nyuuzyou/wikis huggingface dataset:
Quaternion Group/Subgroups

== Subgroups of the Quaternion Group ==
Let $Q$ denote the quaternion group, whose group presentation is given as:
The subsets of $Q$ which form subgroups of $Q$ are:
From Quaternion Group is Hamiltonian we have that all of these subgroups of $Q$ are normal.
== Proof ==
Consider the Cayley table for $Q$:
We have that:
:$a^4 = e$
and so $\gen a = \set {e, a, a^2, a^3}$ forms a subgroup of $Q$ which is cyclic.
We have that:
:$b^2 = a^2$
and so $\gen b = \set {e, b, a^2, a^2 b}$ forms a subgroup of $Q$ which is cyclic.
We have that:
:$\paren {a b}^2 = a^2$
and so $\gen {a b} = \set {e, a b, a^2, a^3 b}$ forms a subgroup of $Q$ which is cyclic.
We have that:
:$\paren {a^2}^2 = e$
and so $\gen {a^2} = \set {e, a^2}$ forms a subgroup of $Q$ which is also a subgroup of $\gen a$, $\gen b$ and $\gen {a b}$.
That exhausts all elements of $Q$.
Any subgroup generated by any $2$ elements of $Q$ which are not both in the same subgroup as described above will generate the whole of $Q$.
== Also see ==
* Quaternion Group is Hamiltonian
== Sources ==
* : Chapter $7$: Normal subgroups and quotient groups: Exercise $3$
Category:Quaternion Group
Quaternion Group has Normal Subgroup without Complement

== Theorem ==
Let $Q$ denote the quaternion group.
There exists a normal subgroup of $Q$ which has no complement.
== Proof ==
From Subgroups of Quaternion Group:
From Quaternion Group is Hamiltonian, all these subgroups are normal.
For two subgroups to be complementary, they need to have an intersection which is trivial.
However, apart from $\set e$ itself, all these subgroups contain $a^2$.
Hence none of these subgroups has a complement.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $22$
Category:Subgroup Complements
Category:Quaternion Group
Quaternion Group is Hamiltonian

== Theorem ==
The quaternion group $Q$ is Hamiltonian.
== Proof ==
For clarity the Cayley table of $Q$ is presented below:
By definition $Q$ is Hamiltonian :
:$Q$ is non-abelian
and:
:every subgroup of $Q$ is normal.
$Q$ is non-abelian as demonstrated by the counter-example:
:$a b \ne b a$
From Subgroups of Quaternion Group:
From Trivial Subgroup and Group Itself are Normal:
:$Q$ and $\set e$ are normal subgroups of $Q$.
From Center of Quaternion Group, $\gen {a^2} = \set {e, a^2}$ is the center of $Q$.
From Center of Group is Normal Subgroup, $\set {e, a^2}$ is normal in $Q$.
The remaining subgroups of $Q$ are of order $4$, and so have index $2$.
From Subgroup of Index 2 is Normal it follows that all of these order $4$ subgroups of $Q$ are normal.
That accounts for all subgroups of $Q$.
Hence the result.
== Sources ==
* : $\S 1.10$: Example $36$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \iota$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $8 \ \text{(ii)}$
* : Chapter $7$: Normal subgroups and quotient groups: Exercise $3$
Category:Quaternion Group
Category:Hamiltonian Groups
Quaternion Group not Dihedral Group

== Theorem ==
Let $Q$ be the quaternion group.
Then $Q$ is not isomorphic to the dihedral group $D_4$.
== Proof ==
From Group Presentation of Dihedral Group, $D_4$ is generated by two elements of orders $4$ and $2$ respectively.
Let these generators be $\alpha$ and $\beta$ where:
:$\alpha^4 = e$
:$\beta^2 = e$
Hence $\alpha^2$ and $\beta$ are different elements of $D_4$ which both have order $2$.
But the quaternion group:
:$Q = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$
has only one element of order $2$, which is $-\mathbf 1$.
The rest have order one or four.
Thus $Q$ and $D_4$ cannot be isomorphic because there are not the same number of elements of order $2$ in both groups.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \lambda$
Category:Quaternion Group
Category:Dihedral Groups
Quaternion Modulus in Terms of Conjugate

== Theorem ==
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.
Let $\size {\mathbf x}$ be the modulus of $\mathbf x$.
Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.
Then:
:$\size {\mathbf x}^2 \mathbf 1 = \mathbf x \overline {\mathbf x}$
== Proof ==
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$.
Then:
Category:Quaternion Modulus
Quaternion Modulus of Conjugate

== Theorem ==
Let $z = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.
Let $\overline z$ be the conjugate of $z$.
Let $\cmod z$ be the quaternion modulus of $z$.
Then:
:$\cmod {\overline z} = \cmod z$
== Proof ==
Category:Quaternion Modulus
Quaternion Modulus of Product of Quaternions

== Theorem ==
Let $\mathbf x, \mathbf y$ be quaternions.
Let $\size {\mathbf x}$ be the modulus of $\mathbf x$.
Then:
:$\size {\mathbf {x y} } = \size {\mathbf x} \size {\mathbf y}$
== Proof ==
Let $\mathbf x, \mathbf y$ be in their matrix form.
Then:
Category:Quaternion Modulus
Quaternion Multiplication

== Theorem ==
Let $\mathbf x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$ and $\mathbf x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$ be quaternions.
Then their product is given by:
== Proof ==
From Matrix Form of Quaternion we have that:
:$\mathbf x_1 \mathbf x_2 = \begin{bmatrix} a_1 + b_1 i & c_1 + d_1 i \\ -c_1 + d_1 i & a_1 - b_1 i \end{bmatrix} \begin{bmatrix} a_2 + b_2 i & c_2 + d_2 i \\ -c_2 + d_2 i & a_2 - b_2 i \end{bmatrix}$
Let $\mathbf x_1 \mathbf x_2 = \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} a + b i & c + d i \\ -c + d i & a - b i \end{bmatrix}$.
Throughout we use the definition of conventional matrix product.
So:
Hence the result from Matrix Form of Quaternion.
Category:Quaternions
Quaternion Multplication is not Commutative

== Theorem ==
The operation of multplication on the quaternions $H$ is not commutative.
== Proof ==
By definition of multplication:
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
* 
* 
Category:Quaternions
Quaternions Defined by Matrices

== Theorem ==
Let $\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ denote the following four elements of the matrix space $\map {\MM_\C} 2$:
:$\mathbf 1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\qquad
\mathbf i = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
\qquad
\mathbf j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
\qquad
\mathbf k = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
$
where $\C$ is the set of complex numbers.
Then $\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ are related to each other in the following way:
== Proof ==
This is demonstrated by straightforward application of conventional matrix multiplication:
----
----
----
----
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $9$
* : $\S 34$. Examples of groups: $(6) \ \text{(ii)}$
Category:Matrix Algebra
Category:Quaternion Group
Quaternions Defined by Ordered Pairs

== Theorem ==
Consider the quaternions $\Bbb H$ as numbers in the form:
: $a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
where:
: $a, b, c, d$ are real numbers;
: $\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ are entities related to each other in the following way:
 
Now consider the quaternions $\Bbb H$ defined as ordered pairs $\tuple {x, y}$ where $x, y \in \C$ are complex numbers, on which the operation of multiplication is defined as follows:
Let $w = a_1 + b_1 i, x = c_1 + d_1 i, y = a_2 + b_2 i, z = c_2 + d_2 i$ be complex numbers.
Then $\tuple {w, x} \tuple {y, z}$ is defined as:
:$\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$
where $\overline w$ and $\overline x$ are the complex conjugates of $w$ and $x$ respectively.
These two definitions are equivalent.
== Proof ==
First we identify the following:
We can see that:
Let:
:$w = a_1 + b_1 i$
:$x = d_1 + c_1 i$
:$y = a_2 + b_2 i$
:$z = d_2 + c_2 i$
and so:
:$\overline w = a_1 - b_1 i$
:$\overline x = d_1 - c_1 i$
Then substituting for $(1)$ to $(4)$ above, we have:
Notice the way $\mathbf j$ and $\mathbf k$ are configured. See that they are what appears to be in the wrong order.
We can then demonstrate the equivalence by showing that:
which is equivalent to:
where:
:$x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$
:$x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$
in accordance with Quaternion Multiplication.
Thus:
So:
as required.
== Sources ==
* 2002: : The Octonions: 2.2 The Cayley-Dickson Construction
Category:Quaternions
Quaternions Subring of Complex Matrix Space

== Theorem ==
The Ring of Quaternions is a subring of the matrix space $\map {\MM_\C} 2$.
== Proof ==
From Matrix Form of Quaternion it is clear that the quaternions $\H$ can be expressed in matrix form, as elements of $\map {\MM_\C} 2$.
Thus $\H \subseteq \map {\MM_\C} 2$.
As the quaternions form a ring, the result follows by definition of subring.
== Sources ==
* : $\S 2.1$: Subrings: Examples $2$
Category:Quaternions
Category:Matrix Algebra
Quaternions form Algebra

== Theorem ==
The set of quaternions $\Bbb H$ forms an algebra over the field of real numbers.
This algebra is:
:$(1): \quad$ An associative algebra, but not a commutative algebra.
:$(2): \quad$ A normed division algebra.
:$(3): \quad$ A nicely normed $*$-algebra.
== Proof ==
The quaternions $\Bbb H$ are formed by the Cayley-Dickson Construction from the complex numbers $\C$.
From Complex Numbers form Algebra, we have that $\C$ forms:
:$(1): \quad$ An associative algebra
:$(2): \quad$ A commutative algebra
:$(3): \quad$ A normed division algebra
:$(4): \quad$ A nicely normed $*$-algebra.
From Cayley-Dickson Construction forms Star-Algebra, $\Bbb H$ is a $*$-algebra.
From Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, $\Bbb H$ is a nicely normed $*$-algebra.
From Cayley-Dickson Construction from Commutative Associative Algebra is Associative, $\Bbb H$ is an associative algebra.
Now suppose $\Bbb H$ formed a commutative algebra.
Then from Cayley-Dickson Construction from Real Star-Algebra is Commutative, that would mean $\C$ is a real $*$-algebra.
But from Complex Numbers form Algebra it is explicitly demonstrated that $\C$ is not a real $*$-algebra.
So $\Bbb H$ can not be a commutative algebra.
=== Proof of Normed Division Algebra ===
Consider the element $\tuple {1, 0}$ of $\C^2$.
We have:
As $\times$ is commutative on $\C$, it follows that $\tuple {1, 0} \times \tuple {x_1, x_2} = \tuple {x_1, x_2}$.
So $\tuple {1, 0} \in \C^2$ functions as a unit.
That is, $\tuple {\C^2, \times}$ is a unitary algebra.
We define a norm on $\left({\C^2, \times}\right)$ by:
:$\forall \mathbf a = \tuple {a_1, a_2} \in \C^2: \norm {\mathbf a} = \sqrt {a_1^2 + a_2^2}$
This is a norm because:
:$(1): \quad \norm {\mathbf x} = 0 \iff \mathbf x = \mathbf 0$
:$(2): \quad \norm {\lambda \mathbf x} = \size \lambda \norm x$
:$(3): \quad \norm {x - y} \le \norm {x - z} + \norm {z - y}$
It also follows that:
:$\norm {x \times y} = \size {x \times y} = \size x \times \size y = \norm x \times \norm y$
and so $\struct {\C^2, \times}$ is a normed division algebra.
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Quaternions
Category:Algebras
Quaternions form Skew Field

== Theorem ==
The set $\H$ of quaternions forms a skew field under the operations of addition and multiplication.
== Proof ==
From Ring of Quaternions is Ring we have that $\H$ forms a ring.
From Multiplicative Identity for Quaternions‎ we have that $\mathbf 1$ is the identity for quaternion multiplication.
From Multiplicative Inverse of Quaternion we have that every element of $\H$ except $\mathbf 0$ has an inverse under quaternion multiplication.
So $\H \setminus \set 0 = \H^*$ is a group.
Hence $\H$ forms a division ring.
But quaternion multiplication is specifically not commutative, for example:
:$\mathbf i \mathbf j = \mathbf k, \ \mathbf j \mathbf i = -\mathbf k$
So $\H$ forms a skew field under addition and multiplication.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $3$. FIELD
* : $\S 55$. Special types of ring and ring elements: $(3)$
* 
Category:Quaternions
Category:Examples of Skew Fields
Quaternions form Vector Space over Reals

== Theorem ==
Let $\R$ be the set of real numbers.
Let $\H$ be the set of quaternions.
Then the $\R$-module $\H$ is a vector space.
== Proof ==
Recall that Real Numbers form Field.
Thus $\R$ is a division ring .
Thus we only need to show that $\R$-module $\H$ is a unitary module, by demonstrating the unitary (left) module axioms:
As $\lambda, \mu \in \R$ it follows that $\lambda, \mu \in \H$.
Thus from Quaternion Multiplication Distributes over Addition, $(1)$ and $(2)$ immediately follow.
$(3)$ follows from Quaternion Multiplication is Associative.
$(4)$ follows from Multiplicative Identity for Quaternions, as $\mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$ is the unity of $\H$.
== Also see ==
* Properties of Quaternions
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $2$
* 
* 
Category:Quaternions
Category:Examples of Vector Spaces
Quaternions form Vector Space over Themselves

== Theorem ==
The set of quaternions $\H$, with the operations of addition and multiplication, forms a vector space.
== Proof ==
Let the set of quaternions be denoted $\struct {\H, +, \times}$.
From Quaternions form Skew Field, the algebraic structure $\struct {\H, +, \times}$ is a skew field.
By definition, a skew field is a division ring.
== Also see ==
* Properties of Quaternions
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $2$
Category:Linear Algebra
Category:Quaternions
Category:Examples of Vector Spaces
Queen's Tour/Examples

== Examples of Queen's Tours ==
=== $3 \times 3$ Chessboard ===
=== $4 \times 4$ Chessboard ===
=== $5 \times 5$ Chessboard ===
=== $7 \times 7$ Chessboard ===
=== $8 \times 8$ Chessboard ===
Category:Queen's Tours
Queen's Tour/Examples/3 x 3

== Example of Queen's Tour ==
:
As can be seen, the queen's tour on a $3 \times 3$ chessboard can be achieved in $2 \times 3 - 2 = 4$ moves.
However, the queen needs to go outside the $3 \times 3$ chessboard in order to do it.
== Sources ==
* : Answers: $416$. Sinking the Fishing-Boats
Category:Queen's Tours
Queen's Tour/Examples/5 x 5

== Example of Queen's Tour ==
:
As can be seen, the queen's tour on a $5 \times 5$ chessboard can be achieved in $2 \times 5 - 2 = 8$ moves.
Notice how the queen does not need to go outside the $5 \times 5$ chessboard in order to do it.
== Sources ==
* : Answers: $416$. Sinking the Fishing-Boats
Category:Queen's Tours
Queen's Tour/Examples/8 x 8

== Example of Queen's Tour ==
: $\qquad$ $\qquad$ 
Here are three examples of a re-entrant queen's tour on a $8 \times 8$ chessboard, achieved in $2 \times 8 - 2 = 14$ moves.
== Historical Note ==
== Sources ==
* : The Trick Donkeys
* : Answers: $416$. Sinking the Fishing-Boats
Category:Queen's Tours
Queen's Tour/Examples/8 x 8/Historical Note

== Historical Note on Queen's Tour Example: $8 \times 8$ ==
The $14$-move re-entrant queen's tour on the $8 \times 8$ chessboard was first achieved by .
He considered it one of his best puzzles.
== Sources ==
* : Answers: $416$. Sinking the Fishing-Boats
Category:Historical Notes
Quintic Equation/Examples/z^5 - 2z^4 - z^3 + 6z - 4 = 0

== Example of Quintic Equations ==
The quintic equation:
:$z^5 - 2 z^4 - z^3 + 6 z - 4 = 0$
has solutions:
:$1, 1, 2, -1 \pm i$
== Proof ==
The integer divisors of $4$ are respectively:
:$\pm 1, \pm 2, \pm 4$
Thus from Conditions on Rational Solution to Polynomial Equation, the possible rational solutions are:
:$\pm 1, \pm 2, \pm 4$
Performing some trial divisions:
 z^4 - z^3 - 2 z^2 - 2 z + 4
 ---------------------------------------
z - 1 ) z^5 - 2 z^4 - z^3 + 0 z^2 + 6 z - 4
 z^5 - z^4
 -----------
 -z^4 - z^3
 -z^4 + z^3
 -------------
 -2 z^3 + 0 z^2
 -2 z^3 + 2 z^2
 -------------
 -2 z^2 + 6 z
 -2 z^2 + 2 z
 -----------
 4 z - 4
 4 z - 4
 -------
Hence we have that $z - 1$ is a factor of $z^5 - 2 z^4 - z^3 + 6 z - 4$, and so $z = 1$ is a solution.
Trying the same thing again:
 z^3 - 2 z - 4
 -----------------------------
z - 1 ) z^4 - z^3 - 2 z^2 - 2 z + 4
 z^4 - z^3
 ---------
 0 -2 z^2 - 2 z
 -2 z^2 + 2 z
 ------------
 -4 z + 4
 -4 z + 4
 -------- 
Hence we have that $z - 1$ is another factor of $z^5 - 2 z^4 - z^3 + 6 z - 4$, and so $z = 1$ is another solution.
Trying another trial division:
 z^2 + 2 z + 2
 -----------------------------
z - 2 ) z^3 + 0 z^2 - 2 z - 4
 z^3 - 2 z^2
 -----------
 2 z^2 - 2 z
 2 z^2 - 4 z
 -----------
 2 z - 4
 2 z - 4
 -------
Hence we have that $z - 2$ is another factor of $z^5 - 2 z^4 - z^3 + 6 z - 4$, and so $z = 2$ is another solution.
We are left with:
Hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $101$
Category:Examples of Quintic Equations
Quintuple Angle Formulas

== Theorem ==
=== Quintuple Angle Formula for Sine ===
=== Quintuple Angle Formula for Cosine ===
=== Quintuple Angle Formula for Tangent ===
where $\sin, \cos, \tan$ denote sine, cosine and tangent respectively.
== Also see ==
* Quadruple Angle Formulas
* Sextuple Angle Formulas
Category:Quintuple Angle Formulas
Category:Multiple Angle Formulas
Category:Trigonometric Identities
Quintuple Angle Formulas/Cosine

== Theorem ==
:$\cos 5 \theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$
where $\cos$ denotes cosine.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.51$
Category:Cosine Function
Category:Quintuple Angle Formula for Cosine
Category:Quintuple Angle Formulas
Quintuple Angle Formulas/Sine

== Theorem ==
:$\sin 5 \theta = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta$
where $\sin$ denotes sine.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.50$
Category:Sine Function
Category:Quintuple Angle Formula for Sine
Category:Quintuple Angle Formulas
Quintuple Angle Formulas/Sine/Corollary

== Theorem ==
For all $\theta$ such that $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$
:$\dfrac {\sin 5 \theta} {\sin \theta} = 16 \cos^4 \theta - 12 \cos^2 \theta + 1$
where $\sin$ denotes sine and $\cos$ denotes cosine.
== Proof ==
First note that when $\theta = 0, \pm \pi, \pm 2 \pi \ldots$:
:$\sin \theta = 0$
so $\dfrac {\sin 5 \theta} {\sin \theta}$ is undefined.
Therefore for the rest of the proof it is assumed that $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $21 \ \text{(b)}$
Category:Quintuple Angle Formula for Sine
Category:Sine Function
Quintuple Angle Formulas/Tangent

== Theorem ==
:$\tan 5 \theta = \dfrac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}$
where $\tan$ denotes tangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.52$
Category:Tangent Function
Category:Quintuple Angle Formulas
Quintuplets of Consecutive Integers which are not Divisor Sum Values

== Theorem ==
The elements of the following $5$-tuples of consecutive integers have the property that they are not values of the divisor sum function $\map {\sigma_1} n$ for any $n$:
:$\tuple {49, 50, 51, 52, 53}$
:$\tuple {115, 116, 117, 118, 119}$
:$\tuple {145, 146, 147, 148, 149}$
== Proof ==
== Sources ==
* : $49$
Category:Divisor Sum Function
Quotient Algebra is Algebra

== Theorem ==
Let $R$ be a ring. 
Let $A$ be a $R$-algebra.
Let $I$ be an ideal of $A$.
Let $A/I$ be the quotient module of $A$ modulo $I$.
For $a, b \in A$, define $\ast : \paren {A/I}^2 \to A/I$ by:
:$\paren {a + I} \ast \paren {b + I} = a b + I$ 
Then $\ast$ is well-defined and $\struct {A/I, \ast}$ is an $R$-algebra.
== Proof ==
=== $\ast$ is Well-Defined ===
It suffices to show that if:
:$a - a' \in I$
and:
:$b - b' \in I$
then:
:$ab - a' b' \in I$ 
Suppose that:
:$a - a' = i_1 \in I$
and:
:$b - b' = i_2 \in I$
Then we have:
:$a' b' = \paren {a + i_1} \paren {b + i_2} = a b + i_1 b + a i_2 + i_1 i_2$
From the definition of an ideal, we have $-\paren {i_1 b + a i_2 + i_1 i_2} \in I$.
Hence $a b - a' b' = -\paren {i_1 b + a i_2 + i_1 i_2} \in I$, which is what we wanted to show. 
=== $\ast$ is $R$-Bilinear ===
Let $x, y, z \in A$ and $\lambda \in R$.
We have: 
Almost identically we have:
:$\paren {x + I} \ast \paren {\paren {y + I} + \lambda \paren {z + I} } = \paren {x + I} \ast \paren {y + I} + \lambda \paren {\paren {y + I} \ast \paren {z + I} }$
Hence $\struct {A/I, \ast}$ is an algebra.
Category:Quotient Algebras
Quotient Algebra is Unital iff Quotienting Ideal is Modular

== Theorem ==
Let $R$ be a ring. 
Let $A$ be a $R$-algebra.
Let $I$ be an proper ideal of $A$.
Let $A/I$ be the quotient algebra of $A$ modulo $I$.
Then $A/I$ is unital $I$ is modular.
== Proof ==
=== Necessary Condition ===
Suppose that $A/I$ is unital.
That is, there exists $u \in I$ such that:
:$\paren {a + I} \paren {u + I} = \paren {u + I} \paren {a + I} = a + I$
for each $a \in A$.
That is:
:$a u + I = u a + I = a + I$
Hence, we have:
:$a - a u \in I$
and:
:$a - u a \in I$
for each $a \in A$.
So $I$ is a modular ideal.
=== Sufficient Condition ===
Let $u \in I$ be such that:
:$a - a u \in I$
and:
:$a - u a \in I$
for each $a \in A$.
Then we have:
:$a + I = a u + I$
and:
:$a + I = u a + I$
In other words:
:$a + I = \paren {a + I} \paren {u + I}$
and:
:$a + I = \paren {u + I} \paren {a + I}$
== Sources ==
* : $1.1$: Banach Algebras
Category:Quotient Algebras
Category:Modular Ideals of Algebras
Quotient Epimorphism Condition for Normal Subgroup Product to be Internal Group Direct Product

== Theorem ==
Let $\struct {G, \odot}$ be a group
Let $\struct {H, \odot}$ and $\struct {K, \odot}$ be normal subgroups of $\struct {G, \odot}$.
Then:
:$\struct {G, \odot}$ is the internal group direct product of $\struct {H, \odot}$ and $\struct {K, \odot}$
:the restriction of the quotient epimorphism $q_H$ to $K$ is an isomorphism from $K$ onto the quotient group $G / H$
and:
:the restriction of the quotient epimorphism $q_K$ to $H$ is an isomorphism from $H$ onto the quotient group $G / K$
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.9$
Category:Internal Group Direct Products
Category:Quotient Epimorphism Condition for Normal Subgroup Product to be Internal Group Direct Product
Quotient Epimorphism Condition for Normal Subgroup Product to be Internal Group Direct Product/Necessary Condition

== Theorem ==
Let $\struct {G, \odot}$ be a group
Let $\struct {H, \odot}$ and $\struct {K, \odot}$ be normal subgroups of $\struct {G, \odot}$.
Let:
:the restriction of the quotient epimorphism $q_H$ to $K$ be an isomorphism from $K$ onto the quotient group $G / H$
and:
:the restriction of the quotient epimorphism $q_K$ to $H$ be an isomorphism from $H$ onto the quotient group $G / K$
Then $\struct {G, \odot}$ is the internal group direct product of $\struct {H, \odot}$ and $\struct {K, \odot}$,
== Proof ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.9 \ \text{(b)}$
Category:Quotient Epimorphism Condition for Normal Subgroup Product to be Internal Group Direct Product