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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:

- Upper bounds for the number of intermediate subgroups
- number of subgroups index p equals number of subgroups order p
- Exactly one nontrivial proper subgroup
- Subgroups of finite abelian groups.
- Group theory proof of Euler's theorem ($a^{\phi(m)} \equiv 1\mbox{ }(\mbox{mod }m)$ if $\gcd(a,m)=1$)
- The rule of three steps for a cyclically ordered group

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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- $\textbf{Theorem.}$ 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 $\textbf{Theorem 7.9}$ in Prof. Keith Conrad blurb here:

By Cauchy’s theorem, $G$ contains an element of order 2. How does this act in the regular representation?

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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