# Math Help - Prove that Disjoint cycles commute

1. ## Prove that Disjoint cycles commute

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?
Thanks!

2. Originally Posted by temp31415
Question: if f and g in Sn are disjoint cycles fg = gf
It is trivial. Say f=(a1,a2,a3) and g=(b1,b2) for simplicity sake. For fg = gf we require that they are the same function. We want to show fg (x) = gf (x). Now if x ! = a_1,a_2,a_3,b_1,b_3 then it says fixed. If x=a1 then fg(a1)=a_2 and gf(a_1)=a_2 because they are disjoint. And the same idea in general.

3. 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.
Example
$f = \left( {\begin{array}{*{20}c} 1 & 2 & 3 & 4 & 5 \\
1 & 4 & 3 & 5 & 2 \\\end{array}} \right)\quad \& \quad g = \left( {\begin{array}{*{20}c}
1 & 2 & 3 & 4 & 5 \\ 3 & 2 & 1 & 4 & 5 \\ \end{array}} \right)$
.

The idea being that if an element is active in f it is not in g, and visa versa.

4. Originally Posted by Plato
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.
Example
$f = \left( {\begin{array}{*{20}c} 1 & 2 & 3 & 4 & 5 \\
1 & 4 & 3 & 5 & 2 \\\end{array}} \right)\quad \& \quad g = \left( {\begin{array}{*{20}c}
1 & 2 & 3 & 4 & 5 \\ 3 & 2 & 1 & 4 & 5 \\ \end{array}} \right)$
.

The idea being that if an element is active in f it is not in g, and visa versa.
Sorry I don't think it's true. Just take permutations $f,g$ such that $f=g$.