Answer to Question #163217 in Abstract Algebra for K

Question

Let S be the set of all real numbers except −1. Define * on S by

a * b = a + b + ab

Show that〈 S,*〉is a group and solve 2 * x * 3 = 7 in S

.

Answer 1

Given that $a \neq-1, b \neq -1 \in \mathbb{R}$ , we know that $(\mathbb{R},+)$ and $(\mathbb{R},.)$ are groups. Hence $S$ is closed.

Next, we show that S is associative

Let $a,b, c \in S$

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

$= a+b+c +ab +ac + bc +abc= a+ab+ac+abc +b+c+bc = a + a(b+c+bc) + (b+c+bc)$

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

Next we show that S has an identity element. Inorder to do this, we suppose there is an $e \in S ~ \ni a*e =a$ i.e $a + e +ae = a \\ \implies e +ae = 0\\ \implies e(1+a)=0$

Either $e =0$ or $1+a = 0$

If $1+a =0$ ,then $a \notin S \therefore e= 0$

(Since $(\mathcal{R},+)$ and $(\mathcal{R}, .)$ are Abelian groups, we have that

$a*e=a=e*a)$

Next we show that every non zero element of $S$ has an inverse.

Let $a,b \in S$

Consider,

$a+b+ab =e=0\\ b+ab = -a\\ b(1+a) = -a\\ b = \frac{-a}{1+a}$

By definition of elements of $S,$ we have that $a\neq -1$ . Hence for every non-zero $a\in S$ , $a^{-1} = \frac{-a}{1+a}$

Hence $\langle S, * \rangle$ is a group

$(2*x)*3 = 7\\ (2+x+2x)*3 =7\\ (2+3x)*3=7\\ 2+3x +3+(2+3x)3 = 7\\ 5+ 3x + 6 + 9x =7\\12x = 7-11\\ 12x = -4 \\ x = \frac{-4}{12}\\ x = \frac{-1}{3}$