Question #283075

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.


1
Expert's answer
2021-12-27T18:33:57-0500

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

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


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

Taking into account that

(ab)c=(a+b+1)c=(a+b+1)+c+1=a+(b+c+1)+1=a+(bc)+1=a(bc)(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,bGa, b \in G, we conclude that the operation * is associative on the set Z.\Z.

Since for e=1e=-1 we get that

1a=1+a+1=a-1*a=-1+a+1=a and a(1)=a1+1=aa*(-1)=a-1+1=a

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

Taking into account that for any aZa\in \Z and a1=a2Za^{-1}=-a-2\in \Z we have that

aa1=aa2+1=1=ea*a^{-1}=a-a-2+1=-1=e and a1a=a2+a+1=1=e,a^{-1}*a=-a-2+a+1=-1=e,

we conclude that a1a^{-1} is the inverse of a.a.

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

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


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