Let $f: G \to H$ be a homomorphism of groups. Assume that $a\in G$ and $\operatorname{ord}(a)=n$. Prove that the order of $f(a)$ is a divisor of $n$.
I know that if $H$ is a subgroup of $G$, then $$|G|=|H|\cdot|\text{distinct cosets of $H$}|$$ so $|H|$ must be divisor of $|G|$, but I don't understand how to apply that logic to homomorphisms (or if that will even work).
Any help is appreciated!
Best Answer
Hint $$ \left( f(a) \right)^n=f(a^n)=f(e)=e \,.$$
You don 't need to calculate the order of $f$, you need to calculate the order of $f(a)$ which is an element in $H$.