Question #213734

Determine whether or not ∗ gives a group structure on the set. If it is not a group, say which

axioms fail to hold.

Define ∗ on Z by a ∗ b = ab.


1
Expert's answer
2021-07-05T17:18:22-0400

Consider the operation \ast on ZZ defined by

ab=aba\ast b=ab

Require to determine whether or not \ast gives a group structure on the set ZZ

Let us verify the group axioms:

(1) Closure: a,bZabZa,b\in Z\Rightarrow a*b\in Z

Now a,bZabZa,b\in Z\Rightarrow ab\in Z

abZ\Rightarrow a* b\in Z

So, a,bZabZa,b\in Z\Rightarrow a*b\in Z

Therefore, ZZ satisfies the closure axiom under the operation \ast

(2) Associativity: a,b,cZa(bc)=(ab)ca,b,c\in Z\Rightarrow a* (b* c)=(a* b)\star c

Now a,b,cZa(bc)=a(bc)=a(bc)=abca,b,c\in Z\Rightarrow a* (b* c)=a* (bc)=a(bc)=abc

And

(ab)c=(ab)c=(ab)c=abc(a* b)* c=(ab)* c=(ab)c=abc

So, a,b,cZa(bc)=(ab)ca,b,c\in Z\Rightarrow a* (b* c)=(a*b)* c

Therefore, ZZ satisfies the associativity axiom under the operation \ast

(3) Identity: For each aZa\in Z there exists eZe\in Z such that ae=a=eaa*e=a=e*a

Now ae=aae=aa*e=a\Rightarrow ae=a

e=1Z\Rightarrow e=1\in Z

And a1=a(1)=a=1(a)=1aa*1=a(1)=a=1(a)=1*a

Therefore, 11 is the identity in ZZ

(4) Inverse: For each non zero aZa\in Z there exists bZb\in Z such that ab=e=baa*b=e=b*a

Let aa be a non zero integer

Now ab=1ab=1a*b=1\Rightarrow ab=1

b=1aZ\Rightarrow b=\frac{1}{a}\notin Z

Therefore, inverse axiom fail to hold

Hence, (Z,)(Z,*) is not a group.


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