Is Comunitative Group a Group?

[physicsjock]

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?

 

[Alesak]

Hmm, and how do you know it's associative?

Give us more precise name of your starting structure.

 

[Norwegian]

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.

 

[physicsjock]

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?

 

[Alesak]

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.