hcju/proofwikips huggingface dataset:
Sequence Lemma
  == Lemma == Let $A$ be a subset of a topological space $X$. If there is a sequence of points of $A$ converging to $x$, then $x \in \bar A$. The converse holds if $X$ is first-countable. == Proof == Assume the sequence of points of $A$ that converges to $x$ is $\sequence {x_i}$. Then for any open set $U$ of $x$, there exists a positive natural number $N$ such that when $i > N$, $x_i \in U$. Thus $U \cap A$ is nonempty, $x \in \bar A$. Let the topological space $X$ be first-countable. Then there is a countable collection of open neighbourhood $\family {U_i}_{i \mathop \in \Bbb Z_+}$ of $x$ such that any open neighbourhood $U$ of $x$ contains at least one of the sets $U_i$. Because $x \in \bar A$, $U_1 \cap A$ is nonempty, we can select a point $x_1$ in it. In a similar manner, $U_1 \cap U_2 \cap A$ is nonempty. Hence we can select a point $x_2$ in it. The point $x_i$ is selected from: :$U_1 \cap U_2 \cap \cdots \cap U_i \cap A$ We then obtain a sequence $\sequence {x_i}$. For any open neighbourhood $U$ of $x$, it contains at least one of the set $U_N$, $N \in \Bbb Z_+$ of $\family {U_i}_{i \mathop \in \Bbb Z_+}$. Thus it contains the set: :$U_1 \cap U_2 \cap \cdots \cap U_N \cap A$ :$U_1 \cap U_2 \cap \cdots \cap U_N \cap U_{N + 1}\cap A$ :$U_1 \cap U_2 \cap \cdots \cap U_N \cap U_{N + 1} \cap U_{N + 2} \cap A$ :$\ldots$ or the set $U_1 \cap U_2 \cap \cdots \cap U_i \cap A$ with $i > N$, hence the points $x_i$ with $i > N$. The sequence $\sequence {x_i}$ converges to $x$.  == Sources == * {{BookReference|Topology|2004|James R. Munkres}} P130, Lemma 21.2
Group Direct Product/Examples/C2 x C2
== Example of Group Direct Product ==  The direct product of $C_2$, the cyclic group of order $2$, with itself is as follows. Let us represent $C_2$ as the group $\struct {\set {1, -1}, \times}$: :$\begin {array} {r|rr} \struct {\set {1, -1} , \times} & 1 & -1 \\ \hline 1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{array}$ Then the Cayley table for $C_2 \times C_2$ can be portrayed as: :$\begin {array} {c|cccc} C_2 \times C_2  & \tuple { 1,  1} & \tuple { 1, -1} & \tuple {-1,  1} & \tuple {-1, -1} \\ \hline \tuple { 1,  1} & \tuple { 1,  1} & \tuple { 1, -1} & \tuple {-1,  1} & \tuple {-1, -1} \\ \tuple { 1, -1} & \tuple { 1, -1} & \tuple { 1,  1} & \tuple {-1, -1} & \tuple {-1,  1} \\ \tuple {-1,  1} & \tuple {-1,  1} & \tuple {-1, -1} & \tuple { 1,  1} & \tuple { 1, -1} \\ \tuple {-1, -1} & \tuple {-1, -1} & \tuple {-1,  1} & \tuple { 1, -1} & \tuple { 1,  1} \\ \end{array}$  This is seen by inspection to be an instance of the Klein $4$-group. == Subgroups == {{:Group Direct Product/Examples/C2 x C2/Subgroups}} == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = External Direct Product of Groups is Group|next = Definition:Multiplicative Notation}}: Chapter $1$: Definitions and Examples: Example $1.10$
Dihedral Group D4/Matrix Representation/Formulation 2/Cayley Table
== Cayley Table for Dihedral Group $D_4$ == The Cayley table for the dihedral group $D_4$: :$D_4 = \set {\mathbf I, \mathbf A, \mathbf B, \mathbf C, \mathbf D, \mathbf E, \mathbf F, \mathbf G}$ under the operation of conventional matrix multiplication, where: :$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad \mathbf A = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \qquad \mathbf B = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \qquad \mathbf C = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix}$ :$\mathbf D = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \qquad \mathbf E = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} \qquad \mathbf F = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \qquad \mathbf G = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$ can be presented as:  :$\begin{array}{r|rrrrrrrr}           & \mathbf I & \mathbf A & \mathbf B & \mathbf C & \mathbf D & \mathbf E & \mathbf F & \mathbf G \\ \hline \mathbf I & \mathbf I & \mathbf A & \mathbf B & \mathbf C & \mathbf D & \mathbf E & \mathbf F & \mathbf G \\ \mathbf A & \mathbf A & \mathbf B & \mathbf C & \mathbf I & \mathbf E & \mathbf F & \mathbf G & \mathbf D \\ \mathbf B & \mathbf B & \mathbf C & \mathbf I & \mathbf A & \mathbf F & \mathbf G & \mathbf D & \mathbf E \\ \mathbf C & \mathbf C & \mathbf I & \mathbf A & \mathbf B & \mathbf G & \mathbf D & \mathbf E & \mathbf F \\ \mathbf D & \mathbf D & \mathbf G & \mathbf F & \mathbf E & \mathbf I & \mathbf C & \mathbf B & \mathbf A \\ \mathbf E & \mathbf E & \mathbf D & \mathbf G & \mathbf F & \mathbf A & \mathbf I & \mathbf C & \mathbf B \\ \mathbf F & \mathbf F & \mathbf E & \mathbf D & \mathbf G & \mathbf B & \mathbf A & \mathbf I & \mathbf C \\ \mathbf G & \mathbf G & \mathbf F & \mathbf E & \mathbf D & \mathbf C & \mathbf B & \mathbf A & \mathbf I \end{array}$  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Square Matrices with +1 or -1 Determinant under Multiplication forms Group|next = Symmetry Group of Square/Cayley Table}}: Chapter $1$: Definitions and Examples: Exercise $2$
Dihedral Group D4/Group Presentation
== Group Presentation of Dihedral Group $D_4$ == The group presentation of the dihedral group $D_4$ is given by:  :$D_4 = \gen {a, b: a^4 = b^2 = e, a b = b a^{-1} }$  == Proof == We have that the group presentation of the dihedral group $D_n$ is: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Setting $n = 4, \alpha = a, \beta = b$, we get: :$D_4 = \gen {a, b: a^4 = b^2 = e, b a b = a^{−1} }$ from which the result follows.
Dihedral Group D3/Group Presentation
== Group Presentation of Dihedral Group $D_3$ == The group presentation of the dihedral group $D_3$ is given by:  :$D_3 = \gen {a, b: a^3 = b^2 = e, a b = b a^{-1} }$  == Proof == We have that the group presentation of the dihedral group $D_n$ is: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Setting $n = 3, \alpha = a, \beta = b$, we get: :$D_3 = \gen {a, b: a^3 = b^2 = e, b a b = a^{−1} }$ from which the result follows.
Dihedral Group D3/Cayley Table
Dihedral Group D3/Cayley Table 0 73563 386386 2019-01-05T09:08:29Z Prime.mover 59 Created page with "== Cayley Table of Dihedral Group $D_3$ ==  The Cayley table of the Definition:Dihed..." wikitext text/x-wiki == Cayley Table of Dihedral Group $D_3$ == The Cayley table of the dihedral group $D_3$ can be written:  :$\begin{array}{c|cccccc}       & e     & a     & a^2   & b     & a b   & a^2 b \\ \hline e     & e     & a     & a^2   & b     & a b   & a^2 b \\ a     & a     & a^2   & e     & a b   & a^2 b & b     \\ a^2   & a^2   & e     & a     & a^2 b & b     & a b   \\ b     & b     & a^2 b & a b   & e     & a^2   & a     \\ a b   & a b   & b     & a^2 b & a     & e     & a^2   \\ a^2 b & a^2 b & a b   & b     & a^2   & a     & e     \\ \end{array}$  where the group presentation of $D_3$ is given as: {{:Group Presentation of Dihedral Group D3}}
Standard Parity Check Matrix/Examples/(6, 3) code in Z2
== Example of Standard Parity Check Matrix ==  Let $C$ be the linear $\tuple {6, 3}$-code in $\Z_2$ whose standard generator matrix $G$ is given by: :$G := \begin{pmatrix} 1 & 0 & 0 & 1 & 1 & 0 \\  0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix}$ Its standard parity check matrix $P$ is given by: :$P := \begin{pmatrix} 1 & 1 & 0 & 1 & 0 & 0 \\  1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \end{pmatrix}$  == Proof == Expressing $G$ in the form: :$G = \paren {\begin{array} {c|c} \mathbf I_k & \mathbf A \end{array} }$ it is seen that: :$\mathbf A = \begin{pmatrix} 1 & 1 & 0 \\  1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ It is noted that $\mathbf A^\intercal$ is: :$\mathbf A^\intercal =  \begin{pmatrix} 1 & 1 & 0 \\  1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ as $\mathbf A$ is symmetrical about the main diagonal. Then each of the elements of $\Z_2$ is self-inverse, so: :$-\mathbf A^\intercal = \mathbf A^\intercal$  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Definition:Standard Parity Check Matrix|next = Definition:Syndrome}}: Chapter $6$: Error-correcting codes: Example $6.18$
Syndrome Decoding/Examples/(6, 3) code in Z2
Syndrome Decoding/Examples/(6, 3) code in Z2 0 73802 387561 2019-01-13T13:16:59Z Prime.mover 59 Created page with "== Example of Syndrome Decoding ==  Let $C$ be the linear code:  :$C = \set {000000, 100110, 010101, 110011, 001011, 101101, 011110..." wikitext text/x-wiki == Example of Syndrome Decoding ==  Let $C$ be the linear code: :$C = \set {000000, 100110, 010101, 110011, 001011, 101101, 011110, 111000}$ Then the Syndrome Decoding table $T$ for $C$ is: :$\begin{array} {cc} 000000 & 000 \\ 100000 & 110 \\ 010000 & 101 \\ 001000 & 011 \\ 000100 & 100 \\ 000010 & 010 \\ 000001 & 001 \\ 100001 & 111 \\ \end{array}$  To find the $8$th row, it is not necessary to hunt directly an element $v$ of $\map V {6, 2}$ of weight $2$ which does not exist anywhere in the other $7$ rows. What you do is identify the remaining syndrome of $3$ digits that has not been used yet (that is: $111$). Then you work out what combinations of coset leaders and their own syndromes which when added together make that last syndrome. The fact that in this case that combination is not unique (here we get $100001$, $010010$, $001100$ and $000111$) means that an element of $\map V {6, 2}$ which is more than $2$ distant from a codeword cannot be uniquely decoded. == Example == {{:Syndrome Decoding/Examples/(6, 3) code in Z2/Example}} == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Syndrome Decoding|next = Syndrome Decoding/Examples/(6, 3) code in Z2/Example}}: Chapter $6$: Error-correcting codes: Example $6.22$
Dihedral Group D4/Cayley Table
== Cayley Table for Dihedral Group $D_4$ == The Cayley table for the dihedral group $D_4$, whose group presentation is: {{:Group Presentation of Dihedral Group D4}} can be presented as:  :$\begin{array}{l|cccccccc}        &     e &     a &   a^2 &   a^3 &     b &   b a & b a^2 & b a^3 \\ \hline  e     &     e &     a &   a^2 &   a^3 &     b &   b a & b a^2 & b a^3 \\  a     &     a &   a^2 &   a^3 &     e & b a^3 &     b &   b a & b a^2 \\  a^2   &   a^2 &   a^3 &     e &     a & b a^2 & b a^3 &     b &   b a \\  a^3   &   a^3 &     e &     a &   a^2 &   b a & b a^2 & b a^3 &     b \\  b     &     b &   b a & b a^2 & b a^3 &     e &     a &   a^2 &   a^3 \\  b a   &   b a & b a^2 & b a^3 &     b &   a^3 &     e &     a &   a^2 \\  b a^2 & b a^2 & b a^3 &     b &   b a &   a^2 &   a^3 &     e &     a \\  b a^3 & b a^3 &     b &   b a & b a^2 &     a &   a^2 &   a^3 &     e \end{array}$  === 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: {{:Dihedral Group D4/Cayley Table/Coset Decomposition of (e, a^2)}} === Coset Decomposition of $\set {e, a, a^2, a^3}$ === Presenting the above Cayley table with respect to the coset decomposition of the normal subgroup $\gen a$ gives: {{:Dihedral Group D4/Cayley Table/Coset Decomposition of (e, a, a^2, a^3)}} == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Dihedral Group D4/Normal Subgroups/Subgroup Generated by a^2/Quotient Group|next = Dihedral Group D4/Cayley Table/Coset Decomposition of (e, a^2)}}: Chapter $7$: Normal subgroups and quotient groups: Example $7.13$
Quaternion Group/Subgroups
== Subgroups of the Quaternion Group == Let $Q$ denote the quaternion group, whose group presentation is given as: {{:Group Presentation of Quaternion Group}}  The subsets of $Q$ which form subgroups of $Q$ are: {{begin-eqn}} {{eqn | o =        | r = Q }} {{eqn | o =        | r = \set e }} {{eqn | o =        | r = \set {e, a^2} }} {{eqn | o =        | r = \set {e, a, a^2, a^3} }} {{eqn | o =        | r = \set {e, b, a^2, a^2 b} }} {{eqn | o =        | r = \set {e, a b, a^2, a^3 b} }} {{end-eqn}} From Quaternion Group is Hamiltonian we have that all of these subgroups of $Q$ are normal.  == Proof == Consider the Cayley table for $Q$: {{:Quaternion Group/Cayley Table}} 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 == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Symmetric Group on 3 Letters/Normal Subgroups|next = Quaternion Group is Hamiltonian}}: Chapter $7$: Normal subgroups and quotient groups: Exercise $3$
Dihedral Group D4/Subgroups
== Subgroups of the Dihedral Group $D_4$ == Let the dihedral group $D_4$ be represented by its group presentation: {{:Group Presentation of Dihedral Group D4}}  The subsets of $D_4$ which form subgroups of $D_4$ are: {{begin-eqn}} {{eqn | o =        | r = D_4 }} {{eqn | o =        | r = \set e }} {{eqn | o =        | r = \set {e, a, a^2, a^3} }} {{eqn | o =        | r = \set {e, a^2} }} {{eqn | o =        | r = \set {e, b} }} {{eqn | o =        | r = \set {e, b a} }} {{eqn | o =        | r = \set {e, b a^2} }} {{eqn | o =        | r = \set {e, b a^3} }} {{eqn | o =        | r = \set {e, a^2, b, b a^2} }} {{eqn | o =        | r = \set {e, a^2, b a, b a^3} }} {{end-eqn}}  == Proof == Consider the Cayley table for $D_4$: {{:Dihedral Group D4/Cayley Table}} We have that: :$a^4 = e$ and so $\gen a = \set {e, a, a^2, a^3}$ forms a subgroup of $D_4$ which is cyclic. We have that: :$\paren {a^2}^2 = e$ and so $\gen {a^2} = \set {e, a^2}$ forms a subgroup of $D_4$ which is cyclic, and also a subgroup of $\gen a$. We have that: :$b^2 = e$ and so $\gen b = \set {e, b}$ forms a subgroup of $D_4$ which is cyclic. We have that: :$\paren {b a}^2 = e$ and so $\gen {b a} = \set {e, b a}$ forms a subgroup of $D_4$ which is cyclic. We have that: :$\paren {b a^2}^2 = e$ and so $\gen {b a^2} = \set {e, b a^2}$ forms a subgroup of $D_4$ which is cyclic. We have that: :$\paren {b a^3}^2 = e$ and so $\gen {b a^3} = \set {e, b a^3}$ forms a subgroup of $D_4$ which is cyclic. Then we have that: :$b a^2 = a^2 b$ and so $\gen {a^2, b} = \set {e, a^2, b, b a^2}$ forms a subgroup of $D_4$ which is not cyclic, but which has subgroups $\set {e, a^2}$, $\set {e, b}$, $\set {e, b a^2}$. Then we have that: :$b a^3 = a^2 b a$ and so $\gen {a^2, b a} = \set {e, a^2, b a, b a^3}$ forms a subgroup of $D_4$ which is not cyclic, but which has subgroups $\set {e, a^2}$, $\set {e, b}$, $\set {e, b a^2}$. That exhausts all elements of $D_4$. Any subgroup generated by any $2$ elements of $Q$ which are not both in the same subgroup as described above generate the whole of $D^4$.
Dihedral Group D4/Center
== Center of the Dihedral Group $D_4$ == Let $D_4$ denote the dihedral group $D_4$, whose group presentation is given as: {{:Group Presentation of Dihedral Group D4}}  The center of $D_4$ is given by: :$\map Z {D_4} = \set {e, a^2}$  == Proof == From Center of Dihedral Group: :$\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, \alpha^{n / 2} } & : n \text { even} \end{cases}$ Hence the result.  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Symmetric Group on 3 Letters/Center|next = Definition:Group Homomorphism}}: Chapter $7$: Normal subgroups and quotient groups: Exercise $5$
Group Action of Symmetric Group on Complex Vector Space/Orbit/Examples/Example 1
== Example of Orbit of Group Action of Symmetric Group on Complex Vector Space == Let $S_n$ denote the symmetric group on $n$ letters. Let $V$ denote a vector space over the complex numbers $\C$. Let $V$ have a basis: :$\mathcal B := \set {v_1, v_2, \ldots, v_n}$ Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as: :$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$ where: :$v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$  Let $n = 4$. Let $v = v_1 + v_2 + v_3 + v_4$. The orbit of $v$ is: :$\Orb v = \set v$  == Proof == {{begin-eqn}} {{eqn | l = \Orb v       | r = \set {w \in V: \exists \rho \in S_4: w = \rho * v}       | c = {{Defof|Orbit (Group Theory)|Orbit}} }} {{eqn | r = \set {w \in V: \exists \rho \in S_4: w = \rho * \sum_{k \mathop = 1}^4 v_k}       | c = Definition of $v$ }} {{eqn | r = \set {w \in V: \exists \rho \in S_4: w = \sum_{k \mathop = 1}^4 v_\map \rho k}       | c = Definition of $*$ }} {{eqn | r = \set {w \in V: \exists \rho \in S_4: w = \sum_{k \mathop = 1}^4 v_k}       | c = Permutation of Indices of Summation }} {{eqn | r = \set {w \in V: \exists \rho \in S_4: w = v}       | c = Definition of $v$ }} {{eqn | r = \set v       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Group Action of Symmetric Group on Complex Vector Space/Stabilizer|next = Group Action of Symmetric Group on Complex Vector Space/Stabilizer/Examples/Example 1}}: Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1 \ \text {(a)}$
Sylow Theorems/Examples/Sylow 3-Subgroups in Group of Order 12
Sylow Theorems/Examples/Sylow 3-Subgroups in Group of Order 12 0 74043 388972 2019-01-21T21:07:43Z Prime.mover 59 Created page with "== Example of Use of Sylow Theorems ==  In a group of order $12$, there are either $1$ or $4$ Definition:..." wikitext text/x-wiki == Example of Use of Sylow Theorems ==  In a group of order $12$, there are either $1$ or $4$ Sylow $3$-subgroups.  == Proof == Let $G$ be a group of order $12$. Let $n_3$ be the number of Sylow $3$-subgroups in $G$. From the Fourth Sylow Theorem, $n_3$ is congruent to $1$ modulo $3$, that is, in $\set {1, 4, 7, \ldots}$ Let $H$ be a Sylow $3$-subgroup of $G$. We have that: :$12 = 4 \times 3$ and so the order of $H$ is $3$. Thus: {{begin-eqn}} {{eqn | l = \index G H       | r = \dfrac {12} {3}       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} From the Fifth Sylow Theorem: :$n_3 \divides 4$ where $\divides$ denotes divisibility. Thus there may be $1$ or $4$ Sylow $3$-subgroups of $G$.  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Sylow Theorems/Examples/Sylow 2-Subgroups in Group of Order 12|next = Direct Product of Normal Subgroups is Normal}}: Chapter $11$: The Sylow Theorems: Exercise $1 \ \text{(e)}$
Pullback is Subgroup
 * Pullback of Quotient Group Isomorphism is Subgroup
Pullback of Quotient Group Isomorphism/Examples/Subgroups of Index 2
== Example of Pullback of Quotient Group Isomorphism ==  Let $G$ and $H$ be groups. Let $N$ and $K$ be normal subgroups of $G$ and $H$ respectively such that: :their quotient groups $G / N$ and $H / K$ are isomorphic :their indices are $2$: ::$\index G N = \index H K = 2$ Let $\theta: G / N \to H / K$ be an isomorphism. The pullback of $G$ and $H$ by $\theta$ is a subset of $G \times H$ of the form: :$G \times^\theta H = \set {\tuple {g, h}: \paren {g \in N, h \in K} \text { or } \paren {g \notin N, h \notin K} }$  == Proof == As $\index G N = \index H K = 2$, it follows that: :$\order {G / N} = \order {H / K} = 2$ and they are the cyclic group of order $2$. Let: : $x \in G: x \notin N$ : $y \in H: y \notin K$ Then: :$G / N = \gen {x N}$ :$H / K = \gen {y K}$ and we have: {{begin-eqn}} {{eqn | l = \map \theta N       | r = K }} {{eqn | l = \map \theta {x N}       | r = y K }} {{end-eqn}} Let $\tuple {g, h} \in G \times^\theta H$. By definition: :$G \times^\theta H = \set {\tuple {g, h}: \map \theta {g N} = h K}$ Let $g \in N$. Then $g N = N$ and so: : $\map \theta {g N} = h K = K$ So $g \in N \implies h \in K$. Let $g \notin N$. Then $g N = x N$ and so: : $\map \theta {g N} = h K = x K$ So $g \notin N \implies h \notin K$. Hence the result.  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Pullback of Quotient Group Isomorphism is Subgroup|next = Pullback of Quotient Group Isomorphism/Examples/Alternating Subgroups of Symmetric Groups}}: Chapter $13$: Direct products: Example $13.12$
Arcsin as an Integral/Lemma 1
Arcsin as an Integral/Lemma 1 0 74302 390719 390718 2019-01-30T14:50:47Z Pelliott 3564 sin_A inverse arcsin_a wikitext text/x-wiki == Lemma ==  Let $sin_A$ be the analytic sine function for real numbers, the one defined by Definition:Sine/Real Numbers. $\arcsin_A$ is the inverse of this function. :$\displaystyle \map {\arcsin_A} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$  == Proof == For this proof only, let $\sin_A$ be the analytic sine function from Definition:Sine/Real Numbers. Consider: :$\displaystyle \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$ Let: :$x = \sin_A \theta \iff x = \map {\arcsin_A} \theta$ Then: {{begin-eqn}} {{eqn | l = \d x       | r = \cos_A \theta \rd \theta       | c = Derivative of Sine Function }} {{eqn | l = \int \frac {\d x} {\sqrt {1 - x^2} }       | r = \int \frac {\d x} {\cos_A \theta} \cos_A \theta \rd \theta       | c = Integration by Substitution }} {{eqn | r = \int 1 \rd \theta }} {{eqn | r = \theta + C }} {{eqn | r = \map {\arcsin_A} x + C }} {{eqn | ll= \leadsto       | l = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }       | r = \map {\arcsin_A} x       | c = Fundamental Theorem of Calculus: Second Part }} {{end-eqn}} {{qed|lemma}} chldcq0gp73np0qqonhd2zb3nt5sh13
Arcsin as an Integral/Lemma 2
Arcsin as an Integral/Lemma 2 0 74304 390774 390773 2019-01-31T06:27:22Z Prime.mover 59 Reverted edits by Prime.mover (talk) to last revision by Pelliott wikitext text/x-wiki == Lemma ==  Let, $\sin_G$ be the Geometric Sine from Definition:Sine/Definition from Circle. $\arcsin_G$ is the inverse of this function. :$\displaystyle \map {\arcsin_G} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$  == Proof == This result will be used in proving Derivative of Sine Function in the geometric case. So we can not use the same reasoning as Arcsin as an Integral/Lemma 1 because our logic would be circular. :640px {{improve|Might be worth revisiting the diagram to make all the text the same size}} Let $\theta$ be the length of the arc associated with the angle on the circle of radius $1$. By definition of arcsine: :$y = \sin \theta \iff \theta = \arcsin y$ We have that arc length is always positive.  For negative $y$, the $\arcsin$ function is defined as being the negative of the arc length. This makes the $\arcsin$ function and the $\sin$ function odd, and puts us in line with mathematical convention: :Inverse Sine is Odd Function. :Sine Function is Odd Without this convention, the derivative of the $\sin$ function would not be continuous.  Now: {{begin-eqn}} {{eqn | l = x^2 + y^2       | r = 1       | c = Equation of Circle       | n = 1 }} {{eqn | l = \dfrac {\d x} {\d y}       | r = -\dfrac y x       | c = Implicit Differentiation }} {{eqn | r = -\dfrac y {\sqrt {1 - y^2} }       | c = substituting for $x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \arcsin_G y       | r = \int_0^y \sqrt {1 + \paren {\dfrac {\d x} {\d y} }^2} \rd y       | c = {{Defof|Arc Length}} }} {{eqn | r = \int_0^y \sqrt {1 + \paren {-\dfrac y x}^2}       | c = substituting for $\dfrac {\d x} {\d y}$ }} {{eqn | r = \int_0^y \sqrt {1 + \dfrac {y^2} {x^2} } \rd y }} {{eqn | r = \int_0^y \sqrt {\dfrac {x^2} {x^2} + \dfrac {y^2} {x^2} } \rd y       | c = rewriting $1$ to create common denominator }} {{eqn | r = \int_0^y \sqrt {\dfrac {x^2 + y^2} {x^2} } \rd y       | c = combining terms with common denominator }} {{eqn | r = \int_0^y \sqrt {\dfrac 1 {x^2} } \rd y       | c = Equation of Circle $(1)$ }} {{eqn | r = \int_0^y \dfrac 1 x \rd y       | c = in Quadrant $\text I$ and Quadrant $\text {IV}$ }} {{eqn | r = \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y       | c = substituting for $x$ in Quadrant $\text I$ and Quadrant $\text {IV}$ }} {{end-eqn}} {{qed|lemma}} l2gxb81370tgmmd6hllwvmdl8u05l8q