Let $G$ and $H$ be two abelian groups. Is it true that $\text{Aut}(G\times H)\cong\text{Aut}(G)\times\text{Aut}(H)$?
I am trying to figure out $\text{Aut}(G)$ where $G=\mathbb{Z}/5\mathbb{Z}\bigoplus \mathbb{Z}/25\mathbb{Z}$.
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The answer is implied by https://math.stackexchange.com/questions/595598/edited-t-f-the-automorphism-group-textaut-mathbb-z-2-times-mathbb-z?rq=1 – Mr. Chip Jul 07 '17 at 18:57
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The "natural strategy" is to start with $G = H$ (since that affords the possibility of "swapping the factors"), and then to take $G$ the simplest non-trivial group, i.e., the cyclic group of order $2$. Aha...this resolves the question (as Dietrich's answer shows). <> As a general matter, when seeking counterexamples, always try to choose them as simply as possible. – Andrew D. Hwang Jul 07 '17 at 19:01
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2For your example, see this question with $p=5$ and $m=n=1$. – Dietrich Burde Jul 07 '17 at 19:02
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No, we have $\rm{Aut}(C_2\times C_2)\cong S_3$, see Show $\operatorname{Aut}(C_2 \times C_2)$ is isomorphic to $D_6$, but ${\rm Aut}(C_2)$ is trivial.
Dietrich Burde
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