If $G/N\cong H$ then $G=NH$?

Question

I am curious if $G/N\cong H$ then $G=NH$?

($N$ is a normal subgroup of $G$, and $H$ is a subgroup of $G$.)

With this setup, we get that $NH$ is a subgroup of $G$ so $NH\subset G$. I am not sure about the other inclusion.

Thanks!

Answer 1

No. Let $G=C_8$, $N=C_4$, $H=C_2 \subset N$. Then $NH=N \ne G$.

Answer 2

For any group $S$, let $G = S \times S$ (direct product). Then take $N = H = S \times \{e_S\}$. Then $N$ is normal, $G/N \cong S \cong H$, and $NH = N = H$ is a strict subset of $G$ unless $S$ is the trivial group (with only one element, $e_S$).

Answer 3

If $G$ is finite, since $|HN|=\frac{|H| \cdot |N|}{|H \cap N|}$, it is true iff $H \cap N=1$.