Let $G$ be a finite group, $H$ and $N$ are normal subgroups of $G$ such that $\gcd(|H|,|N|)=1$. Prove $H \times N \cong H \cdot N$.
$\gcd(|H|,|N|)=1 \implies H \cap N=\{e\} $.
$H,N$ are normal subgroups of $G \implies $$HN \leq G$.
$|HN|=\frac{|H|\cdot|N|}{|H \cap N|}=|H|\cdot|N|$
Denote $q: HN \to H \times N, hn\mapsto (h,n)$ and I'll show that $q$ is well-defined and injective.
Well-defined:
Let $h_1n_1,h_2n_2\in HN$ such that $h_1n_1=h_2n_2.$
I have to show $q(h_1n_1)=q(h_2n_2)=(h_1,n_1)=(h_2,n_2)$.
$h_1n_1=h_2n_2\implies h_2^{-1}h_1n_1=n_2\implies h_3n_1=n_2\implies h_3=n_2n_1^{-1}\implies h_3\in N$.
Since $h_3\in H \cap N\implies h_3=\{e\} \implies n_1=n_2,h_1=h_2.$
Hence $q$ is well-defined.
$q$ is injective:
By definition of $q$,$(h_1,n_1)=(h_2,n_2) \iff h_1=h_2$ and $n_1=n_2$ hence $q$ is injective.
Since $q$ is injective and $|HN|=|H|\cdot|N| \implies H\times N \cong H \cdot N$.
Is my solution correct?
Thanks!