correspondence of normal subgroups and group congruences
We start with a definition.
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

•
$\forall a,b,{a}^{\prime},{b}^{\prime}\in G,(a\sim {a}^{\prime}\text{and}b\sim {b}^{\prime})\Rightarrow ab\sim {a}^{\prime}{b}^{\prime}$

•
$\forall a,b\in G,a\sim b\Rightarrow {a}^{1}\sim {b}^{1}.$
So a group congruence is a http://planetmath.org/node/3403semigroup^{} congruence that additionally preserves the unary operation of taking inverse^{}.
It turns out that group congruences correspond to normal subgroups^{}:
Theorem 2.
An equivalence relation $\mathrm{\sim}$ is a group congruence if and only if there is a normal subgroup such that
$$\forall a,b\in G,a\sim b\u27faa{b}^{1}\in H.$$ 
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 ${a}^{\prime}{\sim}_{H}a$ and ${b}^{\prime}{\sim}_{H}b$ then there are elements ${h}_{1}$ and ${h}_{2}$ of $H$ such that ${a}^{\prime}=a{h}_{1}$ and ${b}^{\prime}=b{h}_{2}$. Furthermore since $H$ is normal in $G$ there is an element ${h}_{3}\in H$ such that ${h}_{1}b=b{h}_{3}$. Then we have
${a}^{\prime}{b}^{\prime}$  $=a{h}_{1}b{h}_{2}$  
$=ab{h}_{3}{h}_{2}$ 
which gives that ${a}^{\prime}{b}^{\prime}\sim ab$.
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 $a{b}^{1}\sim e{e}^{1}$, that is $a{b}^{1}\sim e$; so that $H$ is a subgroup of $G$. Now if $g\in G$ and $h\in H$ we have
$h\sim e$  $\Rightarrow gh{g}^{1}\sim ge{g}^{1}$  
$\Rightarrow gh{g}^{1}\sim e$  
$\Rightarrow gh{g}^{1}\in H.$ 
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
$a{\sim}_{H}b$  $\Rightarrow a{b}^{1}\in H$  
$\Rightarrow a{b}^{1}\sim e$  
$\Rightarrow (a{b}^{1})b\sim eb$  
$\Rightarrow a\sim b$ 
and
$a\sim b$  $\Rightarrow a{b}^{1}\sim b{b}^{1}$  
$\Rightarrow a{b}^{1}\sim e$  
$\Rightarrow a{\sim}_{H}b.$ 
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Title  correspondence of normal subgroups and group congruences 

Canonical name  CorrespondenceOfNormalSubgroupsAndGroupCongruences 
Date of creation  20130322 15:32:52 
Last modified on  20130322 15:32:52 
Owner  Dr_Absentius (537) 
Last modified by  Dr_Absentius (537) 
Numerical id  7 
Author  Dr_Absentius (537) 
Entry type  Theorem 
Classification  msc 2000 
Defines  group congruence 