Let $G$ be a group, $X$ a generating set of $G$. Suppose $X$ is $\operatorname{Aut}(G)$-invariant, i.e. $\sigma(X)\subseteq X$ for all $\sigma \in \operatorname{Aut}(G)$. When is the restriction homomorphism $$ \begin{gathered} \operatorname{Aut}(G) \to \operatorname{Sym}(X)\\ \sigma \mapsto \sigma|_X \end{gathered}$$ an isomorphism?

Example: $A_4$ with $X=\{\text{3-cycles}\}$ satisfies the above properties. $C_2\times C_2$ works, and so do all cyclic groups of order up to 4 (Thanks Jeremy Rickard for the correction). Are there any other examples?

  • $\begingroup$ I don't think $A_4$ is an example. Cyclic groups of order up to $4$ and $C_2\times C_2$ are. $\endgroup$ – Jeremy Rickard Oct 2 '14 at 20:58
  • $\begingroup$ @JeremyRickard $\operatorname{Aut}(A_4)=S_4$, and each automorphism takes takes 3-cycles to 3-cycles. See groupprops.subwiki.org/wiki/… $\endgroup$ – Avi Steiner Oct 2 '14 at 21:01
  • $\begingroup$ But there are eight $3$-cycles. $\endgroup$ – Jeremy Rickard Oct 2 '14 at 21:03
  • $\begingroup$ @JeremyRickard ... oops! You're right. $\endgroup$ – Avi Steiner Oct 2 '14 at 21:04
  • 1
    $\begingroup$ Going back to the question, I think that for $n\ge 5$ there's nothing not abelian, because then every orbit of $Alt(n)$ on $Sym(n)-\{1\}$ has cardinal $>n$. Hence I guess that the answer for your question is a short finite list. $\endgroup$ – YCor Oct 2 '14 at 21:56

For $G$ abelian, the complete list is $C_2$, $C_3$, $C_4$, $C_6$ and $C_2\times C_2$.

Proof: Suppose that $X$ contains a non-involution $x$. Then $x^{-1}\in X$ (since inversion in an automorphism of $G$) but then $\{x,x^{-1}\}$ is a block of size 2 for the action of Aut(G) on $X$ and thus $|X|=2$ and $X=\{x,x^{-1}\}$ and $G$ is cyclic. Since $X$ is $Aut(G)$-invariant, $G$ must have only two element of order $|G|$ and this happens only when $G$ is $C_3$, $C_4$ or $C_6$. We may thus assume that $G$ is generated by involutions and hence is an elementary abelian $2$-group. It's easy to check that $C_2^n$ has this property only for $n=1$ and $n=2$.

(I was assuming $G$ is finite, but the proof can probably be adapted.)

For $G$ nonabelian, I think the only example is $G=Sym(3)$ with $X$ the set of involutions. You could probably prove this following YCor's comments for example.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.