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edited | 126 views
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3?
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I think only one (i).
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4th option is a cyclic group ? @kumar.dilip

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1 and 4
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I'm sure about 1 but not 4. ( I need to revise this topic).

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how you got 4th
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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.
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2) closer not satisfy 1+1=2 which is not in set