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

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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Hint: by Cauchy's theorem, $G$ contains an element of order 2. How does this act in the regular rep? –  Chris Eagle Aug 6 '11 at 12:47
@Chris: your comment above would make a nice answer... –  Pete L. Clark Aug 6 '11 at 12:51
@Chirs: thank you much for the hint. This element is an odd permutation in the regular representation. –  ShinyaSakai Aug 7 '11 at 3:54

• $\textbf{Theorem.}$ 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:

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Thank you very much. I will rewrite and post the proof here. I hope you will not mind... –  ShinyaSakai Aug 7 '11 at 3:48
When $n=0$, the result is clear. Now consider the case when $n \geq 1$. First, there is a normal subgroup of $G$ of order $2^{n-1}m$. In order to proof this, consider the permutation of $G$ on itself by left multiplication, $\rho: G \rightarrow$ Sym$G$. As $G$ has a cyclic $2$-Sylow subgroup, $G$ has an element $g$ of order $2^n$, and $\rho(g)$ is the composition of $m$ $2^n$-cycles. Thus, $\rho(g)$ is an odd permutation. So, the composition map sgn$\circ \rho: G \rightarrow \{ \pm 1 \}$ is onto. The kernel of this map is a normal subgroup $H$ of $G$, of order $2^{n-1}m$. –  ShinyaSakai Aug 7 '11 at 3:49
Now proceed by induction on $n$. For $n=1$, we are done. For $n >1$. $G$ has a normal subgroup $H$ of order $2^{n-1}m$. The intersection of $H$ and the cyclic $2$-Sylow subgroup of $G$ is a cyclic $2$-Sylow subgroup of $H$. Thus, by the induction hypothesis, $H$ has a normal subgroup $N$ of order $m$. Then, $N$ is the only subgroup of $H$ of order $m$. If not, suppose that $N'$ is another such group, as $N$ is normal in $H$, the multiplication $NN'$ containing $N$ is also a subgroup of $N$, of odd order, but this order is divisible by $m$, the order of $N$. So, $NN'=N$, and $N' \subseteq N$. –  ShinyaSakai Aug 7 '11 at 3:50
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 $m$. –  ShinyaSakai Aug 7 '11 at 3:50
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.