The problem: If $f:G \to H$ is an injective homomorphism of groups and $a \in G$, prove that $\vert f(a) \vert = \vert a \vert$.
The solution in the back of the text first shows that if the order of $a$ is $n$, then the order of $f(a)$ must be $n$. Then it shows that if the order of $f(a)$ is $n$ then the order of $a$ must be $n$, and in this second part it uses the fact that $f$ is an injection.
In any case, in attempting to do the problem on my own, I got hung up on the fact that the order of $a$ or $f(a)$ could be infinite, but I believe the text's solution addresses this. By proving that $a$ has finite order $n$ iff $f(a)$ has finite order $n$, the text has also proved that that if one has infinite order, then so does the other.