# Let $H \leq G$ be a subgroup with finite index $n$. Show for every $g \in Z(G)$, $g^n \in H$.

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.

• 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. – Derek Holt Aug 17 '17 at 21:12
• 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. – AMD Aug 17 '17 at 22:03
• @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 $|G/H|$. – Bungo Aug 17 '17 at 23:55
• @Bungo I see, thank you for clearing that up. – AMD Aug 18 '17 at 1:10
• 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 $g^n \in \text{Ker}(q)=H$. – Nex Aug 18 '17 at 3:18

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.