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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?

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  • $\begingroup$ See here $\endgroup$ – Richard D. James Jun 9 '14 at 18:02
  • $\begingroup$ $HN/N\cong H/(H\cap N)$ via $hN\mapsto h(H\cap N)$ $\endgroup$ – blue Jun 9 '14 at 18:12
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It follows immediately from the isomorphism theorems that $G/N = HN/N \cong H/H \cap N \cong H.$

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  • $\begingroup$ How do we know that the intersection $H\cap N$ is trivial? $\endgroup$ – yoyostein Jul 9 '16 at 2:26
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    $\begingroup$ Because it was stated in the question. $\endgroup$ – Geoff Robinson Jul 9 '16 at 8:17
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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 $H \hookrightarrow G \twoheadrightarrow G/N$.

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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 $G/N\cong H$.

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