Let $G$, $H$ be groups. Prove: $e_G \times H$ is normal in $G \times H$ 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.
 A: $\{e_G\}\times H$ is (obviously?) the kernel of the homomorphism $\pi_1\colon G\times H\to G$, $(g,h)\mapsto g$.
A: 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.
