Congruence Relation induces Normal Subgroup

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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:

$(1): \quad \struct {H, \circ \restriction_H}$ is a normal subgroup of $G$
$(2): \quad \RR$ is the equivalence relation $\RR_H$ defined by $H$
$(3): \quad \struct {G / \RR, \circ_\RR}$ is the subgroup $\struct {G / H, \circ_H}$ of the semigroup $\struct {\powerset G, \circ_\PP}$.


Proof

Proof of Normal Subgroup

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

From Congruence Relation 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.

$\blacksquare$


Proof of Equality of Relations

Let $\RR_N$ be the equivalence defined by $N$.

Then:

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


But from Congruence Class Modulo Subgroup is Coset:

$x \mathrel {\RR_N} y \iff x^{-1} \circ y \in N$

Thus:

$\RR = \RR_N$

$\blacksquare$


Proof that Quotient Structure is Quotient Group

Let $\eqclass x \RR \in G / \RR$.

By Congruence Relation on Group induces Normal Subgroup:

$\eqclass x \RR = x N$

where $x N$ is the (left) coset of $N$ in $G$.


Similarly, let $y N \in G / N$.

Then from Normal Subgroup induced by Congruence Relation defines that Congruence:

$y N = \eqclass x \RR$

where:

$\eqclass x \RR$ is the equivalence class of $y$ under $\RR$
$\RR$ is the equivalence relation defined by $N$.


Hence the result.

$\blacksquare$


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