Group of $2n$ elements, $n$ odd, is not simple

Question

Problem

Let $n \geq 3$ be an odd number and let $G=\{1,...,2n\}$ be a group of order $2n$. Let $\phi:G \to S_{2n}$ be the morphism defined by $\phi(g_i)(g_j)=g_ig_j$ and let $H=\phi^{-1}(A_{2n})$. Show that $H$ is a proper normal subgroup of $G$.

So basically the result of this problem is that every group of order $2n$, with $n$ odd, is not simple.

The morphism $\phi$ can be seen as the action of $G$ on itself by left multiplication. Since $A_{2n} \lhd S_{2n}$, then $H=\phi^{-1}(A_{2n}) \lhd G$. I am stuck at showing that $H \neq \{1\},G$. All I could think of is that $G$ has an element $g$ of order $2$, then $\phi(g)$ is of order $1$ or $2$, but it is easy to see that $\phi(g)=1$ if and only if $g=1$, so $ord(\phi(g))=2$. I don't know which are the elements of order two in $S_{2n}$. I would appreciate some help to solve the problem, thanks in advance.

Answer 1

A hint would be to take that element of order $2$, and decompose $\phi(g)$ as a permutation into disjoint cycles. If you know how many cycles and the length of each cycle, you can use parity to determine if that permutation is even or odd.

Answer 2

I denote the symmetric group on the set $G$ with $S_G$

Consider the following homomorphism $\varphi$:

$G\rightarrow S_G\rightarrow Z_2$, where the first homomorphism is the left regular action (the homomorphism that sends $g$ to the permutation $f$ on $G$ defined by $f(a)=ga$ ) and the second homomorphism is the one that sends each permutation to its parity.

The kernel of this homomorphism consists of the elements of odd order. By cauchy there is an element of order $2$ and an element of order $p$ where $p$ is an odd prime dividing $n$, so the kernel is not $\{e\}$ and is not $G$. Since kernels are normal we are done.

How do we prove the kernel consists of the elements of odd order?

If $g$ has order $m$ then the permutation defined by $f(a)=ga$ consists of $n/m$ cycles of order $n$. when $n$ is odd there are no even cycles, so the permutation is even. when $n$ is even $n/m$ is odd and so there is an odd number of even cycles, hence the permutation is odd.

Answer 3

I don't have enough reputation to comment, but this is for @user16924

$\phi(y)$ can be written as a product of p-cycles. We don't need to know how many of them we have! Let's decompose them in transpositions. A p-cycle is $p-1 \equiv 0 (2)$ transpositions, so $t(p-1) \equiv 0 (2)$. And we are done.