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asked in Set Theory & Algebra by Loyal (5.4k points)
edited by | 47 views

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I think it should be C.

a) Every commutative operation need not be associative. For instance, NAND and NOR operations both are commutative but not associative.

b) With the given operation, it will only be ale to generate the multiples of 6 and 3 and not all the elements of the group, hence it is not a generator.

d) Multiplication is not a group on Z as 0 does not have an inverse.

However in option C, it is given that a,b and c are all in the group. Let us multiply both sides with a-1

a-1*a*x*b = a^-1*c

e*x*b = a^-1*c

x*b = a^-1*c

Now since the group follows the property of closure, x belongs to the group.
answered by (365 points)

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