# Binary operations on groups

• Oct 10th 2009, 10:05 AM
spearfish
Binary operations on groups
Hey guys,

I am trying to prove that a set with the specified binary operation below is a group? Here is the problem

on R* x R, (a, b) * (c,d) = (ac, bc + d)

I know that I have to prove that it is has
1.) an identity: a * x = a
2.) an inverse: a * x = e
3.) associative

The thing that throws me off is the (ordered pair). How do I start it off?

This is how I started off trying to prove the identity
a*x = a
(a,b) * (x1, x2) = (a,b) ?? DOn't know if I can use x1,x2 ??
(a*x1, b*x1 + x2 ) = (a,b)
then I get two equations:

a *x1 = a --> x1 = 1
b*x1 + x2 = b --> b*1 + x2 = b --> x2 = 0

NOw I plug x1 and x2 back into original eqn:

(a *1, b*1 + 0 ) = (a, b)
(a,b) = (a,b)

haha, I started off not knowing how to go about and the mere fact of coming to this site and trying to ask a question got me going in the right direction and I ended up solving my own problem, (Clapping). My thanks for this website!
• Oct 10th 2009, 10:06 AM
spearfish
Ok guys, so I just thought I would go ahead and post if anybody else needed the help. I can do the rest, so thanks for watching out!
• Oct 10th 2009, 10:58 AM
Plato
Quote:

Originally Posted by spearfish
I am trying to prove that a set with the specified binary operation below is a group? Here is the problem
on R* x R, (a, b) * (c,d) = (ac, bc + d)
I know that I have to prove that it is has
1.) an identity: a * x = a
2.) an inverse: a * x = e
3.) associative

The identity is $(1,0)$.
But you have to show $(x,y)*(1,0)=(1,0)*(x,y)=(x,y).$

The inverse of $(a,b)$ is $\left(\frac{1}{a},\frac{-b}{a}\right)$.
Again you have to show that inverse properties hold.
• Oct 11th 2009, 07:13 AM
spearfish
Thanks for the clarification Plato! It's kind of funny, but before I went to bed yesterday, it just popped into my head that I still needed to do the (x,y) * (1,0) = (1,0) * (x,y) = (x,y). I ll work on that and the rest of the problem today. Again, thanks!