I'm working through an Abstract Algebra book to teach myself, and came across the problem:

Prove: If $G$ is a finite group of even order, then there exists a $g\in G$ such that $g^2 = e$ and $g \ne e$.

(In this book, $e$ is used as the identity element. I don't know if that's standard or not...)

I have a proof outline, but don't really know how to write it in a formal way. My idea is as follows:

First, note that this problem is equivalent to saying that "there exists a $g\in G$ such that $g = g^{-1}$. Also note that the identity element is its own inverse.

Since the identity element is its own inverse, we have an odd number of elements remaining in the group that need their inverse "assigned." Assign all but one of the remaining elements an inverse so that none satisfy $g = g^{-1}$.

You have one remaining element left; the rest of the elements already have inverses. As inverses are unique, this element must be its own inverse.

My question is twofold:

  1. Am I even close to being on the right track as far as a proof outline goes?
  2. If so, what can I do to make this proof rigorous, and not just an outline?
  • 1
    $\begingroup$ You are close. Note that $\iota \colon g \mapsto g^{-1}$ is an involution, i.e. $\iota \circ \iota = \operatorname{id}$. The number of fixed points of an involution of a finite set must be... $\endgroup$ Jun 27, 2013 at 23:37
  • $\begingroup$ See also this post $\endgroup$
    – amWhy
    Jun 27, 2013 at 23:38
  • $\begingroup$ @amWhy I do believe this is the first time that I've voted to close one of my own questions... :) I promise that I did search, though... $\endgroup$
    – apnorton
    Jun 27, 2013 at 23:40
  • $\begingroup$ Oh, I know, @anorton. Searching MSE is not easy. I usually find possible duplicates appearing on the right column as "related", but only after a question's already been posted! $\endgroup$
    – amWhy
    Jun 27, 2013 at 23:41

2 Answers 2


You've got it. There is an issue that more than one element may be its own inverse. But your idea of pairing those $g$ satisying $g^2\ne e$ with their inverses is on the money.


You've done the hard part (of recognizing what's happening). The linked questions will help you nail it, though. The key idea is noting the existence of a partition of finite $G$ into two groups: those elements which are their own inverses, and those which are not.

Indeed, what we can conclude is more general: that a finite group $G$ of even order must have an odd number of elements $g\in G$ such that $g^{2} = e, \;g\neq e$. (Put differently, a finite group of even order must have an even number of elements such that $g^2 = e$, and one of those elements of course includes $e$).

  • $\begingroup$ Well, yes, that too! ;-) $\endgroup$
    – amWhy
    Jun 27, 2013 at 23:56
  • $\begingroup$ @user14111: If the group is infinite, the number of such elements might also be infinite, in which case "even" and "odd" have no meaning. You might notice that the infinite group $\mathbb{Z}$ of integers under addition does not have a nonzero element which is its own inverse. $\endgroup$ Jun 28, 2013 at 2:00

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