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first of all it is not a Group,because for a group,in cayley table no repetition in any coloumn and row allowed.

* is not defined what kind of operation it is,considering multiplication:

1.it has an identitiy element see 1st column 1st row.

2.it is not commutative,because if commutative transpose of table NOT EQUAL to table.(not symmetric)

1) false : 2*(3*4) <> (2*3)*4

2)false : 4*2 <> 2*4

3)true : 1*{1,2,3,4} = {1,2,3,4}

so only option iii) is true..

The correct answer is choice (c).

* is not commutative. Take this counterexample.

$2 * 4 = 4$, and $4 * 2 = 1$.

Clearly $2 * 4$ is not equal to $4 * 2$.

* is not associative. Let's see if $2 * (3 * 4)$ is equal to $(2 * 3) * 4$.

LHS: $2 * (3 * 4)$ = $2 * 2$ = 2.

RHS: $(2 * 3) * 4$ = $1 * 4$ = 4.

Clearly both are not equal, hence * is not associative either.

However * does have an identity element, and we can clearly see that the identity element is 1.

Option c is right.

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