Is a Commutative Group with a^2=1 Always a Group?

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Discussion Overview

The discussion revolves around whether a set defined as H = can be considered a group, given that it is commutative. Participants explore the implications of the properties of the elements in H and the requirements for a set to qualify as a group, including identity, inverses, and associativity.

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  • Debate/contested

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Areas of Agreement / Disagreement

Participants express differing views on whether the properties given are sufficient to establish H as a group. There is no consensus on the necessity of associativity or the implications of the binary operation.

Contextual Notes

There are unresolved questions regarding the definition of the binary operation and the implications of the properties of the elements in H. The discussion highlights the importance of these definitions in determining group status.

physicsjock
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Hey,

I have a small question about groups,

If you have a comunitative 'group' H = <a in H : a2=1>,

Is that enough information to show that it is a group, without knowing the binary operation?

say b is also in H

then a*b=b*a

(a*b)*(b*a) = (a*a)*(b*b) = 1 (since its comunitative)

So that shows there is an identity, and each element is it's own inverse

It's also associative so everything is satisfied for H to be a group,

So only knowing these two properties of this group can show that it is indeed a group?
 
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Hmm, and how do you know it's associative?

Give us more precise name of your starting structure.
 
physicsjock said:
So only knowing these two properties of this group can show that it is indeed a group?
No, if you have some set H, with a binary operation HxH → H, such that all (a,a) maps to same element (that we may call 1), then this will in general not be a group. You have that every element is its own inverse wrt to the element 1, but 1 is not a unit element (which requires that a*1=1*a=a), and you do not have associativity as there are absolutely no conditions set to the value of a*b with a≠b.

I didnt understand the word "comunitative", but you may have meant commutative, which btw is not taking you a lot further to making this into a group.
 
So I have a set <a in H : a^2=1> where H is a commutative group,
1 being the identity, and aa means a*a, sorry should have specified before

I have been told it is a group but I just want to see if I can justify it myself, having no clear binary operation confused me.

So I know each element is it's own inverse, and it does contain an identity element, 1

To show associativity , for a,b and c in the set

(a*b)*c=a*(b*c)

(a*c)*b= b*(a*c)

(a*c)*b= (a*c)*b

so all boxes ticked?
 
Last edited:
Maybe you want to show that subset of H <a in H : a^2=1> is a subgroup of H?

In this case, you don't need to worry about associativity etc., only under closure of operations. There are three operation you need to verify, one nullary - taking unit, one unary - taking inverse and one binary - multiplication.
 

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