# Congruence Relation on Group induces Normal Subgroup

## 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:

\(\ds x, y\) | \(\in\) | \(\ds H\) | ||||||||||||

\(\ds \implies \ \ \) | \(\ds e\) | \(\mathcal R\) | \(\ds x\) | |||||||||||

\(\, \ds \land \, \) | \(\ds e\) | \(\mathcal R\) | \(\ds y\) | by definition of $H$ | ||||||||||

\(\ds \implies \ \ \) | \(\ds \left({e \circ e}\right)\) | \(\mathcal R\) | \(\ds \left({x \circ y}\right)\) | $\mathcal R$ is compatible with $\circ$ | ||||||||||

\(\ds \implies \ \ \) | \(\ds x \circ y\) | \(\in\) | \(\ds H\) | by definition of $H$ |

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

\(\ds x\) | \(\in\) | \(\ds H\) | ||||||||||||

\(\ds \implies \ \ \) | \(\ds e\) | \(\mathcal R\) | \(\ds x\) | by definition of $H$ | ||||||||||

\(\ds \implies \ \ \) | \(\ds \left({x^{-1} \circ e}\right)\) | \(\mathcal R\) | \(\ds \left({x^{-1} \circ x}\right)\) | $\mathcal R$ is compatible with $\circ$ | ||||||||||

\(\ds \implies \ \ \) | \(\ds x^{-1}\) | \(\mathcal R\) | \(\ds e\) | Group properties | ||||||||||

\(\ds \implies \ \ \) | \(\ds x^{-1}\) | \(\in\) | \(\ds H\) | by 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:

\(\ds x\) | \(\in\) | \(\ds H\) | ||||||||||||

\(\ds \implies \ \ \) | \(\ds e\) | \(\mathcal R\) | \(\ds h\) | for some $h \in H$, by definition of $H$ | ||||||||||

\(\ds \implies \ \ \) | \(\ds \left({x \circ e}\right)\) | \(\mathcal R\) | \(\ds \left({x \circ h}\right)\) | $\mathcal R$ is compatible with $\circ$ | ||||||||||

\(\ds \implies \ \ \) | \(\ds \left({x \circ e \circ x^{-1} }\right)\) | \(\mathcal R\) | \(\ds \left({x \circ h \circ x^{-1} }\right)\) | $\mathcal R$ is compatible with $\circ$ | ||||||||||

\(\ds \implies \ \ \) | \(\ds e\) | \(\mathcal R\) | \(\ds \left({x \circ h \circ x^{-1} }\right)\) | Group properties | ||||||||||

\(\ds \implies \ \ \) | \(\ds x \circ h \circ x^{-1}\) | \(\in\) | \(\ds H\) | by definition of $H$ |

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

$\blacksquare$

## Also see

- Normal Subgroup induced by Congruence Relation defines that Congruence
- Quotient Structure on Group defined by Congruence equals Quotient Group

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 11$: Theorem $11.5$