Have you done the Sylow theorems yet? Seems like the Corollary to # 1 there might be useful.
Prove that a non-abelian group of order 6 must have at least one element of period 3.
This is what I have so far.
Let in be a non-identity element. can have period 2, 3, or 6.
Period of cannot be 6 because that make cyclic, thus abelian.
If has period 3, we are done.
If has period 2, ... (stuck here)
Maybe looking at an element in and is not in , but not sure what to do with it.
Thanks in advance for any help.
Wait, here you go:
I think your idea is the right idea. Note that Therefore, there exists such that By hypothesis and assumption, and (or your group is cyclic, hence abelian). But, divides because it's a subgroup. Therefore,
Now you'll need some clever argument as to why Thinking...
Well, if there are no elements of order 3, then the five non-identity elements must all have order 2 (because as you noted, an element of order 6 would make the group cyclic, hence abelian).
Can you prove that any group whose non-identity elements all have order 2 must be abelian? This would of course give an immediate contradiction to not being abelian.
EDIT: Sorry, using the word "order" where you use "period".
But the fact that a group of exponent 2 is abelian is nearly trivial. Let be such a group, so that , for all .
Suppose are arbitrary. Consider the relation
But if an element has order 1 or 2, it is clearly its own inverse. So the last relation simplifies to
and, since these elements were arbitrary, this implies that is abelian.
So I understand what topspin1617 is saying, but does it apply? I am working under the assumption that has period 2 and that there exists a and (so we don't know the period of ). I don't think that the proof topspin1617 gave implies is commutative, thus not proving that is abelian and creating a contradiction.