Definition 1   Let $G$ be a group. An equivalence relation $\sim$ on $G$ is called a group congruence if it is compatible with the group structure, ie. when the following holds

• $\forall a,b,a',b'\in G,\quad (a\sim a'\,\,\, {and} \,\,\, b\sim b') \Rightarrow ab \sim a'b'$
• $\forall a,b\in G,\quad a\sim b \Rightarrow a^{-1}\sim b^{-1}\,. $

Theorem 2   An equivalence relation $\sim$ is a group congruence if and only if there is a normal subgroup such that $$\forall a,b \in G, \quad a\sim b \Longleftrightarrow ab^{-1}\in H\,. $$

Proof. Let $H$ be a normal subgroup of $G$ and let $\sim_H$ be the equivalence relation $H$ defines in $G$ . To see that this equivalence relation is compatible with the group operation note that if $a'\sim_H a$ and $b'\sim_H b$ then there are elements $h_1$ and $h_2$ of $H$ such that $a' = ah_1$ and $b' = b h_2$ . Furthermore since $H$ is normal in $G$ there is an element $h_3\in H$ such that $h_1b = bh_3$ . Then we have
which gives that $a'b'\sim ab$ .

To prove the converse, assume that $\sim$ is an equivalence relation compatible with the group operation and let $H$ be the equivalence class of the identity $e$ . We will prove that $\sim \,= \,\sim_H$ . We first prove that $H$ is a normal subgroup of $G$ . Indeed if $a\sim e$ and $b \sim e$ then by the compatibility we have that $ab^{1} \sim ee^{-1}$ , that is $ab^{-1}\sim e$ ; so that $H$ is a subgroup of $G$ . Now if $g\in G$ and $h\in H$ we have

Therefore $H$ is a normal subgroup of $G$ . Now consider two elements $a$ and $b$ of $G$ . To finish the proof observe that for $a,b\in G$ we have
and