# Is there only one internal semidirect group?

I am a bit confused on the concept of internal semidirect group $N\rtimes H$.

For external semidirect group $N\rtimes_\varphi H$, I know that there can be possibly many semidirect groups depending on the homomorphism $\varphi: H\to\text{Aut}(N)$.

Q) For internal semidirect group $N\rtimes H$, is there only one such group, where the homomorphism is taken to be conjugation, i.e. $\varphi_h(n)=hnh^{-1}$?

Thanks for the clarification.

• For $N \unlhd G, H \le G$, $$NH = \{\, nh \mid n \in N, h \in H \,\},$$ so there is no question of dependence on any homomorphism. In other words, $NH$ is just the subgroup of $G$ generated by the two subgroups $N$ and $H$. Any two subgroups $H$ and $K$ of $G$ (neither one necessarily normal) can generate a subgroup $\langle H, K \rangle$, and $\langle H, K \rangle = HK$ if and only if $HK = KH$ (the subgroups "permute"). What is true is that for an internal semidirect product, $NH \cong N \rtimes H$, the external semidirect product of $N$ and $H$ taken as groups. – M. Vinay Jul 21 '16 at 6:52
• In this external semidirect product, the action of $H$ on $N$ is given by $n \mapsto hnh^{-1}$ (with $N$ and $H$ seen as subgroups of $G$ for the purpose of defining this action by conjugation). – M. Vinay Jul 21 '16 at 6:53
• In other words, an internal semidirect product is determined completely by the group operation itself. In the case of an external semidirect product, there is no natural binary operation with one operand from each "factor" group — this necessitates the definition of an action of one group on the other. – M. Vinay Jul 22 '16 at 3:25