Trying to follow a proof from Dummit and Foote Section 4.5 - Sylow's Theorem The proof is as follows:

If $p | q - 1$, we shall see in Chapter 5 that there is a unique non-abelian group of order $pq$ (in which, necessarily $n_p = q$. We can prove the existence of this group now. Let $Q$ be a Sylow q-subgroup of the symmetric group of degree $q$, $S_q$. By Exercise 34 in section 3, $|N_{S_q}(Q)| = q(q - 1)$. By assumption, $p | q- 1$ so by Cauchy's theorem $N_{S_Q}(Q)$ has a subgroup $P$ of order $p$. By Corollary 15 in Section 3.2 (which says that if $P$ is a subgroup of the normalizer $N_{S_Q}(Q)$, then $PQ$ is a subgroup), $PQ$ is a subgroup of order $pq$. Since $C_{S_q}(Q) = Q$ (Example 2, section 3), $PQ$ is non-abelian. The essential ingredient in the uniqueness proof of $PQ$ is Theorem 17, on the cyclicity of $Aut(Z_q)$.

Now, I understand the rest of the proof. What I don't understand of how $C_{S_q}(Q) = Q$. I don't see how example 2 in section 3 (there are two "example twos" in section 3) helps either. For reference, they are:

(#1) If $|G| > 1$ then, unlike action by left multiplication, $G$ does not ac transitively on itself by conjugation because $\{1\}$ is always a conjugacy class (i.e. an orbit for this action). More generally, the one element subset $\{a\}$ is a conjugacy class if and only if $gag^{-1} = a$ for all $g \in G$ if and only if $a$ is in the center of $G$.

The second "example 2" is

In any group $G$ we have $<g>$ is a subgroup of $C_G(g)$. This observation helps to minimize ocmputationg of conjugacy classes. For example, in the quaternion group $Q_8$ we see that $\langle i\rangle$ is a subgroup of $C_{Q_8}(i)$ which is a ssubgroup of $Q_8$. Since $i \not\in Z(Q_8)$, and $|Q_8: \langle i\rangle| = 2$, we must have $C_{Q_8}(i) = \langle i\rangle$. Thus, $i$ has precisely 2 conjugates in $Q_8$, namely $i$ and $-i = kik^{-1}$. The other conjugacy classes in $Q_8$ are determined similarly and are (I omit these since it won't help to list the conjugacy classes of $Q_8$.

I understand what the second "example 2" is trying to say. the index being $2$ implies $|\langle i\rangle|$ is $4$, so by Lagrange's theorem 4 divides the centralizer of i and the centeralizer divides $Q_8$, which is has order $8$. So the only options for its order are $4$ and $8$ Since $i$ is not in the center of $Q_8$, then its order must be $4$.
Now, since $Q$ is of prime order $q$, then we know $Q = \langle x\rangle$ for some element $x \in S_q$. So if we tried to do a similar tactic, we get $Q = \langle x\rangle$ is a subgroup of $C_{S_q}(Q)$ which is a subgroup of $S_q$. But for one, this gives us much weaker bounds: namely q divides $|C_{S_q}(Q)|$ and $|C_{S_q}(Q)|$ divides $q!$.
We could try to use the fact that $C_{S_q}(Q)$ is a subgroup of $N_{S_q}(Q)$ instead. Then $q$ divides $|C_{S_q}(Q)|$ and $|C_{S_q}(Q)|$ divides $q(q - 1)$. Much better! And then the index of $|N_{S_q}(Q):C_{S_q}(Q)|$ is....actually what is it? We know from another theorem of the book that the quotient group is isomorphic to a subgroup of $\mathrm{Aut}(Q)$, and since $Q$ is cyclic and prime, its order if $q - 1$. So the index divides $q - 1$. If the index is $q - 1$, then $|C_{S_q}(Q)| = q$ and we are done. But we would still have everything work out if the index was $1$. $1$ still divides $q - 1$.
So basically, my problem comes down to the fact that I don't know how to calculate $|N_{S_q}(Q):C_{S_q}(Q)|$. Or if there's some other approach to proving that $C_{S_q}(Q) = Q$, then that would be much appreciated too.
I also know $x$ is not in the center since for $n \geq 3$ the only element in the center of $S_n$ is the identity. If we showed that there was an element in $N_{S_q}(Q)$ not in $C_{S_q}(Q)$ then we would be done also.
 A: I realized that the book has actually provided us with a direct way to calculate the order of centralizers for some $q$-cycles, which $x$ would be since it has order $q$ which is prime. And the center of $C_{S_q}(Q)$ is equivalent to the center of $C_{S_q}(x)$ because $Q = \langle x\rangle$. Anyways, the formula for an arbitrary m cycle in $S_n$ is $m \cdot (n - m)!$, so for $q$ this would be....$q$.
The derivation of this formula is as follows:
The number of conjugates of an $m$-cycle in $S_n$ is $\frac{n \cdot (n - 1) \dotsm (n - m + 1)}{m}$. This is due to first permuting $m$ items out of $n$, and then dividing by $m$ because one "turn" in the cycle provides the same permutation, and there are $m$ possible "turns" before you're back to the original permutation.
Then due to a theorem in the book, this is the index of the centralizer with respect to $S_n$. Basically, two elements in $S_n$ are conjugate if and only if they have the same cycle type. So the number of $m$-cycles is equal to the conjugacy class of the conjugacy action for $m$-cycles. In general, there is a bijection between an orbit of an action and the index of $|G: G_a|$, where $G$ is an arbitrary group and $G_a$ is the stabilizer of $a$, where $a$ is a representative of the orbit. For conjugacy classes, $G_a = C_G(a)$, since you have that it's in the stabilizer if and only if $gag^{-1} = a$.
So anyways,
$\frac{|S_n|}{|C_{S_n}(\sigma)|}$ equals the number above, and since $|S_n| = n!$, after some algebra we get $|C_{S_n}(\sigma)| = m \cdot (n - m)!$.
