nyuuzyou/wikis huggingface dataset:
Common Sections of Parallel Planes with other Plane are Parallel

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be two planes which are parallel.
Let $AB$ and $CD$ be cut by the plane $EFGH$.
Let:
:$EF$ be the common section of $EFGH$ and $AB$
and:
:$GH$ be the common section of $EFGH$ and $CD$.
It is to be shown that $EF$ is parallel to $GH$.
Suppose that $EF$ and $GH$ are not parallel.
Then when produced, they will meet, either in the direction of $F$ and $H$, or in the direction of $E$ and $G$.
, suppose they meet at $K$ when produced in the direction of $F$ and $H$.
We have that $EFK$ is in the plane $AB$.
Therefore from :
:all the points of $EFK$ are in the plane $AB$.
But $K$ is one of the points of $EFK$.
Therefore $K$ is in the plane $AB$.
For the same reason, $K$ is also in the plane $CD$.
Therefore the planes $AB$ and $CD$ will meet when produced.
But they do not meet, because they are parallel.
Therefore $EF$ and $GH$ will not meet when produced in the direction of $F$ and $H$.
By the same argument, $EF$ and $GH$ will not meet when produced in the direction of $E$ and $G$.
But from :
:$EF$ is parallel to $GH$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Planes
Common Sum of 3 Distinct Amicable Pairs

== Theorem ==
The integer $64 \, 795 \, 852 \, 800$ is the sum of $3$ distinct amicable pairs:
:$29 \, 912 \, 035 \, 725$ and $34 \, 883 \, 817 \, 075$
:$31 \, 695 \, 652 \, 275$ and $33 \, 100 \, 200 \, 525$
:$32 \, 129 \, 958 \, 525$ and $32 \, 665 \, 894 \, 275$
all of them odd.
== Proof ==
We have that:
From $29 \, 912 \, 035 \, 725$ and $34 \, 883 \, 817 \, 075$ are amicable:
From $31 \, 695 \, 652 \, 275$ and $33 \, 100 \, 200 \, 525$ are amicable:
From $32 \, 129 \, 958 \, 525$ and $32 \, 665 \, 894 \, 275$ are amicable:
== Sources ==
* 
* : $64,795,852,800$
Category:Amicable Pairs
Commutation Property in Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then $x$ and $y$ commute $x \circ y \circ x^{-1} = y$.
== Proof ==
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $6 \ \text{(i)}$
Category:Group Theory
Category:Commutativity
Commutation of Inverses in Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $x, y \in S$ such that $x$ and $y$ are both invertible.
Then $x$ commutes with $y$ $x^{-1}$ commutes with $y^{-1}$.
== Proof ==
=== Necessary Condition ===
Let $x$ commute with $y$.
Then:
So $x^{-1}$ commutes with $y^{-1}$.
=== Sufficient Condition ===
Now let $x^{-1}$ commute with $y^{-1}$.
From the above, $\paren {x^{-1} }^{-1}$ commutes with $\paren {y^{-1} }^{-1}$.
From Inverse of Inverse in Monoid, $\paren {x^{-1} }^{-1} = x$ and $\paren {y^{-1} }^{-1} = y$.
Thus $x$ commutes with $y$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.5: \ 3^\circ$
Category:Inverse Elements
Category:Commutativity
Category:Monoids
Commutation with Group Elements implies Commuation with Product with Inverse

== Theorem ==
Let $G$ be a group.
Let $a, b, c \in G$ such that $a$ commutes with both $b$ and $c$.
Then $a$ commutes with $b c^{-1}$.
== Proof ==
== Sources ==
* : $\S 3.2$: Groups; the axioms: Exercise $(11)$
Category:Group Theory
Category:Commutativity
Commutation with Inverse in Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid.
Let $x, y \in S$ such that $y$ is invertible.
Then $x$ commutes with $y$ $x$ commutes with $y^{-1}$.
== Proof ==
=== Necessary Condition ===
Let $x$ commute with $y$.
Then:
So $x$ commutes with $y^{-1}$.
=== Sufficient Condition ===
Now let $x$ commute with $y^{-1}$.
From the above, it follows that $x$ commutes with $\paren {y^{-1} }^{-1}$.
From Inverse of Inverse in Monoid:
:$\paren {y^{-1} }^{-1} = y$
Thus $x$ commutes with $y$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.5: \ 2^\circ$
Category:Inverse Elements
Category:Monoids
Category:Commutativity
Commutative B-Algebra Implies (zy)(zx)=xy

== Theorem ==
Let $\struct {B, \circ}$ be a commutative $B$-algebra. 
Then:
:$\forall x, y, z \in X: \paren {z \circ y} \circ \paren {z \circ x} = x \circ y$
== Proof ==
Let $x, y, z \in X$.
Then:
Category:B-Algebras
Commutative B-Algebra Induces Abelian Group

== Theorem ==
Let $\left({X, \circ }\right)$ be a commutative $B$-algebra.
Let $*$ be the binary operation on $X$ defined as:
:$\forall a, b \in X: a * b := a \circ \left({0 \circ b}\right)$
Then the algebraic structure $\left({X, *}\right)$ is an abelian group such that:
:$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$.
That is:
:$\forall a, b \in X: a * b^{-1} := a \circ b$
== Proof ==
From B-Algebra Induces Group, the algebraic structure $\left({X, *}\right)$ is a group such that:
:$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$.
It remains to show that $*$ is a commutative operation.
Let $x, y \in X$:
Hence the result.
== Also see ==
* Abelian Group Induces Commutative $B$-Algebra
Category:B-Algebras
Commutative B-Algebra is Entropic Structure

== Theorem ==
Let $\struct {G, *}$ be a commutative $B$-algebra.
Then $\struct {G, *}$ is an entropic structure.
== Proof ==
From Commutative $B$-Algebra Induces Abelian Group we have that there exists an abelian group $\struct {G, \circ}$ such that:
:$\forall a, b \in G: a \circ b^{-1} = a * b$ 
where $a * b$ is defined by the binary operation in $\struct {G, *}$.
From Abelian Group Induces Entropic Structure, we have directly that $\struct {G, *}$ is an entropic structure.
Category:B-Algebras
Category:Entropic Structures
Commutative Diagram/Examples/Arbitrary Example 1

== Example of Commutative Diagram ==
Let $A, B, X, Y$ be arbitrary sets.
Let:
be mappings such that:
:$\beta \circ f = g \circ \alpha = k$
where $\circ$ denotes composition of mappings.
This can be depicted using the following commutative diagram:
$\quad\quad \begin{xy} \xymatrix@L+2mu@+1em{
A \ar[r]^*{\alpha} \ar[d]_*{f} \ar[rd]^*{k} & B \ar[d]^*{g}
\\
X \ar[r]^*{\beta} & Y
}\end{xy}$
== Sources ==
* 
* 
Category:Examples of Commutative Diagrams
Commutative Diagram/Examples/Square Function with Square Root

== Example of Commutative Diagram ==
Let $g$ and $h$ be the real functions defined as:
:$\forall x \in \R: \map g x = x^2$
:$\forall x \in \R_{\ge 0}: \map h x = \sqrt x$
The composition $h \circ g$ can be depicted using a commutative diagram as follows:
$\quad\quad\begin{xy} \xymatrix@L+2mu@+1em{
\R \ar[r]^*{g}
\ar@{-->}[rd]_*{h \circ g}
&
\R_{\ge 0} \ar[d]^*{h}
\\ &
\R_{\ge 0}
}\end{xy}$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 8$: Composition of Functions and Diagrams: Exercise $1$
Category:Examples of Commutative Diagrams
Commutative Law of Addition

== Theorem ==
Let $\mathbb F$ be one of the standard number sets: $\N, \Z, \Q, \R$ and $\C$.
Then:
:$\forall x, y \in \mathbb F: x + y = y + x$
That is, the operation of addition on the standard number sets is commutative.
 
=== Natural Number Addition is Commutative ===
 
=== Integer Addition is Commutative ===
 
=== Rational Addition is Commutative ===
 
=== Real Addition is Commutative ===
 
=== Complex Addition is Commutative ===
== Also see ==
* Commutative Law of Multiplication
* Associative Law of Addition
* Associative Law of Multiplication
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $1$. Scalar and Vector Quantities
* : $\S 1.1$. Number Systems: $\text{I}.$
* : $\S 4.2$. Commutative and associative operations: Example $61$
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.1$
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers: $(\text P 4)$
* 
* 
* 
* 
* 
Category:Commutative Law of Addition
Category:Numbers
Category:Addition
Category:Examples of Commutative Operations
Commutative Law of Multiplication

== Theorem ==
Let $\mathbb F$ be one of the standard number sets: $\N, \Z, \Q, \R$ and $\C$.
Then:
:$\forall x, y \in \mathbb F: x + y = y + x$
That is, the operation of multiplication on the standard number sets is commutative.
 
=== Natural Number Multiplication is Commutative ===
 
=== Integer Multiplication is Commutative ===
 
=== Rational Multiplication is Commutative ===
 
=== Real Multiplication is Commutative ===
 
=== Complex Multiplication is Commutative ===
== Also see ==
* Commutative Law of Addition
* Associative Law of Multiplication
* Associative Law of Addition
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $1$. Scalar and Vector Quantities
* : $\S 1.1$. Number Systems: $\text{IV}.$
* : Chapter $\text {I}$: The Group Concept: $\S 1$: Introduction: $(1.1)$
* : $\S 4.2$. Commutative and associative operations: Example $61$
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.1$
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers: $(\text P 8)$
* 
* 
* 
* 
Category:Commutative Law of Multiplication
Category:Numbers
Category:Multiplication
Category:Examples of Commutative Operations
Commutative Linear Transformation is G-Module Homomorphism

== Theorem ==
Let $\rho: G \to \GL V$ be a representation.
Let $f: V \to V$ be a linear transformation.
Let:
:$\forall g \in G: \map \rho g \circ f = f \circ \map \rho g$
Then $f: V \to V$ is a $G$-module homomorphism.
== Proof ==
Let:
:$\forall g \in G: \map \rho g \circ f = f \circ \map \rho g$
Let $v$ be a vector $v \in V$.
Then:
:$\map {\map \rho g} {\map f v} = \map f {\map {\map \rho g} v}$
Using the properties from Correspondence between Linear Group Actions and Linear Representations:
:there exists a $G$-module $\struct {V, \phi}$ associated with $\rho$ such that:
::$\map \phi {g, v} = \map {\map \rho g} v$
Applying the last formula:
:$\map {\map \rho g} {\map f v} = \map \phi {g, \map f v}$
and:
:$\map f {\map \phi {g, v} } = \map f {\map {\map \rho g} v}$
Thus our assumption is equivalent to:
:$\map f {\map \phi {g, v} } = \map \phi {g, \map f v}$
Hence, by definition of $G$-module homomorphism, $f: V \to V$ is a $G$-module homomorphism.
Category:Representation Theory
Commutative Semigroup is Entropic Structure

== Theorem ==
A commutative semigroup is an entropic structure.
== Proof ==
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $a, b, c, d \in S$.
Then:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(a)}$
Category:Semigroups
Category:Entropic Structures
Category:Commutativity
Commutative and Associative Product on Space of Distributions does not Exist

== Theorem ==
Let $\map {\DD'} \R$ be the distribution space.
Let $\alpha \in \map {C^\infty} \R$ be a smooth function.
Let $\circ$ be a product operation on $\map {\DD'} \R$.
Suppose:
:$\forall T \in \map {\DD'} \R : \forall \alpha \in \map {C^\infty} \R : \alpha \circ T := \alpha \cdot T$
where $\cdot$ stands for multiplication of distribution by a smooth function.
Suppose $\circ$ is commutative and associative.
 
Then $\circ$ does not exist.
== Proof ==
 there is such a product.
Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.
We also have that $\ds \PV \frac 1 x$ is a distribution.
Let $\phi, \psi \in \map \DD \R$ be test functions.
Then:
Here $\mathbf 1$ is the distribution associated to the real function defined by:
:$\mathbf 1 : \R \to 1$
Consider the product $\ds \delta \cdot x \cdot \PV \frac 1 x$ in the distributional sense, where $\delta$ and $\PV \frac 1 x$ are distributions, and $x$ is a smooth function.
Altogether, we have that:
:$\ds \paren {\delta \circ x} \circ \PV \frac 1 x \ne \delta \circ \paren {x \circ \PV \frac 1 x}$
Hence, the associativity is violated, and we have reached a contradiction.
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Distributions
Category:Commutativity
Category:Associativity
Commutative and Unitary Ring with 2 Ideals is Field

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$.
Let $\struct {R, +, \circ}$ be such that the only ideals of $\struct {R, +, \circ}$ are:
:$\set {0_R}$
and:
$\struct {R, +, \circ}$ itself.
That is, such that $\struct {R, +, \circ}$ has no non-null proper ideals.
Then $\struct {R, +, \circ}$ is a field.
== Proof ==
From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_R}$ and $\struct {R, +, \circ}$ are ideals of $\struct {R, +, \circ}$.
Let $a \in R^*$, where $R^* := R \setminus \set {0_R}$.
Let $\ideal a$ be the principal ideal of $R$ generated by $a$.
We have that $\ideal a$ is a non-null ideal and hence $\ideal a = R$.
Thus $1_R \in \ideal a$.
Thus $\exists x \in R: x \circ a = 1_R$ by the definition of principal ideal.
Therefore $a$ is invertible.
As $a$ is arbitrary, it follows that all such $a$ are invertible.
Thus by definition $\struct {R, +, \circ}$ is a division ring such that $\circ$ is commutative.
The result follows by definition of field.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Theorem $37$
* : Chapter $9$: Rings: Exercise $6 \ \text {(i)}$
Category:Commutative Algebra
Category:Field Theory
Category:Ideal Theory
Commutativity of Group Direct Product

== Theorem ==
Let $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$ be groups.
Let $\struct {G \times H, \circ}$ be the group direct product of $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$, where the operation $\circ$ is defined as:
:$\tuple {g_1, h_1} \circ \tuple {g_2, h_2} = \tuple {g_1 \circ_g g_2, h_1 \circ_h h_2}$
Let $\struct {H \times G, \star}$ be the group direct product of $\struct {H, \circ_h}$ and $\struct {G, \circ_g}$, where the operation $\star$ is defined as:
:$\tuple {h_1, g_1} \star \tuple {h_2, g_2} = \tuple {h_1 \circ_h h_2, g_1 \circ_g g_2}$
The group direct product $\struct {G \times H, \circ}$ is isomorphic to the $\struct {H \times G, \star}$.
== Proof ==
The mapping $\theta: G \times H \to H \times G$ defined as:
:$\forall g \in G, h \in H: \map \theta {g, h} = \tuple {h, g}$
is to be shown to be a group homomorphism, and that $\theta$ is bijective, as follows:
=== Injective ===
Thus $\theta$ is injective by definition.
=== Surjective ===
Thus $\theta$ is surjective by definition.
=== Group Homomorphism ===
Let $\tuple {g_1, h_1}, \tuple {g_2, h_2} \in G \times H$.
Then:
thus proving that $\theta$ is a homomorphism.
== Sources ==
* : Chapter $13$: Direct products: Proposition $13.1 \ (2)$
Category:Group Direct Products
Commutativity of Hadamard Product

== Theorem ==
Let $\struct {S, \cdot}$ be an algebraic structure.
Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$.
For $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$, let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.
The operation $\circ$ is commutative on $\map {\MM_S} {m, n}$ $\cdot$ is commutative on $\struct {S, \cdot}$.
== Proof ==
=== Necessary Condition ===
Let the operation $\cdot$ be commutative on $\struct {S, \cdot}$.
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of the $m \times n$ matrix space over $S$.
Then:
That is, $\circ$ is commutative on $\map {\MM_S} {m, n}$.
=== Sufficient Condition ===
Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not commutative.
Then there exists $a$ and $b$ such that:
:$a \cdot b \ne b \cdot a$
Let $\mathbf A$ and $\mathbf B$ be elements of $\map {\MM_S} {m, n}$ such that:
:$a_{i j} = a$, $b_{i j} = b$
where:
:$a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$
:$b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$
Then:
:$a_{i j} \cdot b_{i j} \ne b_{i j} \cdot a_{i j}$
 
That is:
:$\mathbf A \circ \mathbf B \ne \mathbf B \circ \mathbf A$
because they differ (at least) at indices $\tuple {i, j}$.
That is, $\circ$ is not commutative on $\map {\MM_S} {m, n}$.
== Also see ==
* Closure of Hadamard Product
* Associativity of Hadamard Product
Category:Hadamard Product
Category:Examples of Commutative Operations
Commutativity of Incidence Matrix with its Transpose for Symmetric Design

== Theorem ==
Let $A$ be the incidence matrix of a symmetric design.
Then:
:$A A^\intercal = A^\intercal A$
where $A^\intercal$ is the transpose of $A$.
== Proof ==
First note, we have:
:$(1): \quad A J = J A = k J$, so $A^\intercal J = \paren {J A}^\intercal = \paren {k J}^\intercal = k J$, and likewise $J A^\intercal = k J$
:$(2): \quad J^2 = v J$
:$(3): \quad$ If a design is symmetric, then $A A^\intercal = \paren {r - \lambda} I + \lambda J = \paren {k - \lambda} I + \lambda J$
From $(3)$, we get:
We now have that:
:$\ds \frac 1 {k - \lambda} \paren {A + \sqrt {\paren {\frac \lambda v} J} }$
is the inverse of:
:$A^\intercal - \sqrt {\paren {\dfrac \lambda v} J}$
which implies that they commute with each other.
Thus:
whence:
:$A A^\intercal = \paren {k - \lambda} + \lambda J = A^\intercal A$
Category:Design Theory
Commutativity of Parameters of Beta Function

== Theorem ==
Let $\map \Beta {x, y}$ be the Beta function.
Then:
:$\map \Beta {x, y} = \map \Beta {y, x}$
== Proof ==
== Sources ==
* : $\S 17$: The Beta Function: Some Important Results: $17.3$
* : $\S 26$: The Beta Function: Some Important Results: $26.3.$
Category:Beta Function
Commutativity of Powers in Group

== Theorem ==
Let $\left ({G, \circ}\right)$ be a group.
Let $a, b \in G$ such that $a$ and $b$ commute.
Then:
: $\forall m, n \in \Z: a^m \circ b^n = b^n \circ a^m$
This can be expressed in additive notation in the group $\left ({G, +}\right)$ as:
: $\forall m, n \in \Z: m a + n b = n b + m a$
== Proof ==
By definition, all elements of a group are invertible.
Therefore Commutativity of Powers in Monoid‎ can be applied directly.
Category:Group Theory
Category:Commutativity
Commutativity of Powers in Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $a, b \in S$ be invertible elements for $\circ$ that also commute.
Then:
:$\forall m, n \in \Z: a^m \circ b^n = b^n \circ a^m$
== Proof ==
By Powers of Commuting Elements of Semigroup Commute, if $m > 0$ and $n > 0$ then $a^m$ commutes with $b^n$.
By Commutation with Inverse in Monoid, again if $m > 0$ and $n > 0$ then $a^m$ commutes with $\paren {b^n}^{-1} = b^{-n}$.
Similarly $b^n$ commutes with $a^{-m}$.
But as $a^{-m}$ commutes with $b^n$, it also commutes with $\paren {b^n}^{-1} = b^{-n}$, again by Commutation with Inverse in Monoid.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.11 \ (5)$
Category:Monoids
Category:Commutativity
Commutativity of Powers in Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $a, b \in S$ both be cancellable elements of $S$.
Then:
:$\forall m, n \in \N_{>0}: a \circ b = b \circ a \implies a^m \circ b^n = b^n \circ a^m$
but it is not necessarily the case that:
:$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m \implies a \circ b = b \circ a$
== Proof ==
Let $a, b \in S: a \circ b = b \circ a$.
Then from Powers of Commuting Elements of Semigroup Commute:
:$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$
However, consider the dihedral group $D_3$ $= \gen {a, b: a^3 = b^2 = e, b a b = a^{−1} }$.
A group is a semigroup.
Moreover, the Cancellation Laws hold in a group.
Hence both $a$ and $b$ are cancellable.
We have $a^3 \circ b^2 = b^2 \circ a^3 = e \circ e = e$.
However $a \circ b \ne b \circ a$.
Hence the result by Proof by Counterexample.
Category:Semigroups
Category:Commutativity
Commutativity of Ring of Continuous Mappings

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a commutative topological ring.
Let $\struct{\map C {S, R}, +, *}$ be the ring of continuous mappings from $S$ to $R$.
Then:
:$\struct{\map C {S, R}, +, *}$ is a commutative ring
== Proof ==
From Structure Induced by Commutative Operation is Commutative:
:the ring of mappings $\struct{R^S, +, *}$ is commutative
From Ring of Continuous Mappings is Subring of All Mappings:
:$\struct{\map C {S, R}, +, *}$ is a subring of $\struct{R^S, +, *}$
From Subring of Commutative Ring is Commutative:
:$\struct{\map C {S, R}, +, *}$ is commutative
== Also see ==
* Ring of Continuous Mappings is Subring of All Mappings
* Zero of Ring of Continuous Mappings
* Additive Inverse in Ring of Continuous Mappings
* Unity of Ring of Continuous Mappings
Category:Rings of Continuous Mappings
Commutativity of Spectrum in Algebra over Complex Numbers

== Theorem ==
Let $A$ be an algebra over $\C$.
Let $x, y \in A$.
Let $\map {\sigma_A} {x y}$ and $\map {\sigma_A} {y x}$ be the spectrum of $x y$ and $y x$ respectively in $A$.
Then:
:$\map {\sigma_A} {x y} \cup \set 0 = \map {\sigma_A} {y x} \cup \set 0$ 
== Proof ==
Let $\map G A$ be the group of units of $A$.
, by replacing $A$ by its unitization if necessary, that $A$ is unital.
Let $x, y \in A$.
We show that ${\mathbf 1}_A - x y \in \map G A$ ${\mathbf 1}_A - y x \in \map G A$.
By swapping $x$ and $y$, it suffices to show that if ${\mathbf 1}_A - x y \in \map G A$, then ${\mathbf 1}_A - y x \in \map G A$. 
Specifically, we show that $\paren { {\mathbf 1}_A - y x}^{-1} = 1 + y \paren {{\mathbf 1}_A - x y}^{-1} x$.
We have:
and:
Hence we have shown that for $x, y \in A$, we have ${\mathbf 1}_A - x y \in \map G A$ ${\mathbf 1}_A - y x \in \map G A$.
Now fix $x, y \in A$ and take $\lambda \ne 0$.
We have ${\mathbf 1}_A - \paren {\lambda^{-1} x} y \in \map G A$ ${\mathbf 1}_A - y \paren {\lambda^{-1} x} \in \map G A$. 
That is, $\lambda {\mathbf 1}_A - x y \in \map G A$ $\lambda {\mathbf 1}_A - y x \in \map G A$ for $\lambda \ne 0$. 
Hence we have $\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x} \setminus \set 0$.
That is, $\map {\sigma_A} {x y} \cup \set 0 = \map {\sigma_A} {y x} \cup \set 0$. 
== Sources ==
* : $4.5$: Proposition $4.16$
Category:Spectra (Spectral Theory)
Commutator is Identity iff Elements Commute

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $x, y \in G$.
Let $\sqbrk {x, y}$ denote the commutator of $x$ and $y$.
Then $\sqbrk {x, y} = e$ $x$ and $y$ commute.
== Proof ==
As $G$ is a group, it is a monoid.
Hence Product of Commuting Elements with Inverses applies:
:$x y x^{-1} y^{-1} = e = x^{-1} y^{-1} x y$
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $15$
Category:Group Commutators
Commutator of Group Element with Identity is Identity

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.
Then:
:$\sqbrk {g, e} = e = \sqbrk {e, g}$
== Proof ==
Category:Group Commutators
Category:Inverse Elements
Commutator of Quotient Group Elements

== Theorem ==
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
:$\sqbrk {x, y} = x^{-1} y^{-1} x y$
Then:
:$\forall x, y \in G: \sqbrk {x N, y N} = \sqbrk {x, y} N$
where $x N$ and $y N$ are left cosets of $N$, and so elements of the quotient group $G / N$ of $G$ by $N$.
== Proof ==
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $16$
Category:Group Commutators
Category:Quotient Groups
Commutator of x and Distributional Derivative acting on Distribution

== Theorem ==
Let $T \in \map {\DD'} \R$ be a distribution.
Let:
:$\sqbrk {x, \dfrac \d {\d x} } T := x \dfrac {\d T}{\d x} - \dfrac {\d \paren {x T}}{\d x}$
where derivatives are to be understood in the distributional sense.
Then:
:$\sqbrk {x, \dfrac \d {\d x} } T = - T$
== Proof ==
Let $\phi \in \map \DD \R$ be a test function.
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Distributional Derivatives
Commutator on Algebra is Alternating Bilinear Mapping

== Theorem ==
Let $\struct {A_R, \oplus}$ be an algebra over a ring.
Then the commutator on $\struct {A_R, \oplus}$ is an alternating bilinear mapping:
:$\forall a, b \in A_R: \sqbrk {a, b} = -\sqbrk {b, a}$
== Proof ==
== Sources ==
* 2002: : The Octonions: 1.1 Preliminaries
Category:Algebras
Category:Alternating Bilinear Mappings
Category:Commutators