Let $G$ be a simple group of order greater than $2$. Then $Z(Aut(G))$ is trivial if and only if $G$ is not Abelian.

Let $H = Z(G)$ and let $g\in G, h\in H$. Then $gh = hg$ for every $g\in G$. So $ghg^{-1} = h$ for every $g\in G$ and we have $gHg^{-1} = H$. Hence $Z(G)$ is a normal subgroup of $G$. Hence either $Z(G)$ is trivial or $Z(G) = G$.

Ok now we're in business potentially.

Because now we can suppose that $G$ is not Abelian, in which case $Z(G)$ is trivial and let $\sigma\in Z(Aut(G))$. We have $Inn(G) \cong G/Z(G)\cong G$ in this case, i.e. the group conjugations of $G$ (as a subgroup of the automorphism group of $G$) is isomorphic to $G$ itself. Then for every $x,y\in G$ we have $\sigma(xyx^{-1}) = x\sigma(y)x^{-1} = \sigma(x)\sigma(y)\sigma(x)^{-1}$ so that conjugation by $x$ and conjugation by $\sigma(x)$ are the same. Hence then $x = \sigma(x)$, because the map from $G$ to conjugations is injective. Thus $\sigma$ is the identity map and $Z(Aut(G))$ is trivial.

My problem is the other direction. How do I show that $G$ is not Abelian if $Z(Aut(G))$ is trivial?

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    $\begingroup$ If $G$ is a abelian, then inversion is a group automorphism. $\endgroup$ – Lorenzo Najt Dec 11 '16 at 0:00
  • $\begingroup$ Oh right and that will commute with every other automorphism, because they are all homomorphisms. Ugh now I feel like a dork. Thanks a ton! :) $\endgroup$ – Tanner Strunk Dec 11 '16 at 0:10
  • $\begingroup$ Side bar, Stack Exchange is really helping with my qual. studying haha...just a few more days... $\endgroup$ – Tanner Strunk Dec 11 '16 at 0:13
  • $\begingroup$ Good luck! Hope it goes well. $\endgroup$ – Lorenzo Najt Dec 11 '16 at 0:28

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