Group Theory: Alternate proof of the 2nd Isomorphism Theorem I can prove the 2nd Isomorphism Theorem in the usual way, but for fun I'm trying to prove that the map $HN \rightarrow H/(H\cap N)$ given by $\phi(hn)=h(H\cap N)$ is a homomorphism. I've gotten as far as showing that $ \phi(h_1n_1h_2n_2) = h_1h_2n'(H\cap N)$ but can't see my way to forward.  Is this the right map to have chosen? 
Statement of the 2nd Isomorphism Theorem: Let $H\leq G$ and $K\trianglelefteq G$. Then $K\trianglelefteq HK$ and $H\cap K \trianglelefteq H$ and $HK/K \equiv H/(H\cap K)$.
 A: First it's not difficult to show that your $\phi$ is well defined, i.e. if $hn=h'n'$, then $\phi(hn)=\phi(h'n').$
Since $HN$ is a subgroup, this means $h_1n_1h_2n_2\in HN$ for $h_1,h_2\in H,n_1,n_2\in N$. Thus there exists $h'\in H,n'\in N$ such that
$$h_1n_1h_2n_2=h'n',\quad (*).$$
For each $h\in H$, denote $[h]:=h(H\cap N)\in H/(H\cap N)$. To prove that $\phi$ is a homomorphism, it suffices to prove that
$$[h_1][h_2]=[h'].$$
Noting that $[h_1h_2]=[h_1][h_2]$, it suffices to prove that
$$(h_1h_2)^{-1}h'\in H\cap N.$$
To see this fact: since $N$ is normal, there exists $m\in N$ such that
$$n_1=h_2mh_2^{-1},$$
and by $(*)$ we have
$$h_1h_2mn_2=h'n'.$$
It follows that
$$mn_2(n')^{-1}=(h_1h_2)^{-1}h',\quad (**).$$
Since
$$\text{LHS of }(**)\in N,\quad \text{RHS of }(**)\in H,$$
it follows that $(h_1h_2)^{-1}h'\in H\cap N$. This proves that $\phi$ is a homomorphism.
It's trivial that $\phi$ is surjective with $\text{Ker}(\phi)=N$. Thus by the first isomorphism theorem
$$(HN)/N\cong H/(H\cap N).$$
