TITLE: Let $H \leq G$ be a subgroup with finite index $n$. Show for every $g \in Z(G)$, $g^n \in H$.
QUESTION [7 upvotes]: Let $H \leq G$ be a subgroup with finite index $n$. Show for every $g \in Z(G)$, $g^n \in H$.

I am given a hint: 

Consider $C = \langle g \rangle$ and show the left multiplication action of $C$ on $G/H$ has orbits of size $|C/(C\cap H)|$.

Proof of hint: Let $g \in Z(G)$ and $C = \langle g \rangle$ act on $G/H$ by left multiplication.  For any $xH \in G/H$, $(g^kx)H = x(g^kH)$ which shows $g^k \in \text{Stab}_C(xH)$ if and only if $g^k \in H$.  By the orbit stabilizer theorem, $|\text{Orb}_C(xH)| = [C:\text{Stab}_C(xH)] = |C/(C\cap H)|$.$\square$
I am having trouble proving the result from here.  I have not yet used the assumption that $H$ has finite index in $G$ which seems essential for finishing this proof.  Any help would be appreciated.

REPLY [2 votes]: Suppose $G \leq H$, $[G:H] = n$, and $g \in Z(G)$. Now, suppose $K = \langle H \cup \{g\} \rangle$. Because $g$ commutes with everything, we see, that $H \triangleleft K$. Also, $[K: H]|n$ by Lagrange theorem, and that leads to $exp(\frac{K}{H})|n$. Now, suppose $\phi$ is the natural homomorphism from $K$ to $\frac{K}{H}$. We see that $\phi(g^n) = \phi(g)^n = e$, which results in $g^n \in Ker(\phi) = H$, Q.E.D.