TITLE: Proving that a group homomorphism preserves the identity element
QUESTION [6 upvotes]: Assume that $(G,*)$ and $(H,o)$ are groups and that $f:(G,*) \rightarrow (H,o)$ is a homomorphism.

Let $e_G$ and $e_H$ denote the identity elements of $G$ and $H$, respectively. Show that $f(e_G)=e_H$.

Approach: $f(e_G)=f(a*a^{-1})$ for $a,a^{-1} \in G$,
so $f(a*a^{-1})=f(a)of(a^{-1})$.
If that’s true, then how do we know that  $f(a^{-1})$ is the inverse of $f(a)$?

REPLY [7 votes]: The answers provided here are excellent, but here is something to just widen your mind for diversity.
So if $a\in G$ $$f(a) = f(a* e_G) = f(a)\circ f(e_G).$$
Left cancel by $f(a)^{-1}$ to get $e_H$