I'm trying to prove that if $G$ is a group with $|G|=2m$ and $m$ odd, then $G$ has exactly one element of order two. What I have is that there actually exist at least one element of order two, and that by Sylow theorems every two involutions are conjugate. I also proved that if a group has odd order then there are no elements of order two. I'm trying to glue all this info together with no success.

  • 6
    $\begingroup$ This is not true: Take $G = S_3$, then $|G| = 6$, but $G$ has both $(12)$ and $(13)$ which are of order 2. $\endgroup$ – Prahlad Vaidyanathan Sep 24 '15 at 1:45
  • $\begingroup$ This is true if $G$ is abelian, though. $\endgroup$ – pjs36 Sep 24 '15 at 1:51
  • $\begingroup$ @PrahladVaidyanathan I see, thanks. There is then a mistake in my teacher's document. $\endgroup$ – Jose Paternina Sep 24 '15 at 1:56
  • $\begingroup$ As @pjs36 points out, it is likely that $G$ is meant to be abelian. $\endgroup$ – Prahlad Vaidyanathan Sep 24 '15 at 1:58

This is false, a dihedral group of order $2m$ with $m$ odd provides a counterexample. If $\rho$ is the rotation and $\tau$ is the reflection then $\rho^n\tau$ has order $2$ regardless of the choice of $n$.


In the abelian case, the proof is a one-liner:

Two different elements of order $2$ generate a subgroup isomorphic to $C_2 \times C_2$, hence $4$ divides the order of the group.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.