# Even order of G

1. Sep 16, 2008

### fk378

1. The problem statement, all variables and given/known data
If G is a group of even order, prove it has an element a=/ e satisfying a^2=e.

3. The attempt at a solution
I showed that a=a^-1, ie a is its own inverse.
So, can't every element in G be its own inverse? Why does G have to be even ordered?

2. Sep 16, 2008

### Dick

The problem says "IF G is group of even order", that's why G has even order. The question of whether there is a nontrivial group of odd order such that a^2=e for all a is a completely different question. Do you want to try and solve it? You don't have to since you already solved the original question. Didn't you? Though you didn't really say how.

Last edited: Sep 16, 2008
3. Sep 16, 2008

### Dick

If you want to know, then if a^2=e for a not equal to e, the G has a subgroup of order 2. Doesn't it? What does that mean?

4. Sep 16, 2008

### fk378

Subgroup of order 2, then the subgroup has 2 elements. a and a-inverse. But can't there exist a b with a b-inverse in the subgroup as well? Why are there only 2 elements?

5. Sep 17, 2008

### Dick

{a,a^(-1)} isn't a subgroup. And it doesn't even have two elements. It's certainly not what I was considering. I don't think you are really thinking about this.

6. Sep 17, 2008

### fk378

I'm sorry, I'm new to proofs. Can you explain your last post? Why are there not 2 elements in {a, a^-1} and why is this not a subgroup?

7. Sep 17, 2008

### fk378

I got some hints from the TA:

Make 2 equivalence relations, one where a~a and another a~a-inverse
Show that every class size is 2 or 1
Show that the sum of the sizes of the equivalence classes=the size of the group

If what the question is asking is false, prove it is equivalent to "for all a =/ 0, the size of a equals 2." This gives a contradiction to the size of G being even.

8. Sep 17, 2008

### Dick

You assumed a^2=e for any a in G. That means a=a^(-1). That means {a,a^(-a)} has only one element. It's not a subgroup since if a!=e, e isn't in it. The subgroup is was referring to is {e,a}.

9. Sep 17, 2008

### Dick

I've begun to realize I'm confusing the original question with one you subsequently posed. Did you get the original question? Yes, pair a with a^(-1) for all elements in the group. e pairs with itself. If no other element pairs with itself then the order of the group must be odd. Right?

10. Sep 17, 2008

### fk378

Not quite following...in fact I don't even know how to use the hints the TA gave me.

11. Sep 17, 2008

### Dick

Ok, let's go back to the original question. I thought you said you had it and you don't. Forget all of the stuff in between, please? Think of all of the sets {a,a^(-1)} for a in G. Every element is in ONE of those sets, yes? And G is the union of all of those sets, also ok? If you don't agree, say so now.

12. Sep 17, 2008

### fk378

I agree.

13. Sep 17, 2008

### Dick

I hope so. Now if a=a^(-1), then the set {a,a^(-1)} has one element, otherwise it has two. Still with me?

14. Sep 17, 2008

### fk378

Yes, still following.

15. Sep 17, 2008

### Dick

Great! So G is the union of a bunch of sets. Some have two elements, some have one. Now {e} is one of those sets, since e=e^(-1). And {e} only has one element. Suppose n1 is the number of sets that have one element and n2 is the number that have two elements. Then the number of elements in G is n1*1+n2*2. Given that the number of elements in G is even, is it possible n1=1?

16. Sep 17, 2008

### fk378

Oh, I'm lost here. I thought G was made up of sets of 2 elements, not two and one. I understand that e is one of the sets of one element, though, you implied that there are more.

Where did you get n2*2 from? How do you know there are 2 sets with 2 elements in them?

17. Sep 17, 2008

### Dick

Sorry. I meant n1 and n2 to be different numbers. Try this "Suppose a is the number of sets that have one element and b is the number that have two elements. Then the number of members of G is a*1+b*2. Given that the number of members of G is even, is it possible a=1?" If G has an member such that g=g^(-1), then {g,g^(-1)} has only one element. There may be other members that belong to one element sets.

18. Sep 18, 2008

### fk378

it is possible for a=1, right? e could be the only element which is its own multiplicative inverse.

19. Sep 18, 2008

### Dick

Absolutely, then the order of G is 1 plus an even number isn't it? What could be wrong with that?

20. Sep 18, 2008

### fk378

That would imply that the order of G is odd.