Quotient Structure is Similar to Structure
Theorem
Let $\RR$ be a congruence relation on a algebraic structure $\struct {G, \circ}$.
 | This article needs to be linked to other articles. In particular: "similar" structure? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
Then the quotient structure $\struct {G / \RR, \circ_\RR}$ is a similar structure to $\struct {G, \circ}$.
Proof
Quotient Structure of Semigroup is Semigroup
From Quotient Structure is Well-Defined we have that $\circ_\RR$ is closed on $S / \RR$.
Let $\eqclass x \RR, \eqclass y \RR, \eqclass z \RR \in S / \RR$.
We shall prove that $\circ_\RR$ is associative:
| \(\ds \paren {\eqclass x \RR \circ_{S / \RR} \eqclass y \RR} \circ_{S / \RR} \eqclass z \RR\) | \(=\) | \(\ds \eqclass {x \circ y} \RR \circ_{S / \RR} \eqclass z \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass {\paren {x \circ y} \circ z} \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass {x \circ \paren {y \circ z} } \RR\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass x \RR \circ_{S / \RR} \eqclass {y \circ z} \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass x \RR \circ_{S / \RR} \paren {\eqclass y \RR \circ_{S / \RR} \eqclass z \RR}\) | Definition of Operation Induced on $S / \RR$ by $\circ$ |
Hence $\struct {S / \RR, \circ_\RR}$ is a semigroup.
$\blacksquare$
Quotient Structure of Monoid is Monoid
From Quotient Structure of Semigroup is Semigroup $\struct {S / \RR, \circ_\RR}$ is a semigroup.
Let $\eqclass x {\RR} \in S / \RR$.
Consider $\eqclass e \RR$:
| \(\ds \eqclass x \RR \circ_{S / \RR} \eqclass e \RR\) | \(=\) | \(\ds \eqclass {x \circ e} \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass x \RR\) | Definition of Identity Element |
Furthermore:
| \(\ds \eqclass e \RR \circ_{S / \RR} \eqclass x \RR\) | \(=\) | \(\ds \eqclass {e \circ x} \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass x \RR\) | Definition of Identity Element |
Hence $\eqclass e \RR$ is an identity.
Hence $\struct {S / \RR, \circ_\RR}$ is a monoid.
$\blacksquare$
Quotient Structure of Group is Group
From Quotient Structure of Monoid is Monoid $\struct {G / \RR, \circ_\RR}$ is a monoid with $\eqclass e \RR$ as its identity.
Let $\eqclass x \RR \in S / \RR$.
Consider $\eqclass {-x} \RR$ where $-x$ denotes the inverse of $x$ under $\circ$.
| \(\ds \eqclass x \RR \circ_{S / \RR} \eqclass {-x} \RR\) | \(=\) | \(\ds \eqclass {x \circ -x} \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass e \RR\) | Definition of Inverse Element |
Furthermore:
| \(\ds \eqclass {-x} \RR \circ_{S / \RR} \eqclass x \RR\) | \(=\) | \(\ds \eqclass {-x \circ x} \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass e \RR\) | Definition of Inverse Element |
Hence $\eqclass {-x} \RR$ is the inverse of $\eqclass x \RR$.
Hence $\struct {G / \RR, \circ_\RR}$ is a group.
$\blacksquare$
Quotient Structure of Abelian Group is Abelian Group
From Quotient Structure of Group is Group we have that $\struct {G / \RR, \circ_\RR}$ is a group.
Let $\eqclass x \RR, \eqclass y \RR \in S / \RR$.
| \(\ds \eqclass x \RR \circ_{S / \RR} \eqclass y \RR\) | \(=\) | \(\ds \eqclass {x \circ y} \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass {y \circ x} \RR\) | $\circ$ is Commutative | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass y \RR \circ_{S / \RR} \eqclass x \RR\) | Definition of Operation Induced on $S / \RR$ by $\circ$ |
Hence $\circ_{S / \RR}$ is commutative.
Hence $\struct {G / \RR, \circ_\RR}$ is an abelian group.
$\blacksquare$