search
Log In
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
0 votes
110 views

 

in Set Theory & Algebra
edited by
110 views

1 Answer

1 vote
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.

Related questions

...