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$T/F:$ The automorphism group $\text{Aut} (\mathbb Z/2 \times \mathbb Z/2)$ is abelian.

Keeping the comments for Which group is meant by $\mathbb Z/2 \times \mathbb Z/2.$ in mind I would like to present a solution for the above mathematical problem to get it verified.

My answer to the problem is: FALSE

My Reasons: $\mathbb Z_2\oplus\mathbb Z_2\simeq K_4=\{e,a,b,c\}$ where $K_4$ is the Kelin's $4$-group with $e,a,b,c$ having their usual meaning. I would like to show that any bijection from $K_4$ onto $K_4$ which keeps $e$ fixed, is an isomorphism. Let $\phi$ be one such. Then W.L.G. we are to show that,

$$\phi(ab)=\phi(a)\phi(b)$$

Now, $\phi(ab)=\phi(c)$ and $\phi(a)\phi(b)$ is different from $\phi(a),\phi(b)$ and also from $e=\phi(e)$ since $\phi(a)\ne\phi(b).$

All such $\phi$ forms a group $\simeq S_3.$ Thus $\text{Aut } K_4$ is not abelian.

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  • Looks good to me. At the end of the second to last paragraph you would help the reader by adding ("by process of elimination, $\phi(a)\phi(b)$ must also be $\phi(c)$"). It took me a second to realize that was what you were arguing. – hunter Dec 06 '13 at 15:16
  • It sort of depends how many details your professor makes you check -- you could flesh out the proof more. But your assertion is completely correct: the claim is false, because the group is not abelian, because it is $S_3$, because any set-theoretic bijection of the three non-trivial elements is in fact a homomorphism. – hunter Dec 06 '13 at 15:19

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