Congruence Relation on Group induces Normal Subgroup

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $\RR$ be a congruence relation for $\circ$.

Let $H = \eqclass e \RR$, where $\eqclass e \RR$ is the equivalence class of $e$ under $\RR$.


Then:

$\struct {H, \circ \restriction_H}$ is a normal subgroup of $G$

where $\circ \restriction_H$ denotes the restriction of $\circ$ to $H$.


Proof

We are given that $\RR$ is a congruence relation for $\circ$.

From Equivalence Relation is Congruence iff Compatible with Operation, we have:

$\forall u \in G: x \mathrel \RR y \implies \paren {x \circ u} \mathrel \RR \paren {y \circ u}, \paren {u \circ x} \mathrel \RR \paren {u \circ y}$


Proof of being a Subgroup

We show that $H$ is a subgroup of $G$.


First we note that $H$ is not empty:

$e \in H \implies H \ne \O$


Then we show $H$ is closed:

\(\ds x, y\) \(\in\) \(\ds H\)
\(\ds \leadsto \ \ \) \(\ds e\) \(\RR\) \(\ds x\)
\(\, \ds \land \, \) \(\ds e\) \(\RR\) \(\ds y\) by definition of $H$
\(\ds \leadsto \ \ \) \(\ds \paren {e \circ e}\) \(\RR\) \(\ds \paren {x \circ y}\) $\RR$ is compatible with $\circ$
\(\ds \leadsto \ \ \) \(\ds x \circ y\) \(\in\) \(\ds H\) by definition of $H$


Next we show that $x \in H \implies x^{-1} \in H$:

\(\ds x\) \(\in\) \(\ds H\)
\(\ds \leadsto \ \ \) \(\ds e\) \(\RR\) \(\ds x\) Definition of $H$
\(\ds \leadsto \ \ \) \(\ds \paren {x^{-1} \circ e}\) \(\RR\) \(\ds \paren {x^{-1} \circ x}\) $\RR$ is compatible with $\circ$
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\RR\) \(\ds e\) Group properties
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\in\) \(\ds H\) Definition of $H$

Thus by the Two-Step Subgroup Test, $H$ is a subgroup of $G$.

$\Box$


Proof of Normality

Next we show that $H$ is normal in $G$.

Thus:

\(\ds x\) \(\in\) \(\ds H\)
\(\ds \leadsto \ \ \) \(\ds e\) \(\RR\) \(\ds h\) for some $h \in H$, by definition of $H$
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ e}\) \(\RR\) \(\ds \paren {x \circ h}\) $\RR$ is compatible with $\circ$
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ e \circ x^{-1} }\) \(\RR\) \(\ds \paren {x \circ h \circ x^{-1} }\) $\RR$ is compatible with $\circ$
\(\ds \leadsto \ \ \) \(\ds e\) \(\RR\) \(\ds \paren {x \circ h \circ x^{-1} }\) Group properties
\(\ds \leadsto \ \ \) \(\ds x \circ h \circ x^{-1}\) \(\in\) \(\ds H\) Definition of $H$


Thus from Subgroup is Normal iff Contains Conjugate Elements, we have that $H$ is normal.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Congruence_Relation_on_Group_induces_Normal_Subgroup&oldid=568000"