# centralizer of a k-cycle

###### Theorem 1.

Let $\sigma $ be a $k$-cycle in ${S}_{n}$. Then the centralizer^{} of $\sigma $ is

$${C}_{{S}_{n}}(\sigma )=\{{\sigma}^{i}\tau \mid 0\le i\le k-1,\tau \in {S}_{n-k}\}$$ |

where ${S}_{n\mathrm{-}k}$ is the subgroup^{} of ${S}_{n}$ consisting of those permutations^{} that fix all elements appearing in $\sigma $.

###### Proof.

This is fundamentally a counting argument. It is clear that $\sigma $ commutes with each element in the set given, since $\sigma $ commutes with powers of itself and also commutes with disjoint permutations. The size of the given set is $k\cdot (n-k)!$. However, the number of conjugates of $\sigma $ is the index of ${C}_{{S}_{n}}(\sigma )$ in ${S}_{n}$ by the orbit-stabilizer theorem, so to determine $|{C}_{{S}_{n}}(\sigma )|$ we need only count the number of conjugates of $\sigma $, i.e. the number of $k$-cycles.

In a $k$-cycle $({a}_{1}\mathrm{\dots}{a}_{k})$, there are $n$ choices for ${a}_{1}$, $n-1$ choices for ${a}_{2}$, and so on. So there are $n(n-1)\mathrm{\cdots}(n-k+1)$ choices for the elements of the cycle. But this counts each cycle $k$ times, depending on which element appears as ${a}_{1}$. So the number of $k$-cycles is

$$\frac{n(n-1)\mathrm{\cdots}(n-k+1)}{k}$$ |

Finally,

$$n!=|{S}_{n}|=\frac{n(n-1)\mathrm{\cdots}(n-k+1)}{k}|{C}_{{S}_{n}}(\sigma )|$$ |

so that

$$|{C}_{{S}_{n}}(\sigma )|=\frac{k\cdot n!}{n(n-1)\mathrm{\cdots}(n-k+1)}=k\cdot (n-k)!$$ |

and we see that the elements in the list above must exhaust ${C}_{{S}_{n}}(\sigma )$. ∎

Title | centralizer of a k-cycle |
---|---|

Canonical name | CentralizerOfAKcycle |

Date of creation | 2013-03-22 17:18:00 |

Last modified on | 2013-03-22 17:18:00 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 6 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 20M30 |