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A semi-direct product of two finite groups is determined by the homomorphism from one group to the other's automorphism group. I wonder if it is possible to have two different homomorphism that turns out to have the same resulting semi-direct product.

If yes, what is the general pattern of this? Is there any (equivalent?) condition that we can put on the two different homomorphisms to make isomorphic semi-direct groups?

Guess I'm currently considering semi-direct product of groups like $\Bbb Z_n^k$, which may be easier to formulate than general one but harder than the case with simply $\Bbb Z_n$ involved

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    Yes, it’s possible. The nonabelian group of order $pq$, $p\lt q$ both prime, $p|q-1$, can be realized via any of the available automorphisms of order $p$ in $\mathrm{Aut}(C_q)$; any choice of automorphism will yield groups that are isomorphic to each other. – Arturo Magidin Sep 25 '19 at 06:34

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The tl;dr is yes! As Arturo mentioned in the comments, every (nontrivial) $\varphi : \mathbb{Z}/p \to \text{Aut}(\mathbb{Z}/q)$ (when $p \mid q-1$) gives rise to the same semidirect product (the unique nonabelian group of order $pq$).

As for your second question, involving a "general pattern", the answer is a (somewhat surprising) no. See this MSE question. There are some nice results in special cases, like when $G$ and $H$ are finite with coprime order (cf. the same MSE question), but in general the problem is too hard to say much.

I hope this helps ^_^

Arturo Magidin
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HallaSurvivor
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  • You mean "when $p|q-1$". If $p$ does not divide $q-1$, then the only map $\varphi$ is trivial, and the group of order $pq$ is abelian. – Arturo Magidin Sep 25 '19 at 14:21