This is my first post here, so forgive for any formatting errors or lack of information; please let me know if I can rectify any issue with my post.
I've been stuck on this problem for quite a bit, and I was wondering if my solution had any merit. Here's what I have:
Proof: Let $X = \{x \in G:x^k = b\}$ for some fixed integer $k$ and fixed element $b \in G$ and $Y = \{y \in \bar G|y^k = \phi(b) \in \bar G\}$.
Then $\forall x \in X$, we have that $(\phi(x))^k = \phi(x^k) = \phi(b) \implies \phi(x) \in Y \implies \phi(X) \subseteq Y$.
Similarly, $\forall y \in Y$, we have that $y = \phi(\phi^{-1}(y)) \implies \phi(b) = \phi(\phi^{-1}(y)^k) \implies b = (\phi^{-1}(y))^k \implies y \in \phi(X) \implies Y \subseteq \phi(X)$.
Then $\phi(X) = Y$. Because $\phi$ is an isomorphism, then $|X| = |\phi(X)|$.
Please let me know if I have made any errors as abstract algebra is definitely not my strong suit. Thank you!
Seems like you essentially got it.
If $x^k=b$ then $\phi(x^k)=\phi(b)\Rightarrow (\phi(x))^k=\phi(b)$ so $\phi(x)\in Y$ for all $x\in X$. Thus $|Y|\geq |X|$ (since $\phi$ is injective).
Also, $y^k=\phi(b)\Rightarrow (\phi(a))^k=\phi(b)$ for some $a\in G$ (since $\phi$ is surjective)
$\Rightarrow \phi(a^k)=\phi(b)$
$\Rightarrow \phi^{-1}(\phi(a^k))=\phi^{-1}(\phi(b))$
$\Rightarrow a^k=b$
$\Rightarrow a\in X$ so $|Y|\leq |X|$ (since $\phi$ is injective).