Subcentral Automorphisms

Abstract

A concept of subcentral automorphisms of group G with respect to a characterstic subgroup M of Z(G) along with relevant mathematical paraphernalia has been introduced. With the help of this, a number of results on central automorphisms have been generalized.

1 Introduction

Let G be a group. We shall denote the commutator, center, group of automorphisms, and group of inner automorphisms of G by $G'$, Z(G), $\mathrm{Aut}(G)$, and $\mathrm{Inn}(G)$, respectively. Let exp(G) denote the exponent of G.

For any group H and abelian group K, let $\mathrm{Hom}(H,K)$ denote the group of all homomorphisms from H to K. This is an abelian group with binary operation $fg(x)=f(x)g(x)$ for $f,g\in \mathrm{Hom}(H,K)$.

An automorphism $\alpha$ of G is called central if $x^{-1}\alpha (x)\in Z(G)$ for all $x\in G$. The set of all central automorphisms of G, which is here denoted by Aut$_c(G)$, is a normal subgroup of Aut(G). Notice that $\mathrm{Aut}_c(G)=C_{\mathrm{Aut}(G)}(\mathrm{Inn}(G)),$ the centralizer of the subgroup $\mathrm{Inn}(G)$ in the group Aut(G). The elements of $\mathrm{Aut}_c(G)$ act trivially on $G'$.

There have been number of results on the central automorphisms of a group. M.J. Curran [2] proved that, “For any non abelian finite group G, $\mathrm{Aut}_z^z(G)$ is isomorphic with Hom $(G/G'Z(G),Z(G)),$ where $\mathrm{Aut}_z^z(G)$ is group of all those central automorphisms which preserve the centre Z(G) elementwise.” In [3], Franciosi et al. showed that, If $``$ $Z(G)$ is torsion free and $Z(G)/G'\cap Z(G)$ is torsion, then $\mathrm{Aut}_c(G)$ acts trivially on Z(G). It is an abelian and torsion free group”. They further proved that,“$\mathrm{Aut}_c(G)$ is trivial when Z(G) is torsion free and $G/G'$ is torsion.” In [5], Jamali et al. proved that, “For a finite group G in which $Z(G)\le G'$, $\mathrm{Aut}_c(G)\cong \ \mathrm{Hom}(G/G',Z(G))$.” They also proved that, “If G is a purely non-abelian finite p-group of class two (p odd), then $\mathrm{Aut}_c(G)$ is elementary abelian if and only if $\varOmega _1(Z(G))=\phi (G)$, and $exp(Z(G))=p$ or $exp(G/G')=p$,” where $\phi (G)$ is Frattini subgroup of G and $\varOmega _1(Z(G))=\langle x\in Z(G)| x^p=1 \rangle$ . Note that, a group G is called purely nonabelian if it has no nontrivial abelian direct factor. Adney [1] proved that, “If a finite group G has no abelian direct factor, then there is a one-one and onto map between $\mathrm{Aut}_c(G)$ and $\mathrm{Hom}(G,Z(G))$.”

In this article, we generalize the above results to subcentral automorphisms.

2 Subcentral Automorphisms

Let M and N be two normal subgroups of G.

By $\mathrm{Aut}^N (G)$, we mean the subgroup of $\mathrm{Aut}(G)$ consisting of all automorphisms which induce identity on G / N.

By $\mathrm{Aut}_M (G)$, we mean the subgroup of $\mathrm{Aut}(G)$ consisting of all automorphisms which induce identity on M.

Let $\mathrm{Aut}_M^N(G)=\mathrm{Aut}^N(G)\cap \mathrm{Aut}_M(G)$. From now onward, M will be a characteristic central subgroup, and elements of $\mathrm{Aut}^M(G)$ will be called as subcentral automorphisms of G (with respect to subcentral subgroup M). It can be seen that, $\mathrm{Aut}^M(G)$ is a normal subgroup of $\mathrm{Aut}_c(G)$.

We further, let $C^{*}=\{\alpha \in \mathrm{Aut}_M(G)|\alpha \beta =\beta \alpha , \forall \beta \in \mathrm{Aut}^M(G)\}$.

Clearly, $C^{*}$ is a normal subgroup of $\mathrm{Aut}(G)$. Since every inner automorphism commutes with elements of $\mathrm{Aut}_c(G)$, $\mathrm{Inn}(G)\le C^{*}$. If we take $M=Z(G)$, then $C^{*}$ is same as $\mathrm{Inn}(G)$.

Let $K=<\left\{ [g,\alpha ]|g\in G,\alpha \in C^{*}\right\} >$, where $[g,\alpha ]\equiv g^{-1}\alpha (g)$.

If $M=Z(G)$ then $K=G'$. However, in general, $G'$ is a subgroup of K for every central subgroup M.

In the following, K and $C^{*}$ will always correspond to a central subgroup of M of G as in the above definitions.

Our main results are given by the following theorems.

Theorem 1

For a finite group G, $\mathrm{Aut}_M^M(G)\cong \mathrm{Hom} \left( \frac{G}{\textit{KM}},M\right)$.

Theorem 2

Let G be a group with M torsion free and $M/M\cap K$ torsion. Then $\mathrm{Aut}^M(G)$ is a torsion-free abelian group which acts trivially on M.

Theorem 3

Let G be a purely nonabelian finite group, then $|\mathrm{Aut}^M(G)|=|\mathrm{Hom} (G,M)|$.

Theorem 4

Let G be a purely nonabelian finite p-group (p odd), then $\mathrm{Aut}^M(G)$ is an elementary abelian p-group if and only if $exp(M)=p$ or $exp(G/K)=p$.

Following proposition shows that each element of K is invariant under the natural action of $\mathrm{Aut}^M(G).$

Proposition 1

$\mathrm{Aut}^M(G)$ acts trivially on K.

Proof

Consider an automorphism $\alpha \in \mathrm{Aut}^M(G)$. This implies $x^{-1}\alpha (x)\in M$, for all $x\in G$. So $\alpha (x)=xm$ for some $m\in M$. Let $\beta \in C^{*}$. By definition of $C^{*}$, we have $\alpha ([x,\beta ])=\alpha (x^{-1}\beta (x))=(\alpha (x))^{-1}\beta (\alpha (x))=m^{-1}x^{-1}\beta (xm)=m^{-1}x^{-1}\beta (x)m=x^{-1}\beta (x)=[x,\beta ]$. Hence the results follows. $\square$

Proof of Theorem 1 For any $\mu \in \mathrm{Aut}_M^M(G)$, define the map $\psi _{\mu }\in \mathrm{Hom} \left( \frac{G}{\textit{KM}},M\right)$ as $\psi _{\mu }(g\textit{KM})=g^{-1}\mu (g)$.

We first show that $\psi _{\mu }$ is well defined.

Let $g\textit{KM}=h\textit{KM}$, i.e., $gh^{-1}\in \textit{KM}.$

$\therefore \mu (gh^{-1})=gh^{-1}\Rightarrow g^{-1}\mu (g)=h^{-1}\mu (h)\Rightarrow \psi _{\mu }(g\textit{KM})=\psi _{\mu }(h\textit{KM}).$

For proving $\psi _{\mu }$ is a homomorphism, consider $\psi _{\mu }(g\textit{KM}hKM)=\psi _{\mu }(gh\textit{KM})$ $=(gh)^{-1}\mu (gh)=h^{-1}g^{-1}\mu (g)\mu (h)$ $=g^{-1}\mu (g)h^{-1}\mu (h)=\psi _{\mu }(g\textit{KM}).\psi _{\mu }(h\textit{KM})$

Now define a map $\psi : \mathrm{Aut}_M^M(G)\longrightarrow \mathrm{Hom} \left( \frac{G}{\textit{KM}},M\right)$, as $\psi (\mu )=\psi _{\mu }$.

We show that $\psi$ is the required isomorphism.

For $f,g \in \mathrm{Aut}_M^M(G)$ and $h\in G$, $\psi (fg)(h\textit{KM})=\psi _{fg}(h\textit{KM})=h^{-1}fg(h)=h^{-1}f(hh^{-1}g(h))=h^{-1}f(h)h^{-1}g(h)=\psi _f(h\textit{KM}) \psi _g(h\textit{KM})=\psi _{f}.\psi _{g}(h\textit{KM})$. Hence $\psi (fg)=\psi (f)\psi (g)$.

Consider $\psi (\mu _1)=\psi (\mu _2)$, i.e., $\psi _{\mu _1}(g\textit{KM})=\psi _{\mu _2}(g\textit{KM}), g\in G$. This implies $g^{-1}\mu _1(g)=g^{-1}\mu _2(g)\Rightarrow \mu _1=\mu _2$, as g is an arbitrary element of G. Thus $\psi$ is a monomorphism.

We next show that $\psi$ is onto. For any $\tau \in \mathrm{Hom} \left( \frac{G}{\textit{KM}},M\right)$, define a map $\mu :G\rightarrow G$ as $\mu (g)=g\tau (g\textit{KM}), g\in G$.

Now we show that $\mu \in \mathrm{Aut}_M^M(G)$. For $g_1,g_2\in G, \mu (g_1g_2)=g_1g_2\tau (g_1g_2\textit{KM})$ $=g_1\tau (g_1\textit{KM})g_2\tau (g_2\textit{KM})$ $=\mu (g_1)\mu (g_2). \therefore \mu$ is a homomorphism on G.

Further, let $\mu (g)=1.$ This implies $g\tau (g \textit{KM})=1 \Rightarrow$ $\tau (g\textit{KM})=g^{-1}\Rightarrow g^{-1}\in M$ $\therefore g\textit{KM}=\textit{KM}$ $\Rightarrow \tau (g\textit{KM})=1\Rightarrow g=1$. Hence $\mu$ is one-one.

As G is finite, $\mu$ must be onto. So $\mu \in \mathrm{Aut}(G).$ Further, as $g^{-1}\mu ( g)=g^{-1}g\tau (g\textit{KM}) =\tau (g\textit{KM})\in M$, so $\mu \in \mathrm{Aut}^M(G)$. Also if $g\in M$, then $\mu (g)=g(\tau (g\textit{KM}))=g\tau (\textit{KM})=g.$ Thus, $\mu \in \mathrm{Aut}_M^M(G)$ and $\psi (\mu )=\tau .$

Hence the theorem follows. $\square$

Corollary 1

Let G be finite group with $M\le K$, then $\mathrm{Aut}^M(G)\cong \mathrm{Hom} (G/K, M)$.

Proof

Since $M\le K$, $\frac{G}{\textit{KM}}=G/K$. The result follows directly from Theorem 1 and Proposition 1. $\square$

Proof of Theorem 2 Let $\alpha \in \mathrm{Aut}^M(G)$. If x is an element of M, then by the hypothesis $x^n\in M\cap K$ for some positive integer n. By Proposition 1, we have $x^n=\alpha (x^n)=(\alpha (x))^n$, and hence $x^{-n}(\alpha (x))^n=1$. Since $x^{-1}\alpha (x)\in M$, this implies $(x^{-1}\alpha (x))^n=1$. As M is torsion free, this implies that $x^{-1}\alpha (x)=1$, i.e., $\alpha (x)=x$. Therefore, $\mathrm{Aut}^M(G)$ acts trivially on M.

Let $\alpha ,\beta \in \mathrm{Aut}^M(G)$ and $x\in G$. So $\alpha \beta (x)=\alpha (\beta (x))=\alpha (xx^{-1}\beta (x))=\alpha (x)x^{-1}\beta (x) =xx^{-1}\alpha (x)x^{-1}\beta (x)$ $=\beta (x)x^{-1}\alpha (x)=\beta (x)\beta (x^{-1}\alpha (x))=\beta \alpha (x)$. Thus, $\mathrm{Aut}^M(G)$ is an abelian group.

Now, consider $\alpha \in \mathrm{Aut}^M(G)$, and suppose there exists $k\in N$ such that $\alpha ^k=1$. Since $x^{-1}\alpha (x)\in M$ for all $x\in G$, there exists $g\in M$ such that $\alpha (x)=xg$. Further, $\alpha ^2(x)=\alpha (\alpha (x))=\alpha (xg)=\alpha (x)\alpha (g)=xg^2$($\because \alpha$ acts trivially on M). Hence, by induction, $\alpha ^n(x)=xg^n.$ But $\alpha ^k=1 \Rightarrow x=xg^k$, i.e., $g^k=1$. As M is torsion free, we must have $g=1$. Thus $\alpha (x)=x$ for every x, i.e., $\alpha =1$.

Therefore, $\mathrm{Aut}^M(G)$ is torsion free, and the theorem follows.$\square$

Proposition 2

Let G be a group in which M is torsion free and G / K is torsion, then $\mathrm{Aut}^M(G)=1$.

Proof

Let $\alpha \in \mathrm{Aut}^M(G)$ and $x\in G$. Then by the assumption, $x^n\in K$ for some $n\in N$. As $\alpha$ fixes K elementwise, we have $(\alpha (x))^n=\alpha (x^n)=x^n$. So $x^{-n}(\alpha (x))^n=1$. But $\alpha \in \mathrm{Aut}^M(G)$ and hence $x^{-1}\alpha (x)\in M\le Z(G)$. This implies that $(x^{-1}\alpha (x))^n=1$. Since M torsion free, it follows that $x^{-1}\alpha (x)=1$, i.e., $\alpha (x)=x, \forall x\in G$. So $\mathrm{Aut}^M(G)=1$. $\square$

Proof of Theorem 3 For $f \in \mathrm{Aut}^M(G)$, we let $\alpha (f)\equiv \alpha _f$ defined as $\alpha (f)(g)\equiv \alpha _f(g)=g^{-1}f(g), g\in G$. It can be shown that $\alpha _f\in \mathrm{Hom}(G,M).$ We thus have $\alpha : \mathrm{Aut}^M(G)\rightarrow \mathrm{Hom} (G,M)$.

One can easily see that $\alpha$ is injective.

It just remains to show that $\alpha$ is onto.

For $\sigma \in \mathrm{Hom} (G,M)$, consider the map $f:G\rightarrow G$ given by $f(g)=g\sigma (g)$. f is an endomorphism and also $g^{-1}f(g)=\sigma (g)\in M$, which implies that f is subcentral endomorphism of G, and hence f is normal endomorphism(i.e., f commutes with all inner automorphisms). So, clearly $\mathrm{Im}(f)$ is a normal subgroup of G.

It is easy to see that $f^n$ is also normal endomorphism and hence $\mathrm{Im} f^n$ is a normal subgroup of G, for all $n\ge 1$. Since G is a finite group, the two series

$$\mathrm{Ker} f\le \mathrm{Ker} f^2\le \dots \dots $$

$$\mathrm{Im}\ f\ge \mathrm{Im}\ f^2\ge \ldots \ldots $$

will terminate.

So there exists $k\in N$ such that

$$\mathrm{Ker} f^k=\mathrm{Ker} f^{k+1}=\ldots =A$$

$$\mathrm{Im} f^k=\mathrm{Im} f^{k+1}=\ldots \ldots =B$$

Now, we prove that $G=AB$.

Let $g\in G,\ f^k(g)\in \mathrm{Im} f^{k}=\mathrm{Im} f^{2k}$, and so $f^k(g)=f^{2k}(h)$, for some $h\in G$. Therefore $f^k(g)=f^k(f^k(h)).$ This implies $f^{k}(g^{-1})f^k(g)=f^k(g^{-1})f^k(f^k(h))$. Thus $(f^k(h))^{-1}g\in \mathrm{Ker} f^k=A$. Thus $g\in AB$ and hence $G=AB$.

Clearly $A\cap B=<1>$ and therefore $G=A\times B$. If $f(g)=1$, then $g^{-1}\sigma (g)=1.$ This implies $\mathrm{Ker} f\le M$. Similarly, if $f^2(g)=1$, i.e., $f(f(g))=1$. Thus $f(g)\in \mathrm{ker} f \le M.$ Therefore, $g\sigma (g) \in M \Rightarrow g\in M$. Hence $\mathrm{ker} f^2\le M.$ Repetition of this argument gives, $A\equiv \mathrm{ker} f^k \le M\le Z(G)$. This implies A is an abelian group. By assumption, G is purely nonabelian and hence, we must have $A\equiv \mathrm{Ker} f^k=1$. This further implies $\mathrm{Ker} f=1$, i.e., f is injective. So $G=B\equiv \mathrm{Im}\ f^k=\mathrm{Im} f.$ Thus f surjective. Hence, $f\in \mathrm{Aut}^M(G)$. From the definition of $\alpha$, it follows that $\alpha (f)=\sigma$. $\alpha$ is thus surjective. Therefore, $\alpha$ is the required bijection. Hence the result follows. $\square$

Proposition 3

Let G be a purely nonabelian finite group, then for each $\alpha \in \mathrm{Hom}(G,M)$ and each $x\in K$, we have $\alpha (x)=1$. Further $\mathrm{Hom}(G/K,M)\cong \mathrm{Hom}(G,M)$.

Proof

Whenever G is purely nonabelian group, then by Theorem 3, $|\mathrm{Aut}^M(G)|=|\mathrm{Hom}(G,M)|$. For every $\sigma \in \mathrm{Aut}^M(G)$, it follows that $f_{\sigma }:x\rightarrow x^{-1}\sigma (x)$ is a homomorphism from G to M. Further the map $\sigma \rightarrow \ f_{\sigma }$ is one–one and thus a bijection because $|\mathrm{Aut}^M(G)|=|\mathrm{Hom}(G,M)|$. So every homomorphism from G to M can be considered as an image of some element of $\mathrm{Aut}^M(G)$ under this bijection. Let $\alpha \in \mathrm{Hom}(G,M)$. Since $K=\left\{ [g,\alpha ]|g\in G, \alpha \in C^{*}\right\}$, a typical generator of K is given by $g^{-1}\beta (g)$ for some $g\in G$, and $\beta \in C^{*}$. So $\alpha (g^{-1}\beta (g))=f_{\sigma }(g^{-1}\beta (g))=(g^{-1}\beta (g))^{-1}\sigma (g^{-1}\beta (g))=\beta ^{-1}(g)gg^{-1}\beta (g)=1 (\because g^{-1}\beta (g)\in K$). It follows that $\alpha (x)=1$, for every $x\in K$.

Now consider the map $\phi : \mathrm{Hom}(G,M)\longrightarrow \mathrm{Hom}(G/K,M)$ such that $\phi (f)=\bar{f}$, where $\bar{f}(gK)=f(g)$ for all $g\in G$. Clearly this map $\phi$ is an isomorphism. $\square$

Proposition 4

Let G be a purely nonabelian finite group, then $|\mathrm{Aut}^M(G)|=|\mathrm{Hom}(G/K,M)|$.

Proof

Proof follows directly from Theorem 3 and Proposition 3. $\square$

Proposition 5

Let p be a prime number. If G is a purely nonabelian finite p-group then $\mathrm{Aut}^M(G)$ is a p-group.

Proof

By the assumption, the subgroup M and hence $\mathrm{Hom}\,(G/K,M)$ are finite p-groups. Hence the result follows directly from Proposition 4. $\square$

Proposition 6

Let G be a purely nonabelian finite group

(i) If $gcd(|G/K|,|M|)=1$, then $\mathrm{Aut}^M (G)=1$ .

(ii) If $\mathrm{Aut}^M(G)=1$, then $M\le K$.

Proof

(i) Follows from Proposition 4.

(ii) Let $|G/K|=a$ and $|M|=b$. Since $\mathrm{Aut}^M(G)=1$, hence by Proposition 4, $(a,b)=1$ . So there exist integers $\lambda$ and $\mu$ such that $\lambda a+\mu b=1.$ Let $x\in M$. Thus $xK=(xK)^{1}=(xK)^{\lambda a+\mu b}$ $=(xK)^{\lambda a}(xK)^{\mu b}$ $=K$ $\Rightarrow x\in K$. $\square$

Remark 1

From Corollary 1, and Proposition 3, we can say that, whenever $M\le K$, $\mathrm{Aut}^M(G)\cong \mathrm{Hom}(G,M)$. Even when $\mathrm{Im} f\le K$, for all $f\in \mathrm{Hom}(G,M),$ this result holds. Thus, if G is a purely nonabelian finite group and if for all $f\in \mathrm{Hom}(G,M)$, $\mathrm{Im} f \le K,$ then $\mathrm{Aut}^M(G)\cong \mathrm{Hom}(G/K,M)$.

Remark 2

For every $f\in \mathrm{Hom}(G,M)$, the map $\sigma _f:x\rightarrow xf(x)$ is a subcentral endomorphism of G. This endomorphism is an automorphism if and only if $f(x)\ne x^{-1}$ for all $1\ne x\in G$ (G is finite).

Following lemma has been proved in [4], we shall use it to prove Theorem 4.

Lemma 1

Let x be an element of a finite p-group G and N a normal subgroup of G containing $G'$ such that $o(x)=o(xN)=p$. If the cyclic subgroup $<x>$ is normal in G such that $ht(xN)=1$, then $<x>$ is a direct factor of G.

In the above statement ht denotes height. Height of an element a of a group G is defined as the largest positive integer n such that for some x in G, $x^n=a.$

Proof of Theorem 4 For the odd prime p, let $\mathrm{Aut}^M(G)$ be an elementary abelian p-group. Assume that the exponent of M and G / K are both strictly greater than p. Since G / K is finite abelian, it has a cyclic direct summand $\langle xK \rangle$ say, of order $p^n(n\ge 2)$ and hence $G/K\cong \langle xK \rangle \times$ L / K. For $f\in \mathrm{Hom}(G,M)$, consider $f(x)=a$ for any $x\in G$. So $\bar{f}(xK)=a$. Since exp (M) is strictly greater than p, the order of a is $p^m$, for some m, $2\le m\le n$.

We can use the homomorphism $\bar{f}$ to get corresponding homomorphism ( also denoted by same notation ) $\bar{f}$ as $\bar{f}:\langle xK \rangle \times L/K\rightarrow M$ with $(x^iK,lK)\rightarrow a^i$. The map $\bar{f}$ on $<xK> \times L/K$ is well defined, since o(a)|o(xK) (as $m\le n$).

If $aK=(x^sK,lK)$ then we show that p|s. Assume $p\not {|}s$, then $<xK>=<x^sK>$ and hence $G/K=<aK>L/K$. Now we have $o(a)\ge o(aK)\ge o(x^sK)=o(xK)\ge o(\bar{f}(xK))=o(a)$. This implies that $o(a)=o(aK)$. Thus $<a>\cap K=1$. As $o(aK)=o(xK)$, we get $G/K\cong <aK>\times L/K$ and hence $G\cong <a>\times L$. This is a contradiction, as G is a purely nonabelian group. Thus p|s.

By Remark 2 and Theorem 3, $\sigma _f\in \mathrm{Aut}^M(G)$ and by assumption $o(\sigma _f)=p$.

Now, we have $\sigma _f(x)=xf(x)=xa$. Since $f(a)=\bar{f}((xK)^s,lK)=a^s$, we have $\sigma _f^2(x)=xa^{s+2}=xa^{\frac{(s+1)^2-1}{s}}.$

Also, $\sigma _f^3(x)=xa^{\left( \frac{(s+1)^3-1}{s}\right) }$.

Generalizing this,

we get $\sigma _f^{t}(x)=xa^{\left( \frac{(s+1)^t-1}{s}\right) }$, for every $t\in N$.

As the order of $\sigma _f$ is p, we have $a^{\frac{(s+1)^p-1}{s}}=1$. Since p is odd and p|s, we have $p^2|(\left( \frac{(s+1)^p-1}{s}\right) -p)$.

$\therefore qp^2+p=\frac{(s+1)^p-1}{s}$ for some $q\in Z$. Thus $(a^p)^{qp+1}=1$. But $o(a)=p^m \Rightarrow$ $o(a^p)=p^{m-1}$.

Now

(1) if $a^p\ne 1$, then $p^{m-1}|(qp+1)$. But this is impossible as $m\ge 2$.

(2) $a^p=1$ is also not possible as $o(a)=p^m$ and $m\ge 2$.

So, the assumption that exp(M) and exp(G / M) are stricly greater than p is wrong.

Conversly, assume that $exp(G/K)=p$ and $f\in \mathrm{Hom}(G,M)$. Then by proposition 3, $\bar{f}\in \mathrm{Hom}(G/K,M)$. So for $x\in G$, put $\bar{f}(xK)=a$. If $aK\ne 1$, then it follows that $o(aK(G))=o(a)=p$. Clearly $<a>\le M(G)\le Z(G)$ and hence the cyclic subgroup $<a>$ is normal in G. We also have $ht(aK)=1$. Now by the Lemma 1, the cyclic subgroup $<a>$ is an abelian direct factor of G, and this contradicts the assumption. Therefore $a\in K$. This implies that $\mathrm{Im}(f)\le K$. Hence by Remark 1 $\mathrm{Aut}^M(G)\cong \mathrm{Hom} (G/K, M).$ But as M is abelian, $\mathrm{Hom}(G/K,M)$ is abelian. Thus $\mathrm{Aut}^M(G)$ is abelian. Since $exp(G/M)=p$, this implies that $\mathrm{Aut}^M(G)$ is an elementary abelian p-group.

Now assume that $exp(M)=p$. Consider $f,g\in \mathrm{Hom}(G,M)$. We first show that $g\circ f(x)=1$, for all $x\in G$. Assume that $\bar{f}(xK)=b\in M$, for $x\in G$. Since $exp(M)=p$, it implies that o(b)|p. If $b=1$ then $g\circ f(x)=g(\bar{f}(xK(G)))=1$. Now take, $o(b)=p$. If $b\in K$ then we have $g(f(x))=g(\bar{f}(xK(G)))=g(b)=1$. Assume b does not belong to K. As $b^p=1$, it follows that $o(bK)=p$. Also, as $b\in M\le Z(G)$, $<b>$ is normal in G. Now if $ht(bK(G))=1$, then by the Lemma 1, the cyclic subgroup $<b>$ is an abelian direct factor of G, giving a contradiction. So assume $ht(bk(G))=p^m$ for some $m\in N$. By the definition of height, there exists an element yK in G / K such that $bK=(yK)^{p^m}.$ But $exp(M)=p$. Therefore $g\circ f(x)=g(b)=\bar{g}(bK)=\bar{g}(yK)^{p^m}=1$. Thus, for all $f,g\in \mathrm{Hom}(G,M)$ and each $x\in G, g(f(x))=1$. We can similarly show that $f(g(x))=1$ and hence $f\circ g=g\circ f$. From Remark 2, $\sigma _f\circ \sigma _g=\sigma _g\circ \sigma _f$. This shows that $\mathrm{Aut}^M(G)$ is abelian.

Now we show that each nontrivial element of $\mathrm{Aut}^M(G)$ has order p. So if $\alpha \in \mathrm{Aut}^M(G)$, then by Remark 2, there exists a homomorphism $f\in \mathrm{Hom}(G,M)$ such that $\alpha =\sigma _f$. Therefore, we have to show that $o(\sigma _f)|p$. Clearly, taking $f=g$ and using $f(f(x))=1, x\in G,$ we have $x\in G$, we have $\sigma _f^2(x)=\sigma _f(xf(x))$ $=x(f(x))^2$. In general for $n\ge 1$, $\sigma _f^n(x)=x(f(x))^n$. As exp $(M)=p$ and $f(x)\in M$ we have, $\sigma _f^p(x)=x$ which implies $\sigma _f^p=1_{\mathrm{Aut}^M(G)}$.

Hence $o(\sigma _f)|p$. Thus, $o(\alpha )|p\ \forall \alpha \in \mathrm{Aut}_M(G).$ $\therefore$ $\mathrm{Aut}^M(G)$ is an elementary abelian group.

$\square$