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6 years ago in Algorithms By Ravneet Singh
Cayley’s theorem proof
I want a proof for Cayley’s theorem. I have checked out many websites but each has quite a different proof. So please provide an appropriate proof.
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By Sonam Choephel Answered 6 years ago
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.
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By Niranjan Singh Answered 6 years ago
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.
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By Thamizhini Nagarajan Answered 6 years ago
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
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