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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!

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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.

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  • $\begingroup$ Would you mind extending your answer to include the second part of the question? I think I might be missing the point with my answer. $\endgroup$ – user146650 May 7 '14 at 15:40

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