Group Structure

Printable View

• Apr 4th 2006, 07:05 AM
TexasGirl
Group Structure
Show that a*b=a+b+ab defines a group structure on G=R\{-1} using that 1+a*b=(1+a)(1+b).

not that G=R\{-1} means that G is all real numbers excluding -1.
• Apr 4th 2006, 03:13 PM
ThePerfectHacker
Quote:

Originally Posted by TexasGirl
Show that a*b=a+b+ab defines a group structure on G=R\{-1} using that 1+a*b=(1+a)(1+b).

not that G=R\{-1} means that G is all real numbers excluding -1.

Binary Operation is Closed: The only case where it would not be closed is when, $a,b \not = -1$
$a*b=-1$
Thus,
$a+b+ab=-1$
Thus,
$b(a+1)=-a-1=-(a+1)$
Thus,
$b=-1$
Impossible, thus the binary operation is closed.

Associative: We have,
$(a*b)*c=(a+b+ab)*c=a+b+ab+c+ac+bc+abc$
And,
$a*(b*c)=a*(b+c+bc)=a+b+c+bc+ab+ac+abc$
Thus, $a*(b*c)=(a*b)*c$

Identity: We have, that,
$0*a=0+a+a0=a$ and,
$a*0=a+0+00=a$.

Inverse: Notice that,
$a*-\frac{a}{a+1}=-\frac{a}{a+1}*a=0$
Finally, we need to show if, $a\not = -1$ then, the element $-\frac{a}{a+1}\not = -1$
Assume, it were then,
$-\frac{a}{a+1}=-1$
Thus,
$\frac{a}{a+1}=1$
Thus,
$a=a+1$
Thus,
$1=0$
Impossible, thus, $-\frac{a}{a+1}\in G$