Answer to Question #133610 in Abstract Algebra for PRATHIBHA ROSE C S

Question

On Q/{-1} define * by a*b=a+b+an for all a,b element of Q/{-1} show that (Q/{-1},*) is an abelian group

Answer 1

$a\in Q/\{-1\}$

$b∈Q/\{−1\}$

$a*b=a+b+ab\in Q/\{-1\}$

$\therefore *, is$  a binary operation in$(Q/\{-1\}, *)$

Associativity: Suppose $a,b,c \in Q/\{-1\}$

$(a*b)*c=(a+b+ab)*c$

$=(a+b+ab)+c+(a+b+ab)c$

$$=a+b+c+ab+bc+ca+abc$$

So,

$a*(b*c)=a*(b+c+bc)$

$=a+(b+c+bc)+a(b+c+bc)$

$=a+(b+c+bc)+a(b+c+bc)$

$=a+b+c+ab+bc+ca+abc$

$(a*b)*c=a*(b*c)$

* is associative.

$Existence$ $of$ $Identity:$

$0\in Q/\{-1\}$

$is$ the identity element.

Because $a\in Q/\{-1\}$

$0*a=0+a+0a=a$

$a*0=a+0+a0=a$

$Existence$ of $Inverse$ :

Every $a\in$ $Q/\{-1\}$

For -a/a+1 $\in$ Q/{-1}

$a\not=-1,$ which is inverse of a

Because

$\bigg(-a/(a+1) \bigg) *a$ $=$

$=\bigg(-a/(a+1)\bigg) +a+\bigg(-a/(a+1)\bigg) *a=0$

$a*\bigg(-a/(a+1)\bigg) =$

$=a+(-a)/(a+1)+a\bigg(-a/(a+1)\bigg) =0$

$Commutativity:-$ $a, b \in Q/\{-1\}$

$a*b=a+b+ab$

$b*a=b+a+ba$

So,

$a*b=b*a$

Operation * is commutative.

The above discussion proves that the group is abelian.