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Given a finite simple group $S$, we can consider its automorphism group ${\rm Aut}(S)$. Since ${\rm Inn}(S) \lhd {\rm Aut}(S)$, and $S \cong {\rm Inn}(S)$, we can ask whether $S$ has a complement in ${\rm Aut}(S)$. I was very surprised to learn that this is not always the case.

The reason I am surprised is as follows: if I take a simple group $S$ and an automorphism $\phi:S \to S$, I can compute the order of $\phi$ (say it is $t$) and then form the semidirect product $S \rtimes C_t$ where a generator of $C_t$ acts as $\phi$ on $S$.

Since automorphism groups of simple groups are solvable, why can't I just do this process a certain number of times and obtain the whole ${\rm Aut}(S)$? I understand that if I just include a set of generators of all the automorphisms then I'd get the holomorph of $S$, which is indeed $S \rtimes {\rm Aut}(S)$... but why can't I just take a set of generators of outer automorphisms? Is this the obstruction?

I tried to look at ${\rm Aut}(A_6)$, which I know to be not split, but it wasn't helpful. Any insight on the right way to think about this would be appreciated.

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    $\begingroup$ I think you mean "Since outer automorphism groups of simple groups are solvable." $\endgroup$ – Jeremy Rickard Sep 27 at 13:50
  • $\begingroup$ Indeed I do, thanks! $\endgroup$ – FifteenPointOne Sep 27 at 14:39
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The outer automorphism group $\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G)$ of a group $G$ may not act as automorphisms of $G$, so forming the semidirect product $G\rtimes\operatorname{Out}(G)$ may not even make sense.

For example, $\operatorname{Out}(A_6)\cong C_2\times C_2$ has one element (of order $2$) that isn't represented by any element of order $2$ in $\operatorname{Aut}(A_6)$. It is represented by an element $\sigma\in\operatorname{Aut}(A_6)$ that has order $4$, where $\sigma^2$ is an inner automorphism, which is why the order of $\sigma$ in $\operatorname{Out}(A_6)$ is $2$.

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