# Determining if a group

• Jun 8th 2009, 11:39 PM
Maccaman
Determining if a group
Hi there,
I'm doing some revision for my exam and I cannot see how this one particular solution has come about. The question is:

Is $\displaystyle (G, *)$ a semigroup, or group in the following?
(a) $\displaystyle G = \mathbb{R} \$\$\displaystyle \{\frac{1}{2}\}$, $\displaystyle a * b = a - 2ab +b$

I have solved a lot of questions like this, so I know its a group because i have been able to show that it has an identity and inverse, and i have the solutions to this problem so i have seen the associative property proved, but that is where I am getting stuck (Shake)

The solution I have says (I'll number each line)....
(1) $\displaystyle (a * b) * c = (a - 2ab + b) * c$
(2) $\displaystyle \ \ \ \ \ \ \ = (a - 2ab + b) - 2(a - 2ab + b)c + c$
(3) $\displaystyle \ \ \ \ \ \ \ = a + b + c - 2ab - 2ac - 2bc + 4abc$
(4) $\displaystyle \ \ \ \ \ \ \ = a - 2a(b - 2bc + c) + (b - 2bc + c)$
(5) $\displaystyle \ \ \ \ \ \ \ = a * (b - 2bc + c)$
(6) $\displaystyle \ \ \ \ \ \ \ = a * (b * c)$

Now normally what I like to do when looking to see if a particular operation is associative is do steps (1), (2), (3), and then do (6), (5), (4), (3) and see if i get the same result for (3) both times. However in this case I cannot see how (2) was obtained from (1). Initially I thought that step 2 would be:
$\displaystyle = (a - 2ab + b) + c - 2c$. I cannot see why (2) is $\displaystyle \ \ \ \ \ \ \ = (a - 2ab + b) - 2(a - 2ab + b)c + c$. Can anyone please explain why this is the case?
Thanks in advance to those who can help.
Cheers,
Maccaman
• Jun 9th 2009, 12:13 AM
Swlabr
To easy confision let us write multiplication as $\displaystyle x * y = x - 2xy +y$. So when going from line 1 to line 2 we have that $\displaystyle x=a*b, y=c$. $\displaystyle a * b = a - 2ab +b$, and the result follows.

The $\displaystyle a$ and $\displaystyle b$ in the definition of multiplication and associativity are just arbitrary elements from the group.
• Jun 9th 2009, 03:38 PM
Maccaman
I see. I must have been tired or something to miss that. Thanks very much Swlabr. You (Rock)