6

Are there any groups (not necessarily finite) that have a non-zero even number of elements of order two?

My attempt: since any finite group containing an element of order 2 must be of even order, hence the number of elements of order 2 must be odd, it suffices to find groups of infinite order.

I browsed through those familiar infinite groups but non of them seem to fit in the construction. How to construct such a group?

ThetaOmega
  • 419
  • 2
  • 6
  • You could take the free product of two copies of $\Bbb{Z}_2$. – Theo Bendit Jun 28 '21 at 02:33
  • 3
    @TheoBendit This fails, since such a group has infinitely many elements of order two (for instance, if $\langle a, b \mid a^2 = b^2 = 1\rangle$ is your presentation, then $aba$ has order two as $abaaba = abba = aa = 1,$ and similarly any odd length word will have order two). –  Jun 28 '21 at 02:35
  • 2
    @MichaelBarz Ah, I thought it was too good to be true. I'll leave my comment up so that others don't make the same error. – Theo Bendit Jun 28 '21 at 02:36
  • The same thing happens on the Möbius group. – ThetaOmega Jun 28 '21 at 02:37
  • Now I'm intrigued to know if there are even any groups of nonzero even number of elements of even order... From what you said, such a group has to be infinite too. – oleout Jun 28 '21 at 02:55
  • 1
    This issue has already been discussed here:https://math.stackexchange.com/questions/740397/the-number-of-elements-of-order-2-in-an-infinite-group – kabenyuk Jun 28 '21 at 12:50

0 Answers0