# Proving normality of a subgroup under componentwise multiplication

The question:

Let $G\times H$= {$(g,h) : g \in G, h\in H$}, where G and H are groups. Let $G_1$ = {$(g,e_H): g \in G$}. Prove that $G_1$ is a normal subgroup of $G\times H$ and that G is isomporphic to G.

My Understanding of it:

Knowing that $G \times H$ is a group under component multiplication my first instinct is to think that it's elements will look something like : {$g_1h_1, g_2gh_2, ... g_nh_n$} (assuming it is finite. Maybe I shouldn't be imagining it as finite, but anyway...)

My first thought is to prove (or rather show) that it is a subgroup. By showing that $G_1$ follows the basic rules:

1. It is non-empty
2. It is closed under multiplication
3. It contains the inverses of all it's elements

But my concern is I may not be sure what $G_1$ looks like. Is it something like this: $G_1$ = {$g_1e_h, g_2e_h, g_3e_h....$}?

And then from there - what?

Any hints would be great. Maybe even a step-by-step explanation of the question and what it's asking. This isn't a homework question or anything. I'm doing some self-study for an exam monday.

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Your first thoughts are correct. Have you tried performing the three basic rules? – Julian Kuelshammer Nov 17 '12 at 14:13
@JulianKuelshammer I couldn't start while I wasn't sure what $G_1$ looks like. But after some thought I couldn't quite figure out how I could use a group that contained the identity of another group. – Siyanda Nov 18 '12 at 9:49

Write the elements of the group and of $\,G_1\,$ as you did:

$$G_1:=\{(g,1)\;\;;\;\;g\in G\}\,$$

and thus for any $\,(g,h)\in G\times H\,$ , we have for any $\,x\in G\,$:

$$(g,h)^{-1}(x,1)(g,h)=(g^{-1},h^{-1})(x,1)(g,h)=(g^{-1}xg,h^{-1}\cdot 1\cdot h)=(g^{-1}xg,1)\in G_1$$

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+1 and I think it would be a good result that $G_1\cong G$. – Babak S. Nov 17 '12 at 14:30
that usually's impossible, @BabakSorouh, unless $\,H\,$ is trivial. Of course, with some infinite groups that can happen... – DonAntonio Nov 17 '12 at 14:59
@BabakSorouh I actually think you're right. Because I suspect the lecturer made a typo in our practice sheet when he wrote "and show that $G$ is an isomorphism to $G$... – Siyanda Nov 18 '12 at 9:50
I think $H$ just might be trivial. – Siyanda Nov 18 '12 at 9:51