Let $G$ be a group of order $2n$, $n$ odd. Prove that there is a unique subgroup $H$ of $G$ of order $n$.

Question

Question: Let $G$ be a group of order $2n$, $n$ odd. Prove that there is a unique subgroup $H$ of $G$ of order $n$.

By Lagrange theorem, I know that there exist a subgroup $H$ of $G$ of order $n$ but I don't know how to prove that $H$ is unique.

Please help me, thank you for considering my request.

Answer 1

Let $H$ and $K$ be subgroups of $G$ of order $n$. Then since both $H$ and $K$ have index $2$, they are both normal subgroups of $G$. In particular, $HK$ is a subgroup of $G$.

Consider $|HK|=\frac{|H|\cdot|K|}{|H\cap K|}$. It is straightforward to see that $|HK|$ is odd. Moreover, $H\le HK\le G$. The only possible case is $HK=H$. That is $H=K$.

By the way, the subgroup of order $n$ does exist, but not by Lagrange. There should be a more elementary way to see that. But the first thing in my mind is the fact that $G$ is soluble and so has a Hall subgroup (of order $n$).

Answer 2

There is a canonical embedding of $G$ into a subgroup of $S_{2n}$ with $2n$ elements, given by associating $g$ to $\Phi_g:x\to gx$. With the only exception of $\Phi_e$, $\Phi_g$ has no fixed points. If we consider the kernel of the $\text{Sign}$ homomorphism we get a subgroup of $G$ with exactly $|G|/2$ elements. Indeed by Cauchy's theorem there is some element $h\in G$ with order $2$, and $\Phi_h$ is the product of an odd number of disjoint transpositions, so $\Phi_h$ is an odd permutation and $\ker\text{Sign}\neq G$.