Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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
$[a]=\{x\in X|a\sim x\}$

share|improve this question
Equivalence relations are symmetric. –  Andreas Blass Aug 4 '14 at 14:12
Right, from this?: $\forall a,b\in X; a\sim b \Rightarrow b\sim a$ –  imME 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

2 Answers 2

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


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 $b \sim a$.

share|improve this answer

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.