MHB Is There a Simple Group of Order $2n$ for Odd $n \geq 3$?

[Euge]

Here is this week's POTW:

-----
Prove that there is no simple group of order $2n$ where $n$ is an odd number $\ge 3$.

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

This week's problem was solved correctly by Olinguito and castor28. You can read their solutions below.

1. Olinguito's solution.

Spoiler

Let $G$ be a group of order $2n$, $n\ge3$ and odd. Consider $G$ as acting on itself by left multiplication. This gives, for each $g\in G$, a permutation $\varphi_g$ of $G$ where $\varphi_g(x)=gx$ for $x\in G$, and the set of all $\varphi_g$ forms a group isomorphic to a subgroup of the symmetric group $S=S_{2n}$ (and may be considered as the subgroup itself).

Now the mapping $\vartheta:G\to S$ with $\vartheta(g)=\varphi_g$ is a monomorphism (injective homomorphism); therefore $T=\vartheta(G)$ has order $2n$. Any Sylow 2-subgroup of $T$ has order $2$ and so contains an element $\tau$ of order $2$. Furthermore $\varphi_g:G\to G$ has no fixed point unless $g=e$. It follows that $\tau$ is a product of disjoint transpositions (since it has order $2$) and there are $\dfrac{2n}2=n$ of these transpositions (since it has no fixed point); as $n$ is odd, $\tau$ is an odd permutation. Therefore $T\ne A$ where $A$ is the alternating group $A_{2n}$.

As $\dfrac{|S|}{|A|}=2$, $AT=S$. By the second isomorphism theorem,
$$\frac{|T|}{|A\cap T|}\ =\ \frac{|AT|}{|A|}\ =\ \frac{|S|}{|A|}\ =\ 2.$$
It follows that $\vartheta^{-1}(A\cap T)$ is a subgroup of $\vartheta^{-1}(T)=G$ of index $2$ and so is normal. So $G$ has a proper normal subgroup, and as $n>1$ it is not the trivial subgroup. This shows that $G$ is not simple.

2. castor28's solution.

Spoiler

Let $G$ be a group of order $2n$, where $n>1$ is odd. The action of $G$ on itself by left multiplication defines a homomorphism $\varphi:G\to S_G$ such that $\varphi(g).x = gx$ ($S_G$ is the symmetric group on the set $G$).

If $\psi:S_G\to\mathbb{Z}_2$ is the parity map, we have a composite homomorphism:
$$
G\stackrel{\varphi}{\to}S_G\stackrel{\psi}{\to}\mathbb{Z}_2
$$
The kernel $K$ of that homomorphism $\psi\circ\varphi$ is a normal subgroup of $G$. To prove that $G$ is not simple, we must prove that $K$ is neither trivial not equal of $G$, i. e. that $G$ contains elements $a$ and $b$ such that $\varphi(a)$ is odd and $\varphi(b)$ is even.

Let us take $a$ as an element of order $2$, whose existence is guaranteed by Cauchy's theorem. The permutation $\varphi(a)$ only contains transpositions. As this permutation contains no fixed points, there are $n$ transpositions. As $n$ is odd, $\varphi(a)$ is an odd permutation.

We can take $b=c^2$ where $c$ is an element of odd order (since $n$ is divisible by at least one odd prime, the existence of such an element is again guaranteed by Cauchy's theorem). This means that $b\ne1$, and $\varphi(b)=\varphi(c)^2$ is an even permutation.

The conclusion is that $K=\ker(\psi\circ\varphi)$ is a proper non-trivial normal subgroup of $G$, and $G$ is not simple.