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I searched in the existing post and didn't find this problem. I am sorry if someone else have already posted.

Let $G$ be a group of order $2m$ where $m$ is odd. Prove that $G$ contains a normal subgroup of order $m$.

There is a hint:

Denote by $\rho$ the regular represetation of $G$: find an odd permutation in ${\rho}(G)$.

I don't know how to find an odd permutation in the regular representation. I am wondering whether all the elements of $G$ of odd order form this subgroup in this case.

Thanks.

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    @Chirs: thank you much for the hint. This element is an odd permutation in the regular representation.2011-08-07

3 Answers 3

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There is a nice generalization of this fact due to John Thompson, known as the "Thompson transfer Lemma". It goes as follows: let $G$ be a finite group which has a subgroup $M$ such that $[G:M] = 2d$ for some odd integer $d$, and suppose that $G$ has no factor group of order $2$. Then every element of order $2$ in $G$ is conjugate to an element of $M$. I will not give the full proof as it reveals too much of the solution of the original question, but the idea is the same: any element of order $2$ in $G$ which does not lie in any conjugate of $M$ must act as an odd permutation in the permutation action of $G$ of the (say, right) cosets of $M$. As a sample application, consider a finite non-Abelian simple group $G$ whose Sylow $2$-subgroup $S$ has a cyclic subgroup $M$ of index $2$. Then $G$ certainly has no factor group of order $2$, so every element of order $2$ (involution) of $G$ is conjugate to an involution of $M$. But $M$ only has one involution as $M$ is cyclic, so $G$ has one conjugacy class of involutions. In case anyone is wondering, the Thompson Transfer Lemma is a true generalzation of the question, because the case $M = 1$ can be applied to the question to conclude that there must be a normal subgroup of index $2$ for $G$, because no element of order $2$ lies in any conjugate of the trivial group.

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    Thank you very much for the generalization.2011-08-07
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By Cauchy's theorem, $G$ contains an element of order 2. How does this act in the regular representation?

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    I'm proving something similar. I'm proving that if $H\le G$ such that $|G:H|=2$ then H contains all of the elements of G of odd order. It's easy to show that H is normal in G. The problem I'm having is showing that H contains the elements of G of odd order. Can I, without loss of generality, assume that the order of $G=n$ where $n$ is odd to show that $xH$ divides $n$. Therefore $xH$ is odd? I'm just having problems hashing out this proof.2016-03-21
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  • $\textbf{Theorem.}$ Let $|G| = 2^{n} \cdot m$ where $2 \nmid m$. If $G$ has a cyclic $2$- Sylow subgroup, then $G$ has a normal subgroup of order $m$.

Your question is just a corollary to this theorem. Please see $\textbf{Theorem 6.9}$ in Prof. Keith Conrad blurb here:

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    The two are of the same order, thus equal. Now, for any $x \in G$, $xNx^{-1} \subseteq xHx^{-1} =H$, so $xNx^{-1}$ is another subgroup of $H$ of order $m$. So, $xNx^{-1}=N$. $N$ is normal in $G$ of order $m$.2011-08-07