>>12765262After S4 it works.
For Sn (n>4), the only normal subgroups are the trivial group, An, and Sn.
First, I'll show conjugation fixes k-cycles. Then I'll show the group generated by k-cycles is normal. Then using the fact above, you can conclude this normal subgroup is Sn.
1) Conjugation will map k-cycles to k-cycles.
This is easy to show. Given any permutation in Sn, write it as a product of disjoint cycles.
https://en.wikipedia.org/wiki/Cycle_index#Disjoint_cycle_representation_of_permutationsIf you think about what conjugation does, it just re-labels the entries of the cycle representation without modifying the parentheses (the number of and sizes of cycles remains unchanged).
Pic related.
2) Now to prove the group generated by k-cycles, N, is normal.
Let n =k1*k2*...*ki by any element of N. Conjugate by any element, g, of Sn to get g*n*g' = g*k1*k2*...*ki*g'.
Multiply between the k's with g'*g = e to get g*k1*g'*g*k2*g'*...*g*ki*g'.
Grouping terms gives (g*k1*g')*(g*k2*g')*...*(g*ki*g') which is just a product of k-cycles and therefore an element of N.
So N is normal.
Clearly N isn't trivial. N can't be An from parity. So N is Sn.
