Can someone give an example when this (particular construction of a ) normal subgroup is not characteristic?

Question

Suppose $G$ is a group, and that in the action of $G$ on itself by right multiplication, we find an element $g$ of odd order. This tells us that $G$ intersects nontrivial with $A_{|G|}$ when we embed $i: G \to S_{|G|}$ by this action.

If the preimage of this intersection $i^{-1}(i(G) \cap A_n)$ is denoted $A$ then it is surely normal in $G$. But is it characteristic? I don't think so, since I can precompose the right action with any automorphism of $G$ to get different embeddings, which means that $A$ is probably not invariant under automorphisms. However, I'm at a loss for an example - can someone provide one?

Incidentally, I guess this would mean that saying "the" right action of a group on its elements is misleading.

Answer 1

It is characteristic.

Let $g \in G$ have order $n$. By Lagrange's theorem, $n$ divides $|G|$, that is $|G|=nk$ for some $k$. Then $i(g)$ is a product of $k$ disjoint $n$-cycles. Hence the parity of $i(g)$ depends only on the order of $g$, not on which automorphisms have been applied.

More concretely: $i^{-1}(i(G) \cap A_n) = \{ g \in G : |g| \text{ is odd, or } |G|/|g| \text{ is even} \}$. This means that $i^{-1}(i(G) \cap A_n) = G$ unless $G$ has a cyclic Sylow 2-subgroup, and is the semi-direct product of the index 2-subgroup of the Sylow with the normal Sylow 2-complement in that case.

I don't think it helps in this case, but it is often helpful to know any automorphism of $G$ is also a permutation of $|G|$, so induced by an inner automorphism of $S_{|G|}$. The normalizer of $i(G)$ is called the holomorph, and the stabilizer of $i(1_G)$ is the automorphism group of $G$ viewed as permutations of $|G|$.

Answer 2

Here's another proof that it's characteristic.

It's well known that the alternating group $A_n$ is the only index 2 subgroup of $S_n$, hence $A_n$ is characteristic in $S_n$. Now let $n := |G|$. Then we may view $G$ as a subgroup of $S_n$ as you described. You're then asking if $A_n\cap G$ is characteristic in $G$. In other words, is $A_n\cap G$ fixed by every automorphism of $G$?

It's easy to prove that under this embedding of $G$ inside $S_n$, every automorphism of $G$ extends to an (inner) automorphism of $S_n$, which necessarily leaves $G$ invariant (as a set). This follows from noting that an automorphism of $G$ is in fact a permutation of its elements, hence an element of $S_n$. This corresponding element of $S_n$ should then be the thing you're conjugating by.

Having noted this, let $\sigma\in\text{Aut}(G)$, then we may extend $\sigma$ to $\tilde{\sigma}\in\text{Aut}(S_n)$, with the property that $\tilde{\sigma}(G) = G$. Then since $\tilde{\sigma}(A_n) = A_n$ (since $A_n$ is characteristic), you have $\tilde{\sigma}(A_n\cap G)\subset A$ and $\tilde{\sigma}(A_n\cap G)\subset G$, hence $\tilde{\sigma}(A_n\cap G) \subset A_n\cap G$, hence $\tilde{\sigma}(A_n\cap G) = A_n\cap G$ since $\tilde{\sigma}$ is an automorphism, but since every automorphism of $G$ can be viewed as a $\tilde{\sigma}\in\text{Aut}(S_n)$ this proves that $A_n\cap G$ is characteristic in $G$.