Condition for a group G to be the product of subgroups

Question

I'm trying to prove that if $H,N<G$ with $N$ a normal subgroup such that $H\cap N=\lbrace 1\rbrace$ and $H\cong G/N$ then $G=NH$. This will tell us that $G$ is the internal semidirect product of $N$ and $H$ with the canonical morphism of conjugation but I've not been able to show that $G=NH$.

I realized that $NH/N\cong G/N$ but that doesn't help. On the other way I thought that if the isomorphism of $f:G/N\to H$ is such that $f(hN)=h$ for all $h\in H$ then I could have finished but is not possible since we don't have information of the isomorphism.

Answer 1

If $G$ is finite, this follows by a counting argument: $|H|=|G/N|=[G:N]$, so $$|HN| = \frac{|H||N|}{|H\cap N|} = |H||N| = [G:N]|N| = |G|,$$ and since all quantities are finite and $HN\subseteq G$, we get $G=HN$.

If $G$ is not finite, then the claim need not hold. Let $G=\mathbb{Z}\times\mathbb{Z}$, $N=\mathbb{Z}\times\{0\}$. and $H=\{0\}\times2\mathbb{Z}$. Then $G/N\cong \mathbb{Z}\cong H$, $N\cap H=\{e\}$. but $NH$ is a proper subgroup of $G$.

(Or just $G$ any group that has a proper subgroup $H$ with $H\cong G$, and take $N=\{e\}$.)

It will become true for infinite groups if instead of asking that $H$ be isomorphic to $G/N$, you specify that the isomorphism is induced by the canonical projection. That is, if you ask that $\pi\colon G\to G/N$ satisfy that $\pi|_H \colon H\to G/N$ is an isomorphism.

With that additional assumption, to prove that $HN=G$, let $g\in G$. Then there exists $h\in H$ such that $\pi(h) = \pi(g)$, so $h^{-1}g\in \ker(\pi) = N$. Therefore, there exists $n\in N$ such that $h^{-1}g=n$, so $g=hn$, proving that $G\subseteq HN$. The reverse inclusion is trivial.

Answer 2

For finite groups this is trivially true: That $NH/N \cong G/N$ gives $|NH|/|N| = |G|/|N| \implies |NH| = |G|$, which, as $NH \subseteq G$ gives you $G=NH$.

For infinite groups, this is not true. Indeed, let $G = \mathbb{Z}^2$ $= \{(x,y); x,y \in \mathbb{Z}\}$, where the operation is component-wise addition. Let $N= \langle (0,1) \rangle$, and let $H= \langle (2,0) \rangle$. Then $G/N \cong \langle (1,0) \rangle \cong H$, but $NH$ does not generate $G$.