correspondence of normal subgroups and group congruences

We start with a definition.

Definition 1.

Let $G$ be a group. An equivalence relation $\sim$ on $G$ iscalled a group congruence if it is compatible with thegroup 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 congruencethat 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 congruenceif and only if there is a normal subgroup such that

$$\forall a,b\in G,a\sim b⟺a{b}^{-1}\in H.$$

Proof.

Let $H$ be a normal subgroup of $G$ and let ${\sim }_H$ be theequivalence relation $H$ defines in $G$. To see that thisequivalence relation is compatible with the group operation notethat 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 anequivalence relation compatible with the group operation and let$H$ be the equivalence class of the identity $e$. We will provethat $\sim ={\sim }_H$. We first provethat $H$ is a normal subgroup of $G$. Indeed if $a\sim e$ and $b\sim e$ then by the compatibility we have that $a{b}^1\sim e{e}^{-1}$, thatis $a{b}^{-1}\sim e$; 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 twoelements $a$ 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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