15
$\begingroup$

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?

  • 5
    "uneven"="non-even"="even $\cup$ infinite"?...2012-06-08

1 Answers 1

35

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.)

  • 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