Question: if f and g in Sn are disjoint cycles fg = gf
f = (a1 a2 a3... an)
g = (b1 b2 b3 ... bm)
where ai != bj for any i or j
gf(ai) where i = 1 to n-1
gf(ai) = g(ai+1) = ai+1
fg(ai) = f(ai) = ai+1
thus f and g commute for ai where i = 1 to n-1 this same method can be used to show that f and g commute for an and for bi.
Is there a better way to prove this or to think about it?
This not intended to answer the original question. It is an interesting observation I think. Several years ago I saw this question posed a different way. Prove that two permutations commute iff their active sets are disjoint.
The idea being that if an element is active in f it is not in g, and visa versa.