Let $\Phi: G \to G'$ be a group homomorphism. Suppose $x \in G$ has order $n$, so $x^n = e_G$. Since $\Phi$ must carry identity to identity, we have $$\Phi(x^n) = \Phi(e_G) = e_{G'},$$ but for any $n \in \mathbb{Z}$, $\Phi(x^n) = (\Phi(x))^n$, so $$(\Phi(x))^n = e_{G'}.$$ The order of $\Phi(x)$ is by definition the smallest positive integer $m$ such that $(\Phi(x))^m = e_{G'}$. Clearly, Since $n$ is positive, the order of $\Phi(x)$ cannot exceed $n$, i.e., $|\Phi(x)| \leq n$. Furthermore, it could be the case that the order of $\Phi(x)$ is a multiple, $k$, of $m$, the order of $\Phi(x)$. I am trying to rule out the possibility, though, that there could exist some other integer, $m$, with the property that $\Phi(x)^m = e_{G'}$ and $m$ does not divide $n$.
Would it be fair to say that this would violate uniqueness of the inverse in $G'$?
Editing with a possible proof of this:
Suppose for a contradiction that this is the case. That is, $\Phi(x)$ has order $m$ where $m \leq n$ (which must be the case from above) but $m$ does not divide $m$. Then $$\Phi(x)^m = e_{G'}$$ by the definition of order. Since $m$ doesn't divide $n$, clearly $m \neq n$, so $m + k = n$ for some positive integer $k$. We then have $$\Phi(x)^n = \Phi(x)^{m+k} = \Phi(x)^m \Phi(x)^k = e_{G'} \Phi(x)^k = \Phi(x)^k = e_{G'}.$$ So I found another $k$ such that $\Phi(x)^k = e_{G'}$. If $k < m$, we have a contradiction to the fact that $m$ is the order of $\Phi(x)$. If not, we still have $k < n$, so we can repeat this process again and again, ad infinitum, so by infinite regress, we have a contradiction. (I suppose the argument is that we are positing the existence of positive integers, for which the principle of infinite descent is valid: we most certainly cannot construct an infinitely decreasing sequence of natural numbers.)
I'm trying to think of a way to formalize the last bit of this argument, but it seems to me that this is in line with what we need.
Divide $n$ by $m$. Get $n=qm+r$, with $r\lt m$. Then $e=g^n=g^{qm+r}=g^{mq}g^r=(g^m)^qg^r=e^qg^r=g^r$. This is a contradiction, since $m$ is the order of $g$ and $r\lt m$.