# How can non-split automorphism extensions exist?

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.

• I think you mean "Since outer automorphism groups of simple groups are solvable." Sep 27, 2020 at 13:50
• Indeed I do, thanks! Sep 27, 2020 at 14:39

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 $$2$$.