# Determining the group homomorphism in semidirect product

We know that if $N$ is a normal subgroup, $H$ is a subgroup, and $\varphi$ is the group homomorphism such that $\varphi:H\to$Aut$(N)$. And this gives a unique group, called the outer semidirect product of $N$ and $H$.

But when a group and two of its subgroups are known, how to determine the homomorphisms? For instance, from here we know $$D_3\times \mathbb{Z}_3=(\mathbb{Z}_3\times \mathbb{Z}_3) \rtimes \mathbb{Z}_2$$ And from here $$\langle a,b,c| a^3=c^3=b^2=1, ac=ca,ab=ba^{-1},cb=bc^{-1}\rangle=(\mathbb{Z}_3\times \mathbb{Z}_3)\rtimes \mathbb{Z}_2$$ The group homomorphism $\varphi:\mathbb{Z}_2 \to$Aut$(\mathbb{Z}_3\times \mathbb{Z}_3)$ are surely different. But I have no idea on how to determine this homomorphism. Any hint or solutions are welcomed! Thanks!

• The action of $h\in H$ on $N$ inside $N\rtimes H$ is conjugation. That is, "$h(n)=hnh^{-1}$." – anon May 25 '13 at 2:42
• @anon You are right! Thanks for your comment, you can write an answer and I will accept it. – Golbez May 25 '13 at 2:47

## 1 Answer

If $N\trianglelefteq G$ and $H$ is a subgroup such that $HN=G$ and $H\cap N=1$, then $G$ is an internal semi-direct product $N\rtimes H$. This means it can be constructed as an external semidrect product $N\rtimes H$ with the data of a homomorphism $H\to{\rm Aut}(N)$. If we want to characterize this homomorphism when we already know $G$ full well, we think of conjugation. That is, for $h\in H$ and $n\in N$ and homomorphism $\alpha:H\to{\rm Aut}(N)$, then $\alpha(h)$ applied to $n$ yields the element $hnh^{-1}\in N\subseteq G$.

I like to think of semidirect products as special quotients of free products. Recall that relations are equations of the form $\rm blah=blah$, and we can quotient by relations $\rm \color{Blue}{blah}=\color{Purple}{blah},\color{Teal}{blah}=\color{Magenta}{blah},$... by quotienting a group by the normal subgroup generated by $\rm \color{Blue}{blah}\color{Purple}{blah}^{-1}$, $\rm \color{Teal}{blah}\color{Magenta}{blah}^{-1}$, etc. (The normal subgroup generated by a set $S$ is the normal closure of $\langle S\rangle$, i.e. the group generated by the union of all conjugates of $S$.) Then $N\rtimes H$ is isomorphic to $N*H$ modulo $h(n)=hnh^{-1}$ for all elements $h\in H$ and $n\in N$ (the expression $h(n)$ is interpreted as the result of $h$'s action on $n$).