# Semidirect product of subgroups

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 $$1\rightarrow N\rightarrow G \rightarrow K\rightarrow 1.$$
• 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. – user661541 Apr 19 at 9:59
• 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. – user661541 Apr 19 at 10:13
• 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. – Dietrich Burde Apr 19 at 10:14
• 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/… – user661541 Apr 19 at 10:51