For a finite group of order $2n$ does there exist $x$ such that $x\ast x=e$?

Question

Let $ (G,\ast)$ be a group with identity $e$ and cardinality $2n$ for some $n\in\omega$. Then, does there exist $x\in G$ such that $x\ast x=e$ and $x\neq e$?

Answer 1

To expand on Daniel Fischer's hint:

Let $f : G \to G$ be defined by $f(x) = x^{-1}$. This is a bijection. An element $x \in G$ verifies $x * x = e$ iff $f(x) = x$, so you want to find the fixed points of $f$, or at least find one that is not $e$. Note that $f(e) = e$ already, so that's one fixed point.

Now $f$ is an involution ($f(f(x)) = x$), and $|G|$ is even, so its number of fixed points has to be even too: let $F \subset G$ be the fixed points, then every element $x \in G \setminus F$ can be paired with $f(x)$, thus $|G \setminus F|$ is even, so $|F|$ is even too. And since we already have one fixed point ($e$) then there is at least one other (because $1$ is odd), say $x \neq e$. Then $x$ has order $2$ and we're done.

Answer 2

(This is a variation on Daniel Fischer's hint in the comments.)

Let's partition the elements of $G$ into subsets: For each $x\in G$, there will be a subset $S_x$ consisting of exactly $\{x, x^{-1}\}$. Note that when $x =x^{-1}$, this subset will have one element, not 2. For example, $S_e$, the subset containing $e$ is $\{e\}$.

Now prove the following:

1. This division into subsets is a partition: each element of $G$ is in exactly one of the subsets.

2. If you add up the sizes of the subsets, you get $2n$.

3. Some of the subsets have 2 elements, some have 1. But there is at least one subset with only 1 element.

By (3), there is at least one subset with only one element. Is it possible that there are no others?

Answer 3

Depending on what you know, the answer can be almost painfully trivial: yes, as any group of even order has an element of order two by Cauchy's Theorem.

If you haven't yet covered the above theorem things are going to be lengthier: pair up as many as possible of the $\;2n\;$ elements of the group by the rule $\;(x\,,\,x^{-1})\;$ . Taking into account that the unit element gets paired with itself, and since the number of elements of the group is even, there must be another, non-unit!, element in the group which is paired with itself, and we're done.