Finding the centralizer of a permutation

I need to find the centralizer of the permutation $\sigma=(1 2 3 ... n)\in S_n$.
I know that:
$C_{S_n}(\sigma)=\left\{\tau \in S_n|\text{ } \tau\sigma\tau^{-1}=\sigma\right\}$
In other words, that the centralizer is the set of all the elements that commute with $\sigma$, and I also know that if two permutations have disjoint cycles it implies that they commute, but the thing is; there are no $\tau\in S_n$ s.t. $\tau$ and $\sigma$ have disjoint cycles, since $\sigma=(1 2 3...n)$.
So can I conclude that $\sigma$ does not commute with any other $\tau$ in $S_n$ (besides $id$ of course)?
I guess my question reduces to: is the second direction of the implication mentioned above also true? meaning, if two permutation commute, does it imply that they have disjoint cycles?
If the answer is no, how else can I find the $C_{S_n}(\sigma)$?

By the way, on related subject, I noticed that if an element $g$ of a group $G$ is alone it its conjugacy class, it commutes with all elements in $G$.
What does it mean, intuitively, for an element to share its conjugacy class with another element? does it mean it "almost" commute with everyone in the group?
Is it true that the bigger the conjugacy class, the lesser its members commutes with others in the group?

Solution 1:

Hint: conjugacy in $S_n$ leaves cycle types in tact: $\tau^{-1}(1 2 3 \dots n)\tau=(\tau(1) \tau(2)\tau(3) \cdots \tau(n))$.

Solution 2:

Let $S_n$ act on itself by conjugation. Let $g = (1,2,\ldots,n)$. The size of the orbit of $g$ in this action is its conjugacy class $\{ h^{-1} gh: h \in S_n\}$. The stabilizer subgroup of $g$ in this action is the set of elements $h$ in $S_n$ such that $h^{-1}gh=g$, i.e. the centralizer $C_{S_n}(g)$. By the orbit-stabilizer lemma, the size of the orbit equals the index of the stabilizer. The size of the orbit is the number of elements that have the same cycle structure as $g$, which is $(n-1)!$. Thus, the index of the centralizer is $(n-1)!$, whence the centralizer has $n! / (n-1)!=n$ elements. Thus the powers of $g$ exhaust all of $C_{S_n}(g)$.

A second proof is as follows. For the special case where $g=(1,2,\ldots,n)$, we can determine its centralizer in $S_n$ without using the orbit-stabilizer lemma. If $h^{-1}gh = g$, then $(h(1),h(2),\ldots,h(n)) = (1,2,\ldots,n)$. Now, $h(1)$ can be chosen in $n$ ways to be any of $1,2,\ldots,n$, but once $h(n)$ is chosen, the remaining elements $h(2),\ldots,h(n)$ are uniquely determined. In fact, if $h(1)=i$, then $h(2)=i+1$, and so on, and so $h$ is just a power of $g$. Thus, there are exactly $n$ different elements $h$ such that $h^{-1}gh=g$.

Solution 3:

Note that $|\sigma| = n$ so that $\sigma^n = 1$. So $$ \langle \sigma \rangle = \{1, \sigma, \sigma^2, \ldots, \sigma^{n-1}\} $$ forms a cyclic commutative subgroup of $S_n$ with order $|\langle \sigma \rangle| = n$. For any permutation $x$ of a symmetric group $S_n$ we have the very convenient formula: $$ |C(x)| = 1^{\alpha_1}2^{\alpha_2}\cdots n^{\alpha_n}\alpha_1!\alpha_2!\ldots\alpha_n!, $$ where $\alpha_i$ is the number of cycles in $x$ of length $i$. For your permutation $\alpha_1 = \alpha_2 = \cdots \alpha_{n-1} = 0$ and $\alpha_n = n$ so $|C(\sigma)| = n$. We have $|C(\sigma)| = |\langle \sigma \rangle| = n$, implying $C(\sigma) = \langle \sigma \rangle$.