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

Question

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)

score 11 · Answer 1 · answered Nov 04 '19 at 09:39

11

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

answered Nov 04 '19 at 09:39

Tsemo Aristide

87,475

score 8 · Accepted Answer · answered Nov 04 '19 at 10:03

8

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 04 '19 at 10:03

Dietrich Burde

130,978

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 $Aut(\mathbb{Z}/p_i^j \mathbb{Z}) \times..$?
– Lucas Tonon Nov 04 '19 at 16:34
1

Yes, see this question. Here $GL(2,\Bbb F_2)\cong S_3$. – Dietrich Burde Nov 04 '19 at 16:38
1

Also this question. – Arturo Magidin Nov 04 '19 at 17:26

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

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