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how you got 4th
Option 1 is correct coz it follows closure, identity is 1, invere of every element exist and multiplication follows associativity.

Option 2 is also correct coz it follows closure, identity is 0, each element has inverse and associativity is applicable on addition.

Option 3 does not follows associativity.

Option 4 is a cyclic group i think....every cyclic group is abelian group and thus it is also a group.
2) closer not satisfy 1+1=2 which is not in set

1 vote
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1 vote