Injective homomorphism $\phi : H\times N\to G$ defined by $\phi (h, n) = hn$ Show that $\phi : H\times N\to G$ defined by $\phi (h, n) = hn$ is a injective homomorphism. $H$ and $N$ are normal subgroups of $G$, and $H\cap N = \{e\}$.
I know that if $hn = e$, then $h=e=n$, and $hn=nh$, for $h\in H$ and $n\in N$. I know the definition of homomorphism but I'm having trouble showing in this case.

To show  $\phi $ is a homomorphism, we need to prove that  $\phi ((h_1,n_1)\cdot  (h_2,n_2))=\phi ((h_1,n_1))\phi ((h_2,n_2)). $
Note that $$\phi ((h_1,n_1)\cdot  (h_2,n_2))=\phi((h_1h_2,n_1n_2))=h_1h_2n_1n_2=(h_1n_1)(h_2n_2)\\=\phi ((h_1,n_1))\phi ((h_2,n_2)).$$
