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Let $G$ be a group with $N \unlhd G$ such that there is $K \leq G$ with $KN = G$ and $K \cap N = 1$.

It is well known that in this case $$G \cong N \rtimes_\phi K$$

where $\phi: K \to Aut(N): k \mapsto (\phi_k: N \to N: n \mapsto knk^{-1})$


In exercises on classifying groups, we look for subgroups $K, H$ as above and then we conclude that $G \cong N \rtimes H$ for some $\phi: K \to Aut(N)$. We then proceed to find out what possibilities there are for such a map.

Why don't we just say that it is the $\phi$ as above? Is it because we don't want our description of $\phi$ to depend on conjugation in $G$?

Because otherwise, $G \cong N \rtimes K$ isn't useful because $\phi$ depends on how we calculate conjugation in $G$ and we want to describe $G$ independently of its subgroups?

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In general, $\phi$ is just any group homomorphism from $K$ to $Aut(N)$. If we have $G=KN$, $K\cap N=1$, we have an inner semidirect product, with the $\phi$ you have given.

"Why don't we just say that it is the $\phi$ as above?" Because there are also outer semidirect products, $G\cong N\rtimes_{\phi} K$ with an arbitrary homomorphism $\phi\colon K\rightarrow Aut(N)$. This is equivalent to the fact that there is a split short exact sequence of groups $$ 1\rightarrow N\rightarrow G \rightarrow K\rightarrow 1. $$

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  • $\begingroup$ Thanks for your answer. I'm a bit confused. Doesn't such a $\psi$ always exist if we can find subgroups as above in my post? I wrote down the explicit function. $\endgroup$ – user661541 Apr 19 at 9:59
  • $\begingroup$ If I understand correctly, you ask about the difference of an inner and an outer semidirect product, right? $\endgroup$ – Dietrich Burde Apr 19 at 10:06
  • $\begingroup$ Suppose we are classifying groups. At a given point, I have found $N,K$ as above.Then $G = N \rtimes_\phi K$ with $\phi$ as defined above. Why exactly do we want to find $\phi$ expxplicitely? Is it because we want to determine $G$ and $\phi$ is defined using calculations in $G$? Sorry if the question isn't clear. $\endgroup$ – user661541 Apr 19 at 10:13
  • $\begingroup$ We don't want to find $\phi$ explicitly. The classification, say, of groups of order $12$ has a group, called $C_3\rtimes C_4$, which is not the direct product, so is not abelian. There are other, more interesting properties about this group than $\phi$, for example a presentation. But we can use the semidirect product to classify all groups of order $12$. For details, see the classification by K. Conrad here. $\endgroup$ – Dietrich Burde Apr 19 at 10:14
  • $\begingroup$ I was in the impression that knowing $\phi$ explicitely allows us to to completely understand $G$ because then we have an independent representation $G \cong N \rtimes K$ (if we know $N,K$). Maybe here is an example to demonstrate my problem: math.stackexchange.com/questions/3193420/… $\endgroup$ – user661541 Apr 19 at 10:51

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