# An abelian group $G$ with ${\rm Aut}(G)$ non-abelian

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)

Take $$G=\mathbb{Z}^2$$ its automorphism group $$Gl(2,\mathbb{Z})$$ is not commutative.

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.

• 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 $Aut(\mathbb{Z}/p_i^j \mathbb{Z}) \times..$? Commented Nov 4, 2019 at 16:34
• Yes, see this question. Here $GL(2,\Bbb F_2)\cong S_3$. Commented Nov 4, 2019 at 16:38
• Also this question. Commented Nov 4, 2019 at 17:26