# A problem involving centralizer and cycles

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 $\langle a\rangle$.

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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!) – Alexander Gruber Nov 12 '12 at 0:20

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 $(2)$.