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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.

Shaun
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Liêu Long Hồ
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    Welcome to MSE. How do you deduce from the Lagrange theorem that $G$ has a subgroup of order $n$? – José Carlos Santos Jul 12 '20 at 11:17
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    "By Lagrange theorem, i know that there exist a subgroup $H$ of $G$ of order $n$" No, you don't. Lagrange's theorem limits possible subgroup orders, but doesn't in any way guarantee the existence of anything. – Arthur Jul 12 '20 at 11:18
  • Abstract duplicate of https://math.stackexchange.com/questions/225987/show-group-of-order-4n-2-has-a-subgroup-of-index-2/226000#226000 – Jack D'Aurizio Jul 12 '20 at 12:30

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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 $\ker\text{Sign}\neq G$.

Jack D'Aurizio
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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 $n$).

Groups
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    An elementary approach is to consider which elements give even permutations in the regular representation. – Angina Seng Jul 12 '20 at 11:31
  • See Fact 2 in this answer: https://math.stackexchange.com/questions/3752162/why-normal-subgroups-occur-as-kernels-of-homomorphisms-is-a-big-deal/3752250#3752250 – halrankard Jul 12 '20 at 12:07
  • The proof of existence uses Cauchy's Theorem which could be thought of as a rough approximation to an inverse of Lagrange's Theorem. Here we only need Cauchy for $p=2$, and my answer also gives a really short proof of that. – halrankard Jul 12 '20 at 12:09
  • Yes this approach is more elementary to see the existence.@halrankard – Groups Jul 12 '20 at 13:19