Relation of Congruence Modulo a Subgroup in a Group

Let H be a subgroup of a group G. If the element a of G belong 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 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.