# If $G = HN$, $N\unlhd G$, $H\cap N = 1$ - is $G/N \cong H$?

This might be a stupid question, but right now it does not seem obvious to me if for a finite group $G$ with normal subgroup $N$ and subgroup $H$ such that $G = HN$ -- is it then true that $G/N \cong H$? How does an isomorphism look?

• See here – Richard D. James Jun 9 '14 at 18:02
• $HN/N\cong H/(H\cap N)$ via $hN\mapsto h(H\cap N)$ – blue Jun 9 '14 at 18:12

It follows immediately from the isomorphism theorems that $G/N = HN/N \cong H/H \cap N \cong H.$
• How do we know that the intersection $H\cap N$ is trivial? – yoyostein Jul 9 '16 at 2:26
Yes. In fact, under these circumstances $G \cong N \rtimes H$ for some semidirect product of $N$ and $H$. The isomorphism is then the composition of $H \hookrightarrow G \twoheadrightarrow G/N$.
The fact that $G=HN$ means precisely that for every $g\in G$ there is a unique pair $(h,n) \in H\times N$ such that $g=hn$, so you can define the projections $$\pi_H:G \to H, g=hn \mapsto h,\quad \pi_N:G \to N, g=hn \mapsto n.$$ Consider the homomorphism $$\Phi: G/N \to H, \Phi(gN)=\pi_H(g).$$ We have $$\Phi(gN)=1 \iff\pi_H(g)=1 \iff h=1.$$ It follows that $$\ker\Phi=\{g=hn\in G: h=1\}=N.$$
Since every $h \in H$ can be regarded as $h=h\cdot1\in HN=G$, we have $$h=\pi_H(h)=\Phi(hN) \quad \forall h \in H,$$ i.e. $\Phi$ is surjective. Hence $G/N\cong H$.