# 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.

## 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.

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

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?

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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?

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

(0,0), (3,3)

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

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.

(0,0), (3,3)

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