Question

Let $G$ be the set of all rational numbers except $1$ and $\ast$ be defined on $G$ by $a\ast b=a+b-ab$ for all $a,b\in G$. Show that $(G,\ast )$ is an infinite Abelian group.

Answer 1

Let $G=Q-\left\{1\right\}$
Let $a,b\in G$. Then $a$ and $b$ are rational numbers and $a\ne 1$, $b\ne 1$.
(i) Closure axiom: Clearly $a\ast b=a+b-ab$ is a rational number. But to prove $a\ast b\in G$, we have to prove that $a\ast b\ne 1$.
On the contrary, assume that $a\ast b=1$ then
$a+b-ab=1$
$\Rightarrow b-ab=1-a$
$\Rightarrow b(1-a)=1-a$
$\Rightarrow b=1$ (since $a\ne 1$, $1-a\ne 0$)
This is impossible, because $b\ne 1$.
$\therefore$ Our assumption is wrong.
$\therefore a\ast b\ne 1$ and hence $a\ast b\in G$.
$\therefore$ Closure axiom is true.
(ii) Associative axiom:
Let $a,b,c\in G$
$a\ast (b\ast c)=a\ast (b+c-bc)$
$=a+(b+c-bc)-a(b+c-bc)$
$=a+b+c-bc-ab-ac+abc$
$(a\ast b)\ast c=(a+b-ab)\ast c$
$=(a+b-ab)+c-(a+b-ab)c$
$=a+b+c-ab-ac-bc+abc$
$\therefore a\ast (b\ast c)=(a\ast b)\ast c\forall a,b,c\in G$.
$\therefore$ Associative axiom is true.
(iii) Identity axiom: Let $e$ be the identity element.
By definition of $e$, $a\ast e=a$
By definition of $\ast$, $a\ast e=a+e-ae$
$\Rightarrow a+e-aw=a$
$\Rightarrow e(1-a)=0$
$\Rightarrow e=0$ since $a\ne 1$
$e=0\in G$
$\therefore$ Identity axiom is satisfied.
(iv) Inverse axiom: Let $a^{-1}$ be the inverse of $a\in G$.
By the definition of inverse, $a\ast a^{-1}=e=0$
By the definition of $\ast$, $a\ast a^{-1}=a+a^{-1}-a{a}^{-1}$
$\Rightarrow a+a^{-1}-a{a}^{-1}=0$
$\Rightarrow a^{-1}(1-a)=-a$
$\Rightarrow a^{-1}=\frac{a}{a-1}\in G$ since $a\ne 1$
$\therefore$ Inverse axiom is satisfied. $\therefore (G,\ast )$ is a group.
(v) Commutative axiom: For any $a,b\in G$
$a\ast b=a+b-ab=b+a-ba=b\ast a$
$\therefore \ast$ is commutative in $G$ and hence $(G,\ast )$ is an abelian group. Since $G$ is infinite, $(G,\ast )$ is an infinite abelian group.