Equivalence Classes

Notes

Group Theory

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} ⟹ 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} ⟹ 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{aligned} g_{a} \sim g_{b} ⟹ 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} (g^{- 1} g = g g^{- 1}) \\ g_{c} g_{b} g_{c}^{- 1} & = g_{a} ⟹ g_{b} \sim g_{a} \end{aligned}

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} ⟹ g_{b} = g_{1} g_{a} g_{1}^{- 1} and g_{b} \sim g_{c} ⟹ g_{c} = g_{2} g_{b} g_{2}^{- 1}

So, $g_{c} = g_{2} g_{b} g_{2}^{- 1} = (g_{2} g_{1}) g_{a} (g_{1}^{- 1} g_{2}^{- 1}) = g_{3} g_{a} g_{3}^{- 1} ⟹ g_{a} \sim g_{c}$ 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: $C_{i}$ , $i = {1, 2, \dots, k}$ .

Theorem 1.5

Consider $C$ , an equivalence class of group $G$ . For any $g \in G$ ,

g^{- 1} C g = C

Furthermore, any subgroup $S$ of $G$ , for which

g^{- 1} S g = S \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 $S g$ or $g S$ is equal to the order $| S |$ .

So, the number of elements in $g^{- 1} S g$ is the same as $| S |$ .

Therefore, each element of $g^{- 1} C g$ is also in $C$ , implying that $g^{- 1} C g = C$ .

Second Part

Consider some subset $S$ , such that $g^{- 1} S g = S$ $\forall g \in G$ , 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 $g_{2} \notin S$ .

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.

Subgroups and Cosets Invariant Subgroups