Finding Elements of Order 2 in Z6

[POtment]

Homework Statement

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.

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.

 

[quantumdude]

The Attempt at a Solution

The elements of Z6 are 1,4,5.

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.

 

[POtment]

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

 

[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?

 

[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?

 

[Dick]

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

 

[POtment]

(0,0), (3,3)

 

[POtment]

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

 

[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.

 

[Dick]

(0,0), (3,3)

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