Finite groups and order of their elements

[Chen]

Hi,

This time around I need to prove that a finite group of order 10 must contain an element of order 2 and an element of order 5.

If the group is cyclic then this is trivial. So assuming the group is not cyclic, it's easy to show that there exists an element of order 2 in the group. And it is just as trivial to show that any element, except for e, must be of either order 2 or order 5. But I'm having a hard time proving that there must be an element of order 5 in the group. I think I just need to show that it's impossible to have every element in the group be of order 2, only problem is I don't know how... I do know that if a group consists only of elements of order 2, then it must be abelian. So perhaps it would be just as good to show that a non-cyclic group of order 10 cannot be abelian?

I'd much appreciate hints and nudges in the right direction. :)

Chen

 

[Chen]

Ok, I think I have an idea. Let's assume that there are two elements of order 2 in G, let them be a and b. If I can prove that the subset {e, a, b, ab, ba} forms a subgroup within G, I'm done. So is it? And is this a good idea?

Another thought: if G is abelian then it's easy to prove that that subset is indeed a subgroup. So now I solved the question for the case that G is abelian. Now how I do proceed from here, proving that it's also true if G is not abelian?

Got it.

 

[Galileo]

Sometimes it's easier to solve a slightly more general problem.

[Nonsense]
There is always exactly one element of order 1 (the identity) in any group, so if all other elements are of order 2, then the order of the group must be odd.
[/Nonsense]

 

[Chen]

Huh? How can a group contain an element of order 2 and have an odd order itself? Isn't that in contradiction with Lagrange's theorem?

 

[matt grime]

I thought I posted an answer?

Anyway, if every element is of order 2 then G is abelian, but we know the abelian groups of order 10. There is exactly one of them. So, not abelian implies there is an element of order 5 (since all elements of order 2 implies abelian), and abelian implies the group is C_2xC_5, and again there is an element of order 5.

 

[Hurkyl]

There is always exactly one element of order 1 (the identity) in any group, so if all other elements are of order 2, then the order of the group must be odd.

That doesn't follow. Take Z₂ as a trivial counterexample.

 

[Galileo]

Ack. I wrote A, but I meant B, while the answer is C.
I wanted to show that a finite group of even group always has an element of order 2, but it can't be used directly to show there has to be an element of order 5 in your case.

Sowry'bout that.

 

[Chen]

Thanks, matt. We've not studied this material yet, so I can't use that proof (and even if I could, I don't understand one bit of it ). But thanks.

 

[mathwonk]

well what do youi know? certainly there is either an element of order 2 or an element of roder 5, by cauchy.

So if all elements have order 2, then G is abelian. Then take one of those elements, say x, and mod out by it. I.e. construct a new group of order 5, whose elements are all 5 pairs of form {y,yx} for all y in G. The product of the pair {y,yx} with {z,zx} is of course {yz,yzx}.

the fact this group has order 5 implies every y in G, except e and x, has order greater than 2.

so you cannot have a groupo of order 10 with all elements of order 2.


now suppose all elements have order 5, except e. then do something similar.




is it ok that