Suppose $(G,\cdot)$ is a finite group of uneven order such that $abab=baba$ for any $a,b\in G$. Does this mean that $G$ is commutative?
Does $abab=baba$ imply commutativity in a Group of uneven order?
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group-theory
finite-groups
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5"uneven"="non-even"="even $\cup$ infinite"?... – 2012-06-08
1 Answers
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Yes. Let $|G|=2k-1$ be the order of the group and $a,b\in G$. Then: $ab=ab(ab)^{2k-1}=(ab)^{2k}=(abab)^k=(baba)^k=(ba)^{2k}=ba(ba)^{2k-1}=ba.$
(Added: I should probably mention that here we use the following fact twice: if $G$ is a finite group of order $n$ and $a\in G$, then $a^n=e$, where $e$ is the identity element.)
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1@DougSpoonwood: I am using the fact Babak mentions. The fact itself follows from elementary divisibility properties of natural numbers. It may also be seen as an easy consequence of [Lagrange's theorem](http://en.wikipedia.org/wiki/Lagrange%27s_theorem_%28group_theory%29). You're quite right, I should have mentioned this from the beginning. – 2012-06-09