Clarificación en el ejemplo de la sección 4.3 Dummit and Foote (3Ed) Hola tengo una problema para entender el ejemplo 2 de la sección 4.3 que dice lo siguiente:
(Preferí no traducirlo para evitar ambigüedades)
In any group $G$ we have $\langle g \rangle \leq C_{G}(g)$; this observation helps to minimize computations
of conjugacy classes. For example, in the quatemion group $Q_{8}$ we see
that $\langle i \rangle \leq C_{Q_8}(i) \leq  Q_{8}$. 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:
$$\{ 1\}, \{- 1 \}, \{\pm i\}, \{\pm j\}, \{\pm k\}$$
The first two classes form $Z(Q_{8})$ and the class equation for this group is
$$|Q_{8}| =  2 + 2 + 2 + 2$$
Me duda radica en el paso que dicen que $C_{Q_{8}}(i) = \langle i \rangle$. Supongo que usan el hecho de que $\langle g \rangle \leq C_{G}(g)$ pero no me queda clara la consecuencia. Estaría muy agradecido si me pueden ayudar a entender este paso

[EDIT] - Translation to English:

Hi, I have a problem understanding Example 2 in Section 4.3 that says the following:
(I preferred not to translate it to avoid ambiguities)
In any group $G$ we have $\langle g \rangle \leq C_{G}(g)$; this observation helps to minimize computations
of conjugacy classes. For example, in the quatemion group $Q_{8}$ we see
that $\langle i \rangle \leq C_{Q_8}(i) \leq  Q_{8}$. 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:
$$\{ 1\}, \{- 1 \}, \{\pm i\}, \{\pm j\}, \{\pm k\}$$
The first two classes form $Z(Q_{8})$ and the class equation for this group is
$$|Q_{8}| =  2 + 2 + 2 + 2$$
My question lies in the step where they say that $ C_ {Q_ {8}} (i) = \langle i \rangle $. I suppose they use the fact that $ \langle g \rangle \leq C_ {G} (g) $ but the consequence is not clear to me. I would be very grateful if you can help me understand this step.
 A: $C_{Q_8}(i)$ means the set of all elements $x \in Q_8$ such that $ix = xi$.  As you said, $C_{Q_8}(i)$ contains $\langle i \rangle$ (because any element commutes with all its powers).  We need to check there are no other elements in $C_{Q_8}(i)$.  The elements of $Q_8$ outside $\langle i \rangle$ are $j, k, -j, -k$ and you can easily check that they do not satisfy $ix = xi$.
A: We know that $\langle i\rangle\le C_{Q_8}(i)$. On the other hand, we know that $i$ is not in the centre of $Q_8$, so there are elements of $Q_8$ that do not commute with $i$; those elements cannot be in $C_{Q_8}(i)$, so $C_{Q_8}(i)\ne Q_8$. The fact that $[Q_8:\langle i\rangle]=2$ is prime means that there cannot be any group $G$ such that $\langle i\rangle<G<Q_8$, so we must have $\langle i\rangle=C_{Q_8}(i)$.
Essentially the same argument without reference to the index of $\langle i\rangle$ in $Q_8$ could go like this. Let $n=|C_{Q_8}(i)|$; we have $\langle i\rangle\le C_{Q_8}(i)\le Q_8$, so $4\mid n$ and $n\mid 8$. Moreover, we know that $C_{Q_8}(i)\ne Q_8$, so $n<8$, and the only possibility is that $n=4$. Since we already know that $\langle i\rangle\le C_{Q_8}(i)$, this implies that $\langle i\rangle=C_{Q_8}(i)$.
