Given a one to one group homomorphism $f: G\to H$ and an element $b\in G$ has order $m$, how can I prove $f(b)\in H$ has order $m$ as well?
My idea was that a cyclic group is only homomorphic to another cyclic group.
I feel like this could be used somehow but still not sure where to start.
Suppose $f:G\to H$ is a group homomorphism and it is one-to-one. Let $b\in G$ has order $m$. $f^m(b)=f(b^m)=f(e_G)=e_H$.
Now we know that $f(b)$ has finite order. But what if it has order $k$ such that $k<m$? We claim that this cannot be the case. Suppose this is true, then $f^k(b)=e_H$, which implies that $f(b^k)=e_H$. Since $f$ is an injective homomorphism, $\ker{f}={e_G}$. So we claim that we must have $b^k=e_G$. Notice that $k<m$, so actually $b$ has order less than $m$, which is impossible.