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:


*

*It is non-empty

*It is closed under multiplication 

*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.

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