# Groups whose elements have order 2

1. Oct 19, 2009

### halvizo1031

1. The problem statement, all variables and given/known data

suppose that G is a group in which every non-identity element has order two. Show that G is commutative.

2. Relevant equations

3. The attempt at a solution

Suppose that a,b and ab all have order two. we will show that a and b commute. By assumption, e=(ab)^2
=abab
As a and b are their own inverses, multiplying on the left by a and then b,
we get
ba=ab.

2. Oct 19, 2009

### VeeEight

Looks good to me.
e = abab implies
b = ababb = aba which implies
ba = abaa = ab

3. Oct 19, 2009

### Dick

Yes, that's right.

4. Oct 20, 2009

### halvizo1031

Thanks! Is there another way i can show this?

5. Oct 20, 2009

### Dick

Why do you need another way? What did you have in mind?

6. Oct 20, 2009

### halvizo1031

well i just don't want to have the same answer as someone else.

7. Oct 20, 2009

### Dick

It's a simple problem. There's a simple answer. There's a few different permutations on the expression of that answer, but they are all really the same. Wouldn't it be better to move on to the next problem?

8. Oct 20, 2009

### halvizo1031

That's true. I'm having trouble showing that an element k is a generator of Zn if and only if k and n are relatively prime.

9. Oct 20, 2009

### Dick

The order 2 part of the problem has nothing to do with proving k is a generator of Zn if k and n are relatively prime. Try thinking about them independently.

10. Oct 20, 2009

### halvizo1031

Sorry, i forgot to mention that this question has nothing to do with the order 2 problem. Here's the new question:

Consider Zn={0,1,...,n-1}. show that an element k is a generator of Zn if and only if k and n are relatively prime.