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True of False? This is a question from a past exam that I'm practicing on. Thanks.

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    One can use a theorem of Frobenius to see that the number of solutions to $x^3=1$ in $G$ is 2 more than a multiple of 3. Combine this with the answers, and you see that in fact it must be $\equiv 2\pmod{6}$.2012-12-11

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True: If $g$ has order three, then so does $g^2$ so you can only have an even number of elements of order three.

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Suppose that $x\in G$ has order $3$. Then $x^2=x^{-1}$ has order $3$ as well. Thus, elements of order $3$ come in pairs, and you can’t have an odd number of them.