Let $\displaystyle G$ be a group of order 6. Show that if $\displaystyle G$ is not abelian then $\displaystyle G$ has an element of order 3.

I believe that if $\displaystyle g, h \in G $ then $\displaystyle gh \neq hg $ in general, and this means that $\displaystyle G$ cannot be a cyclic. Also $\displaystyle g^6 = 1 $ as a corollary of Lagrange's Theorem. I'm not sure if any of this is helpful to the problem at hand though... I'm quite stuck as you can see.

Thanks,