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Show that the set G of all rational numbers except − 1 forms an abelian group with respect to the operation $*$ given by $a * b = a + b + ab$ for all $a, b \in G$.

1 Answer

  • A non-empty set G, together with an operation $*$ i.e., $(G, *)$ is said to be a group if it satisfies the following axioms
  • (1) Closure axiom : $a,b\in G\Rightarrow a*b\in G$
  • (2) Associative axiom : $\forall a, b, c \in G, (a * b) * c = a * (b * c)$
  • (3) Identity axiom : There exists an element $e \in G$ such that $a * e = e * a = a, \forall a ^{-1} G.$
  • (4) Inverse axiom : $\forall a \in G$ there exists an element $a^{-1}\in G$ such that $a^{-1}*a=a*a^{-1}=e.$
  • e is called the identity element of $G$ and $a^{-1}$ is called the inverse of a in $G.$
Step 1:
Let $G=\{x\in Q-\{1\}\}$
The operation $*$ is defined by $a*b$ is defined by $a*b=a+b=a+b+ab$ for $a,b\in G$
Closure :
Let $a,b\in G$.
Suppose $a*b=-1$
$\Rightarrow a+b+ab=-1$
$\Rightarrow a+b+ab+1=0$
$\Rightarrow (a+1)+b(a+1)=0$
$\Rightarrow (a+1)(b+1)=0$
$\Rightarrow a=-1$ or $b=-1$ which is not true since $a,b\in G$
$\therefore a*b\neq -1\Rightarrow a*b\in G$
Closure is satisfied.
Step 2:
Associativity :
Let $a,b,c\in G$
It can be seen that $a*(b*c)=(a*b)*c$
$\therefore$ associative property is satisfied.
Step 3:
Existence of identity :
Let $a,e\in G$
Consider $a*e=a$
$\Rightarrow e(1+a)=0$
$\Rightarrow e=0$ or $1+a=0$
Since $a\neq -1\Rightarrow e=0$
$\therefore 0\in G$ is the identity.
(It can be seen that $0*a=a)$
Step 4:
Existence of inverse :
Let $a\in G$
Consider $a*a'=0$
$\Rightarrow a'(1+a)=-a$
$\Rightarrow a'=\large\frac{-a}{1+a}$
$(1+a\neq 0)$
$\large\frac{a}{1+a}$$\in G(\neq -1)$
$a'=\large\frac{-a}{1+a}$ is the inverse of $'a'$
(It can be seen that $a'*a=0)$
$\therefore$ every element in $G$ has an inverse.
The four group axioms being satisfied,$(G,*)$ is a group.
answered Sep 16, 2013 by sreemathi.v

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