$ord(f(a)) \leqslant ord(a) $ for every a in G

Question

Let $f: G \to H$ be a group homomorphism.

I want to show that:

(i) $ord(f(a)) \leqslant ord(a) $ for every a in G

(ii) If $ord(a)$ is finite, then $ord(f(a))$ divides $ord(a)$

(ii) If $f$ is an isomorphism, then $ord(f(a)) = ord(a)$

I tried showing these things for several hours now but I can't come up with an elegant solution. Help is appreciated!

Answer 1

Homomorphism means that given $a\in G$, you can write $f(a^n) = f(a)f(a)...f(a)$ multiplicatively. So if the order of $a$ is n, above statement is just that $f(a)f(a)...f(a) = f(a^n) = f(1) = 1$. It should be clear now.