Let $G$ be a group and $C_G(g)$ denote the conjugacy class of $g$. In chapter 4.3 of Dummit and Foote, it is noted often that $\langle g\rangle \leq C_G(g)$ and this fact is used throughout some of the exercises (as a tool/hint) in the text.
The following is an example (chapter 4.3 page 124)
But $C_{Q_8}(i) = \{\pm i \}$, it is not possible to reach $\pm 1$ in the conjugation $gig^{-1}$.
In this example $C_G(\cdot)$ is the centralizer (page 59 chapter 2.3)
Can someone clarify the first example for me?
Best Answer
You are simply mistaken; $C_G(g)$ is denoting the centralizer of $g$ in $G$ throughout: $$C_G(g) = \{x\in G\mid xg=gx\}.$$
The reason this subgroup shows up in the consideration of conjugacy classes is that there is a bijection between the cosets of $C_G(g)$ in $G$ and the conjugates of $g$ in $G$.
Indeed, if $xC_G(g) = yC_G(g)$, then there exists $h\in C_G(g)$ such that $x=yh$. Then $$\begin{align}xgx^{-1} & =(yh)g(yh)^{-1}\\ &= yhgh^{-1}y^{-1}\\ & = yghh^{-1}y^{-1} &\text{(since }hg=gh\text{)}\\ &= ygy^{-1}. \end{align}$$ And if $zgz^{-1}=wgw^{-1}$, then $w^{-1}zg = gw^{-1}z$, hence $w^{-1}z\in C_G(g)$, so $zC_G(g) = wC_G(g)$.
Thus, the size of the conjugacy class of $g$ is the index of $C_G(g)$ in $G$.
If we let $C_g = \{xgx^{-1}\mid x\in G\}$ be the conjugacy class of $g$, this amounts to $$|C_g| = [G:C_G(g)]$$ which is a special case of the Orbit-Stabilizer Theorem.
And of course, every power of $g$ commutes with $g$, so $\langle g\rangle \subseteq C_G(g)$ for all $g\in G$.