correspondence of normal subgroups and group congruences
ProofThatTheEquivalenceRelationRelationDefinedByANormalSubgroupIsCompatibleWithTheGroupOperation
2005-10-22 00:15:39
2006-02-01 07:30:19
Theorem
quotient group
group congruence
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We start with a definition.
\begin{defn}
Let $G$ be a group. An equivalence relation $\sim$ on $G$ is
called a \emph{group congruence} if it is compatible with the
group structure, ie. when the following holds
\begin{itemize}
\item $\forall a,b,a',b'\in G,\quad (a\sim a'\,\,\, \text{and}
\,\,\, b\sim b') \Rightarrow ab \sim a'b'$
\item $\forall a,b\in G,\quad a\sim b \Rightarrow a^{-1}\sim b^{-1}\,. $
\end{itemize}
\end{defn}
So a group congruence is a \PMlinkid{semigroup congruence}{3403}
that additionally preserves the unary operation of taking inverse.
It turns out that group congruences correspond to normal subgroups:
\begin{thm}
An equivalence relation $\sim$ is a group congruence
if and only if there is a normal subgroup such that
$$\forall a,b \in G, \quad a\sim b \Longleftrightarrow ab^{-1}\in H\,. $$
\end{thm}
\begin{proof}
Let $H$ be a \emph{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
\begin{align*}
a'b'&=ah_1bh_2 \\
&=abh_3h_2
\end{align*}
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
\begin{align*}
h\sim e &\Rightarrow ghg^{-1} \sim geg^{-1}\\
&\Rightarrow ghg^{-1} \sim e\\
&\Rightarrow ghg^{-1} \in H\,.
\end{align*}
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
\begin{align*}
a\sim_H b &\Rightarrow ab^{-1}\in H\\
&\Rightarrow ab^{-1} \sim e\\
&\Rightarrow (ab^{-1})b \sim eb\\
&\Rightarrow a \sim b
\end{align*}
and
\begin{align*}
a\sim b &\Rightarrow ab^{-1} \sim bb^{-1}\\
&\Rightarrow ab^{-1} \sim e\\
&\Rightarrow a\sim_H b\,.
\end{align*}
\end{proof}