hcju/proofwikips huggingface dataset:
Order of Subgroup Product/Lemma
Order of Subgroup Product/Lemma 0 71989 372771 2018-10-23T14:20:28Z Prime.mover 59 Created page with "== Lemma for Order of Subgroup Product ==  Let $h_1, h_2 \in H$.  Then: :$h_1 K = h_2 K$ : :$h_1$ and $h_2$ are in the same Definition:Left Coset|lef..." wikitext text/x-wiki == Lemma for Order of Subgroup Product ==  Let $h_1, h_2 \in H$. Then: :$h_1 K = h_2 K$ : :$h_1$ and $h_2$ are in the same left coset of $H \cap K$.  == Proof == Let $h_1, h_2 \in H$. Then: {{begin-eqn}} {{eqn | l = h_1 K       | r = h_2 K       | c =  }} {{eqn | ll= \iff       | l = h_1^{-1} h_2       | o = \in       | r = K       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \iff       | l = h_1^{-1} h_2       | o = \in       | r = H \cap K       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \iff       | l = h_1 \paren {H \cap K}       | r = h_2 \paren {H \cap K}       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{end-eqn}}  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Intersection of Left Cosets of Subgroups is Left Coset of Intersection|next = Order of Subgroup Product/Proof 2}}: Chapter $7$: Cosets and Lagrange's Theorem: Exercise $10$
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 $\HH$. :520px 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 == * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Definition:Symmetry Group of Regular Hexagon|next = Symmetry Group of Regular Hexagon/Examples/Subgroup that Fixes C}}: Chapter $2$: Subgroups and Cosets: $\S 35 \zeta$
Composition of Distance-Preserving Mappings is Distance-Preserving
Composition of Distance-Preserving Mappings is Distance-Preserving 0 72279 375364 2018-11-08T10:30:08Z Leigh.Samphier 3031 Created page with " == Theorem ==  Let: : $\struct{X_1, d_1}$ : $\struct{X_2, d_2}$ : $\struct{X_3, d_3}$ be metric spaces.  Let: : $\phi: \..." wikitext text/x-wiki  == Theorem ==  Let: : $\struct{X_1, d_1}$ : $\struct{X_2, d_2}$ : $\struct{X_3, d_3}$ be metric spaces. Let: : $\phi: \struct{X_1, d_1} \to \struct{X_2, d_2}$ : $\psi: \struct{X_2, d_2} \to \struct{X_3, d_3}$ be distance-preserving mappings. Then the composite of $\phi$ and $\psi$ is also a distance-preserving mapping.  == Proof == Let $x,y \in X_1$ then: {{begin-eqn}} {{eqn|l= d_1 \paren {x,y}      |r= d_2 \paren {\map \phi x, \map \phi y }      |c= $\phi$ is a distance-preserving mapping }} {{eqn|r= d_3 \paren {\map \psi {\map \phi x}, \map \psi {\map \phi y} }      |c= $\psi$ is a distance-preserving mapping }} {{eqn|r= d_3 \paren {\map {\psi \circ \phi} x, \map {\psi \circ \phi} y }      |c= Definition of composite mappings }} {{end-eqn}} By the definition of a distance-preserving mapping then $\psi \circ \phi$ is distance-preserving.
First Sylow Theorem/Corollary
== Corollary to First Sylow Theorem ==  Let $p$ be a prime number. Let $G$ be a group. Let: :$p^n \divides \order G$ where: :$\order G$ denotes the order of $G$ :$n$ is a positive integer. Then $G$ has at least one subgroup of order $p$.  == Proof 1 == {{:First Sylow Theorem/Corollary/Proof 1}} == Proof 2 == This result can also be proved directly: {{:First Sylow Theorem/Corollary/Proof 2}} == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 1|next = Sylow Theorems/Historical Note}}: $\S 44$. Some consequences of Lagrange's Theorem
Real Square Function is not Surjective
== Example of Mapping which is Not a Surjection ==  Let $f: \R \to \R$ be the real square function: :$\forall x \in \R: \map f x = x^2$ Then $f$ is not a surjection.  == Proof == For $f$ to be a surjection, it would be necessary that: :$\forall y \in \R: \exists x \in \R: \map f x = y$ However from Square of Real Number is Non-Negative: :$\forall y \in \R_{< 0}: \nexists x \in \R: \map f x = y$ Hence $f$ is not a surjection.
Monoid/Examples/x+y+xy on Reals
== Example of Monoid ==  Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as: :$\forall x, y \in \R: x \circ y := x + y + x y$ Then $\struct {\R, \circ}$ is a monoid whose identity is $0$.  == Proof == We have that: :$\forall x, y \in \R: x \circ y \in \R$ and so $\struct {\R, \circ}$ is closed. Now let $x, y, z \in \R$. We have: {{begin-eqn}} {{eqn | l = x \circ \paren {y \circ z}       | r = x + \paren {y \circ z} + x \paren {y \circ z}       | c = Definition of $\circ$ }} {{eqn | r = x + \paren {y + z + y z} + x \paren {y + z + y z}       | c = Definition of $\circ$ }} {{eqn | r = x + y + z + y z + x y + x z + x y z       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ z       | r = \paren {x \circ y} + z + \paren {x \circ y} z       | c = Definition of $\circ$ }} {{eqn | r = \paren {x + y + x y} + z + \paren {x + y + x y} z       | c = Definition of $\circ$ }} {{eqn | r = x + y + x y + z + x z + y z + x y z       | c =  }} {{end-eqn}} As can be seen by inspection: :$x \circ \paren {y \circ z} = \paren {x \circ y} \circ z$ and so $\circ$ is associative. Then we have: {{begin-eqn}} {{eqn | l = x \circ 0       | r = x + 0 + x \times 0       | c = Definition of $\circ$ }} {{eqn | r = x       | c =  }} {{eqn | r = 0 + x + 0 \times x       | c =  }} {{eqn | r = 0 \circ x       | c = Definition of $\circ$ }} {{end-eqn}} The result follows by definition of monoid.  == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Product of Semigroup Element with Right Inverse is Idempotent|next = Group/Examples/x+y+xy over Reals less -1}}: Chapter $4$. Groups: Exercise $11$
Symmetric Group on 3 Letters/Subgroups
== Subgroups of the Symmetric Group on $3$ Letters == Let $S_3$ denote the Symmetric Group on $3$ Letters, whose Cayley table is given as: {{:Symmetric Group on 3 Letters/Cayley Table}}  The subsets of $S_3$ which form subgroups of $S_3$ are: {{begin-eqn}} {{eqn | o =        | r = S_3 }} {{eqn | o =        | r = \set e }} {{eqn | o =        | r = \set {e, \tuple {123}, \tuple {132} } }} {{eqn | o =        | r = \set {e, \tuple {12} } }} {{eqn | o =        | r = \set {e, \tuple {13} } }} {{eqn | o =        | r = \set {e, \tuple {23} } }} {{end-eqn}}  == Examples == {{:Symmetric Group on 3 Letters/Subgroups/Examples}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Integer Multiples under Addition form Subgroup of Integers|next = Intersection of Subgroups/General Result}}: $\S 5.2$. Subgroups: Example $93$ * {{BookReference|Problems in Group Theory|1967|John D. Dixon|prev = Symmetric Group on 3 Letters/Group Presentation|next = Quaternion Group/Complex Matrices}}: $1$: Subgroups: Problem $1.1$ * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Conjugate of Subgroup is Subgroup|next = Symmetric Group on 3 Letters/Normal Subgroups}}: Chapter $7$: Normal subgroups and quotient groups: Exercise $2$
Alternating Group on 4 Letters
== Group Example == Let $S_4$ denote the symmetric group on $4$ letters.  The '''alternating group on $4$ letters''' $A_4$ is the kernel of the mapping $\sgn: S_4 \to C_2$.  === Cycle Notation === It can be expressed in the form of permutations given in cycle notation as follows: {{:Alternating Group on 4 Letters/Cycle Notation}} === Cayley Table === The Cayley table of $A_4$ can be written: {{:Alternating Group on 4 Letters/Cayley Table}} == Order of Elements == {{:Alternating Group on 4 Letters/Order of Elements}} == Subgroups == {{:Alternating Group on 4 Letters/Subgroups}} == Normality of Subgroups == {{:Alternating Group on 4 Letters/Normality of Subgroups}} == Conjugacy Classes == {{:Alternating Group on 4 Letters/Conjugacy Classes}}
Alternating Group on 4 Letters/Cayley Table
== Cayley Table of Alternating Group on $4$ Letters == The Cayley table of the alternating group on $4$ letters can be written:  :$\begin{array}{c|cccc|cccc|cccc} \circ & e & t & u & v & a & b & c & d & p & q & r & s \\ \hline e & e & t & u & v & a & b & c & d & p & q & r & s \\ t & t & e & v & u & b & a & d & c & q & p & s & r \\ u & u & v & e & t & c & d & a & b & r & s & p & q \\ v & v & u & t & e & d & c & b & a & s & r & q & p \\ \hline a & a & c & d & b & p & r & s & q & e & u & v & t \\ b & b & d & c & a & q & s & r & p & t & v & u & e \\ c & c & a & b & d & r & p & q & s & u & e & t & v \\ d & d & b & a & c & s & q & p & r & v & t & e & u \\ \hline p & p & s & q & r & e & v & t & u & a & d & b & c \\ q & q & r & p & s & t & u & e & v & b & c & a & d \\ r & r & q & s & p & u & t & v & e & c & b & d & a \\ s & s & p & r & q & v & e & u & t & d & a & c & b \\ \end{array}$  where the expression for $A_4$ in cycle notation is given as: {{:Alternating Group on 4 Letters/Cycle Notation}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Modulo Addition/Cayley Table/Modulo 6|next = Definition:Klein Four-Group}}: Tables: $3$. Alternating group $\map A 4$
Alternating Group on 4 Letters/Cycle Notation
== Cycle Notation for Alternating Group on $4$ Letters == The alternating group on $4$ letters can be given in cycle notation as follows:  {{begin-eqn}} {{eqn | l = e       | o = :=       | r = \text { the identity mapping} }} {{eqn | l = t       | o = :=       | r = \tuple {1 2} \tuple {3 4} }} {{eqn | l = u       | o = :=       | r = \tuple {1 3} \tuple {2 4} }} {{eqn | l = v       | o = :=       | r = \tuple {1 4} \tuple {2 3} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a       | o = :=       | r = \tuple {1 2 3} }} {{eqn | l = b       | o = :=       | r = \tuple {1 3 4} }} {{eqn | l = c       | o = :=       | r = \tuple {2 4 3} }} {{eqn | l = d       | o = :=       | r = \tuple {1 4 2} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = p       | o = :=       | r = \tuple {1 3 2} }} {{eqn | l = q       | o = :=       | r = \tuple {2 3 4} }} {{eqn | l = r       | o = :=       | r = \tuple {1 2 4} }} {{eqn | l = s       | o = :=       | r = \tuple {1 4 3} }} {{end-eqn}}  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Symmetric Group on 4 Letters/Subgroups/Examples/Disjoint Transpositions|next = Alternating Group on 4 Letters/Order of Elements}}: Chapter $9$: Permutations: Exercise $3$
Symmetric Group on 3 Letters/Normal Subgroups
== Normal Subgroups of the Symmetric Group on 3 Letters == Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as: {{:Symmetric Group on 3 Letters/Cayley Table}}  Consider the subgroups of $S_3$: {{:Symmetric Group on 3 Letters/Subgroups}} Of those, the normal subgroups in $S_3$ are: :$S_3, \set e, \set {e, \tuple {123}, \tuple {132} }$  == Proof == $S_3$ itself is normal in $S_3$ by Group is Normal in Itself. $\set e$ is normal in $S_3$ by Trivial Subgroup is Normal. $\set {e, \tuple {12} }$: {{begin-eqn}} {{eqn | l = \tuple {123} \tuple {13} \tuple {123}^{-1}       | r = \tuple {123} \tuple {13} \tuple {132}       | c =  }} {{eqn | r = \tuple {123} \tuple {12}       | c =  }} {{eqn | r = \tuple {23}       | c =  }} {{eqn | o = \notin       | r = \set {e, \tuple {12} }       | c =  }} {{end-eqn}} Hence $\set {e, \tuple {12} }$ is not normal in $S_3$. $\set {e, \tuple {23} }$: {{begin-eqn}} {{eqn | l = \tuple {123} \tuple {23} \tuple {123}^{-1}       | r = \tuple {123} \tuple {23} \tuple {132}       | c =  }} {{eqn | r = \tuple {123} \tuple {13}       | c =  }} {{eqn | r = \tuple {12}       | c =  }} {{eqn | o = \notin       | r = \set {e, \tuple {23} }       | c =  }} {{end-eqn}} Hence $\set {e, \tuple {23} }$ is not normal in $S_3$. $\set {e, \tuple {13} }$: {{begin-eqn}} {{eqn | l = \tuple {123} \tuple {13} \tuple {123}^{-1}       | r = \tuple {123} \tuple {13} \tuple {132}       | c =  }} {{eqn | r = \tuple {123} \tuple {12}       | c =  }} {{eqn | r = \tuple {23}       | c =  }} {{eqn | o = \notin       | r = \set {e, \tuple {13} }       | c =  }} {{end-eqn}} Hence $\set {e, \tuple {13} }$ is not normal in $S_3$. $\set {e, \tuple {123}, \tuple {132} }$: We have that $\set {e, \tuple {123}, \tuple {132} }$ is the set of even permutations of $S_3$. Any permutation of the form $\alpha \pi \alpha^{-1}$, for $\pi$ even, is also even. Thus: :$\forall \alpha \in S_3: \alpha \pi \alpha^{-1} \in \set {e, \tuple {123}, \tuple {132} }$ Hence $\set {e, \tuple {123}, \tuple {132} }$ is normal in $S_3$.  == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Subgroup is Superset of Conjugate iff Normal|next = Group is Normal in Itself}}: $\S 6.6$. Normal subgroups: Example $122$ * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Symmetric Group on 3 Letters/Subgroups|next = Subgroups of Quaternion Group}}: Chapter $7$: Normal subgroups and quotient groups: Exercise $2$
Modulo Addition/Cayley Table/Modulo 3
== Cayley Table for Addition Modulo $3$ == The additive group of integers modulo $3$ can be described by showing its Cayley table:  :$\begin{array}{r|rrr} \struct {\Z_3, +_3} & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\ \hline \eqclass 0 3 & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\ \eqclass 1 3 & \eqclass 1 3 & \eqclass 2 3 & \eqclass 0 0 \\ \eqclass 2 3 & \eqclass 2 3 & \eqclass 0 3 & \eqclass 1 3 \\ \end{array}$  It can also be presented: :$\begin{array}{r|rrr} +_3 & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \\ \end{array}$ == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Quotient Group of Integers by Multiples|next = Klein Four-Group is Normal in A4}}: $\S 6.7$. Quotient groups: Example $126$ * {{BookReference|Topology: An Introduction with Application to Topological Groups|1967|George McCarty|prev = Definition:Parity of Integer|next = Definition:Quotient Group/Motivation}}: Chapter $\text{II}$: Groups: A Little Number Theory * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Integers Modulo m under Addition form Abelian Group|next = Mathematician:Arthur Cayley}}: Chapter $2$: Maps and relations on sets: Example $2.33$
Alternating Group on 4 Letters/Subgroups
== Subgroups of the Alternating Group on $4$ Letters == Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as: {{:Alternating Group on 4 Letters/Cayley Table}}  The subsets of $A_4$ which form subgroups of $A_4$ are as follows: Trivial: {{begin-eqn}} {{eqn | o =        | r = \set e       | c = Trivial Subgroup is Subgroup }} {{eqn | o =        | r = A_4       | c = Group is Subgroup of Itself }} {{end-eqn}} Order $2$ subgroups: {{begin-eqn}} {{eqn | o =        | r = \set {e, t}       | c = as $t^2 = e$ }} {{eqn | o =        | r = \set {e, u}       | c = as $u^2 = e$ }} {{eqn | o =        | r = \set {e, v}       | c = as $v^2 = e$ }} {{end-eqn}} Order $3$ subgroups: {{begin-eqn}} {{eqn | o =        | r = \set {e, a, p}       | c = as $a^2 = p$, $a^3 = a p = e$ }} {{eqn | o =        | r = \set {e, b, s}       | c = as $b^2 = s$, $b^3 = b s = e$ }} {{eqn | o =        | r = \set {e, c, q}       | c = as $c^2 = q$, $c^3 = c q = e$ }} {{eqn | o =        | r = \set {e, d, r}       | c = as $d^2 = r$, $d^3 = d r = e$ }} {{end-eqn}} Order $4$ subgroup: {{begin-eqn}} {{eqn | o =        | r = \set {e, t, u, v}       | c = Klein $4$-Group }} {{end-eqn}}  == Examples of Subgroups == {{:Alternating Group on 4 Letters/Subgroups/Examples}} == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Alternating Group on 4 Letters/Order of Elements|next = Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 2}}: Chapter $9$: Permutations: Exercise $3$
Dihedral Group D4/Matrix Representation/Formulation 1
== Matrix Representation of Dihedral Group $D_4$ ==  Let $\mathbf I, \mathbf A, \mathbf B, \mathbf C$ denote the following four elements of the matrix space $\map {\mathcal M_\Z} 2$: :$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad \mathbf A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \qquad \mathbf B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \qquad \mathbf C = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ The set: :$D_4 = \set {\mathbf I, -\mathbf I, \mathbf A, -\mathbf A, \mathbf B, -\mathbf B, \mathbf C, -\mathbf C}$ under the operation of conventional matrix multiplication, forms the '''dihedral group $D_4$'''.  === Cayley Table === Its Cayley table is given by: {{:Dihedral Group D4/Matrix Representation/Formulation 1/Cayley Table}} == Also see == * Dihedral Group D4 Defined by Matrices where it is shown that these have the appropriate properties. == Sources ==
Dihedral Group D4/Matrix Representation/Formulation 1/Cayley Table
== Cayley Table for Dihedral Group $D_4$ == The Cayley table for the dihedral group $D_4$: :$D_4 = \set {\mathbf I, -\mathbf I, \mathbf A, -\mathbf A, \mathbf B, -\mathbf B, \mathbf C, -\mathbf C}$ under the operation of conventional matrix multiplication, where: :$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad \mathbf A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \qquad \mathbf B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \qquad \mathbf C = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ can be presented as:  :$\begin{array}{r|rrrrrrrr}            &  \mathbf I &  \mathbf A &  \mathbf B &  \mathbf C & -\mathbf I & -\mathbf A & -\mathbf B & -\mathbf C \\ \hline  \mathbf I &  \mathbf I &  \mathbf A &  \mathbf B &  \mathbf C & -\mathbf I & -\mathbf A & -\mathbf B & -\mathbf C \\  \mathbf A &  \mathbf A &  \mathbf I & -\mathbf C & -\mathbf B & -\mathbf A & -\mathbf I &  \mathbf C &  \mathbf B \\  \mathbf B &  \mathbf B &  \mathbf C & -\mathbf I & -\mathbf A & -\mathbf B & -\mathbf C &  \mathbf I &  \mathbf A \\  \mathbf C &  \mathbf C &  \mathbf B &  \mathbf A &  \mathbf I & -\mathbf C & -\mathbf B & -\mathbf A & -\mathbf I \\ -\mathbf I & -\mathbf I & -\mathbf A & -\mathbf B & -\mathbf C &  \mathbf I &  \mathbf A &  \mathbf B &  \mathbf C \\ -\mathbf A & -\mathbf A & -\mathbf I &  \mathbf C &  \mathbf B &  \mathbf A &  \mathbf I & -\mathbf C & -\mathbf B \\ -\mathbf B & -\mathbf B & -\mathbf C &  \mathbf I &  \mathbf A &  \mathbf B & -\mathbf C & -\mathbf I & -\mathbf A \\ -\mathbf C & -\mathbf C & -\mathbf B & -\mathbf A & -\mathbf I &  \mathbf C &  \mathbf B &  \mathbf A &  \mathbf I \end{array}$  == Sources ==
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 $\Dic 2$: {{:Quaternion Group/Cayley Table}} 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.  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Dihedral Group D4/Matrix Representation/Formulation 2/Examples of Generated Subgroups/B, F|next = Definition:Left Coset}}: Chapter $4$: Subgroups: Exercise $4$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Generator of Group|next = Definition:Cyclic Group/Definition 2|entry = generator|index = 2}}
Bound for Analytic Function and Derivatives/Analytic Function Bounded on Circle
== Lemma ==  Let $f$ be a complex function. Let $z_0$ be a point in $\C$. Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius in $\R_{>0}$. Let $f$ be analytic on $\Gamma$. Then $f$ is bounded on $\Gamma$.  == Proof == Let: :$\map {f_{\Re} } z = \map \Re {\map f z}$ :$\map {f_{\Im} } z = \map \Im {\map f z}$ Let $\closedint a b$, $a < b$, be a closed real interval. Let $p$ be a continuous complex-valued function defined such that: :$\Gamma = \set {\map p u: u \in \closedint a b}$ $f$ is continuous on $\Gamma$ as $f$ is analytic on $\Gamma$ by the definition of analytic. Also, Real and Imaginary Part Projections are Continuous. Therefore, $f_{\Re}$ and $f_{\Im}$ are continuous by the corollary to Composite of Continuous Mappings is Continuous. Observe that $f_{\Re}$ and $f_{\Im}$ are real-valued functions that are continuous. Also, $p$ is a continuous function defined on a set of real numbers. Therefore $\map {f_{\Re} } {\map p u}$ and $\map {f_{\Im} } {\map p u}$ are continuous real functions by the corollary to Composite of Continuous Mappings is Continuous. $\map {f_{\Re} } {\map p u}$ and $\map {f_{\Im} } {\map p u}$ are bounded on $\closedint a b$ by Continuous Real Function is Bounded. Therefore $\map f {\map p u}$ is bounded on $\closedint a b$ as $\map f {\map p u} = \map {f_{\Re} } {\map p u} + i \map {f_{\Im} } {\map p u}$ where $i = \sqrt {-1}$. Accordingly, $f$ is bounded on $\Gamma$ as $\Gamma = \set {\map p u: u \in \closedint a b}$.
Convergence of Taylor Series of Function Analytic on Disk/Lemma
{{refactor|This (or something very similar) already exists somewhere as a basic null sequence}} == Lemma ==  Let $y > 1$. Then: :$\displaystyle \lim_{n \mathop \to \infty} \frac n {y^n} = 0$  == Proof == Note that $\ln y > 0$ as $y > 1$. {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \frac n {y^n}       | r = \lim_{n \mathop \to \infty} \frac n {\left({e^{\ln y} }\right)^n} }} {{eqn | r = \lim_{n \mathop \to \infty} \frac n {e^{\left({\ln y}\right) n} } }} {{eqn | r = 0       | c = as $\displaystyle \lim_{x \mathop \to \infty} \frac x {e^{\left({\ln y}\right) x} } = 0$ by Limit at Infinity of Polynomial over Complex Exponential as $\ln y > 0$ }} {{end-eqn}}
Sum of Terms of Magic Cube/Sequence
== Sequence of Sums of Terms of Magic Cubes ==  The sequence of the sum totals of all the entries in magic cubes of order $n$ begins: :$1, \paren {36,} \, 378, 2080, 7875, 23 \, 436, 58 \, 996, 131 \, 328, \ldots$ However, note that while $36 = \dfrac {2^3 \paren {2^3 + 1} } 2$, a magic cube of order $2$ does not actually exist.  {{OEIS|A037270}}
Magic Constant of Magic Cube/Sequence
== Sequence of Magic Constants of Magic Cubes ==  The sequence of the magic constants of magic cubes of order $n$ begins: :$1, (9,) \, 42, 130, 315, 651, 1204, 2052, 3285, 5005, 7326, 10 \, 374, 14 \, 287, 19 \, 215, 25 \, 320, 32 \, 776, \ldots$ However, note that while $9 = \dfrac {2 \paren {2^3 + 1} } 2$, a magic cube of order $2$ does not actually exist.  {{OEIS|A027441}}