Prove that $\forall k$ and for any $g \in G$, the sets $A = \{a \in G : a^k = g\}$, $B = \{b \in H : b ^k = f(g)\}$ have the same number of elements.

Question

Let $f : G \to H$ be a group isomorphism. Show that for any integer $k$ and for any $g \in G$, the sets $A = \{a \in G : a^k = g\}$ and $B = \{b \in H : b^k = f(g)\}$ have the same number of elements.

I've been trying this problem for quite a while but to no avail. I don't understand what the saying.

Answer 1

Note that, for any $a \in A$, $$f(a)^k = f(a^k) = f(g)$$ so $f(a) \in B$, and then $f(A) \subseteq B$. Now, apply the same argument with $f^{-1}$ to show that $f^{-1}(B) \subseteq A$. Thus $f(A) = B$.