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 $f(a)$?
Best Answer
Using the fact that $e_G\ast e_G=e_G$ yields $$ f(e_G)=f(e_G\ast e_G)=f(e_G)\circ f(e_G)$$
Now multiply both sides by the inverse of $f(e_G)$ to obtain $f(e_G)=e_H$.