All permutations that commute with a cycle

Question

I know similar questions have been ask, but I can not seem to find answer that is useful for me. My question is, which permutations commute with a cycle $(a_1a_2\dots a_k)(a_{k+1})(a_{k+2}) \dots (a_n)$?

I know that disjoint cycles commute, but those are certainly not the only ones. Since every permutation can be written in form of a product of disjoint cycles, I was advised to examine several cases separately, for example when we have two disjoint cycles and when cycles overlap, and then to further divide the second case into subcases (when they overlap in the middle or at edges).

The problem is, other than the case with disjoint cycles, I don't even have any intuition what happens. We also don't get to use many results about permutations as we only recently started to learn about them, and didn't establish many.

Answer 1

HINT:

An important thing to consider:

If a permutation is written as a product of disjoint cycles

$$ \phi \colon (a_1 a_2 \ldots a_k) (b_1 b_2 \ldots, b_l) (c_1 c_2 \ldots c_m) \cdots$$

and $\sigma$ is another permutation, then $$\sigma \circ \phi \circ \sigma^{-1} \colon ( \sigma(a_1) \sigma(a_2) \ldots \sigma(a_k)) (\sigma(b_1) \sigma(b_2) \ldots \sigma(b_l)) (\sigma(c_1) \sigma(c_2) \ldots \sigma(c_m)) \cdots$$

Indeed, $\sigma \circ \phi \circ \sigma^{-1}$ takes $\sigma(a_i)$ to $\sigma(a_{i+1})$, and the others.

So now, if you want the permutations $\sigma$ that commute with $\phi$, you want the permutations given in cycle form to be the same. When does this happen? The solution is particularly easy when there exists only one cycle of length $>1$ in the decomposition of $\sigma$.

$\bf{Added:}$ From a cycle of length $m$ of $\phi$ we get a cycle of length $m$ of $\sigma \circ \phi\circ \sigma^{-1}$. Therefore:

If $\sigma \phi = \phi \sigma $ ( i.e. $\sigma \phi \sigma^{-1} = \phi$) then $\phi$ takes the set of elements of a cycle of length $m$ of $\phi$ to the set of elements of another cycle of length $m$ of $\phi$ (possibly the same cycle). So the total set $A$ breaks into a disjoint union of $A_m$, and each $A_m$ is invariant under $\sigma$. Let's look at the action of $\sigma$ on an $A_m$. Say $A_m$ consists of $k$ cycles of length $m$. Then we can write $$A_m = \{1, \ldots, k \} \times \{1, \ldots, m\}$$ The action of $\phi$ on $A_m$ is $$(i, j) \mapsto (i, j+1)$$

($+1$ is $\mod m$). Now the permutations of $A_m$ that commute with $\phi$ are

$$\sigma: (i, j) \mapsto ( t(i), j + k_i)$$ where $k_i\in \mathbb{Z}/m$, and $t$ is a permutation of $\{1, \ldots, k\}$.

So the centralizer of $\phi$ restricted to $A_m$ is a wreath product $\mathbb{Z}/m \wr S_k$.

Here is an example:

$\phi\colon (1,2,3,4)(5,6,7,8)\in S_8$. Then we can also write $\phi$ as $\phi\colon (7,8,5,6),(4,1,2,3)$. Therefore $$\sigma = \left( \begin{matrix} 1&2&3&4&5&6&7&8 \\ 7&8&5&6&4&1&2&3\end{matrix}\right)$$ commutes with $\phi$, since $\sigma \circ \phi\circ \sigma^{-1} = \phi$.

Answer 2

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$First compute the number of conjugates of $\sigma = (a_1a_2\dots a_k)(a_{k+1})(a_{k+2}) \dots (a_n)$. This is just the set of $k$-cycles, so there are $$ J = \frac{n \cdot (n-1) \cdot \dots \cdot (n - k + 1)}{k} $$ such conjugates.

Orbit/stabliser tells you that the centraliser of $\sigma$ has order $$ \frac{n!}{J} = k \cdot (n - k)! $$ So you see the centraliser is indeed the usual suspect $\Span{\sigma} \times T$, where $T \cong S_{n-k}$ is the subgroup that fixes $a_{1}, \dots, a_{k}$.