Finding Elements of Order 2 in Z6

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Homework Help Overview

The discussion revolves around identifying elements of order 2 in the additive group of Z6, an abelian group. Participants are tasked with determining which elements satisfy the condition for order 2.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the elements of Z6 and question the definitions and conditions for order 2. There are attempts to clarify whether the group in question is additive or multiplicative, and participants explore the implications of these definitions.

Discussion Status

The discussion has progressed with some participants providing guidance on how to identify elements of order 2. There is an ongoing exploration of the correct elements and their properties, with some participants expressing uncertainty about their previous assertions.

Contextual Notes

There is a focus on the additive group of Z6, and participants are clarifying the definitions and properties related to elements of order 2. Some confusion exists regarding the initial identification of elements and their orders.

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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.
 
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POtment said:

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?
 
Last edited:
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)
 
Last edited:
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.
 
  • #10
POtment said:
(0,0), (3,3)

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

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