Let $|G|=2n$, show that there exists a $g\neq e$ such that $g^2=e$.

Question

Let $G$ be a group with $|G|=2n$ $(n\in \mathbb{N})$.

I have to show that for $g\in G$ with $g\neq e$ and $g^2=e$.

A hint says that I have to use the orbit-stabilizer theorem and interpret $g\mapsto g^{-1}$ as a group operation where $\{-1,1\}$ operates on $G$.

How can I construct that group operation?

I suppose $\gamma : \{-1,1\}\times G\to G, \, (a,g)\mapsto g^{a}$, where $a=1$ for $g\in G$ and $a=-1$ for $g^{-1}\in G$.

The Orbit-stabilizer theorem says $$2=|\{-1,1\}|=|\operatorname{Orb}_{\{-1,1\}}(x))|\cdot |\operatorname{Stab}_{\{-1,1\}}(x))|$$ This is where I am stuck, how does this prove the statement?

Answer 1

If $g\neq g^{-1}, \forall g(\neq e)\in G$,then $|G|=2k+1$,so there be exist $g(\neq e) $ such that $g^2=e$.

Answer 2

For $p=2$, $p$ divides $|G|$, so that $G$ has an element $g$ of order $2$ by Cauchy.

Answer 3

What is says: the orbits are either of size $2$ or $1$. They are of size $1$ if and only if $g=g^{-1}$, that is $g^2=e$. Since $|G|$ is even, the number of orbits of size $1$ is even. There exists at least the orbit of size $1$, consisting of $e$. So there must exist another one. All in all, the number of elements of order $2$ must be odd.

Answer 4

Take all the elements of your group and match them up with their inverses:

$\begin{array}{|c|c|c|c|c} e&g_1&g_2&\cdots&\\ \hline ?&g_1^{-1}&g_2^{-1}&\cdots&\\ \end{array}$

Do you see how this forces at least one element to be its own inverse?