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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.

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    $\begingroup$ The fact that the orbits all have size $|C/(C \cap H)|$ implies that $|C/(C \cap H)|$ divides $n$, and you can deduce the result from that. $\endgroup$ – Derek Holt Aug 17 '17 at 21:12
  • $\begingroup$ I believe this may be true but I can not see why. Since $H$ may not be a normal subgroup of $G$, we can not say $\text{Orb}_C(xH)$ is a subgroup of $G/H$ and apply Lagrange's theorem. $\endgroup$ – AMD Aug 17 '17 at 22:03
  • $\begingroup$ @AMD You don't need Lagrange's theorem, this isn't a statement about subgroups of $G/H$. It's simpler than that. The action of $C$ on $G/H$ partitions $G/H$ into orbits, and your calculation shows that all of these orbits have the same size, $|C/(C \cap H)|$. Therefore $|G/H| = (\text{number of orbits})(\text{orbit size})$, so the orbit size is a divisor of $|G/H|$. $\endgroup$ – Bungo Aug 17 '17 at 23:55
  • $\begingroup$ @Bungo I see, thank you for clearing that up. $\endgroup$ – AMD Aug 18 '17 at 1:10
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    $\begingroup$ A simpler proof? Let $X =Z(G)H$. Note that $H$ is normal in $X$ and has index dividing $n$. Now let $q: X\to X/H$ be the quotient. If $g \in Z(G)$, then $q(g^n)=q(g)^n=1$ and hence $g^n \in \text{Ker}(q)=H$. $\endgroup$ – Nex Aug 18 '17 at 3:18
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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.

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