Permutations are hell for me

How do you prove that whatever permutation from S5 you take, there is n so that σ^n=e?

S5 has order 5!=120. Pick any σ in S5, take the cyclic group <σ>. The order of <σ> must divide 120 by Lagrange's Theorem. Say that order is n. Then σ^n=e by definition of a cycilc group.

For any <math>\sigma</math> in any finite group there must be some <math>k</math>, <math>l</math> such that <math>\sigma^k=\sigma^l</math>, by pigeonhole.  Using <math>(\sigma^l)^{-1}=(\sigma^{-1})^l</math> gives the result, after multiplying by this.

Let G be a finite group and g an element of G.  Consider the infinite sequence g^1, g^2, \ldots.  Since G is finite, there must be two elements of this sequence that are equal, say g^i = g^j with j>i.  Then by cancellation, g^{j-i} = e.

S_5 is a finite group.

Done.

>>4
Man, I wish I had thought of that.