Order of a homomorphism of groups 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!

Hint: in order for $f$ to be a homomorphism, it has to respect powers of $a$. Look at $f(a^{|f(a)|})$.
$|f(a)| = \operatorname{ord}(f(a))$ if you are unfamiliar with the notation.
Edit: I actually believe N.S.'s hint is better. Hopefully mine will get you thinking, anyway.
