Permutations commuting with their powers

1. Dec 1, 2007

I have been asked to show that in Sn the cycle (1,2,....n) only commutes with its powers.
I know that cycles commute when they are disjoint and that every permutation can be written as a product of disjoint cycles but how do i show that this cycle and its powers are disjoint?

PLz help.

Thank you.

2. Dec 1, 2007

Chris Hillman

Is this a homework problem, then? You have been asked to compute the centralizer of an n-cycle in S_n; not to show that the powers of the cycle are all distinct from the cycle itself (which is not even true!).

3. Dec 1, 2007

I see. It is actually not a homework problem, since I'm trying to tackle this on my own. I do appreciate you pointing the right way to think about this. I wonder if you could suggest further reading on the centralizer concept as i am not familiar with it.

4. Dec 2, 2007

I have found some stuff on Centralizers online, and i think i understand the concept. I am however unable yet to make the connection with my problem.
If "The centralizer of a permutation is the set of all permutations
which commute with it" how do i prove that the only set that works are the powers of the cycle?
this still eludes me!!

5. Oct 17, 2011

postylem

Thinking of this in terms of a http://en.wikipedia.org/wiki/Group_action" [Broken] is helpful.

Showing that the centralizer of cycle $a=(1,2,\dots,n) \in S_n$, $C_{S_n}(a)$ is equal to the group generated by the cycle of $\langle a \rangle$, is equivalent to showing that the stabilizer $\text{Stab}_{S_n}(a)=\langle a \rangle$, for $S_n$ acting on itself by conjugation.

It is obvious that every power is in the centralizer group, so then you must show that it contains only those elements. Hint on how to do this: the number of conjugates is the size of the orbit of $a$ in $S_n$. By the orbit-stabilizer theorem, number of orbits equals the index of the stabilizer in $S_n$. So just count the conjugates (the orbits), and solve for the size of the stabilizer.

Last edited by a moderator: May 5, 2017
6. Jan 5, 2012

someriraq

dear "postylem"
plz, example ?
thanks

7. Jan 5, 2012

lavinia

I would do a direct calculation on generators of the group.

For instance, take the permutation 1->2 2->1 all other elements stay fixed. Then this permutation followed by your cycle is 2->1 -> 2 but in the other order 2 ->3 ->3

Last edited: Jan 5, 2012
8. Jan 5, 2012

wisvuze

Remember that powers of an element will always commute with themselves, the cycle decomposition won't have to be disjoint. Furthermore, the "disjoint cycles implies commutativity" implication is only one way, not an "if and only if".
You can just directly compute how the cycle (1 2 .. n ) acts on an arbitrary k-cycle, the fact that the cycle ( 1 2 .. n ) actually moves all symbols should help.

9. Jan 5, 2012

micromass

This thread is 5 years old...