4
$\begingroup$

If I wanted to show that a group of order $66$ has an element of order $33$, could I just say that it has an element of order $3$ (by Cauchy's theorem, since $3$ is a prime and $3 \mid 66$), and similarly that there must be an element of order $11$, and then multiply these to get an element of order $33$? I'm pretty sure this is wrong, but if someone could help me out I would appreciate it. Thanks.

  • 0
    This generally isn't true unless the two elements commute.2012-02-28
  • 0
    I figured this would be the case. Is there a way to repair my "proof"?2012-02-28
  • 5
    Show that they commute!2012-02-28
  • 1
    It's not clear that they do necessarily commute.2012-02-28
  • 4
    I did not say it was *clear*, I said *prove it* :)2012-02-28

3 Answers 3