Soluble Groups with few orbits under automorphisms

TL;DR

This paper investigates soluble groups with finitely many automorphism orbits, showing they decompose into a torsion-free nilpotent part and a finite group, and classifies certain groups with exactly three automorphism orbits.

Contribution

It establishes a structural decomposition for soluble groups with finite automorphism orbits and classifies those with three orbits, advancing understanding of automorphism actions.

Findings

Existence of a torsion-free characteristic nilpotent subgroup in such groups

Decomposition of the group as a semidirect product with a finite group

Classification of soluble groups with exactly three automorphism orbits

Abstract

Let $G$ be a group. The orbits of the natural action of Aut$(G)$ on $G$ are called ``automorphism orbits'' of $G$, and the number of automorphism orbits of $G$ is denoted by $\omega(G)$. We prove that if $G$ is a soluble group with finite rank such that $\omega(G)< \infty$, then $G$ contains a torsion-free characteristic nilpotent subgroup $K$ such that $G = K \rtimes H$, where $H$ is a finite group. Moreover, we classify the mixed order soluble groups of finite rank such that $\omega(G)=3$.

Equations

$$(t_x{t}_y)t_z=t_{xy}c(x,y)t_z=t_{xy}{t}_{z}c(x,y{)}^z=t_{xyz}c(xy,z)c(x,y{)}^z,$$

$$t_x(t_y{t}_z)=t_x{t}_{yz}c(y,z)=t_{xyz}c(x,yz)c(y,z)$$

$$e(zy{)}^{-n}=(e(y{)}^{z}e(z)c(y,z){)}^{-n}.$$

$$s_y{s}_z=t_y{t}_{z}e(y{)}^{z}e(z)=t_{yz}c(y,z)e(y{)}^{z}e(z)=t_{yz}e(yz)=s_{yz}.$$

$$x^p-1=(x-1)(x^{p-1}+x^{p-2}+\dots +x+1).$$

$$(h^{j}a{)}^p=h^{jp}(a{a}^{h^j}\dots a^{h^{(p-1)j}})=1$$

$$\{b_1,b_1^{\alpha },\dots ,b_1^{{\alpha }^{p-2}},\dots ,b_t,b_t^{\alpha },\dots ,b_t^{{\alpha }^{p-2}}\}$$

$$\{c_1,c_1^{\beta },\dots ,c_1^{{\beta }^{p-2}},\dots ,c_t,c_t^{\beta },\dots ,c_t^{{\beta }^{p-2}}\}$$

$$b_i\mapsto c_i\text{and}\alpha \mapsto \beta ,$$

Full text

This work was partially supported by FAPDF - Brazil.

1. Introduction

Let $G$ be a group. The orbits of the natural action of $\operatorname{Aut}(G)$ on $G$ are called “automorphism orbits” of $G$, and the number of automorphism orbits of $G$ is denoted by $\omega(G)$. It is interesting to ask what can we say about “$G$” only knowing $\omega(G)$. It is obvious that $\omega(G)=1$ if and only if $G=\{1\}$, and it is well known that if $G$ is a finite group then $\omega(G)=2$ if and only if $G$ is elementary abelian. In [3], T. J. Laffey and D. MacHale proved that if $G$ is a finite non-soluble group with $\omega(G)\leqslant 4$, then $G$ is isomorphic to $\operatorname{PSL}(2,\mathbb{F}_{4})$. Later, M. Stroppel, in [8], has shown that the only finite non-abelian simple groups $G$ with $\omega(G)\leq 5$ are the groups $\operatorname{PSL}(2,\mathbb{F}_{q})$ with $q\in\{4,7,8,9\}$. In [2], the authors prove that if $G$ is a finite non-soluble group with $\omega(G)\leq 6$, then $G$ is isomorphic to one of $\operatorname{PSL}(2,\mathbb{F}_{q})$ with $q\in\{4,7,8,9\}$, $\operatorname{PSL}(3,\mathbb{F}_{4})$ or $\operatorname{ASL}(2,\mathbb{F}_{4})$ (answering a question of M. Stroppel, cf. [8, Problem 2.5]).

Some aspects of automorphism orbits are also investigated for infinite groups. M. Schwachhöfer and M. Stroppel in [7, Lemma 1.1], have shown that if $G$ is an abelian group with finitely many automorphism orbits, then $G=\operatorname{Tor}(G)\oplus D$, where $D$ is a characteristic torsion-free divisible subgroup of $G$ and $\operatorname{Tor}(G)$ is the set of all torsion elements in $G$. In [1, Theorem A], the authors proved that if $G$ is a $\operatorname{FC}$-group with finitely many automorphism orbits, then the derived subgroup $G^{\prime}$ is finite and $G$ admits a decomposition $G=\operatorname{Tor}(G)\times A$, where $A$ is a divisible characteristic subgroup of $\operatorname{Z}(G)$. For more details concerning automorphism orbits of groups see [8].

If $G$ is a group and $r$ is a positive integer, then $G$ is said to have finite rank $r$ if each finitely generated subgroup of $G$ can be generated by $r$ or fewer elements and if $r$ is the least such integer. The next result can be viewed as a generalization of the above mentioned results from [1] and [7].

Theorem A**.**

Let $G$ be a soluble group of finite rank. If $\omega(G)<\infty$, then $G$ has a torsion-free radicable nilpotent subgroup $K$ such that $G=K\rtimes H$, where $H$ is a finite subgroup.

We do not know whether the hypothesis that $G$ has finite rank is really needed in Theorem A. The proof that we present here uses this assumption in a very essential way.

In [3], T. J. Laffey and D. MacHale showed that $G$ is a finite group in which the order $|G|$ is not prime power and $\omega(G)=3$ if and only if $|G|=pq^{n}$, the Sylow $q$-subgroup $Q$ is a normal elementary abelian subgroup of $G$ and $P$ is a Sylow $p$-subgroup which acts fixed-point-freely on $Q$. See also [5] for groups with $\omega(G)\leqslant 3$ (almost homogeneous groups).

Recall that a group $G$ has mixed order if it contains non-trivial elements of finite order and also elements of infinite order. We obtain the following classification.

Theorem B**.**

Let $G$ be a mixed order soluble group with finite rank. We have $\omega(G)=3$ if and only if $G=A\rtimes H$ where $|H|=p$ for some prime $p$, $H$ acts fixed-point-freely on $A$ and $A=\mathbb{Q}^{n}$ for some positive integer $n$.

2. Proofs

A well-known result in the context of extensions of finite groups, due to I. Schur, states that if $G$ is a finite group and $N$ is a normal abelian subgroup with $(|N|,|G:N|)=1$, then there exists a complement $K$ of $N$. Recall that $K$ is a complement of (a normal subgroup) $N$ in $G$ if $N\cap K=1$ and $G=NK$. In particular, $G=N\rtimes K$. Now, we prove that Schur’s theorem holds under a more general assumption that $G$ contains a divisible abelian subgroup of finite index. The proof presented here is adapted from the ideas of the finite case (cf. [6, 9.1.2]).

Lemma 2.1**.**

Let $A$ be a divisible normal abelian subgroup of finite index of a group $G$. Then there exists a subgroup $H$ of $G$ such that $G=A\rtimes H$.

Proof.

Let $B=G/A$. From each coset $x$ in $B$ we choose a representative $t_{x}$, so that the set $T=\{t_{x}\mid x\in B\}$ is a transversal to $A$ in $G$. Since $t_{x}t_{y}A=t_{xy}A$, there is an element $c(x,y)$ of $A$ such that $t_{x}t_{y}=t_{xy}c(x,y)$. Then

[TABLE]

[TABLE]

and $c(xy,z)c(x,y)^{z}=c(x,yz)c(y,z)$, for each $x,y\in B$. Consider the element $d(y)=\prod_{x\in B}c(x,y)\in A$. As $A$ is an abelian group $d(z)d(y)^{z}=d(yz)c(y,z)^{n}$, where $n=|B|$. We obtain that $d(yz)=d(y)^{z}d(z)c(y,z)^{-n}$. Now, since $A$ is a divisible group, there exists $e(y)\in A$ such that $e(y)^{n}=d(y)^{-1}$ for each $y\in B$. Hence

[TABLE]

Since $A$ is torsion-free, it follows that $e(zy)=(e(y)^{z}e(z)c(y,z))$. Define $s_{x}=t_{x}e(x)$, then

[TABLE]

Thus $x\mapsto s_{x}$ defines a homomorphism $\phi:B\rightarrow G$. Now $s_{x}=1$ implies that $t_{x}\in A$ and $x=A=1_{B}$. From this we conclude that $H=B^{\phi}$ is the desired complement. ∎

Now, we consider groups with finitely many automorphism orbits with a characteristic torsion-free soluble subgroup of finite index (see also Schur-Zassenhaus Theorem [6, 9.1.2]).

Lemma 2.2**.**

Let $n$ be a positive integer. Let $G$ be a group such that $\omega(G)<\infty$ and $A$ a torsion-free characteristic subgroup of $G$ with finite index $n$. If $A$ is soluble, then there exists a subgroup $H$ of $G$ such that $G=A\rtimes H$.

Proof.

If $A$ is abelian, then the result is immediate by Lemma 2.1. Assume that $A$ is non-abelian. Set $d$ the derived length of $A$. First we prove that $A/A^{(d-1)}$ is torsion-free. The subgroup $A^{(d-1)}$ is a torsion-free abelian divisible subgroup since $A$ is torsion-free and has finitely many automorphism orbits. If $A/A^{(d-1)}$ is not torsion-free, then we can find an element $a\in A$ such that $\langle a,A^{(d-1)}\rangle$ has a torsion-free divisible group of finite index. Then by Lemma 2.1 the subgroup $\langle a,A^{(d-1)}\rangle$ has elements of finite order. That is a contradiction. Thus $A/A^{(d-1)}$ is torsion-free.

Now, we complete the proof arguing by induction on the derived length of $A$. Consider the quotient group $\bar{G}=G/A^{(d-1)}$. By induction we deduce that there exists a finite subgroup $\bar{B}$ of order $n$ in $\bar{G}$ such that $\bar{G}=\bar{A}\rtimes\bar{B}$. Set $B$ the inverse image of $\bar{B}$. Clearly $A^{(d-1)}\leq B$ and $A^{(d-1)}$ has finite index $n$ in $B$. Therefore, by Lemma 2.1 $B$ has a subgroup $H$ of order $n$ and so such a subgroup is a complement of $A$ in $G$. The result follows. ∎

The following lemma is well-known. We supply the proof for the reader’s convenience.

Lemma 2.3**.**

Let $G$ be an abelian group of finite rank. If $\omega(G)<\infty$, then the torsion subgroup $\operatorname{Tor}(G)$ is finite.

Proof.

Since $G$ has finitely many automorphism orbits, it follows that the exponent $\exp(\operatorname{Tor}(G))$ is bounded. As $G$ has finite rank we have that $\operatorname{Tor}(G)$ is finitely generated. We deduce that $\operatorname{Tor}(G)$ is finite, which completes the proof. ∎

The following result provides a description of radicable nilpotent groups of finite rank (see [4, Theorem 5.3.6] for more details). Recall that a group $G$ is said to be radicable if each element is an $n$th power for every positive integer $n$.

Lemma 2.4**.**

Let $G$ be a soluble group with finite rank. Then the following are equivalent:

1. (i)

$G$* has no proper subgroups of finite index;* 2. (ii)

$G=G^{m}$* for all $m>0$;* 3. (iii)

$G$* is radicable and nilpotent.*

We are now in a position to prove Theorem A.

Proof of Theorem A.

We argue by induction on derived length of $G$.

Assume that $G$ is abelian. By Schwachhöfer-Stroppel’s result [7], $G=D\oplus T$, where $D$ is characteristic torsion free divisible subgroup and $T$ is the torsion subgroup of $G$. By Lemma 2.3, the torsion subgroup $T=\operatorname{Tor}(G)$ is a finite subgroup of $G$, the result follows.

Now, we assume that $G$ is non-abelian. Set $d$ the derived length of $G$. Arguing as in the previous paragraph, we deduce that $G^{(d-1)}=D_{1}\oplus T_{1}$, where $D_{1}$ is a characteristic torsion-free divisible subgroup and $T_{1}$ is the torsion subgroup of $G^{(d-1)}$ and so, $T_{1}$ is finite. By induction $G/G^{(d-1)}$ has the desired decomposition. More precisely, $G^{(d-1)}=D_{1}\oplus T_{1}$ and $G/G^{(d-1)}=\bar{A}\rtimes\bar{B}$ where $\bar{A}$ is torsion-free and $\bar{B}$ is finite. Note that $\bar{A}^{n}=\bar{A}$ for any positive integer $n$, since the quotient groups $\bar{A}^{(i)}/\bar{A}^{(i+1)}$ are torsion-free divisible groups (we can use Lemma 2.2 to conclude that each quotient is torsion-free).

Note that the centralizer $C_{G}(T_{1})$ is a subgroup of finite index in $G$, because $G/C_{G}(T_{1})$ embeds in the automorphism group of $T_{1}$ which has finite order. Let $A$ be the inverse image of $\bar{A}$. As $\bar{A}$ is torsion-free and $\bar{A}^{n}=\bar{A}$ for any positive integer $n$, we have $A\leqslant C_{G}(T_{1})$. Thus $T_{1}\leq Z(A)$ and $\operatorname{Tor}(A)=T_{1}$. Set $K=A^{e}$, where $e=\exp(T_{1})$. Then $K$ is torsion-free and has finite index in $G$. Therefore, by Lemma 2.2, there exists a finite subgroup $H$ such that $G=K\rtimes H$. According to Lemma 2.4, we deduce that $K$ is a radicable nilpotent group (the subgroup $K$ has no proper subgroups of finite index). The proof is complete. ∎

Now we will deal with Theorem B: Let Let $G$ be a mixed order soluble group with finite rank. We have $\omega(G)=3$ if and only if $G=A\rtimes H$ where $A=\mathbb{Q}^{n}$ for some positive integer $n$, $|H|=p$ for some prime $p$ and $H$ acts fixed-point-freely on $A$.

Proof of Theorem B.

First assume that $G$ is a mixed order soluble group of finite rank and have $\omega(G)=3$. By Theorem A, $G=A\rtimes H$ where $A$ is a torsion-free radicable nilpotent subgroup and $H$ is a finite group. Since $\omega(G)=3$ and $G$ has mixed order, it follows that $A$ must be abelian (so that $A=\mathbb{Q}^{n}$) and $H$ is an elementary abelian $p$-subgroup. On the other hand, since $A$ is characteristic, we deduce that all elements in $G\setminus A$ have order $p$ and then $H$ acts fixed-point-freely on $A$, so that $H$ is cyclic [5, Lemma 2.2].

Conversely, suppose that $G=A\rtimes H$ where $A=\mathbb{Q}^{n}$, $|H|=p$ for some prime $p$, and $H$ acts fixed-point-freely on $A$. First we prove that all elements in $G\setminus A$ have order $p$. Set $h\in G$ such that $H=\langle h\rangle$. Since $h^{p}=1$, we conclude that the minimal polynomials $m_{h}=m_{h}(x)$ of $T_{h}:\mathbb{Q}^{n}\to\mathbb{Q}^{n}$, given by $a\mapsto a^{h}$, divides

[TABLE]

Consequently, $m_{h}(x)=x^{p-1}+x^{p-2}+\ldots+x+1$, because $h$ acts fixed-point-freely on $A$. Thus using the identity $(xy)^{n}=x^{n}(y^{x^{n-1}})\ldots y^{x}y$, we obtain

[TABLE]

for all $a\in\mathbb{Q}^{n}$. So all elements of $G\setminus A$ have order exactly $p$ and act fixed-point-freely on $A$.

Now, let $b,c\in A\setminus\{1\}$ and $\alpha,\beta\in G\setminus A$. By Cyclic Decomposition Theorem, there exist $b_{1},b_{2},\ldots,b_{t},c_{1},c_{2},\ldots,c_{t}\in A$ such that

[TABLE]

and

[TABLE]

are bases of $A$. Without loss of generality we can assume that $b=b_{1}$ and $c=c_{1}$. Thus the map given by

[TABLE]

where $i=1,\ldots,t$ extends to an automorphism of $G$. Hence all non-trivial elements of $A$ belong to the same orbit under the action of $\operatorname{Aut}(G)$, and all elements in $G\setminus A$ are in the same orbit under the action of $\operatorname{Aut}(G)$. The proof is complete. ∎