Some properties on $\mathrm{IA_Z}$-automorphisms of groups

Abstract

Let G be a group and $\mathrm{IA}(G)$ denote the group of all automorphisms of G, which induce identity map on the abelianized group $G_{ab}=G/G'$. Also the group of those $\mathrm{IA}$-automorphisms which fix the centre element-wise is denoted by $\mathrm{IA_Z}(G)$. In the present article, among other results and under some condition we prove that the derived subgroups of finite p-groups, for which $\mathrm{IA_Z}$-automorphisms are the same as central automorphisms, are either cyclic or elementary abelian.

1 Introduction and preliminaries

An automorphism $\alpha$ of a group G is called $\mathrm{IA}$ -automorphism if $x^{-1}\alpha (x) \in G'$, for all $x\in G$. This concept was introduced by Bachmuch [1] in 1965. We remark the letters $\mathrm{I}$ and $\mathrm{A}$ as to remind the reader that are those automorphisms which induce identity on the abelianized group, $G/G'$.

Also, if $x^{-1}\alpha (x) \in Z(G)$ for all $x\in G$, then we say that $\alpha$ is a central automorphism, and if $\alpha$ preserves all conjugacy classes of G, then it is called a class preserving automorphism. The set of all such automorphisms are denoted by $\mathrm{IA}(G)$, $\mathrm{Aut_Z}(G)$ and $\mathrm{Aut_C}(G)$, respectively. These concepts were introduced and studied by Curran [2] in 2001 and Yadav [15, 16] in 2009 and 2013. Clearly, the set of all $\mathrm{IA}$-automorphisms, which fix the centre element-wise, forms a normal subgroup of $\mathrm{IA}(G)$ and is denoted by $\mathrm{IA_Z}(G)$ (see [11, 12] for more information).

For any element x of a group G and automorphism $\alpha$ of G, the autocommutator of x and $\alpha$ is defined as follows:

$$\begin{aligned} {}[x, \alpha ]=x^{-1}x^{\alpha }=x^{-1}\alpha (x). \end{aligned}$$

Now, using the above notation, we have the following:

$$\begin{aligned} \mathrm{Aut_Z}(G)= & {} \lbrace \alpha \in \mathrm{Aut}(G) \mid [x, \alpha ] \in Z(G),\quad \forall x\in G\rbrace , \\ \mathrm{Aut_C}(G)= & {} \lbrace \alpha \in \mathrm{Aut}(G) \mid x^\alpha \in x^{G},\quad \forall x\in G\rbrace , \\ \mathrm{IA_Z}(G)= & {} \lbrace \alpha \in \mathrm{Aut}(G) \mid [x, \alpha ]\in G',~\alpha (z)=z,\quad \forall x\in G~and~z\in Z(G)\rbrace . \end{aligned}$$

One can easily check that any class preserving automorphism is an $\mathrm{IA}$-automorphism, which fixes the centre element-wise and hence

$$\begin{aligned} \mathrm{Inn}(G)\le & {} \mathrm{Aut_C}(G) \le \mathrm{IA_Z}(G) \le \mathrm{IA}(G) \le \mathrm{Aut}(G) , \\ Z(\mathrm{Inn}(G))\le & {} \mathrm{Aut_Z}(G) \le \mathrm{Aut}(G). \end{aligned}$$

The following example shows that every $\mathrm{IA_Z}$-automorphism is not necessarily inner automorphism.

Example 1.1

Consider the group

$$\begin{aligned} G=\langle a,b,x\mid [a,x]=[b,x]=1, [a,b]=x^{s},~s\ne 1 \rangle . \end{aligned}$$

Clearly, $G'=\langle x^{s}\rangle ,~~Z(G)=\langle x\rangle$ and $G/Z(G)=\langle \bar{a},\bar{b}\rangle \cong \mathrm{Inn}(G)$. The $\mathrm{IA_Z}$-automorphism $\alpha$ defined by $\alpha (a)=ax^{s},~~\alpha (b)=bx^{s},~~\alpha (x)=x$ is a non-inner automorphism.

Note that, $\mathrm{Aut_Z}(G)$ fixes the derived subgroup $G'$ element-wise and using this property, we have the following.

Lemma 1.2

For any group G, the central automorphisms commute with $\mathrm{IA_Z}$-automorphisms of G.

Proof

Assume $\alpha$ is a central automorphism of G, then $\alpha (x)=xz$ for any $x\in G$ and some element $z\in Z(G)$. Clearly, every central automorphism $\alpha$ fixes $G^{\prime }$ element-wise, and hence for any $\beta \in \mathrm{IA_Z}(G)$, we have

$$\begin{aligned} \alpha \beta (x)=\alpha (xx^{-1}\beta (x))=\alpha (x) x^{-1}\beta (x)=xzx^{-1}\beta (x)=\beta (\alpha (x))=\beta \alpha (x),\end{aligned}$$

which gives the result. $\square$

2 $\mathrm{IA_Z}$-automorphisms of a group

Considering the converse of Schur’s Theorem, Niroomand [10] proved that if the derived subgroup $G'$ of a given group G is finite and its central factor group, G / Z(G), is generated by d elements, then $|G/Z(G)|\leqslant |G'|^{d}$. Also, Sury [14] generalized this result as follows.

If $G'$ is finite and G / Z(G) is generated by d elements, then $|\mathrm{Inn}(G)|\leqslant |K(G)| ^{d}$, where $K(G)=\langle [x, \alpha ] \mid x\in G, \alpha \in \mathrm{Aut}(G)\rangle$ is the autocommutator subgroup of the group G, see also [4] and [8].

Here, we give a further generalized version of the above result, which improves [10].

Theorem 2.1

Let G be any group with finite derived subgroup. If d is the minimal number of generators of the central factor group of G, then $|\mathrm{IA_Z}(G)|\leqslant |G^{\prime }|^{d}$.

Proof

Assume that G / Z(G) has a minimal set of generators $x_{1}Z(G), \dots , x_{d}Z(G)$ and $\alpha \in \mathrm{IA_Z}(G)$, which fixes Z(G) element-wise. Consider the following map:

$$\begin{aligned} \psi :\mathrm{IA_Z}(G) \longrightarrow \underbrace{G' \times G' \times \cdots \times G'}_{d-times}\ , \end{aligned}$$

given by $\psi (\alpha )=([x_{1}, \alpha ], [x_{2}, \alpha ], \cdots ,[x_{d}, \alpha ])$, for all $\alpha \in \mathrm{IA_Z}(G)$.

One can easily check that $\psi$ is injective, as $\psi (\alpha )=\psi (\beta )$ implies that $[x_i, \alpha ]=[x_i, \beta ]$, for all $1\leqslant i\leqslant d$ and any $\alpha , \beta \in \mathrm{IA_Z}(G)$. Hence, $\alpha = \beta$ and so $\vert \mathrm{IA_Z}(G) \vert \leqslant \vert G' \vert ^d$. $\square$

Remark 2.2

One notes that the above theorem improves the result in [10], as $\mathrm{Inn}(G)\le \mathrm{Aut_C}(G)\le \mathrm{IA_Z}(G)$.

Clearly, $\mathrm{Aut_Z}(G)\cap \mathrm{Inn}(G)=Z(\mathrm{Inn}(G))$, for any group G. Now, we use this property to obtain the following.

Lemma 2.3

Let G be a group such that $Z(G)\le G^{\prime }$. Then $\mathrm{Aut_Z}(G)=\mathrm{IA_Z}(G)$ if and only if $\mathrm{Inn}(G)=Z(\mathrm{Inn}(G))$.

Proof

Assume $\mathrm{Aut_Z}(G)=\mathrm{IA_Z}(G)$, then $\mathrm{Inn}(G)=\mathrm{IA_Z}(G) \cap \mathrm{Inn}(G)=Z(\mathrm{Inn}(G))$.

Conversely, let $\mathrm{Inn}(G)=Z(\mathrm{Inn}(G))$. Then $G' \le Z(G)$ and so the definition of $\mathrm{IA_Z}(G)$ implies that $\mathrm{IA_Z}(G) \le \mathrm{Aut_Z}(G)$. On the other hand, by the assumption $\mathrm{Aut_Z}(G) \le \mathrm{IA_Z}(G)$. Therefore, $\mathrm{Aut_Z}(G)=\mathrm{IA_Z}(G)$. $\square$

The next result shows that the subgroup $Z(\mathrm{IA_Z}(G))$ is between $Z(\mathrm{Inn}(G))$ and $\mathrm{Aut_Z}(G)$.

Proposition 2.4

For any group G,

$$\begin{aligned} \mathrm{Aut_Z}(G) \cap \mathrm{IA_Z}(G) = Z(\mathrm{IA_Z}(G)). \end{aligned}$$

Proof

Suppose that $\alpha \in \mathrm{Aut_Z}(G) \cap \mathrm{IA_Z}(G)$ and $\beta \in \mathrm{IA_Z}(G)$. Lemma 1.2 implies that $[\alpha ,\beta ]=id$., as $\alpha \in \mathrm{Aut_Z}(G)$ and hence $\alpha \in Z(\mathrm{IA_Z}(G))$.

Conversely, for any $\alpha \in Z(\mathrm{IA_Z}(G))$ and $\beta \in \mathrm{Inn}(G)\le \mathrm{IA_Z}(G)$, we have $[\alpha ,\beta ]=id$. This implies that $\alpha \in C_{\mathrm{Aut}(G)}(\mathrm{Inn}(G))=\mathrm{Aut_Z}(G)$, and so $\alpha \in \mathrm{Aut_Z}(G) \cap \mathrm{IA_Z}(G)$. $\square$

In 1940, Hall [3] introduced the concept of isoclinism between two groups and it was extended to n-isoclinism, which is an equivalent relation among all groups and it is weaker than isomorphism. In 1976, Leedham-Green and McKay [7] extended this concept to isologism with respect to a given variety of groups. There have been extensive studies in this area of mathematics (see [5, 6], for more information).

Definition 2.5

Let G and H be arbitrary groups and assume $\alpha :G/Z(G)\rightarrow H/Z(H)$ and $\beta :G' \rightarrow H'$ be isomorphisms such that the following diagram is commutative:

where $f_1(g_1Z(G), g_2Z(G))=[g_1, g_2]$ and $f_2(h_1Z(H), h_2Z(H))=[h_1, h_2]$, for each $h_i\in \alpha (g_iZ(G))$, $i=1, 2$, and $\beta$ is an isomorphism induced by $\alpha$. Then $(\alpha ,\beta )$ is said to be isoclinism, so that G and H are called isoclinic and denoted by $G{\sim }H$.

Here, we characterize all finite p-groups, in which their $\mathrm{IA_Z}$-automorphisms are equal to central automorphisms. Yadav [16] proved that if two finite groups G and H are isoclinic, then $\mathrm{Aut_C}(G) \cong \mathrm{Aut_C}(H)$. With the same assumption, Pradeep [11] showed that $\mathrm{IA_Z}(G) \cong \mathrm{IA_Z}(H)$ and he applied his result to prove the following.

Theorem 2.6

([11], Theorem B(1)) Let G be a finite p-group. Then $\mathrm{Aut_Z}(G)=\mathrm{IA_Z}(G)$ if and only if $Z(G)=G'$.

The following results can be deduced using the above theorem.

Proposition 2.7

Let G be a d-generating finite p-group and $\mathrm{Aut_Z}(G)=\mathrm{IA_Z}(G)$. Then $|\mathrm{IA_Z}(G)|=|G'| ^{d}$.

Proof

The assumption and Theorem 2.6 imply that G / Z(G) is abelian. Hence, assume $G/Z(G)=\langle \bar{x_{1}}\rangle \times \langle \bar{x_{2}}\rangle \times \cdots \times \langle \bar{x_{d}}\rangle$, where $\bar{x_{i}}=x_{i}Z(G)$ is of order $p^{n_{i}}$, for some positive integer $n_i$, $i=1, \ldots , d$. Since $Z(G)=G' \le \Phi (G)$, by [16] Lemma 3.5, $\lbrace x_{1},\ldots ,x_{d}\rbrace$ is the minimal generating set for G. On the other hand, $\mathrm{Hom}(\langle \bar{x_{i}}\rangle ,G^{\prime })=G^{\prime }$, and hence Corollary 2.2 [11] implies that

$$\begin{aligned} \vert \mathrm{IA_Z}(G)\vert =\vert \mathrm{Hom}(G/Z(G) ,G')\vert =\Pi _{i=1}^{d}\vert \mathrm{Hom}(\langle \bar{x_{i}}\rangle ,G')\vert =\Pi _{i=1}^{d}\vert G' \vert =\vert G' \vert ^{d}. \end{aligned}$$

$\square$

Theorem 2.8

Let G and H be isoclinic finite p-groups, and $\mathrm{Aut_Z}(G)=\mathrm{IA_Z}(G)$. Then $\mathrm{Aut_Z}(H)=\mathrm{IA_Z}(H)$ if and only if |G| = |H|.

Proof

Assume that $\mathrm{Aut_Z}(H)=\mathrm{IA_Z}(H)$. The assumption and Theorem 2.6 imply that $G'=Z(G)$. The isoclinic property of G and H implies that

$$\begin{aligned} |Z(G)| =|G'|=|H'|=|Z(H)|. \end{aligned}$$

Therefore,

$$\begin{aligned} |H| =|H/Z(H)| |Z(H)| =|G/Z(G)||Z(G)| =|G|. \end{aligned}$$

Conversely, if $|G| =|H|$ then $|Z(G)|=|Z(H)|$, as G and H are isoclinic. On the other hand, $Z(G)=G'\cong H'$ and hence $\vert H' \vert = \vert Z(H) \vert$. Clearly, G is nilpotent of class 2 and isoclinic with H, which imply that H is also nilpotent of class 2. Then $H'\le Z(H)$ and so $H'=Z(H)$. Now, Theorem 2.6 implies that $\mathrm{Aut_Z}(H)=\mathrm{IA_Z}(H)$. $\square$

The following example gives a class of isoclinic groups, in which their $\mathrm{IA_Z}$-automorphisms are the same as central automorphisms.

Example 2.9

Consider the group

$$\begin{aligned}&G_{i}=\left\langle a, a_{1}, a_{2}, \cdots , a_{2s} \mid a^{p^{r+t}}=1, a_{1}^{p^{r}} = \cdots = a_{2s}^{p^{r}} = a^{p^{i}}, \right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \left. [a_{1}, a_{2}]=[a_{2}, a_{3}]= \cdots =[a_{2s-1}, a_{2s}]=a^{p^{t}} \right\rangle , \end{aligned}$$

where $r, s, t \geqslant 1$ and $1\leqslant i\leqslant r$.

Clearly, the group $G_{i}$ is nilpotent of class 2 and it is of order $p^{r(2s+1)+t}$. Also one can easily see that

$$\begin{aligned} Z(G_{i})\cong C_{p^{r+t}}, G'_{i}\cong C_{p^{r}}, \frac{G_{i}}{Z(G_{i})}\cong (C_{p^{r}})^{2s}. \end{aligned}$$

Now, Proposition 3.2 of [13] and Lemma 2.3 imply that

$$\begin{aligned} \mathrm{Aut_Z}(G_{i})=\mathrm{IA_z}(G_{i})=Z(\mathrm{Inn}(G_{i}))=\mathrm{Inn}(G_{i})\cong (C_{p^{r}})^{2s}, \end{aligned}$$

as the derived subgroup of the group $G_{i}$ is cyclic, for all $1\leqslant i\leqslant r$.

The following lemma of Morigi [9] is useful for our further investigation.

Lemma 2.10

([9], Lemma 0.4) Let G be a finite nilpotent group of class 2. Then $\exp (G^{\prime })=\exp (G/Z(G))$ and in the decomposition of G / Z(G) into direct product of cyclic groups at least two factors of maximal orders must occur.

We recall that the centre of any non-abelian p-group of order $p^{n}$ lies between $p^{2}$ and $p^{n-2}$.

Theorem 2.11

Let G be a non-abelian p-group with $\mathrm{Aut_Z}(G)=\mathrm{IA_Z}(G)$. If $p^{3}\le |G|\le p^{7}$, then the derived subgroups of such groups are either cyclic or elementary abelian p-groups.

Proof

By the assumption and Theorem 2.6, $Z(G)=G^{\prime }$. It is sufficient to show that Z(G) is cyclic or elementary abelian. The result is obvious, when G is of order $p^{3}$ or $p^{4}$. Now assume G is of order $p^{5}$, then $|Z(G)|=p^{2}$ or $p^{3}$. If $|Z(G)|=p^{3}$, then $|G/Z(G)|=p^{2}$. By Morigi’s Lemma, we have

$$\begin{aligned} \exp (Z(G))=\exp (G^{\prime })=\exp (G/Z(G))=p. \end{aligned}$$

Hence, Z(G) is elementary abelian. Suppose that $|G|=p^{6}$. If $|Z(G)|=p^{2}$ or $p^{4}$, then Morigi’s Lemma gives the result. Assume $|Z(G)|=p^{3}$ and $Z(G)\cong C_{p^{2}}\times C_{p}$. Then

$$\begin{aligned} \exp (Z(G))=\exp (G^{\prime })=\exp (G/Z(G))=p^2. \end{aligned}$$

Therefore,

$$\begin{aligned} G/Z(G)\cong C_{p^{2}}\times C_{p}, \end{aligned}$$

which contradicts Morigi’s Lemma and so $Z(G)\cong C_{p^{3}}$ or $C_{p}\times C_{p}\times C_{p}$.

Finally, assume $|G|=p^{7}$. If $|Z(G)|=p^{2}$ or $p^{5}$, then again the result is obtained, by Morigi’s Lemma. Assume Z(G) is of order $p^{3}$ and $Z(G)\cong C_{p^{2}}\times C_{p}$, then

$$\begin{aligned} G/Z(G)\cong C_{p^{2}}\times C_{p^{2}}. \end{aligned}$$

So G has two non-commutative generators and hence $G^{\prime }$ is cyclic, which contradicts the property that $Z(G)=G^{\prime }$.

If $|Z(G)|=p^{4}$ and

$$\begin{aligned} Z(G)\cong C_{p^{2}}\times C_{p}\times C_{p},~ C_{p^{3}}\times C_{p} \quad \mathrm{or} \quad C_{p^{2}}\times C_{p^{2}}, \end{aligned}$$

then

$$\begin{aligned} \exp (Z(G))=\exp (G^{\prime })=\exp (G/Z(G))=p^{2}\quad \mathrm{or} \quad p^{3} \end{aligned}$$

and $|G/Z(G)|=p^{3}$, which contradicts Morigi’s Lemma and so,

$$\begin{aligned} Z(G)\cong C_{p^{4}} \quad \mathrm{or} \quad C_{p}\times C_{p}\times C_{p}\times C_{p}. \end{aligned}$$

$\square$