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.


(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 $$G$$.