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Ignore any trivial cases. Also, even if some examples exist for $|G|$ small, I still want to know if there are simple automorphism groups for arbitrarily large $|G|$.

For such a group to exist, something catastrophically wrong must happen, as there are definitionally many, many normal subgroups of the automorphism group. See here for some examples.

The inner automorphisms are the most famous. Universal power automorphisms are another. I've asked some recent questions about those.

So my question, what groups have a simple automorphism group? And less preciously, what is going on when/if this happens?

abnry
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2 Answers2

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Since $G/Z(G)$ is the inner automorphism group, and it is normal in $Aut(G)$, this has to be either trivial or all of $Aut(G)$. Thus $G$ is either abelian or a simple complete group. In the abelian case, you've already found a nice set of examples: $Z_2^n$ for $n\ge3$. For the simple complete groups, the Mathieu group $M_{23}$ works.

Steve D
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I can now answer half of my question. If $G = Z_2 \times Z_2 \times Z_2$, then $Aut(G) = GL_3(Z_2) = PSL(3,2)$, which is simple according to this page. I have no insight as to why it is simple. I also have no insight if this is just a freak example, as $|G|=8$ is small in this case. So I would really love a better answer.

abnry
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    This should happen for all of the PSL(n, 2) groups (for n > 2), shouldn’t it? Normally you only have easy access to the (non-simple) GL(n, q) as an automorphism group, but it sort of “just so happens” to also be PSL(n, q) in the q = 2 case, since there’s only one nonzero scalar. Aside from that, I don’t know anything about the problem. – pjs36 Mar 22 '18 at 01:46
  • Yes, I was wondering if there is something unique about $GL(n,Z_2)$, due to $Z(G)=G$, if I understand it right. – abnry Mar 22 '18 at 02:49