Let $G_1$ and $G_2$ be groups, and $G := G_1 \times G_2$. Let $e_2$ be the identity element in $G_2$.
Show that $H := G_1 \times {e_2}$ is a normal subgroup of G.
Using the homomorphism theorem without proof, or otherwise, show that $G/H=G_2$
I understand that, if $H$ is normal, then the quotient $G/H$, with the multiplication law on subsets, is a group. And I would check closure etc. to show this.
I'm having trouble with the first bit and I would really appreciate some help!
If $H$ is normal, then for all $g \in G$, $h \in H$ if and only if $ghg^{-1} \in H$.
So let $g = (g_{1}, g_{2}) \in G$. So $g^{-1} = (g_{1}^{-1}, g_{2}^{-1})$. Suppose $h = (g_{i}, e_{2}) \in H$. Now consider $ghg^{-1} = (g_{1}g_{i}g_{1}^{-1}, g_{2}e_{2}g_{2}^{-1}) = (g_{1}g_{i}g_{1}^{-1}, e_{2})$. You can work backwards to prove the converse. And so we have normality.