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Given a finite group $G$ of order an even number, prove that $G$ contains a non-identity element of order $2$.

Proof:

For every non-idenity element $g \in G:g^2 \ne e \iff g \ne g^{-1}$

If such $g$'s exist then the number of them is an even number, since every one of them can be paired with its inverse, follows the number of elements in $G$ that are their own inverse in even, clearly $e=e^{-1}$, so the number of non-identity elements in $G$ that are their own inverse is odd, implies at least $1$ such element does exist.

Another way I tried is as follows:

$G$ has an even order iff $\text{ord}(G)=2n$ for some $n \in \mathbb N^+$ iff $ g^{2n} =e$ for $\langle g \rangle = G$ and $e$ the identity element of the given group.Every element is in the group generated by itself and so $g \in \langle g \rangle$,$G$ is a group and so $g^n \in \langle g \rangle$.

On the other hand $(g^n)^2=g^{2n}=e$.

Then I figured out that this just implies that $\text{ord}(g^n)$ divides $2$ and is not necessarily equal to $2$.

Is my conclusion true? besides is it possible to finish the proof using the second way?

1 Answers1

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The second proof is flawed. For one thing, you seem to assume $G$ is cyclic.

Here is a correct proof: Partition the elements of $G$ into equivalence classes consisting of $\{g,g^{-1}\}$. Since the order of $G$ is even, and the identity forms its own equivalence class, there must be an even number $\ge2$ of classes of size $1$.

Another way is by Cauchy's theorem. A proof of Cauchy's theorem that I like, resembles your second attempt a little. It uses Sylow's theorems. Take a Sylow $2$ subgroup. Then take an element and consider the cyclic group it generates. The order must be $2^m$, for some $m$. But any such cyclic group has a subgroup of order $2$.

  • Isn't ($2|2n$ then $\exists x\in G$ s.t. $x^2=e$) suffices? – C.F.G Oct 26 '20 at 17:03
  • No. Not if $G$ isn't cyclic. That's why Sylow is needed. @C.F.G. –  Oct 26 '20 at 17:10
  • Cauchy's theorem states that if $G$ is a finite group and $p$ is a prime number dividing the order of $G$, then $G$ contains an element of order $p$. Now let $p=2$. – C.F.G Oct 26 '20 at 17:53
  • There's an elegant proof of Cauchy's Theorem sans Sylow Theorems in Roman's "Fundamentals of Group Theory: An Advanced Approach". – Shaun Oct 26 '20 at 17:58
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    Thanks for the reference. @Shaun –  Oct 26 '20 at 18:02
  • You're welcome, @ChrisCuster. – Shaun Oct 26 '20 at 18:05
  • There you go. The smallest noncyclic group is the Klein four group, $\Bbb Z_2\times\Bbb Z_2$. –  Oct 26 '20 at 18:22