If $G$ is a finite group acting on itself via conjugation, Dummit and Foot states in Proposition 6 on page 123 that the number of conjugates of an element $s$ of $G$ is given by $[G:C_G(s)]$, however conceptually I am having trouble believing this fact. Since
$$
C_G(s) = \{ g \in G : gsg^{-1} = s \}
$$
it seems like $C_G(s)$ is the subset of $G$ which acts as the identity on the element $s$, so if I wanted to determine how many elements were in the set $[s]$ I would need to first take the number of elements in $G$ and then subtract away all of the elements from $C_G(s)$, excluding $s$ itself. Doing so would leave me with elements $x \in G$ for which $xsx^{-1} \neq s$, so my thought is that each of these $x$'s would give elements in $[s]$ by conjugation with $s$.
I am sure that there are $x_1$ and $x_2$ in $G$ which would give the same element in the conjugacy class $x_1sx_1^{-1} = x_2sx_2^{-1}$, so it's not accurate to say that the number of elements in $[s]$ would be $|G| – |C_G(s)| + 1$, however I don't understand conceptually how we can say that $[G:C_G(s)]$ gives the number of elements in the conjugacy class of $s$.
Best Answer
This is actually a special case of Orbit-Stabiliser Theorem (You may refer Proposition 2 in page 114).
One of the way to look at it is using the bijection between $[s]$ and $\{gC_G(s):g\in G\}$ defined by $gsg^{-1}\mapsto gC_G(s)$.
Note that for $g,h \in G$, $$gsg^{-1}=hsh^{-1}\iff gC_G(s)=hC_G(s).$$ So the size of $[s]$, which is the number of distinct conjugates of $s$, will be equal to the number of distinct cosets of $C_G(s)$ in $G$, which is $[G:C_G(s)]$.