Let $G$, $H$ be groups. Prove: $e_G \times H$ is normal in $G \times H$

Question

I've proved that $e_G \times H$ is a subgroup of $G\times H$ and I know the various ways of defining normality. I am just having a bit of trouble writing this last part of my proof - any hints, tips etc. for showing that $e_G \times H$ is normal in $G \times H$?

I generally use the form: $aba^{-1} \in A$ where $A$, $B$ groups instead of showing left and right cosets are equal.

Do you mean for $x$ to be $\times$? If so, use \times as in $G\times H$ — Jordan Glen, May 10 '15 at 20:39
@user239484 In my answer I have given an elaboration on Hagen's method as well as a direct proof via the closure-by-conjugation definition you mentioned. — theage, May 10 '15 at 21:21

score 3 · Answer 1 · answered May 10 '15 at 20:43

3

$\{e_G\}\times H$ is (obviously?) the kernel of the homomorphism $\pi_1\colon G\times H\to G$, $(g,h)\mapsto g$.

answered May 10 '15 at 20:43

Hagen von Eitzen

374,180

This actually isn't so obvious for me. I've only just begun studying these things. Would you please elaborate the same way you would when teaching this to someone for the first time? – user239484 May 10 '15 at 20:51
I need to prove this to use the first isomorphic theorem to show that that $G \times H$/${e_G} \times H$ is isomorphic to G. Based on how this question is given on our final review, I assume I can't just state such a thing is obvious and move on. – user239484 May 10 '15 at 20:54
@ Hagen Von Eitzen – user239484 May 10 '15 at 21:10

theage · Accepted Answer · 2015-05-10T21:24:16.700

If you want a direct proof, suppose $b=(e_G,h)\in\{e_G\}\times H$ and $a=(g,k)\in G\times H$. Then $$aba^{-1}=(g,k)(e_G,h)(g^{-1},k^{-1})=(ge_Gg^{-1},khk^{-1}) = (e_G,khk^{-1})$$ recalling that the group structure on $G\times H$ is componentwise, so $(g,h)(g',h'):=(gg',hh')$ and $(g,h)^{-1}=(g^{-1},h^{-1})$. The RHS is clearly in $\{e_G\}\times H$ since $k,h\in H$ implies $khk^{-1}\in H$. Thus your subgroup is normal by the standard definition.

Alternatively if you want to use Hagen's more elegant method, you need to show that $\{e_G\}\times H=\ker\phi$ where $\phi:G\times H\to G$ takes $\phi(g,h)=g$. By definition, \begin{align*}\ker\phi &= \{(g,h)\in G\times H:\phi(g,h)=e_G\}\\&=\{(g,h)\in G\times H:g=e_G\} \\&= \{e_G\}\times H\end{align*} as desired. The first equality is just the definition - the kernel of a homomorphism is the set of all elements that get sent to the identity. Thus since kernels are normal we are done.

Let $G$, $H$ be groups. Prove: $e_G \times H$ is normal in $G \times H$

2 Answers2