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?


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 $$ 1\rightarrow N\rightarrow G \rightarrow K\rightarrow 1. $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.