nyuuzyou/wikis huggingface dataset:
Symmetry Group of Parallelogram/Cayley Table

== Cayley Table of Symmetry Group of Parallelogram ==
=== Definition ===
=== Cayley Table ===
The Cayley table of the symmetry group of the (non-rectangular) parallelogram can be written:
:$\begin{array}{c|cccc}
 & e & r \\
\hline
e & e & r \\
r & r & e \\
\end{array}$
Category:Examples of Cayley Tables
Symmetry Group of Rectangle/Cayley Table

== Cayley Table of Symmetry Group of Rectangle ==
=== Definition ===
=== Cayley Table ===
The Cayley table of the symmetry group of the (non-square) rectangle can be written:
$\quad \begin {array} {c|cccc}
& e & r & h & v \\
\hline
e & e & r & h & v \\
r & r & e & v & h \\
h & h & v & e & r \\
v & v & h & r & e \\
\end {array}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.11$
Category:Examples of Cayley Tables
Category:Symmetry Group of Rectangle
Symmetry Group of Rectangle is Klein Four-Group

== Theorem ==
The symmetry group of the rectangle is the Klein $4$-group.
== Proof ==
Comparing the Cayley tables of the symmetry group of the rectangle with the Klein $4$-group the isomorphism can be seen: 
Thus the required isomorphism $\phi$ can be set up as:
== Sources ==
* : $\S 44$. Some consequences of Lagrange's Theorem: Illustration $1$
* 
Category:Klein Four-Group
Symmetry Group of Regular Hexagon/Examples

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
=== Subgroup of Operations that Fix $C$ ===
=== Subgroup of Operations that Permute $A$, $C$ and $E$ ===
=== Subgroup of Operations Generated by $\alpha^4$ and $\alpha^3 \beta$ ===
=== Elements of Form $\beta \alpha^k$ in Form $\alpha^i \beta^j$ ===
=== Center ===
=== Normalizer of $\alpha$ ===
=== Normalizer of $\beta$ ===
=== Normalizer of $\gen \alpha$ ===
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Examples/Center

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
The center of $D_6$ is:
:$\map Z {D_6} = \set {e, \alpha^3}$
== Proof ==
From Center of Dihedral Group:
:$\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, \alpha^{n / 2} } & : n \text { even} \end{cases}$
for $n \ge 3$.
Here we have that $n = 6$ and so even.
Hence the result.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \zeta \ (5)$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Examples/Elements of Form beta alpha^k

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
Consider elements of $D_6$ of the form $\beta \alpha^k$, for $k \in \Z^{\ge 0}$.
They can be expressed in the form:
:$\beta \alpha^k = \alpha^{6 - k} \beta$
== Proof ==
From Product of Generating Elements of Dihedral Group:
:$\beta \alpha^k = \alpha^{n - k} \beta$
for the dihedral group $D_n$.
The result follows by setting $n = 6$.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \zeta \ (4)$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Examples/Normalizer of Reflection

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
The normalizer of $\alpha$ is:
:$\map {N_{D_6} } {\set \beta} = \set {e, \beta, \alpha^3, \alpha^3 \beta}$
== Proof ==
From Normalizer of Reflection in Dihedral Group:
:$\map {N_{D_n} } {\set \alpha} = \set {e, \beta, \alpha^{n / 2}, \alpha^{n / 2} \beta}$
for even $n$.
Hence the result by setting $n = 6$.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48 \alpha$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Examples/Normalizer of Rotation

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
The normalizer of $\alpha$ is:
:$\map {N_{D_6} } {\set \alpha} = \set {e, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5}$
== Proof ==
From Normalizer of Rotation in Dihedral Group:
:$\map {N_{D_n} } {\set \alpha} = \gen \alpha$
Hence the result by setting $n = 6$.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48 \alpha$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Examples/Normalizer of Subgroup of Rotations

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
Let $\gen \alpha$ denote the subgroup generated by $\alpha$.
The normalizer of $\gen \alpha$ is $D_6$ itself:
:$\map {N_{D_6} } {\gen \alpha} = D_6$
== Proof ==
We have that:
:$\order {\gen \alpha} = 6 = \dfrac {\order {D_6} } 2$
and so:
:$\index {D_6} {\gen \alpha} = 2$
From Subgroup of Index 2 is Normal, $\gen \alpha$ is normal in $D_6$.
The result follows from Normal Subgroup iff Normalizer is Group.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48 \alpha$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Examples/Subgroup Generated by alpha^4 and alpha^3 beta

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
Let $H$ be the subgroup of $D_6$ generated by $\alpha^4$ and $\alpha^3 \beta$.
Then:
:$H = \set {e, \alpha^2, \alpha^4, \alpha \beta, \alpha^3 \beta, \alpha^5 \beta}$
and:
:$H \cong D_3$
== Proof ==
Demonstration by Cayley table:
:$\begin{array}{c|cccccc}
 & e & \alpha^2 & \alpha^4 & \alpha \beta & \alpha^3 \beta & \alpha^5 \beta \\
\hline
e & e & \alpha^2 & \alpha^4 & \alpha \beta & \alpha^3 \beta & \alpha^5 \beta \\
\alpha^2 & \alpha^2 & \alpha^4 & e & \alpha^3 \beta & \alpha^5 \beta & \alpha \beta \\
\alpha^4 & \alpha^4 & e & \alpha^2 & \alpha^5 \beta & \alpha \beta & \alpha^3 \beta \\
\alpha \beta & \alpha \beta & \alpha^5 \beta & \alpha^3 \beta & e & \alpha^4 & \alpha^2 \\
\alpha^3 \beta & \alpha^3 \beta & \alpha \beta & \alpha^5 \beta & \alpha^2 & e & \alpha^4 \\
\alpha^5 \beta & \alpha^5 \beta & \alpha^3 \beta & \alpha \beta & \alpha^4 & \alpha^2 & e \\
\end{array}$
It can be seen by inspection that this is $D_3$.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \zeta \ (3)$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Examples/Subgroup that Fixes C

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
The operations of $D_6$ that fix vertex $C$ form a subgroup of $D_6$ which is isomorphic to the parity group.
== Proof ==
Recall the group action of $D_6$ upon the vertices of $P$:
It is seen by inspection that the only elements of $D_6$ which fix $C$ are $e$ and $\alpha^4 \beta$.
It is further seen that $\alpha^4 \beta$ is the reflection whose axis is $CF$.
The Cayley table of these $2$ elements can be shown to be:
:$\begin{array}{c|cc}
 & e & \alpha^4 \beta \\
\hline
e & e & \alpha^4 \beta \\
\alpha^4 \beta & \alpha^4 \beta & e \\
\end{array}$
whose isomorphism to the parity group is immediate.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \zeta \ (1)$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Examples/Subgroup that Permutes A, C, E

== Examples of Operations on Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
The set of elements of $D_6$ which permute vertices $A$, $C$ and $E$ form a subgroup of $D_6$ which is isomorphic to the dihedral group $D_3$.
== Proof ==
Recall the group action of $D_6$ upon the vertices of $P$:
Let $H$ be the subset of elements of $D_6$ which permute vertices $A$, $C$ and $E$.
It is seen by inspection that $H$ is:
:$H = \set {e, \alpha^2, \alpha^4, \beta, \alpha^2 \beta, \alpha^4 \beta}$
Setting $\gamma = \alpha^2$ we see that $H$ can be written:
:$H = \set {e, \gamma, \gamma^2, \beta, \gamma \beta, \gamma^2 \beta}$
which is seen to be $D^3$.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \zeta \ (2)$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Hexagon/Group Action on Vertices

== Group Action of Symmetry Group of Regular Hexagon ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
$D_6$ acts on the vertices of $\HH$ according to this table:
:$\begin{array}{cccccccccccc}
 e & \alpha & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \beta & \alpha \beta & \alpha^2 \beta & \alpha^3 \beta & \alpha^4 \beta & \alpha^5 \beta \\
\hline
A & B & C & D & E & F & A & B & C & D & E & F \\
B & C & D & E & F & A & F & A & B & C & D & E \\
C & D & E & F & A & B & E & F & A & B & C & D \\
D & E & F & A & B & C & D & E & F & A & B & C \\
E & F & A & B & C & D & C & D & E & F & A & B \\
F & A & B & C & D & E & B & C & D & E & F & A \\
\end{array}$
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \zeta$
Category:Symmetry Group of Regular Hexagon
Symmetry Group of Regular Pentagon/Cayley Table

== Cayley Table of Symmetry Group of Regular Pentagon ==
:
The Cayley table of the symmetry group of the regular pentagon can be written:
:$\begin {array} {c|cccccc}
\circ & e & r & r^2 & r^3 & r^4 & t_A & t_B & t_C & t_D & t_E \\
\hline
e & e & r & r^2 & r^3 & r^4 & t_A & t_B & t_C & t_D & t_E \\
r & r & r^2 & r^3 & r^4 & e & t_C & t_D & t_E & t_A & t_B \\
r^2 & r^2 & r^3 & r^4 & e & r & t_E & t_A & t_B & t_C & t_D \\
r^3 & r^3 & r^4 & e & r & r^2 & t_B & t_C & t_D & t_E & t_A \\
r^4 & r^4 & e & r & r^2 & r^3 & t_D & t_E & t_A & t_B & t_C \\
t_A & t_A & t_D & t_B & t_E & t_C & e & r^2 & r^4 & r & r^3 \\
t_B & t_B & t_E & t_C & t_A & t_D & r^3 & e & r^2 & r^4 & r \\
t_C & t_C & t_A & t_D & t_B & t_E & r & r^3 & e & r^2 & r^4 \\
t_D & t_D & t_B & t_E & t_C & t_A & r^4 & r & r^3 & e & r^2 \\
t_E & t_E & t_C & t_A & t_D & t_B & r^2 & r^4 & r & r^3 & e \\
\end {array}$
where the various symmetries of the regular pentagon $\PP = ABCDE$ are:
:the identity mapping $e$
:the rotations $r, r^2, r^3, r^4$ of $72^\circ, 144^\circ, 216^\circ, 288^\circ$ around the center of $\PP$ anticlockwise respectively
:the reflections $t_A, t_B, t_C, t_D, t_E$ about the lines through the center of $\PP$ and the vertices $A$ to $E$ respectively.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.5$
Category:Examples of Cayley Tables
Category:Symmetry Group of Regular Pentagon
Symmetry Group of Rhombus/Cayley Table

== Cayley Table of Symmetry Group of Rhombus ==
=== Definition ===
=== Cayley Table ===
The Cayley table of the symmetry group of the (non-square) rhombus can be written:
:$\begin{array}{c|cccc}
 & e & r & h & v \\
\hline
e & e & r & h & v \\
r & r & e & v & h \\
h & h & v & e & r \\
v & v & h & r & e \\
\end{array}$
Category:Examples of Cayley Tables
Symmetry Group of Square/Cayley Table

== Cayley Table of Symmetry Group of Square ==
:
The Cayley table of the symmetry group of the square can be written:
:$\begin{array}{c|cccccc} & e & r & r^2 & r^3 & t_x & t_y & t_{AC} & t_{BD} \\ \hline e & e & r & r^2 & r^3 & t_x & t_y & t_{AC} & t_{BD} \\ r & r & r^2 & r^3 & e & t_{AC} & t_{BD} & t_y & t_x \\ r^2 & r^2 & r^3 & e & r & t_y & t_x & t_{BD} & t_{AC} \\ r^3 & r^3 & e & r & r^2 & t_{BD} & t_{AC} & t_x & t_y \\ t_x & t_x & t_{BD} & t_y & t_{AC} & e & r^2 & r^3 & r \\ t_y & t_y & t_{AC} & t_x & t_{BD} & r^2 & e & r & r^3 \\ t_{AC} & t_{AC} & t_x & t_{BD} & t_y & r & r^3 & e & r^2 \\ t_{BD} & t_{BD} & t_y & t_{AC} & t_x & r^3 & r & r^2 & e\\ \end{array}$
where the various symmetries of the square $\SS = ABCD$ are:
:the identity mapping $e$
:the rotations $r, r^2, r^3$ of $90^\circ, 180^\circ, 270^\circ$ around the center of $\SS$ anticlockwise respectively
:the reflections $t_x$ and $t_y$ are reflections about the $x$ and $y$ axis respectively
:the reflection $t_{AC}$ about the diagonal through vertices $A$ and $C$
:the reflection $t_{BD}$ about the diagonal through vertices $B$ and $D$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.5$
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.10$
* : $\S 1.4$: Exercise $2.1$
* : Chapter $1$: Definitions and Examples: Exercise $3$
Category:Examples of Cayley Tables
Category:Symmetry Group of Square
Symmetry Group of Square is Group

== Theorem ==
The symmetry group of the square is a non-abelian group.
=== Definition ===
Recall the definition of the symmetry group of the square:
== Proof ==
Let us refer to this group as $D_4$.
Taking the group axioms in turn:
=== ===
From the Cayley table it is seen directly that $D_4$ is closed.
=== ===
Composition of Mappings is Associative.
=== ===
The identity is $e$ as defined.
=== ===
Each element can be seen to have an inverse:
:$r^{-1} = r^3$ and so $\paren {r^3}^{-1} = r$
:$r^2$, $t_{AC}$, $t_{BD}$, $t_x$ and $t_y$ are all self-inverse.
Thus $D_4$ is seen to be a group.
Note that from the Cayley table it can be observed directly that:
:$r \circ t_x = t_{BD}$
:$t_x \circ r = t_{AC}$
thus illustrating by counterexample that $D_4$ is not abelian.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.3$
Category:Symmetry Group of Square
Symmetry Rule for Binomial Coefficients

== Theorem ==
Let $n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {n - k}$
where $\dbinom n k$ is a binomial coefficient.
=== Complex Numbers ===
== Proof 1 ==
== Proof 2 ==
== Informal Proof ==
== Examples ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.3$: Binomial Coefficients
* : Problems
* : Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis: Theorem $19.10$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.5$
* : $\text {3-1}$ Permutations and Combinations: Exercise $8$
* : $24$
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.29$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.5$
Category:Symmetry Rule for Binomial Coefficients
Category:Binomial Coefficients
Symmetry Rule for Binomial Coefficients/Complex Numbers

== Theorem ==
For all $z, w \in \C$ such that it is not the case that $z$ is a negative integer and $w$ an integer:
:$\dbinom z w = \dbinom z {z - w}$
where $\dbinom z w$ is a binomial coefficient.
== Proof ==
From the definition of the binomial coefficient:
:$\dbinom z w := \ds \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\omega + 1} \map \Gamma {\zeta - \omega + 1} }$
where $\Gamma$ denotes the Gamma function.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $42$ (Solution)
Category:Symmetry Rule for Binomial Coefficients
Symmetry Rule for Binomial Coefficients/Examples/11 choose 8

== Example of Use of Symmetry Rule for Binomial Coefficients ==
Consider the binomial coefficient $\dbinom {11} 8$.
This can be calculated as:
:$\dbinom {11} 8 = \dfrac {11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4} {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$
which is unwieldy.
Or we can use the Symmetry Rule for Binomial Coefficients, and say:
:$\dbinom {11} 8 = \dbinom {11} {11 - 8} = \dbinom {11} 3$
and calculate it as:
:$\dbinom {11} 3 = \dfrac {11 \times 10 \times 9} {3 \times 2 \times 1} = \dfrac {990} 6 = 165$
which is far less trouble.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Two important relations
Category:Symmetry Rule for Binomial Coefficients
Symmetry Rule for Binomial Coefficients/Examples/7 from 10

== Example of Use of Symmetry Rule for Binomial Coefficients ==
The number of ways of choosing:
:$7$ objects from a set of $10$
is the same as the number of ways of choosing:
:$3$ objects from a set of $10$.
Hence:
:$\dbinom {10} 7 = \dbinom {10} 3 = \dfrac {10 \times 9 \times 8} {3 \times 2 \times 1} = \dfrac {720} 6 = 120$
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.5$: Example
Category:Symmetry Rule for Binomial Coefficients
Symmetry Rule for Binomial Coefficients/Examples/8 choose 6

== Example of Use of Symmetry Rule for Binomial Coefficients ==
Let $N$ be the number of ways a team of $6$ people may be selected from a pool of $8$.
Then:
:$N = 28$
== Proof ==
We have:
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $4$
Category:Symmetry Rule for Binomial Coefficients
Symmetry Rule for Binomial Coefficients/Proof 1

== Theorem ==
== Proof ==
Follows directly from the definition of binomial coefficient, as follows.
If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k < 0$.
In both cases:
:$\dbinom n k = \dbinom n {n - k} = 0$
Let $0 \le k \le n$.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Two important relations
* : $\S 1.2.6$: Binomial Coefficients: $\text{B}$
Category:Symmetry Rule for Binomial Coefficients
Symmetry Rule for Binomial Coefficients/Proof 2

== Theorem ==
== Proof ==
From the definition of Cardinality of Set of Subsets, $\dbinom n k$ is the number of subsets of cardinality $k$ of a set with cardinality $n$.
Let $S$ be a set with cardinality $n$.
Let $\powerset S$ denote the power set of $S$.
Let $\AA_k \subseteq \powerset S$ be the set of elements of $\powerset S$ which have cardinality $k$.
For every $T \in \AA_k$ there exists its relative complement:
:$\relcomp S T \in \AA_{n - k}$
Define the mapping $f_k: \AA_k \to \AA_{n - k}$ as:
:$\forall T \in \AA_k: \map f T = \relcomp S T$
From Relative Complement Mapping on Powerset is Bijection it follows that $f_k$ is a bijection.
By definition of set equivalence, that means:
:$\card {A_k} = \card {A_{n - k} }$
That is, from the definition of Cardinality of Set of Subsets:
:$\dbinom n k = \dbinom n {n - k}$
== Sources ==
* : $\S 0.6$: Theorem $8: \ 3$
Category:Symmetry Rule for Binomial Coefficients
Symmetry Rule for Binomial Coefficients/Proof 3

== Theorem ==
== Proof ==
If we choose $k$ objects from $n$, then we leave $n - k$ objects.
Hence for every choice of $k$, we are also making the same choice of $n - k$.
Hence the number of choices of $k$ objects is the same as the number of choices of $n - k$ objects.
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Two important relations
Category:Symmetry Rule for Binomial Coefficients
Symmetry Rule for Gaussian Binomial Coefficients

== Theorem ==
Let $q \in \R_{\ne 1}, n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k_q = \dbinom n {n - k}_q$
where $\dbinom n k_q$ is a Gaussian binomial coefficient.
== Proof ==
If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k < 0$.
In both cases:
:$\dbinom n k_q = \dbinom n {n - k}_q = 0$
Let $0 \le k \le n$.
Consider the case $k \le \dfrac n 2$.
Then $k \le n - k$.
The case $k \ge \dfrac n 2$ can be done by observing:
:$n - k \le \dfrac n 2$
and hence by the above:
:$\dbinom n k_q = \dbinom n {n - \paren {n - k} }_q = \dbinom n {n - k}_q$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $58$
Category:Symmetry Rule for Binomial Coefficients
Category:Gaussian Binomial Coefficients
Symmetry in Space Implies Conservation of Momentum

== Theorem ==
The total derivative of the action $S_{12}$ from states $1$ to $2$ with regard to position is equal to the difference in momentum from states $1$ to $2$:
:$\dfrac {\d S_{1 2} } {\d x} = p_2 - p_1$
== Proof ==
From the definition of generalized momentum and the Euler-Lagrange Equations:
Therefore, via the definition of action, Definite Integral of Partial Derivative and the Fundamental Theorem of Calculus:
Category:Laws of Conservation
Symmetry of Bernoulli Polynomial

== Theorem ==
Let $\map {B_n} x$ denote the nth Bernoulli polynomial.
Then:
:$\map {B_n} {1 - x} = \paren {-1}^n \map {B_n} x$
== Proof ==
Let $\map G {t, x}$ denote the Generating Function for Bernoulli Polynomials:
:$\map G {t, x} = \dfrac {t e^{t x} } {e^t - 1}$
Then:
Thus:
Category:Bernoulli Polynomials
Symmetry of Invariant Metric on Vector Space

== Theorem ==
<onlyinclude>
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $d$ be an invariant metric on $X$. 
Then we have:
:$\map d {x, y} = \map d {-x, -y}$
for each $x, y \in X$.
== Proof ==
Let $x, y \in X$.
We have:
Category:Invariant Metrics on Vector Spaces
Syndrome/Examples/(6, 3) code in Z2

== Example of Syndrome ==
Let $C$ be the linear $\tuple {6, 3}$-code in $\Z_2$ whose standard generator matrix $G$ is given by:
:$G := \begin{pmatrix}
1 & 0 & 0 & 1 & 1 & 0 \\ 
0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix}$
The syndrome of $100000$ is $110$
The syndrome of $110011$ is $000$.
== Proof ==
From Standard Parity Check Matrix on the given linear $\tuple {6, 3}$-code in $\Z_2$:
:$P := \begin{pmatrix}
1 & 1 & 0 & 1 & 0 & 0 \\ 
1 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 \end{pmatrix}$
where $P$ is the standard parity check matrix of $C$.
Thus:
== Sources ==
* : Chapter $6$: Error-correcting codes: Definition $6.19$
Category:Examples of Linear Codes
Syndrome Decoding

== Theorem ==
Let $C$ be a linear $\tuple {n, k}$-code whose master code is $\map V {n, p}$
To decode a given vector $v$ of $\map V {n, p}$, the syndrome of $v$ can be used as follows.
Create an array $T$ of $2$ column consisting of the following:
:The top row contains:
::in column $1$: the zero of $C$ 
::in column $2$: its syndrome.
:The $r$th row subsequent contains:
::in column $1$: any element of $\map V {n, p}$ of minimum weight which is not already included in the first $r - 1$ rows
::in column $2$: its syndrome.
To decode a given vector $v$ of $\map V {n, p}$:
:Calculate its syndrome
:Find it in column $2$ of $T$
:See what is in column $1$ of $T$, and call it $u$, say
:Subtract $u$ from $v$.
== Examples ==
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.22$
Category:Linear Codes