\mathbb{Z}_{p}$. Now by Proposition 2.5, $|X_{2}|=p^{2d-1}$. Let $v_{1},v_{2}\in V$ such that $\{f(v_{1}),f(v_{2})\}$ is a basis of $W$. Observe that $X_{2}$ is generated by the set

[TABLE]

Hence $p^{2d}\geq|X|\geq p^{2d-1}$ and by (2.2),

[TABLE]

$(a)$ Observe that $|X|=p^{2d}$, i.e., the set $\{v_{1}\otimes f(v_{2}),v_{2}\otimes f(v_{1})\}$ is contained in $X$ if and only if $Z^{*}(G)=\operatorname{Z}(G)$, follows by Lemma 2.4. Another possibility is $|X|=p^{2d-1}$, i.e., $X_{1}\subseteq X_{2}$ if and only if $v_{1}\otimes f(v_{2}),v_{2}\otimes f(v_{1})$ are not in $X$. Hence, by Lemma 2.4, $Z^{*}(G)=1$. So $G$ is capable.

$(b)$ By (2.2), $|X|=p^{2d}$ if and only if

[TABLE]

By Theorem 2.3 and Theorem 2.2, taking $Z=\operatorname{Z}(G)$, we see that $\operatorname{M}(G)$ embeds in $\operatorname{M}(G/\operatorname{Z}(G))$ which is elementary abelian. So in this case $\operatorname{M}(G)$ is elementary abelian.

$(c)$ For $x\in\operatorname{M}(G),\;x^{p}\in(V\otimes W)/X$ and $V\otimes W/X$ is elementary abelian, so $x^{p^{2}}=1$.

By (2.2), $X_{1}\subseteq X_{2}$ i.e., $|X|=p^{2d-1}$ if and only if, by (2.2),

[TABLE]

Hence, by Lemma 2.4, $Z^{*}(G)=1$. So $G$ is capable. Converse follows from $(b)$.

$(d)$ Observe that, in both the cases $|\operatorname{M}(G/Z)|=p^{\frac{1}{2}d(d-1)-1}$, follows from (2.1), taking $Z$ a central subgroup of order $p$. Therefore by Theorem 2.8,

[TABLE]

$\Box$

Proof of Theorem 1.3: $(a)$ Assume $G^{p}$ is cyclic of order $p$. Now by Proposition 2.5, $|X_{2}|=p^{d}$. Let $v_{1}\in V$ such that $G^{p}=\langle f(v_{1})\rangle$. Observe that $X_{2}$ is generated by the set

[TABLE]

Hence by Lemma 2.4, $G^{p}\subseteq Z^{*}(G)$. Hence $G$ is not capable with $Z^{*}(G)=G^{p}$ or $Z^{*}(G)=\operatorname{Z}(G)$.

Using Theorem 2.3 and taking $Z=G^{p}$ in (2.1), we have

[TABLE]

Now let $v_{1},v_{2}\in V$ such that $(v_{1},v_{2})\in G^{\prime}\setminus G^{p}$. Then The set

[TABLE]

is linearly independent in $X_{1}$, so $|X_{1}|\geq p^{d-2}$. The set $M^{\prime}\cup N^{\prime}$ is linearly independent in $X$ and $M^{\prime}\cap N^{\prime}=\emptyset$. Thus

[TABLE]

Hence by (2.2),

[TABLE]

$(c)$ Now similarly, as described in the proof of Theorem 1.1, we have $Z^{*}(G)=\operatorname{Z}(G)$ if and only if $|\operatorname{M}(G)|=p^{\frac{1}{2}d(d-1)-2}$ i.e., by (3.1),

[TABLE]

which happens if and only if

[TABLE]

follows from Theorem 2.8.

$(d)$ By Theorem 2.7, it follows that there is no $G/G^{p}$ such that

[TABLE]

Thus by (3.1), $|\operatorname{M}(G)|$ cannot be of order $p^{\frac{1}{2}d(d-1)-1}$. Hence $|\operatorname{M}(G)|=p^{\frac{1}{2}d(d-1)}$ if and only if $Z^{*}(G)=G^{p}$. By (3.1),

[TABLE]

which happens if and only if

[TABLE]

follows by Theorem 2.6.

$(b)$ By $(c)$ and $(d)$ it follows that $p$ must be odd. The group $G/G^{p}$ is of exponent $p$ and $p$ odd, so the homomorphism $\sigma$, described in Section 2, is trivial map and therefore, $\sigma\xi(x)=x^{p}=1$ for $x\in\operatorname{M}(G/G^{p})$. Thus $\operatorname{M}(G/G^{p})$ is elementary abelian. Since $G^{p}\subseteq Z^{*}(G)$, by Theorem 2.3 and Theorem 2.2, $\operatorname{M}(G)$ embeds in $\operatorname{M}(G/G^{p})$. Therefore, $\operatorname{M}(G)$ is also elementary abelian. The proof is complete now. $\Box$

Proof of Theorem 1.4: $(a)$ Since $p$ is odd and $G^{p}=1$, the homomorphism $\sigma$ described in Section 2, is the trivial map and therefore $\sigma\xi(x)=x^{p}=1$. Thus, $\operatorname{M}(G)$ is elementary abelian.

Let $z,z^{\prime}$ be the generators of $\operatorname{Z}(G)$ and $x_{1},x_{2},\ldots,x_{d}$ be the generators of $G$ such that $[x_{1},x_{2}]\in\langle z\rangle$ is non-trivial. Then the set

[TABLE]

consists of $d-2$ linearly independent elements of $X_{1}$.

Now if for some $x_{k}\in\{3,4,\ldots,d\}$, $[x_{1},x_{k}]\in\langle z^{\prime}\rangle$ is non-trivial, then the set

[TABLE]

consists of $d-3$ linearly independent elements of $X_{1}$. Thus, $A\cup B$ consists of $(2d-5)$ linearly independent elements of $X_{1}$.

If $1\neq[x_{2},x_{k}]\in\langle z^{\prime}\rangle$, for some $k\in\{3,4,\ldots,d\}$, then a similar conclusion holds. Suppose then that $[x_{1},x_{k}]$ and $[x_{2},x_{k}]$ are all trivial or in $\langle z\rangle$ for all $k\in\{3,4,\ldots,d\}$. Say, $[x_{3},x_{4}]\in\langle z^{\prime}\rangle$. In this case

[TABLE]

consists of $d-2$ independent elements of $X_{1}$. Thus, $A\cup B_{1}$ consists of $2d-4$ linearly independent elements of $X_{1}$.

Hence in both cases, $2d\geq|X|=|X_{1}|\geq p^{2d-5}$ holds. By (2.2), it follows that,

[TABLE]

which proves $(b)$.

Now if $G$ is not capable, then by (2.1) and [9, Main Theorem] we have

[TABLE]

By Theorem 2.7, there is no $G$ and central subgroup $Z$ such that $|\operatorname{M}(G/Z)|=p^{\frac{1}{2}d(d-1)}$. Hence

[TABLE]

Assume then that $G$ is capable. By [5, Proposition 3], $p^{5}\leq|G|\leq p^{7},$ with $p$ odd. If $|G|=p^{5}$, then looking through the list of groups given in [7], it follows that $G\cong\Phi_{4}(1^{5})$. Since $|X|=p$, it follows by (2.2) that $|\operatorname{M}(G)|=p^{6}=p^{\frac{1}{2}d(d-1)+3}$.

If $|G|=p^{6}$, then looking through the list of groups given in [7], it follows that $G\cong\Phi_{12}(1^{6}),\;\Phi_{13}(1^{6})$ or $\Phi_{15}(1^{6})$. Since $|X|=p^{4}$, it follows by (2.2) that $|\operatorname{M}(G)|=p^{8}=p^{\frac{1}{2}d(d-1)+2}$.

Now consider groups of order $p^{7}$ of exponent $p$. By [6], it follows that there is only one capable group

[TABLE]

up to isomorphism. By (2.2), $|\operatorname{M}(G)|=p^{9}=p^{\frac{1}{2}d(d-1)-1}$.

Now $(c),(d),(e),(f)$ follow by (3.2).

$(g)$ Now by (2.1) we have that $Z^{*}(G)\cong\operatorname{Z}(G)$ if and only if $|\operatorname{M}(G)|=p^{\frac{1}{2}d(d-1)-2}$, taking $Z=\operatorname{Z}(G)$. Hence by (2.1) we have that for every central subgroup $Z$ of order $p$, $|\operatorname{M}(G/Z)|=p^{\frac{1}{2}d(d-1)-1}$, thus

[TABLE]

follows from Theorem 2.8.

$(h)$ Suppose $Z^{*}(G)\cong\mathbb{Z}_{p}$. By (3.2) and $(g)$, it follows that $|\operatorname{M}(G)|=p^{\frac{1}{2}d(d-1)}$.

Conversely suppose $|\operatorname{M}(G)|=p^{\frac{1}{2}d(d-1)}$. From the previous cases it follows that $G$ is not capable since $Z^{*}(G)\cong\mathbb{Z}_{p}$.

In this case, we have

[TABLE]

By Theorem 2.6, it follows that $|\operatorname{M}(G/Z^{*}(G))|=p^{\frac{1}{2}d(d-1)+1}$ if and only if

[TABLE]

The proof is complete now. $\Box$

Proof of Theorem 1.5: By Theorem 1.3, it follows that there is no special $2$-group of rank $2$ with $G^{2}\cong\mathbb{Z}_{2}$.

Assume that $p=2$. As in the proof of Theorem 1.4, we conclude that

[TABLE]

Let $Z$ be a central subgroup of order $2$. If $G$ is not capable then by (2.1),

[TABLE]

By Theorem 2.6, 2.7 and 2.8, there is no group $G$ and central subgroup $Z$ such that $G/Z$ of exponent $2$ and $2^{\frac{1}{2}d(d-1)-1}\leq|\operatorname{M}(G/Z)|\leq 2^{\frac{1}{2}d(d-1)+1}$. Hence $G$ must be capable and $|\operatorname{M}(G/Z)|\leq 2^{\frac{1}{2}d(d-1)-2}$.

Suppose $|\operatorname{M}(G)|=2^{\frac{1}{2}d(d-1)-2+k},\;0\leq k\leq 5$. Then by (2.2), $|\operatorname{Ker}\lambda_{\operatorname{Z}(G)}|=|X|=2^{2d-k}$. Hence $1\leq|\operatorname{Im}\lambda_{Z}|=2^{m}\leq 2^{k}$. By (2.1), it follows that

[TABLE]

Hence $k-m\leq-1$, which is not possible. So there is no special $2$-group of rank $2$ with $G^{2}=1$. The proof is complete now.

$\Box$

Acknowledgement: The research of the author is partly supported by Infosys grant.