Comunitative Group

1. Mar 17, 2012

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?

2. Mar 17, 2012

Alesak

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

Give us more precise name of your starting structure.

3. Mar 17, 2012

Norwegian

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.

4. Mar 17, 2012

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?

Last edited: Mar 17, 2012
5. Mar 18, 2012

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.