(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

if f [itex]\in[/itex] S_{n}show that there is some positive integer k, depending on f, such that f^{k}=i. (from baby Herstein).

3. The attempt at a solution

Suppose that S={x_{1},x_{2},...,x_{n}}. Elements of S_{n}are bijections from S to S. to show that f^{k}=i it's enough to show that f^{k}(x_{m})=x_{m}for every m (1<=m<=n)

If f is in S_{n}, then f may stabilize(not permute) a finite number of elements and permutes the other elements. the elements that are not permuted by f will also be stabilized by f^{k}. (because if f(x_{s})=x_{s}, then f^{2}(x_{s})=f(x_{s})=x_{s}and by induction It's possible to show that f^{k}(x_{s})=x_{s}). So, for the elements that are not permuted by f we're done.

Now, We can suppose that x_{p}is permuted by f and f(x_{p})=x_{q}(x_{p}[itex]\neq[/itex]x_{q}). if we apply f again, f^{2}will again permute x_{p}to another member of S and this process of permutation never halts because if finally x_{p}is mapped to something that is not permuted anymore by f, like x_{s}, then f^{k}(x_{p})=x_{s}and f^{k}(x_{s})=x_{s}which contradicts the fact that f^{k}is bijective. Since this process never halts and S is a finite set and f is a bijection, then finally after k times (2<=k<=n) f^{k}(x_{p})=x_{p}(I'm in someway using the pigeonhole principle). this k depends on x_{p}. (It can be shown by studying permutations of S_{n}for n>2)

It's easy to show that if f^{k}(x_{p})=x_{p}then f^{nk}(x_{p})=x_{p}. since k is dependent to the x_{p}we choose, let's associate a k_{i}with every element of S that is permuted. by the same argument we know that f^{ki}(x_{pi})=x_{pi}. if we set K=LCM(k_{i}) then by our previous arguments, It's obvious that for any x_{p}we'll have f^{K}(x_{p})=x_{p}. Q.E.D.

I come up with this proof after studying permutations in S_{5}. I know that the way I've explained my proof might be a little bit confusing but I hope that it's understandable. Is my proof correct? and if yes, is there any other way to prove it in a simpler way?

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# A problem from Herstein's Abstract algebra

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