Answer to Question #283075 in Abstract Algebra for Lucifer

Question

Prove that the set Z of all integers with binary operation * defined by a*b=a+b+1

for all a, b belonging to G is an Abelian group.

Answer 1

Let us prove that the set $\Z$ of all integers with binary operation * defined by $a*b=a+b+1$

for all $a, b \in G$ is an Abelian group.

Since for any $a, b \in G$ we have that $a*b=a+b+1\in \Z,$ the operation $*$ is defined on the set $\Z.$

Taking into account that

$(a*b)*c=(a+b+1)*c=(a+b+1)+c+1=a+(b+c+1)+1\\=a+(b*c)+1=a*(b*c)$

for all $a, b \in G$, we conclude that the operation $*$ is associative on the set $\Z.$

Since for $e=-1$ we get that

$-1*a=-1+a+1=a$ and $a*(-1)=a-1+1=a$

for each $a\in G,$ we conclude that $e=-1$ is the identity of the semigroup $(\Z,*).$

Taking into account that for any $a\in \Z$ and $a^{-1}=-a-2\in \Z$ we have that

$a*a^{-1}=a-a-2+1=-1=e$ and $a^{-1}*a=-a-2+a+1=-1=e,$

we conclude that $a^{-1}$ is the inverse of $a.$

Therefore, $(\Z,*)$ is a group.

Since $a*b=a+b+1=b+a+1=b*a$ for any $a,b\in G,$ we conclude that $(\Z,*)$ is an Abelian group.