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.
A group of order $66$ has an element of order $33$
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group-theory
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0This generally isn't true unless the two elements commute. – 2012-02-28
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0I figured this would be the case. Is there a way to repair my "proof"? – 2012-02-28
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5Show that they commute! – 2012-02-28
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1It's not clear that they do necessarily commute. – 2012-02-28
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4I did not say it was *clear*, I said *prove it* :) – 2012-02-28