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Every nonabelian group $G$ acts nontrivially on itself by conjugation, and this action is by (inner) automorphisms. Therefore, if $G$ is finite of order $n\in\mathbb N$, then the semidirect product $G\rtimes_{\text{conj}}G$ is a nonabelian group of order $n^2$. Nonetheless, this resource doesn't mention neither $S_3\rtimes S_3$ among the aliases of any nonabelian group of order $36$, nor $D_4\rtimes D_4$ or $Q_8\rtimes Q_8$ among those of any nonabelian group of order $64$. Why are those aliases not mentioned? Maybe $G\rtimes_{\text{conj}}G\cong$ $G\times G$ (is it so?), but if so wouldn't that precisely mean that $G\rtimes_{\text{conj}}G$ is indeed another alias to mention?

Kan't
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    It is isomorphic to the direct product $G \times G$. Try to prove that. – Nicky Hekster Apr 22 '25 at 21:34
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    What does $\rtimes_{\mathrm{conj}}$ mean? – Rob Arthan Apr 22 '25 at 21:49
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    Got it! It's the outer semidirect product where conj is your name for the conjugation homomorphism $G \to \mathrm{Aut}(G)$. I've edited the question so that others won't have to share my puzzlement. – Rob Arthan Apr 22 '25 at 22:19
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    To answer your literal question "Why is [...] not an alias?", I suppose the answer is: Why repeatedly mention many special cases of an elementary general operation? But to be honest, such a question is really about mathematical culture and not so much about mathematics itself. – Lee Mosher Apr 23 '25 at 02:13

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As already mentioned in comments, $G \rtimes_{\text{conj}} G \cong G \times G$. Indeed, it is easy to check that the following map is an isomorphism:

$$G \rtimes_{\text{conj}} G \ni (g, h) \mapsto (gh, h) \in G \times G$$

David Gao
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