Order of a Group and its Element(s)

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,

Re: Order of a Group and its Element(s)

You're on the right track. You've shown that it can't be cyclic, because cyclic groups are abelian.

The way I see is to proceed to show it by contradiction. Assume G has no element of order 3.

Since G isn't cyclic, G has no element of order 6.

By assumption, G has no element of order 3.

With that information, and |G| = 6, you can determine the orders of G's elements (hint: Lagrange.)

Once you've done that, consider the two elements in G, a and b, that don't commute.

That means that ba is not equal to ab.

The contradiction will appear by looking at g = ab and asking what happens when you square g.

Remember that you'll know something about the order of g at this point - and also the orders of a and b.

So consider what you can say about g squared = abab for those non-commuting a and b.

With a little bit of manipulation when considering g squared, the contradiction will appear.

(The contraction will arise by showing that a and b do commute.)

Thus the assumption was false, and so G has an element of order 3.

Re: Order of a Group and its Element(s)

Ah, excellent. When you describe it like that, I see how the proof can be done. Thanks for the help!