If $\mathrm{Aut}(G \times H) \cong \mathrm{Aut}(G) \times \mathrm{Aut}(H)$, then is $\mathrm{gcd}(\left| G \right|, \left| H \right|) = 1$?

Asked Nov 08 '21 at 03:33

Active Nov 08 '21 at 03:33

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I know that the converse is true, i.e., if $\mathrm{gcd}(\left| G \right|, \left| H \right|) = 1$, then $\mathrm{Aut}(G \times H) \cong \mathrm{Aut}(G) \times \mathrm{Aut}(H)$. It is said here.

My question is: Is the converse true? I.e., If $\mathrm{Aut}(G \times H) \cong \mathrm{Aut}(G) \times \mathrm{Aut}(H)$, then is $\mathrm{gcd}(\left| G \right|, \left| H \right|) = 1$?

There is no work done for this and this is just a curious question.

asked Nov 08 '21 at 03:33

Yoyos Tutoring

2

I doubt it. If $G$ and $H$ are simple, I believe this is true if neither is isomorphic to a subgroup of the other. In particular, if $G$ and $H$ are simple and neither of $|G|$ or $|H|$ divides the other, then it is true, I believe. – Thomas Andrews Nov 08 '21 at 03:40
@ThomasAndrews, leave a like! – Yoyos Tutoring Nov 08 '21 at 03:44
1

https://math.stackexchange.com/q/420884/562051 I believe this post would be helpful – Display name Nov 08 '21 at 03:59

If $\mathrm{Aut}(G \times H) \cong \mathrm{Aut}(G) \times \mathrm{Aut}(H)$, then is $\mathrm{gcd}(\left| G \right|, \left| H \right|) = 1$?

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