# Why this equivalence class?

Let $H$ be a subgroup of a group $G$. Why is the equivalence class of $a\in G$ under right congruence, $\{ x\in G | x\equiv_r a\}$?

Shouldn't it be $\{x\in G|a\equiv_r x\}$? Because

The equivalence class of an element $a$ is defined as the set
$[a]=\{x\in X|a\sim x\}$

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Equivalence relations are symmetric. – Andreas Blass Aug 4 '14 at 14:12
Right, from this?: $\forall a,b\in X; a\sim b \Rightarrow b\sim a$ – Mill Aug 4 '14 at 14:13
Andreas's point is that $\{ x\in G | x\equiv_r a\}$ and $\{x\in G|a\equiv_r x\}$ are always equal, because $x\equiv_r a$ holds if and only if $a\equiv_r x$ holds, because $\equiv_r$ is symmetric. – MJD Aug 4 '14 at 16:10

$$x\equiv_r a \text{ is equivalent to: } a\equiv_r x$$

EDIT:

Given a set $X$ and an equivalence relation $\sim$ on $X$:

For every two elements $a$ and $b$ in $X$, if $a \sim b$, then $b \sim a$.

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A relation is equivalent, so $\forall a,b\in X; a\sim b\Rightarrow b\sim a$.

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