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Let U be the set of elements of a group G and let * its operation. If V is the set of one-to-one functions from U to S then such functions are known as permutations of the set S. The set V with function composition (·) will be a group.

Sonam, I don’t think the one you mentioned is right. I guess this is the right proof. Consider an identity element e(y)=y for all y belonging to set S. If f(y)=x then f-1(x)=y. Thus (F, ·) is a group.

Niranjan, whatever is mentioned by name1 is right although it is not the proof name is asking for. And whatever you have mentioned is not the proof of the theorem, it is the proof of the proposition. For any element g of set S consider the function fg(y)=g*y for all y in S. Consider fg*h(y). Since G is a group g*h, an element of S, hence fg*h is an element of F. Since * is associative in G, (g*h)*y = g*(h*y) = g*(fh(y)) = fg(fh(y)) = fg·fh(y) But, (g*h)*y is fg*h(y) So, fg*h = fg·fh