If $G$ is a group such that $g^3=e$ for every $g\in G$, what is the upper bound for its order? I am aware of the Heisenberg group, and I cannot find a group with greater order that has this property. So I conjecture that $|G|\leq 27$. As for proving it, I am sure it has something to do with the fact that the order is bounded by $3^3$. Any help is appreciated.
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6$C_3 \times C_3\times C_3\times\cdots$ – Matthew Towers Oct 03 '19 at 16:35
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1@MatthewTowers I didn't think of that. – Hossmeister Oct 03 '19 at 16:37
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If you add the constraint that the group has to be generated by 2 elements, your guess is right. – Matthew Towers Oct 03 '19 at 16:40
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@MatthewTowers In that case, any element would be of the form $g^ah^b$ and therefore there would be finitely many elements of that form, correct? – Hossmeister Oct 03 '19 at 16:44
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No, not every element has that form (then there was only be 9 elements, but there are 27). You have to consider things like $ghghghghgghh$. – Matthew Towers Oct 05 '19 at 14:36
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Possible duplicate of Is there a formula for $[F_n : V_{{x^3}}(F_n)]$? – Chain Markov Oct 05 '19 at 16:41
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If $G$ has exponent $3$ and is generated by $m$ elements, then $|G| \le 3^c$ with $c = m + \binom{m}{2} + \binom{m}{3}$, and this bound is best possible.
According to this site this was proved in 1933 independently by Levi and van der Waerden.
Derek Holt
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