Question

Show that $(Z,\ast )$ is an infinite abelian group, where ${}^{\prime }{\ast }^{\prime }$ is defined as $a\ast b=a+b+2$ and $Z$ is the set of all integers

Answer 1

(i) Closure axiom : Since $a,b$ and $2$ are integers $a+b+2$ is also an integer
$\therefore a\ast bϵz\forall a,bϵz$
(ii) Associature axiom : Let $a,b,c\ast G$.
$(a\ast b)\ast c=(a+b+2)\ast c=(a+b+2)+c+2=a+b+c+4$
$a\ast (b\ast c)=a\ast (b+c+2)=a+(b+c+2)+2=a+b+c+4$
$\Rightarrow (a\ast b)\ast c=a\ast (b\ast c)$
Thus associature axiam is true.
(iii) Identity axiom:
Let $e$ be the identity element.
By the definition of $e,a\ast e=a$
By the definition of $\ast ,a\ast e=a+e+2$
$\Rightarrow a+e+2=a$
$\Rightarrow e=-2$
Also $-2ϵZ$. Thus identity axiom is true.
(iv) Inverset axiom:
Let $aϵG$ and $a^{-1}$ be the inverse element of $a$
By the definition of $a^{-1},a\ast a^{-1}=e=-2$
By the definition of $\ast ,a\ast a^{-1}=a+a^{-1}+2$
$\Rightarrow a+a^{-1}+2=-2$
$\Rightarrow a^{-1}=-a-4ϵZ$
$\therefore$ Inverse axiom is true
$\therefore (Z,\ast )$ is a group.
(v) Commutature property:
Let $a,bϵG$
$a\ast b=a+b+2=b+a+2=b\ast a$
$\therefore \ast$ is Commulative
$\therefore (Z,\ast )$ is an abelian group also $Z$ is an infinite set.
$\therefore Z$ is an infinite abelian group.