# Relation of Congruence Modulo a Subgroup in a Group

Let $H$ be a subgroup of a group $G$. If the element $a$ of $G$ belongs to the right coset $Hb$, i.e. if $a \in Hb$, i.e., if $a{b^{ – 1}} \in H$, then it is said that $a$ is congruent to $b$ modulo $H$.

Definition: Let $H$ be a subgroup of a group $G$. For $a,b \in G$ we say that $a$ is congruent to $b\bmod H$ if and only if $a{b^{ – 1}} \in H$.

Symbolically, it can be expressed as $a \equiv b\left( {\bmod H} \right)$ if $a{b^{ – 1}} \in H$.

Theorem: The relation of congruency in a group $G$ is defined by $a \equiv b\left( {\bmod H} \right)$ if and only if $a{b^{ – 1}} \in H$ is an equivalence relation.

Proof:
(i) Reflexivity: Let $a \in G$ then $a{a^{ – 1}} = e \in H$ because $H$ is a subgroup of $G$.

Hence $a \equiv a\left( {\bmod H} \right)$ for all $\forall a \in G$. The relation is reflexive.

(ii) Symmetry:
$a \equiv b\left( {\bmod H} \right)$
$\Rightarrow a{b^{ – 1}} \in H$
$\Rightarrow {\left( {a{b^{ – 1}}} \right)^{ – 1}} \in H$
$\Rightarrow b{a^{ – 1}} \in H$
$\Rightarrow b \equiv a\left( {\bmod H} \right)$

Hence the relation is symmetric.

(iii) Transitivity:
$a \equiv b\left( {\bmod H} \right)\,\,\,\,\,{\text{and}}\,\,\,\,\,\,b \equiv a\left( {\bmod H} \right)$
$\Rightarrow a{b^{ – 1}} \in H\,\,\,\,\,\,\,\,\,\,\,{\text{and}}\,\,\,\,\,\,\,b{a^{ – 1}} \in H$
$\Rightarrow \left( {a{b^{ – 1}}} \right)\left( {b{c^{ – 1}}} \right) \in H$
$\Rightarrow a\left( {{b^{ – 1}}b} \right){c^{ – 1}} \in H$
$\Rightarrow a{c^{ – 1}} \in H$
$\Rightarrow a \equiv c\left( {\bmod H} \right)$

Hence the relation is transitive.

Thus the relation congruence $\bmod H$ is an equivalence relation in $G$.