Congruence Relation on Group induces Normal Subgroup

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Theorem

Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $\mathcal R$ be a congruence relation for $\circ$.

Let $H = \left[\!\left[{e}\right]\!\right]_\mathcal R$, where $\left[\!\left[{e}\right]\!\right]_\mathcal R$ is the equivalence class of $e$ under $\mathcal R$.


Then:

$\left({H, \circ \restriction_H}\right)$ is a normal subgroup of $G$

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


Proof


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

From Congruence Relation iff Compatible with Operation, we have:

$\forall u \in G: x \mathop {\mathcal R} y \implies \left({x \circ u}\right) \mathop {\mathcal R} \left({y \circ u}\right), \left({u \circ x}\right)\mathop {\mathcal R} \left({u \circ y}\right)$


Proof of being a Subgroup

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


First we note that $H$ is not empty:

$e \in H \implies H \ne \varnothing$


Then we show $H$ is closed:

\(\displaystyle x, y\) \(\in\) \(\displaystyle H\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle e\) \(\mathcal R\) \(\displaystyle x\) $\quad$ $\quad$
\(\, \displaystyle \land \, \) \(\displaystyle e\) \(\mathcal R\) \(\displaystyle y\) $\quad$ by definition of $H$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({e \circ e}\right)\) \(\mathcal R\) \(\displaystyle \left({x \circ y}\right)\) $\quad$ $\mathcal R$ is compatible with $\circ$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x \circ y\) \(\in\) \(\displaystyle H\) $\quad$ by definition of $H$ $\quad$


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

\(\displaystyle x\) \(\in\) \(\displaystyle H\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle e\) \(\mathcal R\) \(\displaystyle x\) $\quad$ by definition of $H$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({x^{-1} \circ e}\right)\) \(\mathcal R\) \(\displaystyle \left({x^{-1} \circ x}\right)\) $\quad$ $\mathcal R$ is compatible with $\circ$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x^{-1}\) \(\mathcal R\) \(\displaystyle e\) $\quad$ Group properties $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x^{-1}\) \(\in\) \(\displaystyle H\) $\quad$ by definition of $H$ $\quad$

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:

\(\displaystyle x\) \(\in\) \(\displaystyle H\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle e\) \(\mathcal R\) \(\displaystyle h\) $\quad$ for some $h \in H$, by definition of $H$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({x \circ e}\right)\) \(\mathcal R\) \(\displaystyle \left({x \circ h}\right)\) $\quad$ $\mathcal R$ is compatible with $\circ$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({x \circ e \circ x^{-1} }\right)\) \(\mathcal R\) \(\displaystyle \left({x \circ h \circ x^{-1} }\right)\) $\quad$ $\mathcal R$ is compatible with $\circ$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle e\) \(\mathcal R\) \(\displaystyle \left({x \circ h \circ x^{-1} }\right)\) $\quad$ Group properties $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x \circ h \circ x^{-1}\) \(\in\) \(\displaystyle H\) $\quad$ by definition of $H$ $\quad$


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

$\blacksquare$


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