# Automorphisms of powers of finite simple groups

It is a theorem that a finite nontrivial group $$G$$ has no proper nontrivial characteristic subgroups if and only if $$G \cong S^n$$ where $$S$$ is simple and $$n > 0$$ is the number of copies of $$S$$ in a direct product. Call an automorphism $$\varphi \in \operatorname{Aut}(G)$$ irreducible if $$\varphi(H) \not\subseteq H$$ for all proper nontrivial normal $$H \subseteq G$$. (I am just coining this term, please let me know if there is an actual name for this.) When $$S$$ is abelian, irreducible automorphisms are precisely irreps of the group $$\mathbb{Z}$$ over the vector space $$\mathbb{F}_p^n$$ and are precisely characterized by irreducible polynomials in the ring $$\mathbb{F}_p[t]$$ of degree $$n$$, excluding the polynomial $$t$$ because it corresponds to the $$0$$ map. What is the situation like when $$S$$ is non-abelian?

• So I'm sure this is classical but if $S$ is nonabelian it looks to me like the automorphism group is the wreath product $\text{Aut}(S) \wr S_n$; basically the point is that the obvious copies of $S$ are uniquely determined by some property (e.g. I think they are the only copies of $S$ which are normal, or which are direct summands) so an automorphism has to permute them. Then a necessary condition for irreducibility is that the corresponding permutation of the copies of $S$ is a cycle. Commented Aug 6 at 22:53
• Conversely if the permutation is a cycle then I think $\varphi$ is "indecomposable" in the sense that it doesn't preserve a nontrivial direct summand. This is equivalent to your condition in the abelian case. In the nonabelian case I don't know that irreducibility is a good condition; e.g. if $n = 1$ and $S$ has trivial outer automorphism group then every automorphism of $G$ is inner and so preserves some cyclic subgroup. Commented Aug 6 at 23:00
• Oops, that made me realize an error I made. I should have said that they don't fix normal subgroups. Commented Aug 6 at 23:11

Suppose $$S$$ is non-abelian simple. Using https://groupprops.subwiki.org/wiki/Normal_subdirect_product_of_perfect_groups_equals_direct_product we know the normal subgroups of $$S^n$$ are of the form $$S^I$$ for some subset $$I\subset\{1,\dots,n\}$$ of indices. So $$\varphi$$ being irreducible is equivalent to $$\varphi(S^I)\not\subset S^I$$.
Lemma The automorphism group of $$S^n$$ is $$\mathrm{Aut}(S)\wr \Sigma_n$$.
Proof Let $$\varphi\colon S^n\to S^n$$ be an automorphism. For each $$i\in\{1,\dots,n\}$$, the $$i$$-th summand $$S_i\subset S^n$$ is a normal subgroup, hence $$\varphi(S_i)\subset S^n$$ is a normal subgroup. But by the above characterization of normal subgroups of $$S^n$$ we know $$\varphi(S_i)=S_{\sigma(i)}$$ for some index $$\sigma(i)\in\{1,\dots,n\}$$. Thus $$\varphi=\psi\circ \sigma$$ where $$\sigma$$ permutes the factors and $$\psi$$ is an automorphism of $$S^n$$ sending $$S_i$$ to $$S_i$$. But then $$\psi=\psi_1\times\cdots\times\psi_n$$ where $$\psi_i\in\mathrm{Aut}(S)$$.
Now, let $$\varphi\in\mathrm{Aut}(S^n)$$ be an automorphism which induces a permutation $$\sigma\in \Sigma_n$$ on the direct factors. Then the condition that $$\varphi(S^I)\not\subset S^I$$ is equivalent to $$\sigma(I)\not\subset I$$ for any non-trivial subset $$I$$ of $$\{1,\dots,n\}$$. This is equivalent to $$\sigma$$ being a $$n$$-cycle. Thus, we conclude:
Proposition An automorphism $$\varphi$$ of $$S^n$$ is irreducible if and only if the permutation $$\sigma\in\Sigma_n$$ induced on the direct factors is a $$n$$-cycle.