Equivalence Classes
Definition 1.9 – Equivalence
An element $g_{a} \in G$ is equivalent to another element $g_{b} \in G$ ($g_{a} \sim g_{b}$) if there exists a third element $g_{c} \in G$ such that an equivalence transformation can be performed.
$$ g_{a} \sim g_{b} \implies g_{b} = g_{c} g_{a} g_{c}^{-1} $$Important Equivalence Properties
First property : every element is equivalent to itself.
$$ g_{a} \sim g_{a} \implies g_{a} = e g_{a} e^{-1} = e g_{a} e = g_{a} $$Second property : if $g_{a} \sim g_{b}$, then $g_{b} \sim g_{a}$.
$$ \begin{align} g_{a} \sim g_{b} \implies \qquad\qquad g_{b} &= g_{c} g_{a} g_{c}^{-1} \\ g_{c^{-1}} g_{b} &= g_{a} g_{c^{-1}} \\ g_{c}^{-1} g_{b} g_{c} &= g_{a} \qquad (g^{-1}g = gg^{-1}) \\ g_{c} g_{b} g_{c}^{-1} &= g_{a} \qquad\qquad \implies g_{b} \sim g_{a} \end{align} $$Third Property : if $g_{a} \sim g_{b}$ and $g_{b} \sim g_{c}$, then $g_{a} \sim g_{c}$.
$$ g_{a}\sim g_{b} \implies g_{b} = g_{1} g_{a} g_{1}^{-1} \quad \text{and} \quad g_{b} \sim g_{c} \implies g_{c} = g_{2}g_{b}g_{2}^{-1} $$So, $g_{c} = g_{2} g_{b} g_{2}^{-1} = (g_{2}g_{1})g_{a}(g_{1}^{-1}g_{2}^{-1}) = g_{3} g_{a} g_{3}^{-1} \implies g_{a} \sim g_{c}$ with $g_{3} = (g_{2}g_{1})$.
Definition 1.10 – Equivalence Class
An equivalence class is a maximal subset of mutually equivalent elements of some group $G$. We denote the $k$ equivalence classes of a group: $\cal{C}_{i}$, $i = \{1,2,\dots, k\}$.
Theorem 1.5
Consider $\cal{C}$, an equivalence class of group $G$. For any $g \in G$,
$$ g^{-1} \cal{C} g = \cal{C} $$Furthermore, any subgroup $S$ of $G$, for which
$$ g^{-1} S g = S \quad \forall g \in G $$Is composed of complete equivalence classes.
Proof
First Part
We know, through theorem 1.2, that, for a subset $S$ of group $G$, the number of distinct elements in $Sg$ or $gS$ is equal to the order $\lvert S \rvert$.
So, the number of elements in $g^{-1} Sg$ is the same as $\lvert S \rvert$.
Therefore, each element of $g^{-1}\cal{C}g$ is also in $\cal{C}$, implying that $g^{-1} \cal{C} g = \cal{C}$.
Second Part
Consider some subset $S$, such that $g^{-1}Sg = S$ $\forall g \in G$, and that $S$ is not composed of equivalence classes.
This means that there exists two equivalent elements $g_{1},g_{2} \in G$ such that $g_{1} \in S$ and $g_{2} \not\in S$.
Consequently, $g^{-1} g_{1} g = g_{2}$ for some $g \in G$ which implies that $g_{2} \in S$ since $g_{1} \in g^{-1} S g$. This contradicts $g_{2} \in S$. Therefore, if $g^{-1} S g = S$ $\forall g \in G$, $S$ must be composed of complete equivalence classes.