# 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.

• So you know $hn=nh$. Now what is the problem? You can write $h_1h_2n_1n_2$ as $h_1n_1h_2n_2$. Jul 10 '19 at 2:00
• Can you elaborate it for me? Still don't get it. Jul 10 '19 at 2:12
• See This.
– Bach
Jul 10 '19 at 2:17
• You can use MathJax to format your posts. Jul 10 '19 at 3:21

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 $$\phi ((h_1,n_1)\cdot (h_2,n_2))=\phi((h_1h_2,n_1n_2))=h_1h_2n_1n_2=(h_1n_1)(h_2n_2)\\=\phi ((h_1,n_1))\phi ((h_2,n_2)).$$