# Abelian group

1. Apr 28, 2008

### POtment

1. The problem statement, all variables and given/known data
Let G be an abelian group. Show that the set of all elements of G of order 2 forms a subgroup of G. Find all elements of order 2 in Z6.

3. The attempt at a solution
The elements of Z6 are 1,4,5. I'm not sure how to find the set of all elements of order 2. Can someone help with that? I think I can prove from there.

2. Apr 28, 2008

### Tom Mattson

Staff Emeritus
Nope. How did you come up with that? And are you talking about the additive group $\mathbb{Z}_6$, or the multiplicative group $\mathbb{Z}_6$? It does make a difference.

3. Apr 28, 2008

### POtment

We are talking about the additive group of Z6. Can you nudge me in the right direction to find the correct elements?

4. Apr 28, 2008

### Dick

If it's additive then an element g has order 2 if g+g=0 and g is not zero. It should be pretty easy to find them. But why do you say the elements are 1,4,5???

Last edited: Apr 28, 2008
5. Apr 28, 2008

### POtment

OK. I understand what you are saying and know how to construct the table to find the elements of order 2. However, I still don't understand what the elements of Z6 are. Is it as obvious as 0,1,2,3,4,5?

6. Apr 28, 2008

### Dick

It's as obvious as that, yes. So for which ones is g+g=0 mod 6?

7. Apr 29, 2008

### POtment

(0,0), (3,3)

Last edited: Apr 29, 2008
8. Apr 29, 2008

### POtment

This one is solved. Thanks for nudging me in the right direction!

9. Apr 29, 2008

### Dick

I didn't ask for which pairs is (a,b) is a+b=0. Why would I?? Look up the definition of 'order 2'. Now write it on a blackboard ten times. Now step back and look at it. Now tell me why most of the pairs in your 'guess' aren't interesting.

10. Apr 29, 2008

### Dick

Better. So the subgroup is made of the elements 0 and 3.