Could you give me a simple example of $G$ abelian with ${\rm Aut}(G)$ non-abelian? Otherwise how could I prove that $G$ abelian implies ${\rm Aut}(G)$ abelian. (I don't really think that's true)
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2 Answers
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Take $G=\mathbb{Z}^2$ its automorphism group $Gl(2,\mathbb{Z})$ is not commutative.
answered Nov 4, 2019 at 9:39
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There are also examples of finite abelian groups with non-abelian automorphism group, e.g., $$ \operatorname{Aut}(C_2\times C_2)\cong S_3. $$ This seems to be the easiest example.
answered Nov 4, 2019 at 10:03
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$\begingroup$ Thank you! Is there a way to deduce this result for a generic abelian group? For example using the decomposition of $G=\mathbb{Z}/p_i^{j}\mathbb{Z} \times .... $ and then show that because $Aut(H\times K)=Aut(H)\times Aut(K)$ (when the orders of both groups have order that are relatively prime) it's possible to find a group that is isomorphic to these $Aut(\mathbb{Z}/p_i^j \mathbb{Z}) \times..$? $\endgroup$ Commented Nov 4, 2019 at 16:34
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1$\begingroup$ Yes, see this question. Here $GL(2,\Bbb F_2)\cong S_3$. $\endgroup$ Commented Nov 4, 2019 at 16:38
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