I've just run into a little confusion when doing questions on the Homomorphism Theorem.
Say we have a homomorphism $\rho: G \times H \to G$ such that $\rho(g,h) = g,$ where $G \times H$ represents the direct product of groups $G$ and $H$. Then the kernel of $\rho$ will be all elements of $G \times H$ of the form $(e, h)$. So this includes $(e, e)$.
The Homomorphism Theorem states that $\operatorname{image}(\rho)$ is isomorphic to $(G \times H) / \ker(\rho)$.
In this case $\operatorname{image}(\rho) = G$. So $G$ is isomorphic to $(G \times H) / \ker(\rho)$. But if $\ker(\rho)$ contains the identity of $G \times H$ then it seems that $(G \times H) / \ker(\rho)$ does not contain an identity element. How can $(G \times H) / \ker(\rho)$ be isomorphic to $G$ if it doesn't contain an identity element?
Obviously something in my thought process is wrong so I was hoping someone could help me realise my error :)
$G\times H/ker(\rho)$ does contain an identity element - namely, $[(e, e)]_\rho$, the equivalence class of $(e, e)$ modulo "is sent to the same element by $\rho$."
More generally, if $J$ is any group and $\varphi$ is a homomorphism from $J$, then $J/ ker(\varphi)$'s identity element is $[e_J]_\varphi$ - the equivalence class of the identity element $e_J\in J$, modulo (the equivalence relation induced by) $\varphi$.
Elements in the kernel don't cease to exist, they just get mapped to the identity (which, arguably confusingly, is called "vanishing," but it doesn't mean that they're no longer there).