nyuuzyou/wikis huggingface dataset:
Quantum-Charge Ratio

== Theorem ==
The ratio of Planck's constant to the elementary charge is given by:
== Proof ==
We have: 
== Sources ==
* : $2$. Physical Constants and Conversion Factors: Table $2.3$ Adjusted Values of Constants
::which gives the mantissas of these figures as:
:::$4 \cdotp 135 \, 708$ with an uncertainty of $\pm 14$ corresponding to the $2$ least significant figures
:::$1 \cdotp 379 \, 523 \, 4$ with an uncertainty of $\pm 46$ corresponding to the $2$ least significant figures
Category:Planck's Constant
Category:Elementary Charge
Quartic Equation/Examples/6z^4 - 25z^3 + 32z^2 + 3z - 10 = 0

== Example of Quartic Equations ==
The quartic equation:
:$6 z^4 - 25 z^3 + 32 z^2 + 3 z - 10 = 0$
has solutions:
:$-\dfrac 1 2, \dfrac 2 3, 2 + i, 2 - 1$
== Proof ==
The integer divisors of $6$ and $-10$ are respectively:
:$\pm 1, \pm 2, \pm 3, \pm 6$
and
:$\pm 1, \pm 2, \pm 5, \pm 10$
Thus from Conditions on Rational Solution to Polynomial Equation, the possible rational solutions are:
:$\pm 1, \pm \dfrac 1 2, \pm \dfrac 1 3, \pm \dfrac 1 6, \pm 2, \pm \dfrac 2 3, \pm 5, \pm \dfrac 5 2, \pm \dfrac 5 3, \pm \dfrac 5 6, \pm 10, \pm \dfrac {10} 3$
By trial, we find that $z = -\dfrac 1 2$ and $z = \dfrac 2 3$ are solutions.
Thus:
:$\paren {2 z + 1} \paren {3 z - 2} = 6 z^2 - z - 2$
is a factor of $6 z^4 - 25 z^3 + 32 z^2 + 3 z - 10$.
The other factor is found to be:
:$z^2 - 4 z + 5$
Hence:
Hence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polynomial Equations: $34$
Category:Examples of Quartic Equations
Quasicomponent is Intersection of Clopen Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $p \in S$.
Then the quasicomponent containing $p$ equals the intersection of all sets which are both open and closed which contain $p$.
== Proof ==
Let $C$ be the quasicomponent of $p$.
Let $Q$ be the set of clopen sets containing $p$ that are not equal to $S$.
First let $x \in C$.
Let $U \in Q$.
By definition of $Q$:
:$p \in U$
Let $V = \relcomp S U$ be the complement of $U$ relative to $S$.
By Complement of Clopen Set is Clopen $V$ is also a clopen set of $T$.
By Clopen Set and Complement form Separation $U$ and $V$ separate $S$.
By the definition of quasicomponent, $U \mid V$ is a separation of $S$ which contains a single open set $U$ containing both $x$ and $p$.
Thus:
:$x \in U$
As this holds for all such $U$:
:$x \in \ds \bigcap Q$
As this holds for all such $x$:
:$C \subseteq Q$
Let $x \in \ds \bigcap Q$.
Let $U \mid V$ be a separation of $S$.
Let $p \in U$.
Then $U \in Q$.
Thus $x \in U$.
Thus $x$ lies on the same side as $p$ of any separation of $X$.
So $x \in C$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Clopen Sets
Category:Connectedness Between Two Points
Quasicomponent is not necessarily Component

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $Q$ be a quasicomponent of $T$.
Then it is not necessarily the case that $C$ is also a component of $T$.
== Proof ==
From Component of Point is not always Intersection of its Clopen Sets, the set intersection of the clopen sets containing a point $x$ may not always be contained in the component of $x$.
The result follows from Quasicomponent is Intersection of Clopen Sets.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Components
Category:Quasicomponents
Quasicomponent of Compact Hausdorff Space is Connected

== Theorem ==
Let $\struct {X, \tau}$ be a compact Hausdorff space.
Let $C$ be a quasicomponent of $\struct {X, \tau}$.
Then $C$ is connected.
== Proof ==
Let $p \in C$.
 $C$ is not connected.
Therefore, by definition of connected, there exist disjoint closed sets $A, B$ of $\struct {X, \tau}$ such that $C = A \cup B$.
By Compact Hausdorff Space is T4, there exist disjoint open sets $U, V$ of $\struct {X, \tau}$ such that $U \supseteq A$ and $V \supseteq B$.
By Quasicomponent is Intersection of Clopen Sets, $C$ is the intersection of all clopen sets of $\struct {X, \tau}$ containing $p$.
Since $U$ and $V$ are open, $X \setminus \paren {U \cup V}$ is closed.
Hence $X \setminus \paren {U \cup V}$ is compact.
Let $S$ be the set of clopen sets of $\struct {X, \tau}$ containing $p$.
Let $S'$ be the set of complements relative to $\struct {X, \tau}$ of elements of $S$.
Then $S'$ is an open cover of $X \setminus \paren {U \cup V}$.
Thus by compactness has a finite subcover $T'$.
Let $T$ be the set of complements of elements of $T'$.
Then $\ds C \subseteq \bigcap T \subseteq U \cup V$.
Furthermore, since $T$ is a finite set of clopen sets of $\struct {X, \tau}$]], $\bigcap T$ is clopen.
Let $\ds U' = U \cap \bigcap T$ and let $\ds V' = V \cap \bigcap T$.
Then $C \subseteq U' \cup V' = T$.
Since $T$ is clopen, so is $U'$.
But $C$ contains points in $U'$ and points not in $U'$, contradicting the fact that $C$ is a quasicomponent of $\struct {X, \tau}$.
It follows that $C$ is connected.
Category:Compact Spaces
Category:Connectedness Between Two Points
Quasicomponents and Arc Components are Equal in Locally Arc-Connected Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is locally arc-connected.
Then $A \subseteq S$ is an arc component of $T$ $A$ is a quasicomponent of $T$.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Arc-Connected Spaces
Category:Quasicomponents
Category:Arc Components
Quasicomponents and Components are Equal in Compact Hausdorff Space

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a compact Hausdorff space.
Then for each $A \subseteq S$:
$A$ is a component of $S$ $A$ is a quasicomponent of $S$.
== Proof ==
Category:Compact Spaces
Category:Connectedness Between Two Points
Category:Connected Spaces
Quasicomponents and Components are Equal in Locally Connected Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is locally connected.
Then $A \subseteq S$ is a component of $T$ $A \subseteq S$ is a quasicomponent of $T$.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Connected Spaces
Category:Components
Category:Quasicomponents
Quasicomponents and Path Components are Equal in Locally Path-Connected Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is locally path-connected.
Then $A \subseteq S$ is a path component of $T$ $A \subseteq S$ is a quasicomponent of $T$.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Path-Connected Spaces
Category:Quasicomponents
Category:Path Components
Quasicomponents of Integer Reciprocal Space with Zero are Single Points

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then the quasicomponents of $A$ are singletons.
== Proof ==
From Integer Reciprocal Space with Zero is Totally Separated, $A$ is totally separated.
The result follows from Totally Separated iff Quasicomponents are Singletons.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $2 \ \text{(c)}$
Category:Integer Reciprocal Space
Quasigroup is not necessarily B-Algebra

== Theorem ==
Let $\struct {S, \circ}$ be a quasigroup.
Then $\struct {S, \circ}$ is not necessarily a $B$-algebra.
== Proof ==
;Proof by Counterexample
From Group is Quasigroup we take an arbitrary small group.
Consider the Cayley table of the group of order $3$:
:$\begin{array}{c|cccccc}
 & 0 & 1 & 2 \\
\hline
0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1 \\
\end{array}$
By inspection we see that does not hold as $1 \circ 1 \ne 0$.
== Also see ==
* $B$-Algebra is Quasigroup
Category:B-Algebras
Category:Examples of Quasigroups
Quasilinear Differential Equation/Examples/First Order Ordinary

== Example of Quasilinear Differential Equation ==
A first order quasilinear ordinary differential equation is a differential equation which can be written in the form:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $1$ Introduction: $(1')$
Category:Examples of Quasilinear Differential Equations
Quasilinear Differential Equation/Examples/x + y y' = 0

== Theorem ==
The first order quasilinear ordinary differential equation over the real numbers $\R$:
:$x + y y' = 0$
has the general solution:
:$x^2 + y^2 = C$
where:
:$C > 0$
:$y \ne 0$
:$x < \size {\sqrt C}$
with the singular point:
:$x = y = 0$
== Proof 1 ==
== Proof 2 ==
== Explicit Solution ==
== Solution Curves ==
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $1$ Introduction: Example $1$
Category:Examples of First Order ODEs
Category:Examples of Quasilinear Differential Equations
Category:Quasilinear Differential Equation/Examples/x + y y' = 0
Quasilinear Differential Equation/Examples/x + y y' = 0/Explicit Solution

== Theorem ==
The first order quasilinear ordinary differential equation:
:$x + y y' = 0$
has a general solution which can be expressed explicitly as:
:$y = \pm \sqrt {C - x^2}$
over the domain:
:$-\sqrt C < x < \sqrt C$
== Proof ==
The general solution of $x + y y' = 0$ is implicitly over the real numbers.
We have:
But when $y = 0$, the derivative $y'$ of the ordinary differential equation $x + y y' = 0$ needs to be infinite for non-zero $x$.
Hence:
:$-\sqrt C < x < \sqrt C$
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $1$ Introduction: Example $1$
Category:Quasilinear Differential Equation/Examples/x + y y' = 0
Quasilinear Differential Equation/Examples/x + y y' = 0/Proof 1

== Theorem ==
== Proof ==
Let us rearrange the equation in question:
This is in the form:
:$y \rd y = k x \rd x$
where $k = 1$.
From First Order ODE: $y \rd y = k x \rd x$:
:$y^2 = \paren {-1} x^2 + C$
from which the result follows.
Category:Quasilinear Differential Equation/Examples/x + y y' = 0
Quasilinear Differential Equation/Examples/x + y y' = 0/Proof 2

== Theorem ==
== Proof ==
Observe that from Derivative of Power and Chain Rule for Derivatives:
:$\dfrac {\map \d {x^2 + y^2} } {\d y} = 2 x + 2 y \dfrac {\d y} {\d x}$
Hence:
:$\dfrac {\map \d {x^2 + y^2} } {\d y} = 0$
So from Derivative of Constant:
:$x^2 + y^2 = C$
where $C$ is abitrary.
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $1$ Introduction: Example $1$
Category:Quasilinear Differential Equation/Examples/x + y y' = 0
Quasilinear Differential Equation/Examples/x + y y' = 0/Solution Curves

== Solution Curves to $x + y y' = 0$ ==
Consider the first order quasilinear ordinary differential equation:
:$x + y y' = 0$
From Quasilinear Differential Equation: $x + y y' = 0$ we have that the general solution is given by:
:$x^2 + y^2 = C$
Its solution curves can be presented as:
:
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $1$ Introduction: Example $1$
Category:Quasilinear Differential Equation/Examples/x + y y' = 0
Quasiperfect Number is Square of Odd Integer

== Theorem ==
Let $n$ be a quasiperfect number.
Then:
:$n = \paren {2 k + 1}^2$
for some $k \in \Z_{>0}$.
That is, a quasiperfect number is the square of an odd integer.
== Proof ==
By definition of quasiperfect number:
:$\map {\sigma_1} n = 2 n + 1$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
That is, $\map {\sigma_1} n$ is odd.
Then from Divisor Sum is Odd iff Argument is Square or Twice Square:
$n$ is either square or twice a square.
Suppose $n = 2^k m^2$ is a quasiperfect number, where $m$ is odd and $k \in \Z_{\ge 0}$.
Then:
Hence we have:
:$\paren {2^{k + 1} - 1} \divides \paren {2^{k + 1} m^2 + 1}$
Since:
:$\paren {2^{k + 1} - 1} \divides \paren {2^{k + 1} m^2 - m^2}$
we have:
:$\paren {2^{k + 1} - 1} \divides \paren {1 + m^2}$
 $k > 0$.
Write:
:$m^2 \equiv -1 \pmod {2^{k + 1} - 1}$
But by First Supplement to Law of Quadratic Reciprocity (extended to Jacobi symbols):
:$\paren {\dfrac {-1} {2^{k + 1} - 1} } = \paren {-1}^{\frac {2^{k + 1} - 1 - 1} 2} = \paren {-1}^{2^k - 1} = -1$
Hence $-1$ is not a quadratic residue modulo $2^{k + 1} - 1$, contradicting the above.
Therefore we must have $k = 0$.
In this case, $n = m^2$, an odd square.
== Sources ==
* : $16$
* : $16$
Category:Quasiperfect Numbers
Quasiuniformity Induces Topology

== Theorem ==
Let $\UU$ be a quasiuniformity on a set $S$.
Let $\tau \subseteq \powerset S$ be a subset of the power set of $S$, created from $\UU$ by:
:$\tau := \set {\map u x: u \in \UU, x \in S}$
where:
:$\forall x \in S: \map u x = \set {y: \tuple {x, y} \in u}$
That is, where $\map u x$ is the image of $x$ under $u$, where $u$ is considered as a relation on $S$.
Then $\tau$ is a topology on $S$.
That is, the quasiuniform space $\struct {\struct {S, \UU}, \tau}$ is also the topological space $\struct {S, \tau}$.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Uniformities
Category:Uniformities
Quaternion Addition forms Abelian Group

== Theorem ==
Let $\mathbb H$ be the set of quaternions.
Then $\struct {\mathbb H, +}$, where $+$ denotes quaternion addition, is an abelian group.
== Proof ==
Taking the abelian group axioms in turn:
=== ===
Let:
:$\mathbf x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$
:$\mathbf x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$
be quaternions.
By definition of quaternion addition:
:$\mathbf x_1 + \mathbf x_2 = \paren {a_1 + a_2} \mathbf 1 + \paren {b_1 + b_2} \mathbf i + \paren {c_1 + c_2} \mathbf j + \paren {d_1 + d_2} \mathbf k$
So as $a_1, a_2, b_1, b_2$ etc. are all elements of $\R$, then so are $a_1 + a_2, b_1 + b_2$ etc.
So $\paren {a_1 + a_2} \mathbf 1 + \paren {b_1 + b_2} \mathbf i + \paren {c_1 + c_2} \mathbf j + \paren {d_1 + d_2} \mathbf k$ is a quaternion.
Hence $\struct {\mathbb H, +}$ is closed.
=== ===
From Matrix Form of Quaternion, we can express a quaternion $\mathbf x$ in the form of a matrix:
:$\mathbf x = \begin {bmatrix} a + b i & c + d i \\ -c + d i & a - b i \end {bmatrix}$
We have that Matrix Entrywise Addition is Associative.
It follows that quaternion addition is also associative
=== ===
The identity element of $\struct \mathbb H, +}$ is:
:$\mathbf 0 = 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$
as can be seen:
=== ===
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.
The inverse of $\struct {\mathbb H, +}$ is:
:$-\mathbf x = -a \mathbf 1 + -b \mathbf i + -c \mathbf j + -d \mathbf k$
as can be seen:
and similarly for $\mathbf x + \mathbf -x$.
=== Commutativity ===
Commutativity follows from Real Addition is Commutative.
Thus all the abelian group axioms are seen to be fulfilled.
Category:Quaternions
Category:Abelian Groups
Quaternion Conjugation is Involution

== Theorem ==
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.
Let $\overline {\mathbf x}$ denote the quaternion conjugate of $\mathbf x$.
Then the operation of quaternion conjugation is an involution:
:$\overline {\paren {\overline {\mathbf x} } } = \mathbf x$
== Proof ==
Category:Complex Conjugates
Category:Involutions
Quaternion Group/Cayley Table

== Cayley Table for Quaternion Group==
The Cayley table for the quaternion group given with the group presentation:
:$Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
can be presented as:
:$\begin {array} {r|rrrrrrrr}
& e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\
\hline
e & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\
a & a & a^2 & a^3 & e & a b & a^2 b & a^3 b & b \\
a^2 & a^2 & a^3 & e & a & a^2 b & a^3 b & b & a b \\
a^3 & a^3 & e & a & a^2 & a^3 b & b & a b & a^2 b \\
b & b & a^3 b & a^2 b & a b & a^2 & a & e & a^3 \\
a b & a b & b & a^3 b & a^2 b & a^3 & a^2 & a & e \\
a^2 b & a^2 b & a b & b & a^3 b & e & a^3 & a^2 & a \\
a^3 b & a^3 b & a^2 b & a b & b & a & e & a^3 & a^2 
\end {array}$
=== Coset Decomposition of $\set {e, a^2}$ ===
Presenting the above Cayley table with respect to the coset decomposition of the normal subgroup $\gen a^2$ gives:
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \iota$
Category:Quaternion Group
Category:Examples of Cayley Tables
Quaternion Group/Cayley Table/Coset Decomposition of (e, a^2)

== Cayley Table for Quaternion Group $Q$ ==
The Cayley table for the quaternion group $Q$, with respect to the coset decomposition of the normal subgroup $\gen a^2$, is:
can be presented as:
:$\begin {array} {r|rr|rr|rr|rr}
& e & a^2 & a & a^3 & b & a^2 b & a b & a^3 b \\
\hline
e & e & a^2 & a & a^3 & b & a^2 b & a b & a^3 b \\
a^2 & a^2 & e & a^3 & a & a^2 b & b & a^3 b & a b \\
\hline
a & a & a^3 & a^2 & e & a b & a^3 b & a^2 b & b \\
a^3 & a^3 & a & e & a^2 & a^3 b & a b & b & a^2 b \\
\hline
b & b & a^2 b & a^3 b & a b & a^2 & e & a & a^3 \\
a^2 b & a^2 b & b & a b & a^3 b & e & a^2 & a^3 & a \\
\hline
a b & a b & a^3 b & b & a^2 b & a^3 & a & a^2 & e \\
a^3 b & a^3 b & a b & a^2 b & b & a & a^3 & e & a^2 
\end {array}$
which is seen to be an example of the Klein $4$-group.
Category:Quaternion Group
Category:Examples of Cayley Tables
Quaternion Group/Complex Matrices

== Representation of Quaternion Group ==
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.
The set:
:$\Dic 2 = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$
under the operation of conventional matrix multiplication, forms the quaternion group:
=== Cayley Table ===
Its Cayley table is given by:
== Also see ==
* Quaternions Defined by Matrices where it is shown that these have the appropriate properties.
In Matrix Form of Quaternion it is shown that a general element $\mathbf x$ of $\mathbb H$ has the form:
:$\mathbf x = \begin {bmatrix} a + b i & c + d i \\ -c + d i & a - bi \end {bmatrix}$
== Sources ==
* : $\S 34$. Examples of groups: $(6) \ \text{(ii)}$
* 
Category:Quaternion Group
Quaternion Group/Complex Matrices/Cayley Table

== Cayley Table for Quaternion Group==
The Cayley table for the quaternion group:
:$Q = \Dic 2 = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$
under the operation of conventional matrix multiplication, where:
:$\mathbf 1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\qquad
\mathbf i = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
\qquad
\mathbf j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
\qquad
\mathbf k = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$
can be presented as:
:$\begin{array}{r|rrrrrrrr}
 & \mathbf 1 & \mathbf i & -\mathbf 1 & -\mathbf i & \mathbf j & \mathbf k & -\mathbf j & -\mathbf k \\
\hline
 \mathbf 1 & \mathbf 1 & \mathbf i & -\mathbf 1 & -\mathbf i & \mathbf j & \mathbf k & -\mathbf j & -\mathbf k \\
 \mathbf i & \mathbf i & -\mathbf 1 & -\mathbf i & \mathbf 1 & \mathbf k & -\mathbf j & -\mathbf k & \mathbf j \\
-\mathbf 1 & -\mathbf 1 & -\mathbf i & \mathbf 1 & \mathbf i & -\mathbf j & -\mathbf k & \mathbf j & \mathbf k \\
-\mathbf i & -\mathbf i & \mathbf 1 & \mathbf i & -\mathbf 1 & -\mathbf k & \mathbf j & \mathbf k & -\mathbf j \\
 \mathbf j & \mathbf j & -\mathbf k & -\mathbf j & \mathbf k & -\mathbf 1 & \mathbf i & \mathbf 1 & -\mathbf i \\
 \mathbf k & \mathbf k & \mathbf j & -\mathbf k & -\mathbf j & -\mathbf i & -\mathbf 1 & \mathbf i & \mathbf 1 \\
-\mathbf j & -\mathbf j & \mathbf k & \mathbf j & -\mathbf k & \mathbf 1 & -\mathbf i & -\mathbf 1 & \mathbf i \\
-\mathbf k & -\mathbf k & -\mathbf j & \mathbf k & \mathbf j & \mathbf i & \mathbf 1 & -\mathbf i & -\mathbf 1
\end{array}$
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $14$
Category:Quaternion Group
Category:Examples of Cayley Tables
Quaternion Group/Group Presentation

== Group Presentation of Quaternion Group ==
The group presentation of the quaternion group is given by:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
== Proof ==
Let $G = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$.
It is to be demonstrated that $\Dic 2$ is isomorphic to $G$.
Consider the Cayley table for $\Dic 2$:
We have that:
:$a^4 = e$
:$b^2 = a^2$
:$\paren {a b} a = b$
demonstrating that $\Dic 2$ has the same group presentation as $G$.
Hence the result.
== Also presented as ==
== Sources ==
* : Chapter $4$: Subgroups: Exercise $4$
Category:Group Presentation of Quaternion Group
Category:Quaternion Group
Category:Examples of Group Presentations
Quaternion Group/Order 4 Matrices

== Representation of Quaternion Group ==
Let $\mathbf I, \mathbf J, \mathbf K, \mathbf L$ denote the following four elements of the matrix space $\map {\MM_\Z} 4$:
where $\Z$ is the set of integers.
The set:
:$\Dic 2 = \set {\mathbf I, -\mathbf I, \mathbf J, -\mathbf J, \mathbf K, -\mathbf K, \mathbf L, -\mathbf L}$
under the operation of conventional matrix multiplication, forms the quaternion group.
This can be generated by the $2$ elements $\mathbf J$ and $\mathbf K$.
== Proof ==
We note that:
Hence we can construct the Cayley table:
:$\begin{array}{r|rrrrrrrr}
 & \mathbf I & \mathbf J & -\mathbf I & -\mathbf J & \mathbf K & \mathbf L & -\mathbf K & -\mathbf L \\
\hline
 \mathbf I & \mathbf I & \mathbf J & -\mathbf I & -\mathbf J & \mathbf K & \mathbf L & -\mathbf K & -\mathbf L \\
 \mathbf J & \mathbf J & -\mathbf I & -\mathbf J & \mathbf I & \mathbf L & -\mathbf K & -\mathbf L & \mathbf K \\
-\mathbf I & -\mathbf I & -\mathbf J & \mathbf I & \mathbf J & -\mathbf K & -\mathbf L & \mathbf K & \mathbf L \\
-\mathbf J & -\mathbf J & \mathbf I & \mathbf J & -\mathbf I & -\mathbf L & \mathbf K & \mathbf L & -\mathbf K \\
 \mathbf K & \mathbf K & -\mathbf L & -\mathbf K & \mathbf L & -\mathbf I & \mathbf J & \mathbf I & -\mathbf J \\
 \mathbf L & \mathbf L & \mathbf K & -\mathbf L & -\mathbf K & -\mathbf J & -\mathbf I & \mathbf J & \mathbf I \\
-\mathbf K & -\mathbf K & \mathbf L & \mathbf K & -\mathbf L & \mathbf I & -\mathbf J & -\mathbf I & \mathbf J \\
-\mathbf L & -\mathbf L & -\mathbf K & \mathbf L & \mathbf K & \mathbf J & \mathbf I & -\mathbf J & -\mathbf I
\end{array}$
and it can be seen that this is an instance of the quaternion group.
== Also see ==
* Quaternions Defined by Matrices where it is shown that these have the appropriate properties.
In Matrix Form of Quaternion it is shown that a general element $\mathbf x$ of $\mathbb H$ has the form:
:$\mathbf x = \begin {bmatrix} a + b i & c + d i \\ -c + d i & a - b i \end {bmatrix}$
== Sources ==
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* 
Category:Quaternion Group
Quaternion Group/Subgroup Generated by a^2/Quotient Group

== Quotient Group of Subgroup Generated by $a^2$ of the Quaternion Group $Q$==
Let the quaternion group $Q$ be represented by its group presentation:
Consider the subgroup $\gen {a^2}$ of $Q$:
:$\gen {a^2} = \set {e, a^2}$
Let $E := N, A := a N, B := b N, C := a b N$.
Thus the quotient group of $G$ by $N$ is:
:$G / N = \set {E, A, B, C}$
whose Cayley table can be presented as:
:$\begin{array}{c|cccc}
 & E & A & B & C \\
\hline
E & E & A & B & C \\
A & A & E & C & B \\
B & B & C & E & A \\
C & C & B & A & E \\
\end{array}$
which is seen to be an example of the Klein $4$-group.
== Sources ==
* : Chapter $7$: Normal subgroups and quotient groups: Exercise $3$
Category:Quaternion Group