# Finding all positive integers $k$ such that $\sigma$ and $\pi^k$ are conjugated

I'm very new to working with permutations and I have a question that I cannot really understand how to solve.

Suppose that you have the two permutations $\sigma=(17)(264)(35)$ and $\pi=(162)(3574)$ and both belong to the set $\mathrm S_7$. Then the task is to find all the positive integers $k$ such that $\pi^k$ and $\sigma$ are conjugated.

According to the definition of conjugacy between two permutations, both of the permutations must have the same cycle structure. But then I'm trapped. One of the matters I have with this question is that I really do not understand how the permutation $\pi^k$ changes its cycle cycle structure when $k$ goes to infinity. I believe that each possible cycle structure somehow gets repeated depending on the value of the positive integer $k$. But I don't know, I'm simply clueless.

Thank you very much in advance with helping me with this!

• Consider an easy case. Suppose $\pi$ is a 4-cycle in $S_4$. $\pi=(1 2 3 4)$. This can be thought of as a cyclic permutation of 4 vertices of a square. The $\pi^2$ will send every vertex to its diagonally opposite one. That means 1 is sent to 3 (and 3 to 1), 2 is sent to 4 (and 4 to 1). So we get $\pi^2 =(13)(24)$, this is not a 4-cycle it is two disjoint transpositions. Now work on more cases. – P Vanchinathan Oct 22 '16 at 11:09
• I really don't understand the purpose here, because after testing that case you have given, I only get that for all $pi$=$pi^k$ whenever $k=4(n-1)+1$ for some natural integer $n>0$ (and in between I get some different patterns). But how can I apply this to the question in my answer? And what if I am working with bigger permutation sets? – user2566415 Oct 22 '16 at 11:31

The hint in my comment seems to be not enough for you here are more:

First some basic facts about permutations.

FACT 1: Any permutation is a product of disjoint cycles which commute with each other.

FACT 2: Order of a permutation (least power making it identity) is the least common multiple of lengths of all the cycles appearing in the above product.

FACT 3: If $\sigma$ is a cycle of length $r$ then $\sigma ^k$ will be a be a permutation of order $r/\gcd(r,k)$. and product of equal cycles of that length.

Now your $\pi$ is a product of a 3-cycle with a 4-cycle. Hence by Fact 2 it has order 12. So you need to consider only 12 powers of $\pi$. That is $\pi^{12+m}=\pi^m$.

You should be able to now work this out now.

• Just to understand the solution on high level: when I have the permutation $pi^k$ for some positive integer $k$, then I want to reconstruct so that $pi^k$ gets the same cycle structure as $sigma$ and use then the fact 3 to do that by studying each cycle contained in some $pi^k$ and match k to that correct pattern. Is it right thinking? I am, as earlier mentioned, very clueless. So please be patient with my low understanding. – user2566415 Oct 22 '16 at 21:35
• I made a mistake in Fact 3 now corrected it. – P Vanchinathan Oct 23 '16 at 1:06