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

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

I believe that if then in general, and this means that cannot be a cyclic. Also 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,

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.

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