I'm trying to trying to prove this following result:
Let $\Phi: G \to G'$ be a group homomorphism. Prove that for any $n \in \mathbb{Z}$ and $x \in G$, $\Phi(x^n) = (\Phi(x))^n$.
I'm not sure if there is a better, more efficient way to prove this, but the natural strategy to me seems to be induction for the non-negative side and closure under inverses for the negative side.
Solution. For integers $n \geq 0$, we proceed by induction.
Base Case ($n = 0$): If $n = 0$, $x^n = e_G$, and $\Phi(e_G) = e_{G'}$. Similarly, $(\Phi(x))^0 = e_{G'}$.
Induction Hypothesis: Assume for $n = k$: $$\Phi(x^k) = (\Phi(x))^k.$$
Induction Step: Prove for $n = k + 1$: \begin{align*} \Phi(x^{k+1}) & = \Phi((x^k) x) \\ & = \Phi(x^k) \Phi(x) \\ & = (\Phi(x))^k \Phi(x) \\ & = (\Phi(x))^{k+1} \end{align*} So for all $n \in \mathbb{Z}$, $\Phi(x^n) = ((\Phi(x))^n$. By taking inverses (since they are unique), this implies that $\Phi(x^n)^{-1} = ((\Phi(x))^{-n}$, which gives $\Phi(x^{-n}) = ((\Phi(x))^{-n}$.
Just to get the question off the unanswered list: yes, your proof is correct. It is well-written and easy to follow. Good job!