2
$\begingroup$

I have just begun studying the group $S_n$ and I am having trouble with the cycle notation so this problem seems a bit hard. Any help will be deeply appreciated:

If $a \in S_n$ is an $n$-cycle or an $(n-1)$-cycle then the centralizer of $a$ is equal to the cyclic group $\langle a\rangle$.

  • 0
    Hint: if you conjugate$a$cycle $a$ by another cycle $b$, all you have to do is apply the $b$ permutation to the numbers in cycle $a$. For example, $(1234)^{(12)}=(2134)$. (You should prove this if you plan to use it!)2012-11-12

1 Answers 1

1

Let $a$ be an $m$-cycle. The number of conjugates of $a$ is the number of $m$-cycles, which is: $ \frac{n(n-1)\cdots(n-m+1)}{m} $

By the orbit-stabilizer theorem, the number of conjugates of $a$ is: $ \frac{\left|S_n\right|}{\left|C_{S_n}(a)\right|} = \frac{n!}{\left|C_{S_n}(a)\right|} $

Where $C_{S_n}(a)$ is the centralizer of $a$ in $S_n$. It follows that: $ \left|C_{S_n}(a)\right| = m (n-m)! \tag{1} $

Since $a$ commutes with all elements in the cyclic group $\langle a \rangle$, we have: $ \langle a \rangle \le C_{S_n}(a) \tag{2} $

For $m \in \{n-1, n\}$, $(1)$ gives $\left|C_{S_n}(a)\right| = m$. Since $\left|\langle a \rangle\right| = m$, we have equality in $(2)$.