# 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

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

• $\forall a,b\in G,\quad a\sim b\Rightarrow a^{-1}\sim b^{-1}\,.$

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 $\sim$ 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 $a^{\prime}\sim_{H}a$ and $b^{\prime}\sim_{H}b$ then there are elements $h_{1}$ and $h_{2}$ of $H$ such that $a^{\prime}=ah_{1}$ and $b^{\prime}=bh_{2}$. Furthermore since $H$ is normal in $G$ there is an element $h_{3}\in H$ such that $h_{1}b=bh_{3}$. Then we have

 $\displaystyle a^{\prime}b^{\prime}$ $\displaystyle=ah_{1}bh_{2}$ $\displaystyle=abh_{3}h_{2}$

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 $ab^{1}\sim ee^{-1}$, 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

 $\displaystyle h\sim e$ $\displaystyle\Rightarrow ghg^{-1}\sim geg^{-1}$ $\displaystyle\Rightarrow ghg^{-1}\sim e$ $\displaystyle\Rightarrow ghg^{-1}\in H\,.$

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

 $\displaystyle a\sim_{H}b$ $\displaystyle\Rightarrow ab^{-1}\in H$ $\displaystyle\Rightarrow ab^{-1}\sim e$ $\displaystyle\Rightarrow(ab^{-1})b\sim eb$ $\displaystyle\Rightarrow a\sim b$

and

 $\displaystyle a\sim b$ $\displaystyle\Rightarrow ab^{-1}\sim bb^{-1}$ $\displaystyle\Rightarrow ab^{-1}\sim e$ $\displaystyle\Rightarrow a\sim_{H}b\,.$

Title correspondence of normal subgroups and group congruences CorrespondenceOfNormalSubgroupsAndGroupCongruences 2013-03-22 15:32:52 2013-03-22 15:32:52 Dr_Absentius (537) Dr_Absentius (537) 7 Dr_Absentius (537) Theorem msc 20-00 group congruence