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proofwiki-15000
Graph of Nonlinear Additive Function is Dense in the Plane
Let $f: \R \to \R$ be an additive function which is not linear. Then the graph of $f$ is dense in the real number plane.
From Additive Function is Linear for Rational Factors: :$\map f q = q \map f 1$ for all $q \in \Q$. {{WLOG}}, let: :$\map f q = q$ for all $q \in \Q$. Since $f$ is not linear, let $\alpha \in \R \setminus \Q$ be such that: :$\map f \alpha = \alpha + \delta$ for some $\delta \ne 0$. Consider an arbitrary nonempty circle...
Let $f: \R \to \R$ be an [[Definition:Additive Function (Algebra)|additive function]] which is not linear. Then the [[Definition:Graph of Mapping|graph]] of $f$ is [[Definition:Everywhere Dense|dense]] in the [[Definition:Real Number Plane|real number plane]].
From [[Additive Function is Linear for Rational Factors]]: :$\map f q = q \map f 1$ for all $q \in \Q$. {{WLOG}}, let: :$\map f q = q$ for all $q \in \Q$. Since $f$ is not linear, let $\alpha \in \R \setminus \Q$ be such that: :$\map f \alpha = \alpha + \delta$ for some $\delta \ne 0$. Consider an arbitrary nonempt...
Graph of Nonlinear Additive Function is Dense in the Plane
https://proofwiki.org/wiki/Graph_of_Nonlinear_Additive_Function_is_Dense_in_the_Plane
https://proofwiki.org/wiki/Graph_of_Nonlinear_Additive_Function_is_Dense_in_the_Plane
[ "Additive Functions" ]
[ "Definition:Additive Function (Algebra)", "Definition:Graph of Mapping", "Definition:Everywhere Dense", "Definition:Real Number Plane" ]
[ "Additive Function is Linear for Rational Factors", "Rational Numbers are Everywhere Dense in Set of Real Numbers/Topology", "Rational Numbers are Everywhere Dense in Set of Real Numbers/Topology", "Additive Function is Linear for Rational Factors", "Additive Function is Linear for Rational Factors" ]
proofwiki-15001
Order of Elements in Quaternion Group
Let $Q = \Dic 2$ be the quaternion group, whose group presentation is given by: :$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ Then $\Dic 2$ has: :$1$ element of order $2$ and: :$6$ elements of order $4$.
From Identity is Only Group Element of Order 1, the identity element $e$ , and only $e$, is of order $1$. From here, we inspect the Cayley table: {{:Quaternion Group/Cayley Table}} It is immediately seen that: :$\paren {a^2}^2 = e$ and so by definition $a^2$ is of order $2$. As can be seen from inspection of the main d...
Let $Q = \Dic 2$ be the [[Definition:Quaternion Group|quaternion group]], whose [[Definition:Group Presentation|group presentation]] is given by: :$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ Then $\Dic 2$ has: :$1$ [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $2$ and: :$6$ [...
From [[Identity is Only Group Element of Order 1]], the [[Definition:Identity Element|identity element]] $e$ , and only $e$, is of [[Definition:Order of Group Element|order $1$]]. From here, we inspect the [[Quaternion Group/Cayley Table|Cayley table]]: {{:Quaternion Group/Cayley Table}} It is immediately seen that: ...
Order of Elements in Quaternion Group
https://proofwiki.org/wiki/Order_of_Elements_in_Quaternion_Group
https://proofwiki.org/wiki/Order_of_Elements_in_Quaternion_Group
[ "Quaternion Group" ]
[ "Definition:Dicyclic Group/Quaternion Group", "Definition:Group Presentation", "Definition:Element", "Definition:Order of Group Element", "Definition:Element", "Definition:Order of Group Element" ]
[ "Identity is Only Group Element of Order 1", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Order of Group Element", "Quaternion Group/Cayley Table", "Definition:Order of Group Element", "Definition:Matrix/Diagonal/Main", "Definition:Element", "Definition:Order of Group Eleme...
proofwiki-15002
Multiplicative Group of Reduced Residues Modulo 7 is Cyclic
Let $\struct {\Z'_7, \times_7}$ denote the multiplicative group of reduced residues modulo $7$. Then $\struct {\Z'_7, \times_7}$ is cyclic.
From Reduced Residue System under Multiplication forms Abelian Group it is noted that $\struct {\Z'_7, \times_7}$ is a group. It remains to be shown that $\struct {\Z'_7, \times_7}$ is cyclic. It will be demonstrated that: :$\gen {\eqclass 3 7} = \struct {\Z'_7, \times_7}$ That is, that $\eqclass 3 7$ is a generator of...
Let $\struct {\Z'_7, \times_7}$ denote the [[Multiplicative Group of Reduced Residues Modulo 7|multiplicative group of reduced residues modulo $7$]]. Then $\struct {\Z'_7, \times_7}$ is [[Definition:Cyclic Group|cyclic]].
From [[Reduced Residue System under Multiplication forms Abelian Group]] it is noted that $\struct {\Z'_7, \times_7}$ is a [[Definition:Group|group]]. It remains to be shown that $\struct {\Z'_7, \times_7}$ is [[Definition:Cyclic Group|cyclic]]. It will be demonstrated that: :$\gen {\eqclass 3 7} = \struct {\Z'_7, \t...
Multiplicative Group of Reduced Residues Modulo 7 is Cyclic
https://proofwiki.org/wiki/Multiplicative_Group_of_Reduced_Residues_Modulo_7_is_Cyclic
https://proofwiki.org/wiki/Multiplicative_Group_of_Reduced_Residues_Modulo_7_is_Cyclic
[ "Multiplicative Groups of Reduced Residues", "Multiplicative Group of Reduced Residues Modulo 7", "Examples of Cyclic Groups" ]
[ "Multiplicative Group of Reduced Residues/Examples/Modulo 7", "Definition:Cyclic Group" ]
[ "Reduced Residue System under Multiplication forms Abelian Group", "Definition:Group", "Definition:Cyclic Group", "Definition:Cyclic Group/Generator", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Power of Element/Group", "Definition:Element", "Definition:Cyclic Group" ]
proofwiki-15003
Reduced Residues Modulo 5 under Multiplication form Cyclic Group
Let $\struct {\Z'_5, \times_5}$ denote the multiplicative group of reduced residues modulo $5$. Then $\struct {\Z'_5, \times_5}$ is cyclic.
From Reduced Residue System under Multiplication forms Abelian Group it is noted that $\struct {\Z'_5, \times_5}$ is a group. It remains to be shown that $\struct {\Z'_5, \times_5}$ is cyclic. It will be demonstrated that: :$\gen {\eqclass 2 5} = \struct {\Z'_5, \times_5}$ That is, that $\eqclass 2 5$ is a generator of...
Let $\struct {\Z'_5, \times_5}$ denote the [[Multiplicative Group of Reduced Residues Modulo 5|multiplicative group of reduced residues modulo $5$]]. Then $\struct {\Z'_5, \times_5}$ is [[Definition:Cyclic Group|cyclic]].
From [[Reduced Residue System under Multiplication forms Abelian Group]] it is noted that $\struct {\Z'_5, \times_5}$ is a [[Definition:Group|group]]. It remains to be shown that $\struct {\Z'_5, \times_5}$ is [[Definition:Cyclic Group|cyclic]]. It will be demonstrated that: :$\gen {\eqclass 2 5} = \struct {\Z'_5, \t...
Reduced Residues Modulo 5 under Multiplication form Cyclic Group
https://proofwiki.org/wiki/Reduced_Residues_Modulo_5_under_Multiplication_form_Cyclic_Group
https://proofwiki.org/wiki/Reduced_Residues_Modulo_5_under_Multiplication_form_Cyclic_Group
[ "Multiplicative Groups of Reduced Residues", "Multiplicative Group of Reduced Residues Modulo 5", "Examples of Cyclic Groups", "Cyclic Group of Order 4" ]
[ "Multiplicative Group of Reduced Residues/Examples/Modulo 5", "Definition:Cyclic Group" ]
[ "Reduced Residue System under Multiplication forms Abelian Group", "Definition:Group", "Definition:Cyclic Group", "Definition:Cyclic Group/Generator", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Power of Element/Group", "Definition:Element", "Definition:Cyclic Group", "C...
proofwiki-15004
Number of Generators of Cyclic Group whose Order is Power of 2
Let $G$ be a finite cyclic group. Let the order of $G$ be $2^k$ for some $k \in \Z_{>0}$. Then $G$ has $2^{n - 1}$ distinct generators.
From Finite Cyclic Group has Euler Phi Generators, $G$ has $\map \phi {2^n}$ generators. The result follows from {{Corollary|Euler Phi Function of Prime Power}}. {{qed}}
Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Cyclic Group|cyclic group]]. Let the [[Definition:Order of Structure|order]] of $G$ be $2^k$ for some $k \in \Z_{>0}$. Then $G$ has $2^{n - 1}$ [[Definition:Distinct|distinct]] [[Definition:Generator of Group|generators]].
From [[Finite Cyclic Group has Euler Phi Generators]], $G$ has $\map \phi {2^n}$ [[Definition:Generator of Group|generators]]. The result follows from {{Corollary|Euler Phi Function of Prime Power}}. {{qed}}
Number of Generators of Cyclic Group whose Order is Power of 2
https://proofwiki.org/wiki/Number_of_Generators_of_Cyclic_Group_whose_Order_is_Power_of_2
https://proofwiki.org/wiki/Number_of_Generators_of_Cyclic_Group_whose_Order_is_Power_of_2
[ "Cyclic Groups", "Generators of Groups" ]
[ "Definition:Finite Group", "Definition:Cyclic Group", "Definition:Order of Structure", "Definition:Distinct", "Definition:Generator of Group" ]
[ "Finite Cyclic Group has Euler Phi Generators", "Definition:Generator of Group" ]
proofwiki-15005
Equivalent Statements for Congruence Modulo Subgroup/Left
Let $x \equiv^l y \pmod H$ denote that $x$ is left congruent modulo $H$ to $y$. Then the following statements are equivalent: {{begin-eqn}} {{eqn | n = 1 | l = x | o = \equiv^l | r = y \pmod H }} {{eqn | n = 2 | l = x^{-1} y | o = \in | r = H }} {{eqn | n = 3 | q = \exists h \i...
{{begin-eqn}} {{eqn | l = x | o = \equiv^l | r = y \pmod H }} {{eqn | ll= \leadstoandfrom | l = x^{-1} y | o = \in | r = H | c = {{Defof|Left Congruence Modulo Subgroup}} }} {{eqn | ll= \leadstoandfrom | q = \exists h \in H | l = x^{-1} y | r = h | c = {{Defof...
Let $x \equiv^l y \pmod H$ denote that $x$ is [[Definition:Left Congruence Modulo Subgroup|left congruent modulo $H$]] to $y$. Then the following statements are [[Definition:Logical Equivalence|equivalent]]: {{begin-eqn}} {{eqn | n = 1 | l = x | o = \equiv^l | r = y \pmod H }} {{eqn | n = 2 | ...
{{begin-eqn}} {{eqn | l = x | o = \equiv^l | r = y \pmod H }} {{eqn | ll= \leadstoandfrom | l = x^{-1} y | o = \in | r = H | c = {{Defof|Left Congruence Modulo Subgroup}} }} {{eqn | ll= \leadstoandfrom | q = \exists h \in H | l = x^{-1} y | r = h | c = {{Defof...
Equivalent Statements for Congruence Modulo Subgroup/Left
https://proofwiki.org/wiki/Equivalent_Statements_for_Congruence_Modulo_Subgroup/Left
https://proofwiki.org/wiki/Equivalent_Statements_for_Congruence_Modulo_Subgroup/Left
[ "Congruence Modulo Subgroup" ]
[ "Definition:Congruence Modulo Subgroup/Left Congruence", "Definition:Logical Equivalence" ]
[ "Division Laws for Groups" ]
proofwiki-15006
Equivalent Statements for Congruence Modulo Subgroup/Right
Let $x \equiv^r y \pmod H$ denote that $x$ is right congruent modulo $H$ to $y$. Then the following statements are equivalent: {{begin-eqn}} {{eqn | n = 1 | l = x | o = \equiv^r | r = y \pmod H }} {{eqn | n = 2 | l = x y^{-1} | o = \in | r = H }} {{eqn | n = 3 | q = \exists h \...
{{begin-eqn}} {{eqn | l = x | o = \equiv^r | r = y \pmod H }} {{eqn | ll= \leadstoandfrom | l = x y^{-1} | o = \in | r = H | c = {{Defof|Right Congruence Modulo Subgroup}} }} {{eqn | ll= \leadstoandfrom | q = \exists h \in H | l = x y^{-1} | r = h | c = {{Defo...
Let $x \equiv^r y \pmod H$ denote that $x$ is [[Definition:Right Congruence Modulo Subgroup|right congruent modulo $H$]] to $y$. Then the following statements are [[Definition:Logical Equivalence|equivalent]]: {{begin-eqn}} {{eqn | n = 1 | l = x | o = \equiv^r | r = y \pmod H }} {{eqn | n = 2 ...
{{begin-eqn}} {{eqn | l = x | o = \equiv^r | r = y \pmod H }} {{eqn | ll= \leadstoandfrom | l = x y^{-1} | o = \in | r = H | c = {{Defof|Right Congruence Modulo Subgroup}} }} {{eqn | ll= \leadstoandfrom | q = \exists h \in H | l = x y^{-1} | r = h | c = {{Defo...
Equivalent Statements for Congruence Modulo Subgroup/Right
https://proofwiki.org/wiki/Equivalent_Statements_for_Congruence_Modulo_Subgroup/Right
https://proofwiki.org/wiki/Equivalent_Statements_for_Congruence_Modulo_Subgroup/Right
[ "Congruence Modulo Subgroup" ]
[ "Definition:Congruence Modulo Subgroup/Right Congruence", "Definition:Logical Equivalence" ]
[ "Division Laws for Groups" ]
proofwiki-15007
Element of Group is in its own Coset/Left
Let: : $x H$ be the left coset of $x$ modulo $H$. Then: : $x \in x H$
Let $e$ be the identity of $G$. Then: {{begin-eqn}} {{eqn | l = e | o = \in | r = H | c = Identity of Subgroup }} {{eqn | l = x | r = x e | c = {{Defof|Identity Element}} }} {{eqn | ll= \leadsto | q = \exists h \in H | l = x | r = x h | c = Existential Generalisatio...
Let: : $x H$ be the [[Definition:Left Coset|left coset]] of $x$ modulo $H$. Then: : $x \in x H$
Let $e$ be the [[Definition:Identity Element|identity]] of $G$. Then: {{begin-eqn}} {{eqn | l = e | o = \in | r = H | c = [[Identity of Subgroup]] }} {{eqn | l = x | r = x e | c = {{Defof|Identity Element}} }} {{eqn | ll= \leadsto | q = \exists h \in H | l = x | r = x h ...
Element of Group is in its own Coset/Left
https://proofwiki.org/wiki/Element_of_Group_is_in_its_own_Coset/Left
https://proofwiki.org/wiki/Element_of_Group_is_in_its_own_Coset/Left
[ "Element of Group is in its own Coset" ]
[ "Definition:Coset/Left Coset" ]
[ "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Identity of Subgroup", "Existential Generalisation" ]
proofwiki-15008
Element of Group is in its own Coset/Right
Let: : $H x$ be the right coset of $x$ modulo $H$. Then: : $x \in H x$
Let $e$ be the identity of $G$. Then: {{begin-eqn}} {{eqn | l = e | o = \in | r = H | c = Identity of Subgroup }} {{eqn | l = x | r = e x | c = {{Defof|Identity Element}} }} {{eqn | ll= \leadsto | q = \exists h \in H | l = x | r = h x | c = Existential Generalisatio...
Let: : $H x$ be the [[Definition:Right Coset|right coset]] of $x$ modulo $H$. Then: : $x \in H x$
Let $e$ be the [[Definition:Identity Element|identity]] of $G$. Then: {{begin-eqn}} {{eqn | l = e | o = \in | r = H | c = [[Identity of Subgroup]] }} {{eqn | l = x | r = e x | c = {{Defof|Identity Element}} }} {{eqn | ll= \leadsto | q = \exists h \in H | l = x | r = h x ...
Element of Group is in its own Coset/Right
https://proofwiki.org/wiki/Element_of_Group_is_in_its_own_Coset/Right
https://proofwiki.org/wiki/Element_of_Group_is_in_its_own_Coset/Right
[ "Element of Group is in its own Coset" ]
[ "Definition:Coset/Right Coset" ]
[ "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Identity of Subgroup", "Existential Generalisation" ]
proofwiki-15009
Element of Group is in Unique Coset of Subgroup/Left
There exists a exactly one left coset of $H$ containing $x$, that is: $x H$
Follows directly from: * Left Congruence Modulo Subgroup is Equivalence Relation * Element in its own Equivalence Class. {{qed}}
There exists a [[Definition:Unique|exactly one]] [[Definition:Left Coset|left coset]] of $H$ containing $x$, that is: $x H$
Follows directly from: * [[Left Congruence Modulo Subgroup is Equivalence Relation]] * [[Element in its own Equivalence Class]]. {{qed}}
Element of Group is in Unique Coset of Subgroup/Left
https://proofwiki.org/wiki/Element_of_Group_is_in_Unique_Coset_of_Subgroup/Left
https://proofwiki.org/wiki/Element_of_Group_is_in_Unique_Coset_of_Subgroup/Left
[ "Element of Group is in Unique Coset of Subgroup" ]
[ "Definition:Unique", "Definition:Coset/Left Coset" ]
[ "Left Congruence Modulo Subgroup is Equivalence Relation", "Element in its own Equivalence Class" ]
proofwiki-15010
Element of Group is in Unique Coset of Subgroup/Right
There exists a exactly one right coset of $H$ containing $x$, that is: $H x$
Follows directly from: * Right Congruence Modulo Subgroup is Equivalence Relation * Element in its own Equivalence Class. {{qed}}
There exists a [[Definition:Unique|exactly one]] [[Definition:Right Coset|right coset]] of $H$ containing $x$, that is: $H x$
Follows directly from: * [[Right Congruence Modulo Subgroup is Equivalence Relation]] * [[Element in its own Equivalence Class]]. {{qed}}
Element of Group is in Unique Coset of Subgroup/Right
https://proofwiki.org/wiki/Element_of_Group_is_in_Unique_Coset_of_Subgroup/Right
https://proofwiki.org/wiki/Element_of_Group_is_in_Unique_Coset_of_Subgroup/Right
[ "Element of Group is in Unique Coset of Subgroup" ]
[ "Definition:Unique", "Definition:Coset/Right Coset" ]
[ "Right Congruence Modulo Subgroup is Equivalence Relation", "Element in its own Equivalence Class" ]
proofwiki-15011
Equivalence of Definitions of Infinite Order Element
{{TFAE|def = Infinite Order Element}} Let $G$ be a group whose identity is $e_G$. Let $x \in G$ be an element of $G$.
=== $(1)$ implies $(2)$ === Let $x$ be an infinite order element of $G$ by definition 1. Then by definition there exists no $k \in \Z_{>0}$ such that $x^k = e_G$. {{AimForCont}} not all $x^r$ are distinct for all $r \in \Z_{>0}$. Then $x^m = x^n$ for some $m, n \in \Z$ where $m \ne n$. {{WLOG}}, let $m > n$. Then: : $x...
{{TFAE|def = Infinite Order Element}} Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e_G$. Let $x \in G$ be an [[Definition:Element|element]] of $G$.
=== $(1)$ implies $(2)$ === Let $x$ be an [[Definition:Order of Group Element/Infinite/Definition 1|infinite order element of $G$ by definition 1]]. Then by definition there exists no $k \in \Z_{>0}$ such that $x^k = e_G$. {{AimForCont}} not all $x^r$ are [[Definition:Distinct Elements|distinct]] for all $r \in \Z_{...
Equivalence of Definitions of Infinite Order Element
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Infinite_Order_Element
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Infinite_Order_Element
[ "Order of Group Elements" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Element" ]
[ "Definition:Order of Group Element/Infinite/Definition 1", "Definition:Distinct/Plural", "Definition:Contradiction", "Definition:Distinct", "Definition:Order of Group Element/Infinite/Definition 2", "Definition:Order of Group Element/Infinite/Definition 2", "Definition:Distinct/Plural", "Definition:Co...
proofwiki-15012
Equivalence of Definitions of Finite Order Element
{{TFAE|def = Finite Order Element}} Let $G$ be a group whose identity is $e_G$. Let $x \in G$ be an element of $G$.
=== $(1)$ implies $(2)$ === Let $x$ be a finite order element of $G$ by definition 1. Then by definition there exists $k \in \Z_{>0}$ such that $x^k = e_G$. Consider some $m, n \in \Z_{>0}$ such that $m = n + k$. {{begin-eqn}} {{eqn | l = x^m | r = x^{n + k} | c = {{hypothesis}} }} {{eqn | r = x^n x^k ...
{{TFAE|def = Finite Order Element}} Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e_G$. Let $x \in G$ be an [[Definition:Element|element]] of $G$.
=== $(1)$ implies $(2)$ === Let $x$ be a [[Definition:Order of Group Element/Finite/Definition 1|finite order element of $G$ by definition 1]]. Then by definition there exists $k \in \Z_{>0}$ such that $x^k = e_G$. Consider some $m, n \in \Z_{>0}$ such that $m = n + k$. {{begin-eqn}} {{eqn | l = x^m | r = x^{...
Equivalence of Definitions of Finite Order Element
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Finite_Order_Element
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Finite_Order_Element
[ "Order of Group Elements" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Element" ]
[ "Definition:Order of Group Element/Finite/Definition 1", "Powers of Group Elements/Sum of Indices", "Definition:Order of Group Element/Finite/Definition 2", "Definition:Order of Group Element/Finite/Definition 2", "Powers of Group Elements", "Powers of Group Elements/Sum of Indices", "Definition:Order o...
proofwiki-15013
Order of Element in Group equals its Order in Subgroup
Let $G$ be a group. Let $H \le G$, where $\le$ denotes the property of being a subgroup. Let $x \in H$. Then the order of $x$ in $H$ equals the order of $x$ in $G$.
Let $\gen x$ be the subgroup of $G$ generated by $x$. By definition, $\gen x \le G$. All the elements of $\gen x$ are powers of $x$. As $x \in H$ it follows by {{Group-axiom|0}} that all the powers of $x$ are elements of $H$. That is: :$\gen x \le H$ By Order of Cyclic Group equals Order of Generator: :$\order x = \ord...
Let $G$ be a [[Definition:Group|group]]. Let $H \le G$, where $\le$ denotes the property of being a [[Definition:Subgroup|subgroup]]. Let $x \in H$. Then the [[Definition:Order of Group Element|order]] of $x$ in $H$ equals the [[Definition:Order of Group Element|order]] of $x$ in $G$.
Let $\gen x$ be the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $x$. By definition, $\gen x \le G$. All the [[Definition:Element|elements]] of $\gen x$ are [[Definition:Power of Group Element|powers]] of $x$. As $x \in H$ it follows by {{Group-axiom|0}} that all the [[Definition:Power of Group Ele...
Order of Element in Group equals its Order in Subgroup
https://proofwiki.org/wiki/Order_of_Element_in_Group_equals_its_Order_in_Subgroup
https://proofwiki.org/wiki/Order_of_Element_in_Group_equals_its_Order_in_Subgroup
[ "Order of Group Elements", "Subgroups" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Order of Group Element", "Definition:Order of Group Element" ]
[ "Definition:Generated Subgroup", "Definition:Element", "Definition:Power of Element/Group", "Definition:Power of Element/Group", "Definition:Element", "Order of Cyclic Group equals Order of Generator", "Definition:Cyclic Group" ]
proofwiki-15014
Order of Product of Commuting Group Elements of Coprime Order is Product of Orders
Let $G$ be a group. Let $g_1, g_2 \in G$ be commuting elements such that: {{begin-eqn}} {{eqn | l = \order {g_1} | r = n_1 }} {{eqn | l = \order {g_2} | r = n_2 }} {{end-eqn}} where $\order {g_1}$ denotes the order of $g_1$ in $G$. Let $n_1$ and $n_2$ be coprime. Then: :$\order {g_1 g_2} = n_1 n_2$
Let $g_1 g_2 = g_2 g_1$. We have: :$\paren {g_1 g_2}^{n_1 n_2} = e$ Thus: :$\order {g_1 g_2} \le n_1 n_2$ Suppose $\order {g_1 g_2}^r = e$. Then: {{begin-eqn}} {{eqn | l = {g_1}^r | r = {g_2}^{-r} | c = }} {{eqn | o = \in | r = \gen {g_1} \cap \gen {g_2} | c = }} {{eqn | ll= \leadsto | l...
Let $G$ be a [[Definition:Group|group]]. Let $g_1, g_2 \in G$ be [[Definition:Commuting Elements|commuting elements]] such that: {{begin-eqn}} {{eqn | l = \order {g_1} | r = n_1 }} {{eqn | l = \order {g_2} | r = n_2 }} {{end-eqn}} where $\order {g_1}$ denotes the [[Definition:Order of Group Element|order...
Let $g_1 g_2 = g_2 g_1$. We have: :$\paren {g_1 g_2}^{n_1 n_2} = e$ Thus: :$\order {g_1 g_2} \le n_1 n_2$ Suppose $\order {g_1 g_2}^r = e$. Then: {{begin-eqn}} {{eqn | l = {g_1}^r | r = {g_2}^{-r} | c = }} {{eqn | o = \in | r = \gen {g_1} \cap \gen {g_2} | c = }} {{eqn | ll= \leadsto ...
Order of Product of Commuting Group Elements of Coprime Order is Product of Orders
https://proofwiki.org/wiki/Order_of_Product_of_Commuting_Group_Elements_of_Coprime_Order_is_Product_of_Orders
https://proofwiki.org/wiki/Order_of_Product_of_Commuting_Group_Elements_of_Coprime_Order_is_Product_of_Orders
[ "Order of Group Elements" ]
[ "Definition:Group", "Definition:Commutative/Elements", "Definition:Order of Group Element", "Definition:Coprime/Integers" ]
[ "Definition:Divisor (Algebra)/Integer" ]
proofwiki-15015
Right Cosets are Equal iff Left Cosets by Inverse are Equal
Let $G$ be a group whose identity is $e$. Let $H$ be a subgroup of $G$. Let $g_1, g_2 \in G$. Then: :$H g_1 = H g_2 \iff {g_1}^{-1} H = {g_2}^{-1} H$ where: :${g_1}^{-1}$ and ${g_2}^{-1}$ denote the inverses of $g_1$ and $g_2$ in $G$ :$H g_1$ and $H g_2$ denote the right cosets of $H$ by $g_1$ and $g_2$ respectively :$...
{{begin-eqn}} {{eqn | l = H g_1 | r = H g_2 | c = }} {{eqn | ll= \leadstoandfrom | l = g_1 {g_2}^{-1} | o = \in | r = H | c = Right Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom | l = \paren { {g_1}^{-1} }^{-1} {g_2}^{-1} | o = \in ...
Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. Let $g_1, g_2 \in G$. Then: :$H g_1 = H g_2 \iff {g_1}^{-1} H = {g_2}^{-1} H$ where: :${g_1}^{-1}$ and ${g_2}^{-1}$ denote the [[Definition:Inverse Element|inverses]]...
{{begin-eqn}} {{eqn | l = H g_1 | r = H g_2 | c = }} {{eqn | ll= \leadstoandfrom | l = g_1 {g_2}^{-1} | o = \in | r = H | c = [[Right Cosets are Equal iff Product with Inverse in Subgroup]] }} {{eqn | ll= \leadstoandfrom | l = \paren { {g_1}^{-1} }^{-1} {g_2}^{-1} | o = ...
Right Cosets are Equal iff Left Cosets by Inverse are Equal
https://proofwiki.org/wiki/Right_Cosets_are_Equal_iff_Left_Cosets_by_Inverse_are_Equal
https://proofwiki.org/wiki/Right_Cosets_are_Equal_iff_Left_Cosets_by_Inverse_are_Equal
[ "Cosets" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Subgroup", "Definition:Inverse (Abstract Algebra)/Inverse", "Definition:Coset/Right Coset", "Definition:Coset/Left Coset" ]
[ "Right Cosets are Equal iff Product with Inverse in Subgroup", "Inverse of Group Inverse", "Left Cosets are Equal iff Product with Inverse in Subgroup" ]
proofwiki-15016
Subgroup of Subgroup with Prime Index
Let $\struct {G, \circ}$ be a group. Let $H$ be a subgroup of $G$. Let $K$ be a subgroup of $H$. Let: :$\index G K = p$ where: :$p$ denotes a prime number :$\index G K$ denotes the index of $K$ in $G$. Then either: :$H = K$ or: :$H = G$
From the Tower Law for Subgroups: :$\index G K = \index G H \index H K$ As $\index G K = p$ is prime, either $\index G H = p$ or $\index H K = p$. Thus either $\index G H = 1$ or $\index H K = 1$. The result follows from Index is One iff Subgroup equals Group. {{qed}}
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. Let $K$ be a [[Definition:Subgroup|subgroup]] of $H$. Let: :$\index G K = p$ where: :$p$ denotes a [[Definition:Prime Number|prime number]] :$\index G K$ denotes the [[Definition:Index of Subgroup|index]]...
From the [[Tower Law for Subgroups]]: :$\index G K = \index G H \index H K$ As $\index G K = p$ is [[Definition:Prime Number|prime]], either $\index G H = p$ or $\index H K = p$. Thus either $\index G H = 1$ or $\index H K = 1$. The result follows from [[Index is One iff Subgroup equals Group]]. {{qed}}
Subgroup of Subgroup with Prime Index
https://proofwiki.org/wiki/Subgroup_of_Subgroup_with_Prime_Index
https://proofwiki.org/wiki/Subgroup_of_Subgroup_with_Prime_Index
[ "Index of Subgroups" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Subgroup", "Definition:Prime Number", "Definition:Index of Subgroup" ]
[ "Tower Law for Subgroups", "Definition:Prime Number", "Index is One iff Subgroup equals Group" ]
proofwiki-15017
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Reverse Implication
:$\vdash \paren {\paren {p \lor q} \land \paren {p \lor r} } \implies \paren {p \lor \paren {q \land r} }$
{{finish|Use formulation 1 + Modus Ponens}}
:$\vdash \paren {\paren {p \lor q} \land \paren {p \lor r} } \implies \paren {p \lor \paren {q \land r} }$
{{finish|Use formulation 1 + Modus Ponens}}
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Reverse Implication
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Reverse_Implication
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Reverse_Implication
[ "Rule of Distribution" ]
[]
[]
proofwiki-15018
Completeness Theorem for Hilbert Proof System Instance 2 and Boolean Interpretations
Instance 2 of the Hilbert proof systems is a complete proof system for boolean interpretations. That is, for every WFF $\mathbf A$: :$\models_{\mathrm{BI}} \mathbf A$ implies $\vdash_{\mathscr H_2} \mathbf A$
{{ProofWanted|This is going to be fun}}
[[Definition:Hilbert Proof System/Instance 2|Instance 2]] of the [[Definition:Hilbert Proof System|Hilbert proof systems]] is a [[Definition:Complete Proof System|complete proof system]] for [[Definition:Boolean Interpretation|boolean interpretations]]. That is, for every [[Definition:WFF of Propositional Logic|WFF]] ...
{{ProofWanted|This is going to be fun}}
Completeness Theorem for Hilbert Proof System Instance 2 and Boolean Interpretations
https://proofwiki.org/wiki/Completeness_Theorem_for_Hilbert_Proof_System_Instance_2_and_Boolean_Interpretations
https://proofwiki.org/wiki/Completeness_Theorem_for_Hilbert_Proof_System_Instance_2_and_Boolean_Interpretations
[ "Completeness Theorem", "Hilbert Proof System Instance 2", "Named Theorems" ]
[ "Definition:Hilbert Proof System/Instance 2", "Definition:Hilbert Proof System", "Definition:Complete Proof System", "Definition:Boolean Interpretation", "Definition:Language of Propositional Logic/Formal Grammar/WFF" ]
[]
proofwiki-15019
Symmetry Group of Rectangle is Klein Four-Group
The symmetry group of the rectangle is the Klein $4$-group.
Comparing the Cayley tables of the symmetry group of the rectangle with the Klein $4$-group the isomorphism can be seen: {{:Symmetry Group of Rectangle/Cayley Table}} {{:Klein Four-Group/Cayley Table}} Thus the required isomorphism $\phi$ can be set up as: {{begin-eqn}} {{eqn | l = \map \phi e | r = e }} {{eqn |...
The [[Definition:Symmetry Group of Rectangle|symmetry group of the rectangle]] is the [[Definition:Klein Four-Group|Klein $4$-group]].
Comparing the [[Definition:Cayley Table|Cayley tables]] of the [[Symmetry Group of Rectangle/Cayley Table|symmetry group of the rectangle]] with the [[Klein Four-Group/Cayley Table|Klein $4$-group]] the [[Definition:Group Isomorphism|isomorphism]] can be seen: {{:Symmetry Group of Rectangle/Cayley Table}} {{:Klein F...
Symmetry Group of Rectangle is Klein Four-Group
https://proofwiki.org/wiki/Symmetry_Group_of_Rectangle_is_Klein_Four-Group
https://proofwiki.org/wiki/Symmetry_Group_of_Rectangle_is_Klein_Four-Group
[ "Klein Four-Group" ]
[ "Definition:Symmetry Group of Rectangle", "Definition:Klein Four-Group" ]
[ "Definition:Cayley Table", "Symmetry Group of Rectangle/Cayley Table", "Klein Four-Group/Cayley Table", "Definition:Isomorphism (Abstract Algebra)/Group Isomorphism", "Definition:Isomorphism (Abstract Algebra)/Group Isomorphism" ]
proofwiki-15020
Klein Four-Group as Subgroup of S4
Let $G$ be the following subset of the symmetric group on $4$ letters $S_4$, expressed in two-row notation: {{begin-eqn}} {{eqn | l = e | r = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{bmatrix} | c = }} {{eqn | l = a | r = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{bmatrix} | ...
By inspection, the Cayley table is constructed: :<nowiki>$\begin{array}{c|cccc} & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end{array}$</nowiki> Again by inspection this can be seen to be the same as the Cayley table for the Klein $4$-group. {{qed}}
Let $G$ be the following [[Definition:Subset|subset]] of the [[Definition:Symmetric Group|symmetric group on $4$ letters]] $S_4$, expressed in [[Definition:Two-Row Notation|two-row notation]]: {{begin-eqn}} {{eqn | l = e | r = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{bmatrix} | c = }} {{eqn | l...
By inspection, the [[Definition:Cayley Table|Cayley table]] is constructed: :<nowiki>$\begin{array}{c|cccc} & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end{array}$</nowiki> Again by inspection this can be seen to be the same as the [[Klein Four-Grou...
Klein Four-Group as Subgroup of S4
https://proofwiki.org/wiki/Klein_Four-Group_as_Subgroup_of_S4
https://proofwiki.org/wiki/Klein_Four-Group_as_Subgroup_of_S4
[ "Klein Four-Group" ]
[ "Definition:Subset", "Definition:Symmetric Group", "Definition:Permutation on n Letters/Two-Row Notation", "Definition:Klein Four-Group" ]
[ "Definition:Cayley Table", "Klein Four-Group/Cayley Table" ]
proofwiki-15021
Quotient of Cauchy Sequences is Metric Completion
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring. Let $d$ be the metric induced by $\struct {R, \norm {\, \cdot \,} }$. Let $\CC$ be the ring of Cauchy sequences over $R$. Let $\NN$ be the set of null sequences in $R$. Let $\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maxi...
By the definition of the metric induced by a norm then: :a sequence $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\, \cdot \,} }$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, d}$. So $\CC$ is the set of Cauchy sequences in $\struct {R, d}$. Let $\sim$ be the equivalence relation on $\C...
Let $\struct {R, \norm {\, \cdot \,} }$ be a [[Definition:Normed Division Ring|normed division ring]]. Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] by $\struct {R, \norm {\, \cdot \,} }$. Let $\CC$ be the [[Definition:Ring of Cauchy Sequences|ring of Cauchy sequences over $R$]...
By the definition of the [[Definition:Metric Induced by Norm on Division Ring|metric induced by a norm]] then: :a sequence $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\struct {R, \norm {\, \cdot \,} }$ {{iff}} $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in M...
Quotient of Cauchy Sequences is Metric Completion
https://proofwiki.org/wiki/Quotient_of_Cauchy_Sequences_is_Metric_Completion
https://proofwiki.org/wiki/Quotient_of_Cauchy_Sequences_is_Metric_Completion
[ "Normed Division Rings", "Complete Metric Spaces", "Completion of Normed Division Ring" ]
[ "Definition:Normed Division Ring", "Definition:Metric Induced by Norm on Division Ring", "Definition:Ring of Cauchy Sequences", "Definition:Set", "Definition:Null Sequence/Normed Division Ring", "Quotient Ring of Cauchy Sequences is Division Ring", "Null Sequences form Maximal Left and Right Ideal", "...
[ "Definition:Metric Induced by Norm on Division Ring", "Definition:Cauchy Sequence/Normed Division Ring", "Definition:Cauchy Sequence/Metric Space", "Definition:Cauchy Sequence/Metric Space", "Equivalence Relation on Cauchy Sequences", "Definition:Equivalence Class", "Definition:Equivalence Class", "De...
proofwiki-15022
Completion of Normed Division Ring
Let $\struct {R, \norm {\, \cdot \,}_R }$ be a normed division ring. Let $\CC$ be the ring of Cauchy sequences over $R$. Let $\NN$ be the set of null sequences. Let $Q = \CC / \NN$ where $\CC / \NN$ denotes a quotient ring. Let $\norm {\, \cdot \,}_Q: Q \to \R_{\ge 0}$ be the norm on the quotient ring $Q$ defined by:...
From Quotient Ring of Cauchy Sequences is Normed Division Ring: :$\struct {Q, \norm {\, \cdot \,}_Q}$ is a normed division ring. Let $d_R$ be the metric induced by $\struct {R, \norm {\, \cdot \,}_R }$. Let $d_Q$ be the metric induced by $\struct {Q, \norm {\, \cdot \,}_Q}$. From Quotient of Cauchy Sequences is Metric ...
Let $\struct {R, \norm {\, \cdot \,}_R }$ be a [[Definition:Normed Division Ring|normed division ring]]. Let $\CC$ be the [[Definition:Ring of Cauchy Sequences|ring of Cauchy sequences over $R$]]. Let $\NN$ be the set of [[Definition:Null Sequence of Normed Division Ring|null sequences]]. Let $Q = \CC / \NN$ where ...
From [[Quotient Ring of Cauchy Sequences is Normed Division Ring]]: :$\struct {Q, \norm {\, \cdot \,}_Q}$ is a [[Definition:Normed Division Ring|normed division ring]]. Let $d_R$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] by $\struct {R, \norm {\, \cdot \,}_R }$. Let $d_Q$ be the [[D...
Completion of Normed Division Ring
https://proofwiki.org/wiki/Completion_of_Normed_Division_Ring
https://proofwiki.org/wiki/Completion_of_Normed_Division_Ring
[ "Normed Division Rings", "Complete Metric Spaces", "Completion of Normed Division Ring" ]
[ "Definition:Normed Division Ring", "Definition:Ring of Cauchy Sequences", "Definition:Null Sequence/Normed Division Ring", "Definition:Quotient Ring", "Definition:Norm/Division Ring", "Definition:Quotient Ring", "Definition:Completion (Normed Division Ring)" ]
[ "Quotient Ring of Cauchy Sequences is Normed Division Ring", "Definition:Normed Division Ring", "Definition:Metric Induced by Norm on Division Ring", "Definition:Metric Induced by Norm on Division Ring", "Quotient of Cauchy Sequences is Metric Completion", "Definition:Completion (Metric Space)", "Defini...
proofwiki-15023
Group of Order 27 has Subgroup of Order 3
Let $G$ be a group whose identity element is $e$. Let $G$ be of order $27$. Then $G$ has at least one subgroup of order $3$.
Let $x \in G \setminus \set e$. From Identity is Only Group Element of Order 1: :$\order x > 1$ where $\order x$ denotes the order of $x$. From Lagrange's Theorem, $\order x$ is $3$, $9$ or $27$. Thus one of the following applies: {{begin-eqn}} {{eqn | l = \order x | r = 3 }} {{eqn | l = \order {x^3} | r = ...
Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e$. Let $G$ be of [[Definition:Order of Structure|order $27$]]. Then $G$ has at least one [[Definition:Subgroup|subgroup]] of [[Definition:Order of Structure|order $3$]].
Let $x \in G \setminus \set e$. From [[Identity is Only Group Element of Order 1]]: :$\order x > 1$ where $\order x$ denotes the [[Definition:Order of Group Element|order]] of $x$. From [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], $\order x$ is $3$, $9$ or $27$. Thus one of the following applies: {{be...
Group of Order 27 has Subgroup of Order 3
https://proofwiki.org/wiki/Group_of_Order_27_has_Subgroup_of_Order_3
https://proofwiki.org/wiki/Group_of_Order_27_has_Subgroup_of_Order_3
[ "Order of Group Elements", "Groups of Order 27" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Order of Structure", "Definition:Subgroup", "Definition:Order of Structure" ]
[ "Identity is Only Group Element of Order 1", "Definition:Order of Group Element", "Lagrange's Theorem (Group Theory)", "Definition:Generated Subgroup", "Definition:Subgroup", "Definition:Order of Structure" ]
proofwiki-15024
Group does not Necessarily have Subgroup of Order of Divisor of its Order
Let $G$ be a finite group whose order is $n$. Let $d$ be a divisor of $n$. Then it is not necessarily the case that $G$ has a subgroup of order $d$.
Proof by Counterexample: Consider $S_5$, the symmetric group on $5$ letters. By Order of Symmetric Group, $\order {S_5} = 5! = 120$. We have that $120 = 8 \times 15$ and so $15$ is a divisor of $120$. However, Symmetric Group on 5 Letters has no Subgroup of Order 15. {{qed}}
Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Order of Structure|order]] is $n$. Let $d$ be a [[Definition:Divisor of Integer|divisor]] of $n$. Then it is not necessarily the case that $G$ has a [[Definition:Subgroup|subgroup]] of [[Definition:Order of Structure|order]] $d$.
[[Proof by Counterexample]]: Consider $S_5$, the [[Definition:Symmetric Group on n Letters|symmetric group on $5$ letters]]. By [[Order of Symmetric Group]], $\order {S_5} = 5! = 120$. We have that $120 = 8 \times 15$ and so $15$ is a [[Definition:Divisor of Integer|divisor]] of $120$. However, [[Symmetric Group on...
Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 1
https://proofwiki.org/wiki/Group_does_not_Necessarily_have_Subgroup_of_Order_of_Divisor_of_its_Order
https://proofwiki.org/wiki/Group_does_not_Necessarily_have_Subgroup_of_Order_of_Divisor_of_its_Order/Proof_1
[ "Group does not Necessarily have Subgroup of Order of Divisor of its Order", "Subgroups", "Order of Groups" ]
[ "Definition:Finite Group", "Definition:Order of Structure", "Definition:Divisor (Algebra)/Integer", "Definition:Subgroup", "Definition:Order of Structure" ]
[ "Proof by Counterexample", "Definition:Symmetric Group/n Letters", "Order of Symmetric Group", "Definition:Divisor (Algebra)/Integer", "Symmetric Group on 5 Letters has no Subgroup of Order 15" ]
proofwiki-15025
Group does not Necessarily have Subgroup of Order of Divisor of its Order
Let $G$ be a finite group whose order is $n$. Let $d$ be a divisor of $n$. Then it is not necessarily the case that $G$ has a subgroup of order $d$.
Proof by Counterexample: Consider the symmetric group $S_4$. Then the order of the alternating group $A_4$ is $12$. We list the subgroups of $A_4$: {{:Alternating Group on 4 Letters/Subgroups}} Now $6$ divides $12$. But there is no subgroup of $A_4$ of order $6$. {{qed}}
Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Order of Structure|order]] is $n$. Let $d$ be a [[Definition:Divisor of Integer|divisor]] of $n$. Then it is not necessarily the case that $G$ has a [[Definition:Subgroup|subgroup]] of [[Definition:Order of Structure|order]] $d$.
[[Proof by Counterexample]]: Consider the [[Definition:Symmetric Group|symmetric group]] $S_4$. Then the [[Definition:Order of Structure|order]] of the [[Definition:Alternating Group|alternating group]] $A_4$ is $12$. We list the [[Alternating Group on 4 Letters/Subgroups|subgroups of $A_4$]]: {{:Alternating Group ...
Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 2
https://proofwiki.org/wiki/Group_does_not_Necessarily_have_Subgroup_of_Order_of_Divisor_of_its_Order
https://proofwiki.org/wiki/Group_does_not_Necessarily_have_Subgroup_of_Order_of_Divisor_of_its_Order/Proof_2
[ "Group does not Necessarily have Subgroup of Order of Divisor of its Order", "Subgroups", "Order of Groups" ]
[ "Definition:Finite Group", "Definition:Order of Structure", "Definition:Divisor (Algebra)/Integer", "Definition:Subgroup", "Definition:Order of Structure" ]
[ "Proof by Counterexample", "Definition:Symmetric Group", "Definition:Order of Structure", "Definition:Alternating Group", "Alternating Group on 4 Letters/Subgroups", "Definition:Divisor (Algebra)/Integer", "Definition:Subgroup", "Definition:Order of Structure" ]
proofwiki-15026
Characteristic Function of Normal Distribution
The characteristic function of the normal distribution with mean $\mu$ and variance $\sigma^2$ is given by: :$\map \phi t = e^{i t \mu - \frac 1 2 t^2 \sigma^2}$
=== {{Lemma|Characteristic Function of Normal Distribution|1}} === {{:Characteristic Function of Normal Distribution/Lemma 1}}{{qed|lemma}}
The [[Definition:Characteristic Function of Random Variable|characteristic function]] of the [[Definition:Normal Distribution|normal distribution]] with mean $\mu$ and variance $\sigma^2$ is given by: :$\map \phi t = e^{i t \mu - \frac 1 2 t^2 \sigma^2}$
=== {{Lemma|Characteristic Function of Normal Distribution|1}} === {{:Characteristic Function of Normal Distribution/Lemma 1}}{{qed|lemma}}
Characteristic Function of Normal Distribution
https://proofwiki.org/wiki/Characteristic_Function_of_Normal_Distribution
https://proofwiki.org/wiki/Characteristic_Function_of_Normal_Distribution
[ "Characteristic Function of Normal Distribution", "Characteristic Functions of Random Variables", "Normal Distribution" ]
[ "Definition:Characteristic Function of Random Variable", "Definition:Normal Distribution" ]
[]
proofwiki-15027
Product of Subset with Intersection/Equality does not Hold
While it is the case that: :$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$ it is not necessarily the case that: :$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
Proof by Counterexample: Let $a \in G$ such that $a \ne a^{-1}$. Let $X = \set {a, a^{-1} }, Y = \set a, Z = \set {a^{-1} }$. Then: :$X \circ \paren {Y \cap Z} = X \circ \O = \O$ :$\paren {X \circ Y} \cap \paren {X \circ Z} = \set {a^2, e} \cap \set {e, a^{-2} } \ne \O$ so: :$X \circ \paren {Y \cap Z} \ne \paren {X \ci...
While it is the case that: :$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$ it is not necessarily the case that: :$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
[[Proof by Counterexample]]: Let $a \in G$ such that $a \ne a^{-1}$. Let $X = \set {a, a^{-1} }, Y = \set a, Z = \set {a^{-1} }$. Then: :$X \circ \paren {Y \cap Z} = X \circ \O = \O$ :$\paren {X \circ Y} \cap \paren {X \circ Z} = \set {a^2, e} \cap \set {e, a^{-2} } \ne \O$ so: :$X \circ \paren {Y \cap Z} \ne \pare...
Product of Subset with Intersection/Equality does not Hold
https://proofwiki.org/wiki/Product_of_Subset_with_Intersection/Equality_does_not_Hold
https://proofwiki.org/wiki/Product_of_Subset_with_Intersection/Equality_does_not_Hold
[ "Product of Subset with Intersection" ]
[]
[ "Proof by Counterexample" ]
proofwiki-15028
Coset of Subgroup of Subgroup
Let $G$ be a group. Let $H, K \le G$ be subgroups of $G$. Let $K \subseteq H$. Let $x \in G$. Then either: :$x K \subseteq H$ or: :$x K \cap H = \O$ where $x K$ denotes the left coset of $K$ by $x$.
Suppose $x K \cap H \ne \O$. Then: {{begin-eqn}} {{eqn | l = x K \cap H | o = \ne | r = \O | c = }} {{eqn | ll= \leadsto | q = \exists y \in G | l = y | o = \in | r = x K \cap H | c = }} {{eqn | ll= \leadsto | l = y | o = \in | r = x K | c = }} ...
Let $G$ be a [[Definition:Group|group]]. Let $H, K \le G$ be [[Definition:Subgroup|subgroups]] of $G$. Let $K \subseteq H$. Let $x \in G$. Then either: :$x K \subseteq H$ or: :$x K \cap H = \O$ where $x K$ denotes the [[Definition:Left Coset|left coset]] of $K$ by $x$.
Suppose $x K \cap H \ne \O$. Then: {{begin-eqn}} {{eqn | l = x K \cap H | o = \ne | r = \O | c = }} {{eqn | ll= \leadsto | q = \exists y \in G | l = y | o = \in | r = x K \cap H | c = }} {{eqn | ll= \leadsto | l = y | o = \in | r = x K | c = }...
Coset of Subgroup of Subgroup
https://proofwiki.org/wiki/Coset_of_Subgroup_of_Subgroup
https://proofwiki.org/wiki/Coset_of_Subgroup_of_Subgroup
[ "Cosets" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Coset/Left Coset" ]
[ "Left Cosets are Equal iff Element in Other Left Coset", "Definition:Group", "Axiom:Group Axioms", "Definition:Subset Product" ]
proofwiki-15029
Intersection of Left Cosets of Subgroups is Left Coset of Intersection
Let $G$ be a group. Let $H, K \le G$ be subgroups of $G$. Let $a, b \in G$. Let: :$a H \cap b K \ne \O$ where $a H$ denotes the left coset of $H$ by $a$. Then $a H \cap b K$ is a left coset of $H \cap K$.
Let $x \in a H \cap b K$. Then: {{begin-eqn}} {{eqn | l = x | o = \in | r = a H }} {{eqn | ll= \leadsto | l = x H | r = a H | c = Left Cosets are Equal iff Element in Other Left Coset }} {{end-eqn}} and similarly: {{begin-eqn}} {{eqn | l = x | o = \in | r = b K }} {{eqn | ll= \...
Let $G$ be a [[Definition:Group|group]]. Let $H, K \le G$ be [[Definition:Subgroup|subgroups]] of $G$. Let $a, b \in G$. Let: :$a H \cap b K \ne \O$ where $a H$ denotes the [[Definition:Left Coset|left coset]] of $H$ by $a$. Then $a H \cap b K$ is a [[Definition:Left Coset|left coset]] of $H \cap K$.
Let $x \in a H \cap b K$. Then: {{begin-eqn}} {{eqn | l = x | o = \in | r = a H }} {{eqn | ll= \leadsto | l = x H | r = a H | c = [[Left Cosets are Equal iff Element in Other Left Coset]] }} {{end-eqn}} and similarly: {{begin-eqn}} {{eqn | l = x | o = \in | r = b K }} {{eqn ...
Intersection of Left Cosets of Subgroups is Left Coset of Intersection
https://proofwiki.org/wiki/Intersection_of_Left_Cosets_of_Subgroups_is_Left_Coset_of_Intersection
https://proofwiki.org/wiki/Intersection_of_Left_Cosets_of_Subgroups_is_Left_Coset_of_Intersection
[ "Cosets" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Coset/Left Coset", "Definition:Coset/Left Coset" ]
[ "Left Cosets are Equal iff Element in Other Left Coset", "Left Cosets are Equal iff Element in Other Left Coset", "Definition:Coset/Left Coset" ]
proofwiki-15030
Right Cosets are Equal iff Element in Other Right Coset
Let $H x$ denote the right coset of $H$ by $x$. Then: :$H x = H y \iff x \in H y$
{{begin-eqn}} {{eqn | l = H x | r = H y | c = }} {{eqn | ll= \leadstoandfrom | l = x y^{-1} | o = \in | r = H | c = Right Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom | l = x | o = \in | r = H y | c = Element in Right C...
Let $H x$ denote the [[Definition:Right Coset|right coset]] of $H$ by $x$. Then: :$H x = H y \iff x \in H y$
{{begin-eqn}} {{eqn | l = H x | r = H y | c = }} {{eqn | ll= \leadstoandfrom | l = x y^{-1} | o = \in | r = H | c = [[Right Cosets are Equal iff Product with Inverse in Subgroup]] }} {{eqn | ll= \leadstoandfrom | l = x | o = \in | r = H y | c = [[Element in R...
Right Cosets are Equal iff Element in Other Right Coset
https://proofwiki.org/wiki/Right_Cosets_are_Equal_iff_Element_in_Other_Right_Coset
https://proofwiki.org/wiki/Right_Cosets_are_Equal_iff_Element_in_Other_Right_Coset
[ "Cosets" ]
[ "Definition:Coset/Right Coset" ]
[ "Right Cosets are Equal iff Product with Inverse in Subgroup", "Element in Right Coset iff Product with Inverse in Subgroup" ]
proofwiki-15031
Subgroup of Subgroup with Prime Index/Corollary
Let $\struct {G, \circ}$ be a group. Let $H$ and $K$ be subgroups of $G$. Let $K \subsetneq H$. Let: :$\index G K = p$ where: :$p$ denotes a prime number :$\index G K$ denotes the index of $K$ in $G$. Then: :$H = G$
As $K \subsetneq H$ and $K$ is a subgroups of $G$, it follows that $K$ is a proper subgroup of $H$. That is, $K \ne H$ Hence from Subgroup of Subgroup with Prime Index: :$H = G$ {{qed}}
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$. Let $K \subsetneq H$. Let: :$\index G K = p$ where: :$p$ denotes a [[Definition:Prime Number|prime number]] :$\index G K$ denotes the [[Definition:Index of Subgroup|index]] of $K$ in $G$. Then: :...
As $K \subsetneq H$ and $K$ is a [[Definition:Subgroup|subgroups]] of $G$, it follows that $K$ is a [[Definition:Proper Subgroup|proper subgroup]] of $H$. That is, $K \ne H$ Hence from [[Subgroup of Subgroup with Prime Index]]: :$H = G$ {{qed}}
Subgroup of Subgroup with Prime Index/Corollary
https://proofwiki.org/wiki/Subgroup_of_Subgroup_with_Prime_Index/Corollary
https://proofwiki.org/wiki/Subgroup_of_Subgroup_with_Prime_Index/Corollary
[ "Index of Subgroups" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Prime Number", "Definition:Index of Subgroup" ]
[ "Definition:Subgroup", "Definition:Proper Subgroup", "Subgroup of Subgroup with Prime Index" ]
proofwiki-15032
Order of Element divides Order of Centralizer
Let $G$ be a finite group. Let $x \in G$ be an element of $G$. Let $\map {C_G} x$ denote the centralizer of $x$. Then: :$\order x \divides \order {\map {C_G} x}$ where: :$\order x$ denotes the order of $x$ in $G$ :$\divides$ denotes divisibility :$\order {\map {C_G} x}$ denotes the order of $\map {C_G} x$.
From Order of Cyclic Group equals Order of Generator: :$\order x = \order {\gen x}$ where $\gen x$ denotes the subgroup of $G$ generated by $x$. By definition, $\gen x$ is a cyclic group. By Cyclic Group is Abelian, all elements of $\gen x$ commute with $x$. Thus by definition of centralizer: :$\gen x \subseteq \map {...
Let $G$ be a [[Definition:Finite Group|finite group]]. Let $x \in G$ be an [[Definition:Element|element]] of $G$. Let $\map {C_G} x$ denote the [[Definition:Centralizer of Group Element|centralizer]] of $x$. Then: :$\order x \divides \order {\map {C_G} x}$ where: :$\order x$ denotes the [[Definition:Order of Group ...
From [[Order of Cyclic Group equals Order of Generator]]: :$\order x = \order {\gen x}$ where $\gen x$ denotes the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $x$. By definition, $\gen x$ is a [[Definition:Cyclic Group|cyclic group]]. By [[Cyclic Group is Abelian]], all [[Definition:Element|elemen...
Order of Element divides Order of Centralizer
https://proofwiki.org/wiki/Order_of_Element_divides_Order_of_Centralizer
https://proofwiki.org/wiki/Order_of_Element_divides_Order_of_Centralizer
[ "Centralizers", "Order of Group Elements" ]
[ "Definition:Finite Group", "Definition:Element", "Definition:Centralizer/Group Element", "Definition:Order of Group Element", "Definition:Divisor (Algebra)/Integer", "Definition:Order of Structure" ]
[ "Order of Cyclic Group equals Order of Generator", "Definition:Generated Subgroup", "Definition:Cyclic Group", "Cyclic Group is Abelian", "Definition:Element", "Definition:Commutative/Elements", "Definition:Centralizer/Group Element", "Centralizer of Group Element is Subgroup", "Definition:Subgroup"...
proofwiki-15033
Left Coset of Stabilizer in Group of Transformations
Let $S$ be a non-empty set. Let $G$ be a group of permutations of $S$. Let $t \in G$. Let $G_t$ be the set defined as: :$G_t = \set {g \in G: \map g t = t}$ Then each left coset of $G_t$ in $G$ consists of the elements of $G$ that map $t$ to some element of $S$. {{explain|The source work does not discuss group actions,...
Let $x \in G$. Consider the left coset $x G_t$. Let $\map x t = s$. Then: {{begin-eqn}} {{eqn | l = y | o = \in | r = x G_t | c = }} {{eqn | ll= \leadstoandfrom | l = y^{-1} x | o = \in | r = G_t | c = Element in Left Coset iff Product with Inverse in Subgroup }} {{eqn | ll= \...
Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]]. Let $G$ be a [[Definition:Permutation Group|group of permutations]] of $S$. Let $t \in G$. Let $G_t$ be the [[Definition:Set|set]] defined as: :$G_t = \set {g \in G: \map g t = t}$ Then each [[Definition:Left Coset|left coset]] of $G_t$ in...
Let $x \in G$. Consider the [[Definition:Left Coset|left coset]] $x G_t$. Let $\map x t = s$. Then: {{begin-eqn}} {{eqn | l = y | o = \in | r = x G_t | c = }} {{eqn | ll= \leadstoandfrom | l = y^{-1} x | o = \in | r = G_t | c = [[Element in Left Coset iff Product with Inve...
Left Coset of Stabilizer in Group of Transformations
https://proofwiki.org/wiki/Left_Coset_of_Stabilizer_in_Group_of_Transformations
https://proofwiki.org/wiki/Left_Coset_of_Stabilizer_in_Group_of_Transformations
[ "Cosets" ]
[ "Definition:Non-Empty Set", "Definition:Set", "Definition:Permutation Group", "Definition:Set", "Definition:Coset/Left Coset", "Definition:Element", "Definition:Element", "Definition:Stabilizer" ]
[ "Definition:Coset/Left Coset", "Element in Left Coset iff Product with Inverse in Subgroup" ]
proofwiki-15034
Intersection of Coprime Cyclic Subgroups is Trivial
Let $G$ be a group whose identity is $e$. Let $x, y \in G$ such that: :$\order x \perp \order y$ where: :$\order x, \order y$ denotes the orders of $x$ and $y$ in $G$ respectively :$\perp$ denotes the coprimality relation. Then: :$\gen x \cap \gen y = \set e$ where $\gen x, \gen y$ denotes the subgroup of $G$ generated...
From Order of Cyclic Group equals Order of Generator: :$\order x = \order {\gen x}$ and: :$\order y = \order {\gen y}$ where $\order {\gen x}, \order {\gen y}$ denote the orders of $\gen x$ and $\gen y$ respectively. From Intersection of Subgroups is Subgroup: :$\gen x \cap \gen y$ is a subgroup of both $\gen x$ and $\...
Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $x, y \in G$ such that: :$\order x \perp \order y$ where: :$\order x, \order y$ denotes the [[Definition:Order of Group Element|orders]] of $x$ and $y$ in $G$ respectively :$\perp$ denotes the [[Definition:Coprime Intege...
From [[Order of Cyclic Group equals Order of Generator]]: :$\order x = \order {\gen x}$ and: :$\order y = \order {\gen y}$ where $\order {\gen x}, \order {\gen y}$ denote the [[Definition:Order of Structure|orders]] of $\gen x$ and $\gen y$ respectively. From [[Intersection of Subgroups is Subgroup]]: :$\gen x \cap \g...
Intersection of Coprime Cyclic Subgroups is Trivial
https://proofwiki.org/wiki/Intersection_of_Coprime_Cyclic_Subgroups_is_Trivial
https://proofwiki.org/wiki/Intersection_of_Coprime_Cyclic_Subgroups_is_Trivial
[ "Cyclic Groups" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Order of Group Element", "Definition:Coprime/Integers", "Definition:Generated Subgroup" ]
[ "Order of Cyclic Group equals Order of Generator", "Definition:Order of Structure", "Intersection of Subgroups is Subgroup", "Definition:Subgroup", "Lagrange's Theorem (Group Theory)", "Definition:Trivial Subgroup" ]
proofwiki-15035
Normed Division Ring is Field iff Completion is Field
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring. Let $\struct {R', \norm {\, \cdot \,}' }$ be a normed division ring completion of $\struct {R, \norm {\, \cdot \,} }$ Then: :$R$ is a field {{iff}} $R'$ is a field.
By the definition of a normed division ring completion then: :$(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$. :$(2): \quad$ $\struct {R', \norm {\, \cdot \,}' }$ is a complete metric space. :$(3): \quad$ The image $\phi \sqbrk R$ of $\phi$ is a dense subspace in $\struct {R', \norm {...
Let $\struct {R, \norm {\, \cdot \,} }$ be a [[Definition:Normed Division Ring|normed division ring]]. Let $\struct {R', \norm {\, \cdot \,}' }$ be a [[Definition:Completion (Normed Division Ring)|normed division ring completion]] of $\struct {R, \norm {\, \cdot \,} }$ Then: :$R$ is a [[Definition:Field (Abstract Alg...
By the definition of a [[Definition:Completion (Normed Division Ring)|normed division ring completion]] then: :$(1): \quad$ there exists a [[Definition:Distance-Preserving Mapping|distance-preserving]] [[Definition:Ring Monomorphism|ring monomorphism]] $\phi: R \to R'$. :$(2): \quad$ $\struct {R', \norm {\, \cdot \,}' ...
Normed Division Ring is Field iff Completion is Field
https://proofwiki.org/wiki/Normed_Division_Ring_is_Field_iff_Completion_is_Field
https://proofwiki.org/wiki/Normed_Division_Ring_is_Field_iff_Completion_is_Field
[ "Normed Division Rings", "Complete Metric Spaces", "Completion of Normed Division Ring" ]
[ "Definition:Normed Division Ring", "Definition:Completion (Normed Division Ring)", "Definition:Field (Abstract Algebra)", "Definition:Field (Abstract Algebra)" ]
[ "Definition:Completion (Normed Division Ring)", "Definition:Distance-Preserving Mapping", "Definition:Ring Monomorphism", "Definition:Complete Metric Space", "Definition:Image (Set Theory)/Mapping/Mapping", "Definition:Everywhere Dense", "Definition:Topological Subspace", "Ring Homomorphism Preserves ...
proofwiki-15036
Non-Abelian Order 8 Group has Order 4 Element
Let $G$ be a non-abelian group of order $8$. Then $G$ has at least one element of order $4$.
Let $e \in G$ be the identity of $G$. Let $g \in G$ be an arbitrary element of $G$ such that $g \ne e$. From Identity is Only Group Element of Order 1, only $e$ has order $1$. Thus from Order of Element Divides Order of Finite Group: :$\order g \in \set {2, 4, 8}$ Suppose $\order g = 8$. Then $G$ is cyclic. So by Cycli...
Let $G$ be a non-[[Definition:Abelian Group|abelian]] [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $8$. Then $G$ has at least one [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $4$.
Let $e \in G$ be the [[Definition:Identity Element|identity]] of $G$. Let $g \in G$ be an arbitrary [[Definition:Element|element]] of $G$ such that $g \ne e$. From [[Identity is Only Group Element of Order 1]], only $e$ has [[Definition:Order of Group Element|order]] $1$. Thus from [[Order of Element Divides Order o...
Non-Abelian Order 8 Group has Order 4 Element
https://proofwiki.org/wiki/Non-Abelian_Order_8_Group_has_Order_4_Element
https://proofwiki.org/wiki/Non-Abelian_Order_8_Group_has_Order_4_Element
[ "Groups of Order 8" ]
[ "Definition:Abelian Group", "Definition:Group", "Definition:Order of Structure", "Definition:Element", "Definition:Order of Group Element" ]
[ "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Element", "Identity is Only Group Element of Order 1", "Definition:Order of Group Element", "Order of Element Divides Order of Finite Group", "Definition:Cyclic Group", "Cyclic Group is Abelian", "Definition:Abelian Group", "De...
proofwiki-15037
Group of Prime Order p has p-1 Elements of Order p
Let $p$ be a prime number. Let $G$ be a group with identity $e$ whose order is $p$. Then $G$ has $p - 1$ elements of order $p$.
Let $\order g$ denote the order of an element $g$ of $G$. From Order of Element Divides Order of Finite Group: :$\order g \divides p$ where $\divides$ denotes divisibility. By definition of prime number, the only divisors of $p$ are $1$ and $p$. From Identity is Only Group Element of Order 1, only $e$ has order $1$ in ...
Let $p$ be a [[Definition:Prime Number|prime number]]. Let $G$ be a [[Definition:Group|group]] with [[Definition:Identity Element|identity]] $e$ whose [[Definition:Order of Structure|order]] is $p$. Then $G$ has $p - 1$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $p$.
Let $\order g$ denote the [[Definition:Order of Group Element|order]] of an [[Definition:Element|element]] $g$ of $G$. From [[Order of Element Divides Order of Finite Group]]: :$\order g \divides p$ where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]]. By definition of [[Definition:Prime Number|pri...
Group of Prime Order p has p-1 Elements of Order p
https://proofwiki.org/wiki/Group_of_Prime_Order_p_has_p-1_Elements_of_Order_p
https://proofwiki.org/wiki/Group_of_Prime_Order_p_has_p-1_Elements_of_Order_p
[ "Prime Groups", "Order of Group Elements" ]
[ "Definition:Prime Number", "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Order of Structure", "Definition:Element", "Definition:Order of Group Element" ]
[ "Definition:Order of Group Element", "Definition:Element", "Order of Element Divides Order of Finite Group", "Definition:Divisor (Algebra)/Integer", "Definition:Prime Number", "Definition:Divisor (Algebra)/Integer", "Identity is Only Group Element of Order 1", "Definition:Order of Group Element", "D...
proofwiki-15038
Number of Order p Elements in Group with m Order p Subgroups
Let $G$ be a group whose identity is $e$. Let $G$ have $m$ subgroups of order $p$. The total number of elements of $G$ of order $p$ is $m \paren {p - 1}$.
Let $H \le G$ be a subgroup of $G$ of order $p$. From Prime Group is Cyclic, $H$ is a cyclic group. From Group of Prime Order p has p-1 Elements of Order p, $H$ has $p - 1$ elements of order $p$. From Intersection of Subgroups of Prime Order, each of the $m$ sets of $p - 1$ non-identity elements of the $m$ subgroups of...
Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $G$ have $m$ [[Definition:Subgroup|subgroups]] of [[Definition:Order of Structure|order]] $p$. The total number of [[Definition:Element|elements]] of $G$ of [[Definition:Order of Group Element|order]] $p$ is $m \paren ...
Let $H \le G$ be a [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Structure|order]] $p$. From [[Prime Group is Cyclic]], $H$ is a [[Definition:Cyclic Group|cyclic group]]. From [[Group of Prime Order p has p-1 Elements of Order p]], $H$ has $p - 1$ [[Definition:Element|elements]] of [[Definition:Ord...
Number of Order p Elements in Group with m Order p Subgroups
https://proofwiki.org/wiki/Number_of_Order_p_Elements_in_Group_with_m_Order_p_Subgroups
https://proofwiki.org/wiki/Number_of_Order_p_Elements_in_Group_with_m_Order_p_Subgroups
[ "Order of Group Elements" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Subgroup", "Definition:Order of Structure", "Definition:Element", "Definition:Order of Group Element" ]
[ "Definition:Subgroup", "Definition:Order of Structure", "Prime Group is Cyclic", "Definition:Cyclic Group", "Group of Prime Order p has p-1 Elements of Order p", "Definition:Element", "Definition:Order of Group Element", "Intersection of Subgroups of Prime Order", "Definition:Set", "Definition:Ide...
proofwiki-15039
Non-Cyclic Group of Order p^2 has p+3 Subgroups
Let $p$ be a prime number. Let $G$ be a non-cyclic group whose order is $p^2$. Then $G$ has exactly $p + 3$ subgroups.
By Order of Element Divides Order of Finite Group, all elements of $G$ have order in $\set {1, p, p^2}$. But as $G$ is non-cyclic, it can have no element of order $p^2$. By Identity is Only Group Element of Order 1, $G$ has $p^2 - 1$ elements of order $p$. Let $m$ denote the number of subgroups of $G$ of order $p$. Fro...
Let $p$ be a [[Definition:Prime Number|prime number]]. Let $G$ be a non-[[Definition:Cyclic Group|cyclic group]] whose [[Definition:Order of Structure|order]] is $p^2$. Then $G$ has exactly $p + 3$ [[Definition:Subgroup|subgroups]].
By [[Order of Element Divides Order of Finite Group]], all [[Definition:Element|elements]] of $G$ have [[Definition:Order of Group Element|order]] in $\set {1, p, p^2}$. But as $G$ is non-[[Definition:Cyclic Group|cyclic]], it can have no [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $p...
Non-Cyclic Group of Order p^2 has p+3 Subgroups
https://proofwiki.org/wiki/Non-Cyclic_Group_of_Order_p^2_has_p+3_Subgroups
https://proofwiki.org/wiki/Non-Cyclic_Group_of_Order_p^2_has_p+3_Subgroups
[ "Examples of Order of Group Elements" ]
[ "Definition:Prime Number", "Definition:Cyclic Group", "Definition:Order of Structure", "Definition:Subgroup" ]
[ "Order of Element Divides Order of Finite Group", "Definition:Element", "Definition:Order of Group Element", "Definition:Cyclic Group", "Definition:Element", "Definition:Order of Group Element", "Identity is Only Group Element of Order 1", "Definition:Element", "Definition:Order of Group Element", ...
proofwiki-15040
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
Let $p$ be a prime number. Let $G$ be a finite abelian group whose identity is $e$. Let $G$ have at least $p$ elements of order $p$. Then: : $p^2 \divides \order G$ where: :$\divides$ denotes divisibility :$\order G$ denotes the order of $G$.
Let $x \in G$ be of order $p$. Consider $\gen x$, the subgroup generated by $x$. By Group of Prime Order p has p-1 Elements of Order p, the elements of $\gen x$ are all of order $p$ except $e$. Thus, by hypothesis, there must exist another $y \in G$ of order $p$. Consider the subset of $G$: :$S := \set {x^i y^j: 0 \le ...
Let $p$ be a [[Definition:Prime Number|prime number]]. Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$. Let $G$ have at least $p$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $p$. Then: ...
Let $x \in G$ be of [[Definition:Order of Group Element|order]] $p$. Consider $\gen x$, the [[Definition:Generated Subgroup|subgroup generated]] by $x$. By [[Group of Prime Order p has p-1 Elements of Order p]], the [[Definition:Element|elements]] of $\gen x$ are all of [[Definition:Order of Group Element|order]] $p$...
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
https://proofwiki.org/wiki/Order_of_Finite_Abelian_Group_with_p+_Order_p_Elements_is_Divisible_by_p^2
https://proofwiki.org/wiki/Order_of_Finite_Abelian_Group_with_p+_Order_p_Elements_is_Divisible_by_p^2
[ "Order of Group Elements", "Examples of Order of Group Elements", "Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2", "Abelian Groups", "Finite Groups" ]
[ "Definition:Prime Number", "Definition:Finite Group", "Definition:Abelian Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Element", "Definition:Order of Group Element", "Definition:Divisor (Algebra)/Integer", "Definition:Order of Structure" ]
[ "Definition:Order of Group Element", "Definition:Generated Subgroup", "Group of Prime Order p has p-1 Elements of Order p", "Definition:Element", "Definition:Order of Group Element", "Definition:By Hypothesis", "Definition:Order of Group Element", "Definition:Subset", "Finite Subgroup Test", "Defi...
proofwiki-15041
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
Let $p$ be a prime number. Let $G$ be a finite abelian group whose identity is $e$. Let $G$ have at least $p$ elements of order $p$. Then: : $p^2 \divides \order G$ where: :$\divides$ denotes divisibility :$\order G$ denotes the order of $G$.
By hypothesis there are elements $x, y$ of order $3$ in $G$ such that $x, y, x^2$ are all different. Consider the subset of $G$: :$S := \set {x^i y^j: 0 \le i, j \le 2}$ By the Finite Subgroup Test, $S$ is a subgroup of $G$ which has $9$ elements. By Lagrange's theorem: :$\order S \divides \order G$ But $\order S = 9$ ...
Let $p$ be a [[Definition:Prime Number|prime number]]. Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$. Let $G$ have at least $p$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $p$. Then: ...
[[Definition:By Hypothesis|By hypothesis]] there are [[Definition:Element|elements]] $x, y$ of [[Definition:Order of Group Element|order]] $3$ in $G$ such that $x, y, x^2$ are all different. Consider the [[Definition:Subset|subset]] of $G$: :$S := \set {x^i y^j: 0 \le i, j \le 2}$ By the [[Finite Subgroup Test]], $S$...
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2/Examples/Order 3/Proof 1
https://proofwiki.org/wiki/Order_of_Finite_Abelian_Group_with_p+_Order_p_Elements_is_Divisible_by_p^2
https://proofwiki.org/wiki/Order_of_Finite_Abelian_Group_with_p+_Order_p_Elements_is_Divisible_by_p^2/Examples/Order_3/Proof_1
[ "Order of Group Elements", "Examples of Order of Group Elements", "Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2", "Abelian Groups", "Finite Groups" ]
[ "Definition:Prime Number", "Definition:Finite Group", "Definition:Abelian Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Element", "Definition:Order of Group Element", "Definition:Divisor (Algebra)/Integer", "Definition:Order of Structure" ]
[ "Definition:By Hypothesis", "Definition:Element", "Definition:Order of Group Element", "Definition:Subset", "Finite Subgroup Test", "Definition:Subgroup", "Definition:Element", "Lagrange's Theorem (Group Theory)" ]
proofwiki-15042
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
Let $p$ be a prime number. Let $G$ be a finite abelian group whose identity is $e$. Let $G$ have at least $p$ elements of order $p$. Then: : $p^2 \divides \order G$ where: :$\divides$ denotes divisibility :$\order G$ denotes the order of $G$.
An example of Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$, setting $p = 3$. {{qed}}
Let $p$ be a [[Definition:Prime Number|prime number]]. Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$. Let $G$ have at least $p$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $p$. Then: ...
An example of [[Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2|Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$]], setting $p = 3$. {{qed}}
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2/Examples/Order 3/Proof 2
https://proofwiki.org/wiki/Order_of_Finite_Abelian_Group_with_p+_Order_p_Elements_is_Divisible_by_p^2
https://proofwiki.org/wiki/Order_of_Finite_Abelian_Group_with_p+_Order_p_Elements_is_Divisible_by_p^2/Examples/Order_3/Proof_2
[ "Order of Group Elements", "Examples of Order of Group Elements", "Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2", "Abelian Groups", "Finite Groups" ]
[ "Definition:Prime Number", "Definition:Finite Group", "Definition:Abelian Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Element", "Definition:Order of Group Element", "Definition:Divisor (Algebra)/Integer", "Definition:Order of Structure" ]
[ "Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2" ]
proofwiki-15043
Abelian Group of Semiprime Order is Cyclic
Let $p$ and $q$ be distinct prime numbers. Let $G$ be an abelian group such that: :$\order G = p q$ where $\order G$ denotes the order of $G$. Then $G$ is cyclic.
By Order of Element Divides Order of Finite Group, the order of elements of $G$ are all in $\set {1, p, q, p q}$. We have that Identity is Only Group Element of Order 1. Suppose $G$ were to contain more than $p - 1$ elements of order $p$. Then by Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible b...
Let $p$ and $q$ be [[Definition:Distinct|distinct]] [[Definition:Prime Number|prime numbers]]. Let $G$ be an [[Definition:Abelian Group|abelian group]] such that: :$\order G = p q$ where $\order G$ denotes the [[Definition:Order of Structure|order]] of $G$. Then $G$ is [[Definition:Cyclic Group|cyclic]].
By [[Order of Element Divides Order of Finite Group]], the [[Definition:Order of Group Element|order]] of [[Definition:Element|elements]] of $G$ are all in $\set {1, p, q, p q}$. We have that [[Identity is Only Group Element of Order 1]]. Suppose $G$ were to contain more than $p - 1$ [[Definition:Element|elements]] ...
Abelian Group of Semiprime Order is Cyclic
https://proofwiki.org/wiki/Abelian_Group_of_Semiprime_Order_is_Cyclic
https://proofwiki.org/wiki/Abelian_Group_of_Semiprime_Order_is_Cyclic
[ "Abelian Groups", "Finite Groups", "Cyclic Groups" ]
[ "Definition:Distinct", "Definition:Prime Number", "Definition:Abelian Group", "Definition:Order of Structure", "Definition:Cyclic Group" ]
[ "Order of Element Divides Order of Finite Group", "Definition:Order of Group Element", "Definition:Element", "Identity is Only Group Element of Order 1", "Definition:Element", "Definition:Order of Group Element", "Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2", "Definition...
proofwiki-15044
General Morphism Property for Groups
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups. Let $\phi: G \to H$ be a group homomorphism. Then: :$\forall g_k \in H: \map \phi {g_1 \circ g_2 \circ \cdots \circ g_n} = \map \phi {g_1} * \map \phi {g_2} * \cdots * \map \phi {g_n}$
A group is a semigroup. The result then follows from the General Morphism Property for Semigroups. {{Qed}}
Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]]. Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|group homomorphism]]. Then: :$\forall g_k \in H: \map \phi {g_1 \circ g_2 \circ \cdots \circ g_n} = \map \phi {g_1} * \map \phi {g_2} * \cdots * \map \phi {g_n}$
A [[Definition:Group|group]] is a [[Definition:Semigroup|semigroup]]. The result then follows from the [[General Morphism Property for Semigroups]]. {{Qed}}
General Morphism Property for Groups
https://proofwiki.org/wiki/General_Morphism_Property_for_Groups
https://proofwiki.org/wiki/General_Morphism_Property_for_Groups
[ "Morphism Property", "Group Homomorphisms" ]
[ "Definition:Group", "Definition:Group Homomorphism" ]
[ "Definition:Group", "Definition:Semigroup", "General Morphism Property for Semigroups" ]
proofwiki-15045
General Linear Group to Determinant is Homomorphism/Corollary
The kernel of the $\det$ mapping is the special linear group $\SL {n, \R}$.
From General Linear Group to Determinant is Homomorphism: :$\det$ is a group homomorphism. The special linear group $\SL {n, \R}$ is the subset of $\GL {n, \R}$ such that: :$\forall \mathbf A \in \SL {n, \R}: \map \det {\mathbf A} = 1$ From Real Multiplication Identity is One: : $1$ is the identity of the multiplicativ...
The [[Definition:Kernel of Group Homomorphism|kernel]] of the $\det$ [[Definition:Mapping|mapping]] is the [[Definition:Special Linear Group|special linear group]] $\SL {n, \R}$.
From [[General Linear Group to Determinant is Homomorphism]]: :$\det$ is a [[Definition:Group Homomorphism|group homomorphism]]. The [[Definition:Special Linear Group|special linear group]] $\SL {n, \R}$ is the [[Definition:Subset|subset]] of $\GL {n, \R}$ such that: :$\forall \mathbf A \in \SL {n, \R}: \map \det {\m...
General Linear Group to Determinant is Homomorphism/Corollary
https://proofwiki.org/wiki/General_Linear_Group_to_Determinant_is_Homomorphism/Corollary
https://proofwiki.org/wiki/General_Linear_Group_to_Determinant_is_Homomorphism/Corollary
[ "Examples of Group Homomorphisms", "General Linear Group", "Special Linear Group", "Determinants" ]
[ "Definition:Kernel of Group Homomorphism", "Definition:Mapping", "Definition:Special Linear Group" ]
[ "General Linear Group to Determinant is Homomorphism", "Definition:Group Homomorphism", "Definition:Special Linear Group", "Definition:Subset", "Real Multiplication Identity is One", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Multiplicative Group of Real Numbers", "Defini...
proofwiki-15046
Quotient Group of General Linear Group by Special Linear Group
Let $\GL {n, \R}$ denote the general linear group of degree $n$ over $\R$. Let $\SL {n, \R}$ denote the special linear group of degree $n$ over $\R$. Then the quotient group $\GL {n, \R} / \SL {n, \R}$ is the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
Let $\det: \GL {n, \R} \to \struct {\R_{\ne 0}, \times}$ be the group homomorphism: :$\mathbf A \mapsto \map \det {\mathbf A}$ where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$. This is demonstrated to be a homomorphism in General Linear Group to Determinant is Homomorphism From {{Corollary|General Linear...
Let $\GL {n, \R}$ denote the [[Definition:General Linear Group|general linear group]] of degree $n$ over $\R$. Let $\SL {n, \R}$ denote the [[Definition:Special Linear Group|special linear group]] of degree $n$ over $\R$. Then the [[Definition:Quotient Group|quotient group]] $\GL {n, \R} / \SL {n, \R}$ is the [[Defi...
Let $\det: \GL {n, \R} \to \struct {\R_{\ne 0}, \times}$ be the [[Definition:Group Homomorphism|group homomorphism]]: :$\mathbf A \mapsto \map \det {\mathbf A}$ where $\map \det {\mathbf A}$ is the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$. This is demonstrated to be a [[Definition:Group Homomorp...
Quotient Group of General Linear Group by Special Linear Group
https://proofwiki.org/wiki/Quotient_Group_of_General_Linear_Group_by_Special_Linear_Group
https://proofwiki.org/wiki/Quotient_Group_of_General_Linear_Group_by_Special_Linear_Group
[ "General Linear Group", "Special Linear Group", "Examples of Quotient Groups" ]
[ "Definition:General Linear Group", "Definition:Special Linear Group", "Definition:Quotient Group", "Definition:Multiplicative Group of Real Numbers" ]
[ "Definition:Group Homomorphism", "Definition:Determinant/Matrix", "Definition:Group Homomorphism", "General Linear Group to Determinant is Homomorphism", "Definition:Kernel of Group Homomorphism", "Kernel is Normal Subgroup of Domain", "Definition:Normal Subgroup", "First Isomorphism Theorem/Groups", ...
proofwiki-15047
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 1
Let $\psi' = \phi_2 \circ \phi_1^{-1}:\phi_1 \paren R \to \phi_2 \paren R$ be the composition of $\phi_1^{-1}$ with $\phi_2$. Then $\psi': \struct {\map {\phi_1} R, \norm {\, \cdot \,}_1 } \to \struct {\map {\phi_2} R, \norm {\, \cdot \,}_2 }$ is an isometric ring isomorphism.
By Monomorphism Image is Isomorphic to Domain, $\phi_1:R \to \map {\phi_1} R$ and $\phi_2:R \to \map {\phi_2} R$ are ring isomorphisms. By Distance-Preserving Image Isometric to Domain for Metric Spaces, $\phi_1:R \to \map {\phi_1} R$ and $\phi_2:R \to \map {\phi_2} R$ are isometries. By Inverse of Algebraic Structure ...
Let $\psi' = \phi_2 \circ \phi_1^{-1}:\phi_1 \paren R \to \phi_2 \paren R$ be the composition of $\phi_1^{-1}$ with $\phi_2$. Then $\psi': \struct {\map {\phi_1} R, \norm {\, \cdot \,}_1 } \to \struct {\map {\phi_2} R, \norm {\, \cdot \,}_2 }$ is an [[Definition:Isometry (Metric Spaces)|isometric]] [[Definition:Ring ...
By [[Monomorphism Image is Isomorphic to Domain]], $\phi_1:R \to \map {\phi_1} R$ and $\phi_2:R \to \map {\phi_2} R$ are [[Definition:Ring Isomorphism|ring isomorphisms]]. By [[Distance-Preserving Image Isometric to Domain for Metric Spaces]], $\phi_1:R \to \map {\phi_1} R$ and $\phi_2:R \to \map {\phi_2} R$ are [[Def...
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 1
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_1
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_1
[ "Normed Division Ring Completions are Isometric and Isomorphic" ]
[ "Definition:Isometry (Metric Spaces)", "Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism" ]
[ "Monomorphism Image is Isomorphic to Domain", "Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism", "Distance-Preserving Image Isometric to Domain for Metric Spaces", "Definition:Isometry (Metric Spaces)", "Inverse of Algebraic Structure Isomorphism is Isomorphism", "Definition:Isomorphism (Abstra...
proofwiki-15048
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 2
Let $\psi: S_1 \to S_2$ be defined by: :$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'} {x_n}$ where $\ds x = \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$ Then $\psi$ is a well-defined mapping.
Let $x \in S_1$. By the definition of dense subset: :$\map \cl {R_1} = S_1$ By Closure of Subset of Metric Space by Convergent Sequence, there exists a sequence $\sequence {x_n} \subseteq R_1 $ that converges to $x$, that is: :$\ds \lim_{n \mathop \to \infty} x_n = x$ By Isometric Image of Cauchy Sequence is Cauchy Seq...
Let $\psi: S_1 \to S_2$ be defined by: :$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'} {x_n}$ where $\ds x = \lim_{n \mathop \to \infty} x_n$ for some [[Definition:Sequence|sequence]] $\sequence {x_n} \subseteq R_1$ Then $\psi$ is a [[Definition:Well-Defined Mapping|well-defined mapping...
Let $x \in S_1$. By the definition of [[Definition:Dense|dense subset]]: :$\map \cl {R_1} = S_1$ By [[Closure of Subset of Metric Space by Convergent Sequence]], there exists a [[Definition:Sequence|sequence]] $\sequence {x_n} \subseteq R_1 $ that [[Definition:Convergent Sequence in Normed Division Ring|converges]] t...
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 2
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_2
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_2
[ "Normed Division Ring Completions are Isometric and Isomorphic" ]
[ "Definition:Sequence", "Definition:Well-Defined/Mapping" ]
[ "Definition:Dense", "Closure of Subset of Metric Space by Convergent Sequence", "Definition:Sequence", "Definition:Convergent Sequence/Normed Division Ring", "Isometric Image of Cauchy Sequence is Cauchy Sequence", "Definition:Cauchy Sequence/Normed Division Ring", "Definition:Complete Normed Division R...
proofwiki-15049
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 3
:$\psi$ is a surjective mapping.
Let $y \in S_2$. By the definition of dense subset: :$\map \cl {R_2} = S_2$ By Closure of Subset of Metric Space by Convergent Sequence: :there exists a sequence $\sequence {y_n} \subseteq R_2 $ that converges to $y$, that is, $\ds \lim_{n \mathop \to \infty} y_n = y$ By Isometric Image of Cauchy Sequence is Cauchy Seq...
:$\psi$ is a [[Definition:Surjection|surjective mapping]].
Let $y \in S_2$. By the definition of [[Definition:Dense|dense subset]]: :$\map \cl {R_2} = S_2$ By [[Closure of Subset of Metric Space by Convergent Sequence]]: :there exists a [[Definition:Sequence|sequence]] $\sequence {y_n} \subseteq R_2 $ that [[Definition:Convergent Sequence in Normed Division Ring|converges]] ...
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 3
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_3
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_3
[ "Normed Division Ring Completions are Isometric and Isomorphic" ]
[ "Definition:Surjection" ]
[ "Definition:Dense", "Closure of Subset of Metric Space by Convergent Sequence", "Definition:Sequence", "Definition:Convergent Sequence/Normed Division Ring", "Isometric Image of Cauchy Sequence is Cauchy Sequence", "Definition:Cauchy Sequence/Normed Division Ring", "Definition:Complete Normed Division R...
proofwiki-15050
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 4
:$\psi$ is an isometry.
Let $x, y \in S_1$. Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R_1$ such that $\ds \lim_{n \mathop \to \infty} x_n = x, \lim_{n \mathop \to \infty} y_n = y$. Then: {{begin-eqn}} {{eqn | l = x - y | r = \lim_{n \mathop \to \infty} x_n - y_n | c = Difference Rule for Sequences in Normed Divi...
:$\psi$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
Let $x, y \in S_1$. Let $\sequence {x_n}$ and $\sequence {y_n}$ be [[Definition:Sequence|sequences]] in $R_1$ such that $\ds \lim_{n \mathop \to \infty} x_n = x, \lim_{n \mathop \to \infty} y_n = y$. Then: {{begin-eqn}} {{eqn | l = x - y | r = \lim_{n \mathop \to \infty} x_n - y_n | c = [[Difference Rule...
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 4
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_4
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_4
[ "Normed Division Ring Completions are Isometric and Isomorphic" ]
[ "Definition:Isometry (Metric Spaces)" ]
[ "Definition:Sequence", "Combination Theorem for Sequences/Normed Division Ring/Difference Rule", "Modulus of Limit/Normed Division Ring", "Combination Theorem for Sequences/Normed Division Ring/Difference Rule", "Modulus of Limit/Normed Division Ring", "Convergent Sequence in Metric Space has Unique Limit...
proofwiki-15051
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 5
:$\psi$ is a ring isomorphism.
By {{Lemma|Normed Division Ring Completions are Isometric and Isomorphic|4}}, $\psi$ is an isometry. By the definition of an isometry, $\psi$ is a bijection. By the definition of a ring isomorphism, all that remains is to show that $\psi$ is a ring homomorphism. That is: :$(1): \quad \forall x, y \in S_1: \map \psi {x ...
:$\psi$ is a [[Definition:Ring Isomorphism|ring isomorphism]].
By {{Lemma|Normed Division Ring Completions are Isometric and Isomorphic|4}}, $\psi$ is an [[Definition:Isometry (Metric Spaces)|isometry]]. By the definition of an [[Definition:Isometry (Metric Spaces)|isometry]], $\psi$ is a [[Definition:Bijective|bijection]]. By the definition of a [[Definition:Ring Isomorphism|ri...
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 5
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_5
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic/Lemma_5
[ "Normed Division Ring Completions are Isometric and Isomorphic" ]
[ "Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism" ]
[ "Definition:Isometry (Metric Spaces)", "Definition:Isometry (Metric Spaces)", "Definition:Bijection", "Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism", "Definition:Ring Homomorphism", "Definition:Sequence", "Definition:Sequence", "Combination Theorem for Sequences/Normed Division Ring/Sum ...
proofwiki-15052
Normed Division Ring Completions are Isometric and Isomorphic
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring. Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be normed division ring completions of $\struct {R, \norm {\, \cdot \,} }$ Then there exists an isometric isomorphism: :$\psi: \struct {S_1, \norm {\, \cdot \,}_1 } ...
By the definition of a normed division ring completion then: :there exists a distance-preserving ring monomorphisms $\phi_1: R \to S_1$ :$R_1 = \map {\phi_1} R$ is a dense subring of $S_1$ :$S_1$ is a complete metric space :there exists a distance-preserving ring monomorphisms $\phi_2: R \to S_2$ :$R_2 = \map {\phi_2} ...
Let $\struct {R, \norm {\, \cdot \,} }$ be a [[Definition:Normed Division Ring|normed division ring]]. Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be [[Definition:Completion (Normed Division Ring)|normed division ring completions]] of $\struct {R, \norm {\, \cdot \,} }$ Th...
By the definition of a [[Definition:Completion (Normed Division Ring)|normed division ring completion]] then: :there exists a [[Definition:Distance-Preserving Mapping|distance-preserving]] [[Definition:Ring Monomorphism|ring monomorphisms]] $\phi_1: R \to S_1$ :$R_1 = \map {\phi_1} R$ is a [[Definition:Dense|dense]] [[...
Normed Division Ring Completions are Isometric and Isomorphic
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic
https://proofwiki.org/wiki/Normed_Division_Ring_Completions_are_Isometric_and_Isomorphic
[ "Normed Division Ring Completions are Isometric and Isomorphic", "Completion of Normed Division Ring", "Normed Division Rings", "Complete Metric Spaces" ]
[ "Definition:Normed Division Ring", "Definition:Completion (Normed Division Ring)", "Definition:Isometric Isomorphism/Normed Division Ring" ]
[ "Definition:Completion (Normed Division Ring)", "Definition:Distance-Preserving Mapping", "Definition:Ring Monomorphism", "Definition:Dense", "Definition:Subring", "Definition:Complete Metric Space", "Definition:Distance-Preserving Mapping", "Definition:Ring Monomorphism", "Definition:Dense", "Def...
proofwiki-15053
Distance-Preserving Image Isometric to Domain for Metric Spaces
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces. Let $\phi: M_1 \to M_2$ be a distance-preserving mapping. Then: :$\phi: M_1 \to \Img \phi$ is an isometry.
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces. Let $\phi$ be a distance-preserving mapping from $M_1$ to $M_2$. Let $A = \Img \phi$ be the image of $\phi$. By Subspace of Metric Space is Metric Space, $\struct {A, d_2}$ is a metric space. As $\phi$ is a distance-preserving mapping, by D...
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]]. Let $\phi: M_1 \to M_2$ be a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]]. Then: :$\phi: M_1 \to \Img \phi$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]]. Let $\phi$ be a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]] from $M_1$ to $M_2$. Let $A = \Img \phi$ be the [[Definition:Image of Mapping|image]] of $\phi$. By [[Subspace of Metric S...
Distance-Preserving Image Isometric to Domain for Metric Spaces
https://proofwiki.org/wiki/Distance-Preserving_Image_Isometric_to_Domain_for_Metric_Spaces
https://proofwiki.org/wiki/Distance-Preserving_Image_Isometric_to_Domain_for_Metric_Spaces
[ "Isometries (Metric Spaces)", "Metric Spaces" ]
[ "Definition:Metric Space", "Definition:Distance-Preserving Mapping", "Definition:Isometry (Metric Spaces)" ]
[ "Definition:Metric Space", "Definition:Distance-Preserving Mapping", "Definition:Image (Set Theory)/Mapping/Mapping", "Subspace of Metric Space is Metric Space", "Definition:Metric Space", "Definition:Distance-Preserving Mapping", "Distance-Preserving Mapping is Injection of Metric Spaces", "Definitio...
proofwiki-15054
Distance-Preserving Mapping is Injection of Metric Spaces
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces. Let $\phi: M_1 \to M_2$ be a distance-preserving mapping. Then $\phi$ is an injection.
Let $a, b \in A_1$ and suppose that $\map \phi a = \map \phi b$. Then by the definition of a metric space: :$\map {d_2} {\map \phi a, \map \phi b} = 0$ By the definition of a distance-preserving mapping then: :$\map {d_1} {a, b} = 0$ Thus by the definition of a metric space: :$a = b$ Hence $\phi$ is injective. {{qed}} ...
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]]. Let $\phi: M_1 \to M_2$ be a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]]. Then $\phi$ is an [[Definition:Injective|injection]].
Let $a, b \in A_1$ and suppose that $\map \phi a = \map \phi b$. Then by the definition of a [[Definition:Metric Space|metric space]]: :$\map {d_2} {\map \phi a, \map \phi b} = 0$ By the definition of a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]] then: :$\map {d_1} {a, b} = 0$ Thus by th...
Distance-Preserving Mapping is Injection of Metric Spaces
https://proofwiki.org/wiki/Distance-Preserving_Mapping_is_Injection_of_Metric_Spaces
https://proofwiki.org/wiki/Distance-Preserving_Mapping_is_Injection_of_Metric_Spaces
[ "Metric Spaces" ]
[ "Definition:Metric Space", "Definition:Distance-Preserving Mapping", "Definition:Injective" ]
[ "Definition:Metric Space", "Definition:Distance-Preserving Mapping", "Definition:Metric Space", "Definition:Injection", "Category:Metric Spaces" ]
proofwiki-15055
Odd Power Function is Surjective
Let $n \in \Z_{\ge 0}$ be an odd positive integer. Let $f_n: \R \to \R$ be the real function defined as: :$\map {f_n} x = x^n$ Then $f_n$ is a surjection.
From Existence of Positive Root of Positive Real Number we have that: :$\forall x \in \R_{\ge 0}: \exists y \in \R: y^n = x$ From Power of Ring Negative: :$\paren {-x}^n = -\paren {x^n}$ and so: :$\forall x \in \R_{\le 0}: \exists y \in \R: y^n = x$ Thus: :$\forall x \in \R: \exists y \in \R: y^n = x$ and so $f_n$ is a...
Let $n \in \Z_{\ge 0}$ be an [[Definition:Odd Integer|odd]] [[Definition:Strictly Positive Integer|positive integer]]. Let $f_n: \R \to \R$ be the [[Definition:Real Function|real function]] defined as: :$\map {f_n} x = x^n$ Then $f_n$ is a [[Definition:Surjection|surjection]].
From [[Existence of Positive Root of Positive Real Number]] we have that: :$\forall x \in \R_{\ge 0}: \exists y \in \R: y^n = x$ From [[Power of Ring Negative]]: :$\paren {-x}^n = -\paren {x^n}$ and so: :$\forall x \in \R_{\le 0}: \exists y \in \R: y^n = x$ Thus: :$\forall x \in \R: \exists y \in \R: y^n = x$ and so ...
Odd Power Function is Surjective
https://proofwiki.org/wiki/Odd_Power_Function_is_Surjective
https://proofwiki.org/wiki/Odd_Power_Function_is_Surjective
[ "Powers", "Surjections", "Real Functions" ]
[ "Definition:Odd Integer", "Definition:Strictly Positive/Integer", "Definition:Real Function", "Definition:Surjection" ]
[ "Existence of Positive Root of Positive Real Number", "Power of Ring Negative", "Definition:Surjection", "Category:Powers", "Category:Surjections", "Category:Real Functions" ]
proofwiki-15056
Residue at Multiple Pole
Let $f: \C \to \C$ be a function meromorphic on some region, $D$, containing $a$. Let $f$ have a single pole in $D$, of order $N$, at $a$. Then the residue of $f$ at $a$ is given by: :$\ds \Res f a = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac {\d^{N - 1} } {\d z^{N - 1} } \paren {\paren {z - a}^N \map f z}...
By Existence of Laurent Series, there exists a Laurent series: :$\ds \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$ convergent on $D \setminus \set a$. As $f$ has a pole of order $N$ at $a$, we have $c_n = 0$ for $n < -N$. So: :$\ds \paren {z - a}^N \map f z = \sum_{n \mathop = -N}^\infty c_n \...
Let $f: \C \to \C$ be a [[Definition:Complex Function|function]] [[Definition:Meromorphic Function|meromorphic]] on some [[Definition:Region (Complex Analysis)|region]], $D$, containing $a$. Let $f$ have a single [[Definition:Pole|pole]] in $D$, of order $N$, at $a$. Then the [[Definition:Residue (Complex Analysis)|...
By [[Existence of Laurent Series]], there exists a [[Definition:Laurent Series|Laurent series]]: :$\ds \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$ convergent on $D \setminus \set a$. As $f$ has a pole of order $N$ at $a$, we have $c_n = 0$ for $n < -N$. So: :$\ds \paren {z - a}^N \map f...
Residue at Multiple Pole
https://proofwiki.org/wiki/Residue_at_Multiple_Pole
https://proofwiki.org/wiki/Residue_at_Multiple_Pole
[ "Complex Analysis" ]
[ "Definition:Complex Function", "Definition:Meromorphic Function", "Definition:Region/Complex", "Definition:Pole", "Definition:Residue (Complex Analysis)" ]
[ "Existence of Laurent Series", "Definition:Laurent Series", "Definition:Taylor Series", "Definition:Residue (Complex Analysis)", "Taylor Series of Holomorphic Function", "Category:Complex Analysis" ]
proofwiki-15057
Bijection from Cartesian Product of Initial Segments to Initial Segment
Let $\N_k$ be used to denote the set of the first $k$ non-zero natural numbers: :$\N_k := \set {1, 2, \ldots, k}$ Then a bijection can be established between $\N_k \times \N_l$ and $\N_{k l}$, where $\N_k \times \N_l$ denotes the Cartesian product of $\N_k$ and $\N_l$.
Let $\phi: \N_k \times \N_l \to \N_{k l}$ be defined as: :$\forall \tuple {m, n} \in \N_k \times \N_l: \map \phi {m, n} = \paren {m - 1} \times l + n$ First it is confirmed that the codomain of $\phi$ is indeed $\N_{k l}$. {{finish|fiddly and tedious, can't think of an elegant way to prove it}}
Let $\N_k$ be used to denote the [[Definition:Set|set]] of the first $k$ [[Definition:Non-Zero Natural Number|non-zero natural numbers]]: :$\N_k := \set {1, 2, \ldots, k}$ Then a [[Definition:Bijection|bijection]] can be established between $\N_k \times \N_l$ and $\N_{k l}$, where $\N_k \times \N_l$ denotes the [[De...
Let $\phi: \N_k \times \N_l \to \N_{k l}$ be defined as: :$\forall \tuple {m, n} \in \N_k \times \N_l: \map \phi {m, n} = \paren {m - 1} \times l + n$ First it is confirmed that the [[Definition:Codomain of Mapping|codomain]] of $\phi$ is indeed $\N_{k l}$. {{finish|fiddly and tedious, can't think of an elegant way...
Bijection from Cartesian Product of Initial Segments to Initial Segment
https://proofwiki.org/wiki/Bijection_from_Cartesian_Product_of_Initial_Segments_to_Initial_Segment
https://proofwiki.org/wiki/Bijection_from_Cartesian_Product_of_Initial_Segments_to_Initial_Segment
[ "Bijections", "Natural Numbers" ]
[ "Definition:Set", "Definition:Non-Zero Natural Number", "Definition:Bijection", "Definition:Cartesian Product" ]
[ "Definition:Codomain (Set Theory)/Mapping" ]
proofwiki-15058
Bijection between S x T and T x S
Let $S$ and $T$ be sets. Let $S \times T$ be the Cartesian product of $S$ and $T$. Then there exists a bijection from $S \times T$ to $T \times S$.
Let $\phi: S \times T \to T \times S$ be the mapping defined as: :$\forall \tuple {s, t} \in S \times T: \map \phi {s, t} = \tuple {t, s}$ Then $\phi$ is the bijection required, as follows: The domain of $\phi$ is $S \times T$. Let $\tuple {t, s} \in T \times S$. Then there exists $\tuple {s, t} \in S \times T$ such th...
Let $S$ and $T$ be [[Definition:Set|sets]]. Let $S \times T$ be the [[Definition:Cartesian Product|Cartesian product]] of $S$ and $T$. Then there exists a [[Definition:Bijection|bijection]] from $S \times T$ to $T \times S$.
Let $\phi: S \times T \to T \times S$ be the [[Definition:Mapping|mapping]] defined as: :$\forall \tuple {s, t} \in S \times T: \map \phi {s, t} = \tuple {t, s}$ Then $\phi$ is the [[Definition:Bijection|bijection]] required, as follows: The [[Definition:Domain of Mapping|domain]] of $\phi$ is $S \times T$. Let $...
Bijection between S x T and T x S
https://proofwiki.org/wiki/Bijection_between_S_x_T_and_T_x_S
https://proofwiki.org/wiki/Bijection_between_S_x_T_and_T_x_S
[ "Cartesian Product" ]
[ "Definition:Set", "Definition:Cartesian Product", "Definition:Bijection" ]
[ "Definition:Mapping", "Definition:Bijection", "Definition:Domain (Set Theory)/Mapping", "Definition:Surjection", "Definition:Injection", "Definition:Bijection" ]
proofwiki-15059
Bijection between R x (S x T) and (R x S) x T
Let $R$, $S$ and $T$ be sets. Let $S \times T$ be the Cartesian product of $S$ and $T$. Then there exists a bijection from $R \times \paren {S \times T}$ to $\paren {R \times S} \times T$. Hence: :$\card {R \times \paren {S \times T} } = \card {\paren {R \times S} \times T}$
Let $\phi: R \times \paren {S \times T} \to \paren {R \times S} \times T$ be the mapping defined as: :$\forall \tuple {r, \tuple {s, t} } \in R \times \paren {S \times T}: \map \phi {s, t} = \tuple {\tuple {r, s}, t}$ Then $\phi$ is the bijection required, as follows: The domain of $\phi$ is $R \times \paren {S \times ...
Let $R$, $S$ and $T$ be [[Definition:Set|sets]]. Let $S \times T$ be the [[Definition:Cartesian Product|Cartesian product]] of $S$ and $T$. Then there exists a [[Definition:Bijection|bijection]] from $R \times \paren {S \times T}$ to $\paren {R \times S} \times T$. Hence: :$\card {R \times \paren {S \times T} } = \...
Let $\phi: R \times \paren {S \times T} \to \paren {R \times S} \times T$ be the [[Definition:Mapping|mapping]] defined as: :$\forall \tuple {r, \tuple {s, t} } \in R \times \paren {S \times T}: \map \phi {s, t} = \tuple {\tuple {r, s}, t}$ Then $\phi$ is the [[Definition:Bijection|bijection]] required, as follows: ...
Bijection between R x (S x T) and (R x S) x T
https://proofwiki.org/wiki/Bijection_between_R_x_(S_x_T)_and_(R_x_S)_x_T
https://proofwiki.org/wiki/Bijection_between_R_x_(S_x_T)_and_(R_x_S)_x_T
[ "Cartesian Product" ]
[ "Definition:Set", "Definition:Cartesian Product", "Definition:Bijection" ]
[ "Definition:Mapping", "Definition:Bijection", "Definition:Domain (Set Theory)/Mapping", "Definition:Surjection", "Definition:Injection", "Definition:Bijection" ]
proofwiki-15060
Bijection between Power Set of Disjoint Union and Cartesian Product of Power Sets
Let $S$ and $T$ be disjoint sets. Let $\powerset S$ denote the power set of $S$. Then there exists a bijection between $\powerset {S \cup T}$ and $\paren {\powerset S} \times \paren {\powerset T}$. Hence: :$\powerset {S \cup T} \sim \paren {\powerset S} \times \paren {\powerset T}$ where $\sim$ denotes set equivalence.
Let $\phi: \paren {\powerset S} \times \paren {\powerset T} \to \powerset {S \cup T}$ be defined as: :$\forall \tuple {A, B} \in \paren {\powerset S} \times \paren {\powerset T}: \map \phi {A, B} = A \cup B$ In order to show that $\phi$ is a bijection, it needs to be shown that $\phi$ has the following properties: :$\...
Let $S$ and $T$ be [[Definition:Disjoint Sets|disjoint sets]]. Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$. Then there exists a [[Definition:Bijection|bijection]] between $\powerset {S \cup T}$ and $\paren {\powerset S} \times \paren {\powerset T}$. Hence: :$\powerset {S \cup T} \sim \pa...
Let $\phi: \paren {\powerset S} \times \paren {\powerset T} \to \powerset {S \cup T}$ be defined as: :$\forall \tuple {A, B} \in \paren {\powerset S} \times \paren {\powerset T}: \map \phi {A, B} = A \cup B$ In order to show that $\phi$ is a [[Definition:Bijection|bijection]], it needs to be shown that $\phi$ has t...
Bijection between Power Set of Disjoint Union and Cartesian Product of Power Sets
https://proofwiki.org/wiki/Bijection_between_Power_Set_of_Disjoint_Union_and_Cartesian_Product_of_Power_Sets
https://proofwiki.org/wiki/Bijection_between_Power_Set_of_Disjoint_Union_and_Cartesian_Product_of_Power_Sets
[ "Power Set", "Set Union", "Cartesian Product" ]
[ "Definition:Disjoint Sets", "Definition:Power Set", "Definition:Bijection", "Definition:Set Equivalence" ]
[ "Definition:Bijection", "Definition:Property", "Definition:Mapping", "Definition:Left-Total Relation", "Definition:Many-to-One Relation", "Definition:Surjection", "Definition:Injection", "Definition:Left-Total Relation", "Definition:Many-to-One Relation", "Definition:Mapping", "Definition:Power ...
proofwiki-15061
Bijection between Power Set of nth Initial Section and Initial Section of nth Power of 2
Let $\N_n$ be used to denote the first $n$ non-zero natural numbers: :$\N_n = \set {1, 2, \ldots, n}$ Then there exists a bijection between the power set of $\N_n$ and $\N_{2^n}$.
Let $\phi: \powerset {\N_n} \to \N_{2^n}$ be defined as: :$\forall A \in \powerset {\N_n}: \map \phi A = \begin{cases} \ds \sum_{k \mathop \in A} 2^{k - 1} & : A \ne \O \\ 2^k & : A = \O \end{cases}$ Apart from $\O$, every $A \in \powerset {\N_n}$ consists of a set of integers between $1$ and $n$. The expression $\ds \...
Let $\N_n$ be used to denote the [[Definition:Initial Segment of One-Based Natural Numbers|first $n$ non-zero natural numbers]]: :$\N_n = \set {1, 2, \ldots, n}$ Then there exists a [[Definition:Bijection|bijection]] between the [[Definition:Power Set|power set]] of $\N_n$ and $\N_{2^n}$.
Let $\phi: \powerset {\N_n} \to \N_{2^n}$ be defined as: :$\forall A \in \powerset {\N_n}: \map \phi A = \begin{cases} \ds \sum_{k \mathop \in A} 2^{k - 1} & : A \ne \O \\ 2^k & : A = \O \end{cases}$ Apart from $\O$, every $A \in \powerset {\N_n}$ consists of a [[Definition:Set|set]] of [[Definition:Integer|integers]...
Bijection between Power Set of nth Initial Section and Initial Section of nth Power of 2
https://proofwiki.org/wiki/Bijection_between_Power_Set_of_nth_Initial_Section_and_Initial_Section_of_nth_Power_of_2
https://proofwiki.org/wiki/Bijection_between_Power_Set_of_nth_Initial_Section_and_Initial_Section_of_nth_Power_of_2
[ "Power Set", "Integer Powers" ]
[ "Definition:Initial Segment of Natural Numbers/One-Based", "Definition:Bijection", "Definition:Power Set" ]
[ "Definition:Set", "Definition:Integer", "Definition:Summation", "Definition:Set", "Definition:Power (Algebra)/Integer", "Definition:Left-Total Relation", "Definition:Many-to-One Relation", "Definition:Mapping", "Basis Representation Theorem", "Definition:Integer", "Definition:Addition/Sum", "D...
proofwiki-15062
Trivial Group is Group
The trivial group is a group.
Let $G = \struct {\set e, \circ}$ be an algebraic structure.
The [[Definition:Trivial Group|trivial group]] is a [[Definition:Group|group]].
Let $G = \struct {\set e, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Trivial Group is Group
https://proofwiki.org/wiki/Trivial_Group_is_Group
https://proofwiki.org/wiki/Trivial_Group_is_Group
[ "Trivial Group" ]
[ "Definition:Trivial Group", "Definition:Group" ]
[ "Definition:Algebraic Structure/One Operation" ]
proofwiki-15063
Symmetry Group of Line Segment is Group
The symmetry group of the line segment is a group.
Let us refer to this group as $D_1$. Taking the group axioms in turn:
The [[Definition:Symmetry Group of Line Segment|symmetry group of the line segment]] is a [[Definition:Group|group]].
Let us refer to this group as $D_1$. Taking the [[Axiom:Group Axioms|group axioms]] in turn:
Symmetry Group of Line Segment is Group
https://proofwiki.org/wiki/Symmetry_Group_of_Line_Segment_is_Group
https://proofwiki.org/wiki/Symmetry_Group_of_Line_Segment_is_Group
[ "Symmetry Group of Line Segment" ]
[ "Definition:Symmetry Group of Line Segment", "Definition:Group", "Definition:Symmetry Group of Line Segment", "Definition:Symmetry Group of Line Segment" ]
[ "Axiom:Group Axioms" ]
proofwiki-15064
Order of Dihedral Group
The dihedral group $D_n$ is of order $2 n$.
By definition, $D_n$ is the symmetry group of the regular polygon of $n$ sides. :500px 500px Let $P$ be a regular $n$-gon. By inspection, it is seen that: :$(1): \quad$ there are $n$ symmetries of the vertices of $P$ by rotation :$(2): \quad$ there are a further $n$ symmetries of the vertices of $P$ by rotation after r...
The [[Definition:Dihedral Group|dihedral group]] $D_n$ is of [[Definition:Order of Structure|order]] $2 n$.
By definition, $D_n$ is the [[Definition:Symmetry Group|symmetry group]] of the [[Definition:Regular Polygon|regular polygon]] of $n$ [[Definition:Side of Polygon|sides]]. :[[File:SymmetryGroupOddPolygon.png|500px]] [[File:SymmetryGroupEvenPolygon.png|500px]] Let $P$ be a [[Definition:Regular Polygon|regular $n$-gon]...
Order of Dihedral Group
https://proofwiki.org/wiki/Order_of_Dihedral_Group
https://proofwiki.org/wiki/Order_of_Dihedral_Group
[ "Dihedral Groups" ]
[ "Definition:Dihedral Group", "Definition:Order of Structure" ]
[ "Definition:Symmetry Group", "Definition:Polygon/Regular", "Definition:Polygon/Side", "File:SymmetryGroupOddPolygon.png", "File:SymmetryGroupEvenPolygon.png", "Definition:Polygon/Regular", "Definition:Symmetry (Geometry)", "Definition:Polygon/Vertex", "Definition:Rotation (Geometry)/Plane", "Defin...
proofwiki-15065
Klein Four-Group is Group
The Klein $4$-group $K_4$ is a group.
From Klein Four-Group as Subgroup of $S_4$ it is demonstrated that $K_4$ is a subgroup of the $4$th symmetric group. Hence the result. {{qed}}
The [[Definition:Klein Four-Group|Klein $4$-group]] $K_4$ is a [[Definition:Group|group]].
From [[Klein Four-Group as Subgroup of S4|Klein Four-Group as Subgroup of $S_4$]] it is demonstrated that $K_4$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Symmetric Group|$4$th symmetric group]]. Hence the result. {{qed}}
Klein Four-Group is Group
https://proofwiki.org/wiki/Klein_Four-Group_is_Group
https://proofwiki.org/wiki/Klein_Four-Group_is_Group
[ "Klein Four-Group" ]
[ "Definition:Klein Four-Group", "Definition:Group" ]
[ "Klein Four-Group as Subgroup of S4", "Definition:Subgroup", "Definition:Symmetric Group" ]
proofwiki-15066
Dihedral Group is Non-Abelian
Let $n \in \N$ be a natural number such that $n > 2$. Let $D_n$ denote the dihedral group of order $2 n$. Then $D_n$ is not abelian.
From Group of Order less than 6 is Abelian we have that $D_1$ and $D_2$ are abelian, which is why the condition on $n$. From Group Presentation of Dihedral Group we have: :$\beta \alpha = \alpha^{n - 1} \beta$ for some $\alpha, \beta \in D_n$ such that $\alpha \ne \beta$. We also have: :$\alpha^n = e$ But if $D_n$ were...
Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n > 2$. Let $D_n$ denote the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$. Then $D_n$ is not [[Definition:Abelian Group|abelian]].
From [[Group of Order less than 6 is Abelian]] we have that $D_1$ and $D_2$ are [[Definition:Abelian Group|abelian]], which is why the condition on $n$. From [[Group Presentation of Dihedral Group]] we have: :$\beta \alpha = \alpha^{n - 1} \beta$ for some $\alpha, \beta \in D_n$ such that $\alpha \ne \beta$. We also ...
Dihedral Group is Non-Abelian
https://proofwiki.org/wiki/Dihedral_Group_is_Non-Abelian
https://proofwiki.org/wiki/Dihedral_Group_is_Non-Abelian
[ "Dihedral Groups" ]
[ "Definition:Natural Numbers", "Definition:Dihedral Group", "Definition:Order of Structure", "Definition:Abelian Group" ]
[ "Group of Order less than 6 is Abelian", "Definition:Abelian Group", "Dihedral Group/Group Presentation", "Definition:Abelian Group", "Cancellation Laws", "Definition:Dihedral Group" ]
proofwiki-15067
Matrix Entrywise Addition is Commutative
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems. For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$. The operation $+$ is commutative on $\map \MM {m, n}$. That is: :$\mathbf A + \...
From: :Integers form Ring :Rational Numbers form Ring :Real Numbers form Ring :Complex Numbers form Ring the standard number systems $\Z$, $\Q$, $\R$ and $\C$ are rings. Hence we can apply Matrix Entrywise Addition over Ring is Commutative. {{qed|lemma}} The above cannot be applied to the natural numbers $\N$, as they ...
Let $\map \MM {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over one of the [[Definition:Standard Number System|standard number systems]]. For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the [[Definition:Matrix Entrywise Addition|matrix entrywise sum]] of ...
From: :[[Integers form Ring]] :[[Rational Numbers form Ring]] :[[Real Numbers form Ring]] :[[Complex Numbers form Ring]] the [[Definition:Standard Number System|standard number systems]] $\Z$, $\Q$, $\R$ and $\C$ are [[Definition:Ring (Abstract Algebra)|rings]]. Hence we can apply [[Matrix Entrywise Addition over Rin...
Matrix Entrywise Addition is Commutative/Proof 1
https://proofwiki.org/wiki/Matrix_Entrywise_Addition_is_Commutative
https://proofwiki.org/wiki/Matrix_Entrywise_Addition_is_Commutative/Proof_1
[ "Matrix Entrywise Addition is Commutative", "Matrix Entrywise Addition", "Examples of Commutative Operations" ]
[ "Definition:Matrix Space", "Definition:Number", "Definition:Matrix Entrywise Addition", "Definition:Commutative/Operation" ]
[ "Integers form Commutative Ring", "Rational Numbers form Ring", "Real Numbers form Ring", "Complex Numbers form Ring", "Definition:Number", "Definition:Ring (Abstract Algebra)", "Matrix Entrywise Addition over Ring is Commutative", "Definition:Natural Numbers", "Definition:Ring (Abstract Algebra)", ...
proofwiki-15068
Matrix Entrywise Addition is Commutative
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems. For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$. The operation $+$ is commutative on $\map \MM {m, n}$. That is: :$\mathbf A + \...
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be matrices whose order is $m \times n$. Then: {{begin-eqn}} {{eqn | l = \mathbf A + \mathbf B | r = \sqbrk a_{m n} + \sqbrk b_{m n} | c = Definition of $\mathbf A$ and $\mathbf B$ }} {{eqn | r = \sqbrk {a + b}_{m n} | c = {{Defof|Matri...
Let $\map \MM {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over one of the [[Definition:Standard Number System|standard number systems]]. For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the [[Definition:Matrix Entrywise Addition|matrix entrywise sum]] of ...
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be [[Definition:Matrix|matrices]] whose [[Definition:Order of Matrix|order]] is $m \times n$. Then: {{begin-eqn}} {{eqn | l = \mathbf A + \mathbf B | r = \sqbrk a_{m n} + \sqbrk b_{m n} | c = Definition of $\mathbf A$ and $\mathbf B$ }} {{e...
Matrix Entrywise Addition is Commutative/Proof 2
https://proofwiki.org/wiki/Matrix_Entrywise_Addition_is_Commutative
https://proofwiki.org/wiki/Matrix_Entrywise_Addition_is_Commutative/Proof_2
[ "Matrix Entrywise Addition is Commutative", "Matrix Entrywise Addition", "Examples of Commutative Operations" ]
[ "Definition:Matrix Space", "Definition:Number", "Definition:Matrix Entrywise Addition", "Definition:Commutative/Operation" ]
[ "Definition:Matrix", "Definition:Matrix/Order", "Commutative Law of Addition" ]
proofwiki-15069
Special Linear Group is not Abelian
Let $K$ be a field whose zero is $0_K$ and unity is $1_K$. Let $\SL {n, K}$ be the special linear group of order $n$ over $K$. Then $\SL {n, K}$ is not an abelian group.
From Special Linear Group is Subgroup of General Linear Group we have that $\SL {n, K}$ is a group. From Matrix Multiplication is not Commutative it follows that $\SL {n, K}$ is not abelian. {{qed}}
Let $K$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_K$ and [[Definition:Unity of Field|unity]] is $1_K$. Let $\SL {n, K}$ be the [[Definition:Special Linear Group|special linear group of order $n$ over $K$]]. Then $\SL {n, K}$ is not an [[Definition:Abelian Group|abe...
From [[Special Linear Group is Subgroup of General Linear Group]] we have that $\SL {n, K}$ is a [[Definition:Group|group]]. From [[Matrix Multiplication is not Commutative]] it follows that $\SL {n, K}$ is not [[Definition:Abelian Group|abelian]]. {{qed}}
Special Linear Group is not Abelian
https://proofwiki.org/wiki/Special_Linear_Group_is_not_Abelian
https://proofwiki.org/wiki/Special_Linear_Group_is_not_Abelian
[ "Special Linear Group" ]
[ "Definition:Field (Abstract Algebra)", "Definition:Field Zero", "Definition:Multiplicative Identity", "Definition:Special Linear Group", "Definition:Abelian Group" ]
[ "Special Linear Group is Subgroup of General Linear Group", "Definition:Group", "Matrix Multiplication is not Commutative", "Definition:Abelian Group" ]
proofwiki-15070
Summation Formula (Complex Analysis)
:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = - \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
By Summation Formula: Lemma, there exists a constant $A$ such that: :$\cmod {\map \cot {\pi z} } < A$ for all $z$ on $C_N$. Since $f$ has only finitely many poles, we can take $N$ large enough so that no poles of $f$ lie on $C_N$. Let $X_N$ be the set of poles of $f$ contained in the region bounded by $C_N$. From P...
:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = - \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
By [[Summation Formula (Complex Analysis)/Lemma|Summation Formula: Lemma]], there exists a constant $A$ such that: :$\cmod {\map \cot {\pi z} } < A$ for all $z$ on $C_N$. Since $f$ has only [[Definition:Finite Set|finitely many]] [[Definition:Pole (Complex Analysis)|poles]], we can take $N$ large enough so that n...
Summation Formula (Complex Analysis)
https://proofwiki.org/wiki/Summation_Formula_(Complex_Analysis)
https://proofwiki.org/wiki/Summation_Formula_(Complex_Analysis)
[ "Complex Analysis" ]
[]
[ "Summation Formula (Complex Analysis)/Lemma", "Definition:Finite Set", "Definition:Isolated Singularity/Pole", "Definition:Isolated Singularity/Pole", "Definition:Set", "Definition:Isolated Singularity/Pole", "Definition:Region/Complex", "Definition:Boundary (Geometry)", "Poles of Cotangent Function...
proofwiki-15071
Summation Formula (Complex Analysis)/Lemma
Let $N \in \N$ be an arbitrary natural number. Let $C_N$ be the square embedded in the complex plane with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$. Then there exists a constant real number $A$ independent of $N$ such that: :$\cmod {\map \cot {\pi z} } < A$ for all $z \in C_N$.
Let $z = x + iy$ for real $x, y$.
Let $N \in \N$ be an arbitrary [[Definition:Natural Number|natural number]]. Let $C_N$ be the [[Definition:Square (Geometry)|square]] embedded in the [[Definition:Complex Plane|complex plane]] with [[Definition:Vertex of Polygon|vertices]] $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$. Then there exists a [[Definit...
Let $z = x + iy$ for real $x, y$.
Summation Formula (Complex Analysis)/Lemma
https://proofwiki.org/wiki/Summation_Formula_(Complex_Analysis)/Lemma
https://proofwiki.org/wiki/Summation_Formula_(Complex_Analysis)/Lemma
[ "Complex Analysis" ]
[ "Definition:Natural Numbers", "Definition:Quadrilateral/Square", "Definition:Complex Number/Complex Plane", "Definition:Polygon/Vertex", "Definition:Constant", "Definition:Real Number" ]
[]
proofwiki-15072
Product of Generating Elements of Dihedral Group
Let $D_n$ be the dihedral group of order $2 n$. Let $D_n$ be defined by its group presentation: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Then for all $k \in \Z_{\ge 0}$: :$\beta \alpha^k = \alpha^{n - k} \beta$
The proof proceeds by induction. For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition: :$\beta \alpha^k = \alpha^{n - k} \beta$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \beta \alpha^0 | r = \beta e | c = }} {{eqn | r = e \beta | c = }} {{eqn | r = \alpha^n \beta | c = }} {{e...
Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$. Let $D_n$ be defined by its [[Group Presentation of Dihedral Group|group presentation]]: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Then for all $k \in \Z_{\ge ...
The proof proceeds by [[Principle of Mathematical Induction|induction]]. For all $k \in \Z_{\ge 0}$, let $\map P k$ be the [[Definition:Proposition|proposition]]: :$\beta \alpha^k = \alpha^{n - k} \beta$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \beta \alpha^0 | r = \beta e | c = }} {{eqn | r =...
Product of Generating Elements of Dihedral Group
https://proofwiki.org/wiki/Product_of_Generating_Elements_of_Dihedral_Group
https://proofwiki.org/wiki/Product_of_Generating_Elements_of_Dihedral_Group
[ "Dihedral Groups" ]
[ "Definition:Dihedral Group", "Definition:Order of Structure", "Dihedral Group/Group Presentation" ]
[ "Principle of Mathematical Induction", "Definition:Proposition", "Principle of Mathematical Induction", "Principle of Mathematical Induction", "Principle of Mathematical Induction", "Principle of Mathematical Induction" ]
proofwiki-15073
Center of Dihedral Group
Let $n \in \N$ be a natural number such that $n \ge 3$. Let $D_n$ be the dihedral group of order $2 n$, given by: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Let $\map Z {D_n}$ denote the center of $D_n$. Then: :$\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, ...
By definition, the center of $D_n$ is: :$\map Z {D_n} = \set {g \in D_n: g x = x g, \forall x \in D_n}$ For $n \le 2$ we have that $\order {D_n} \le 4$ and so by Group of Order less than 6 is Abelian $D_n$ is abelian for $n < 3$. Hence by definition of abelian group: :$\map Z {D_n} = D_n$ for $n < 3$. So, let $n \ge 3$...
Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n \ge 3$. Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$, given by: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Let $\map Z {D_n}$ ...
By definition, the [[Definition:Center of Group|center]] of $D_n$ is: :$\map Z {D_n} = \set {g \in D_n: g x = x g, \forall x \in D_n}$ For $n \le 2$ we have that $\order {D_n} \le 4$ and so by [[Group of Order less than 6 is Abelian]] $D_n$ is [[Definition:Abelian Group|abelian]] for $n < 3$. Hence by definition of [...
Center of Dihedral Group
https://proofwiki.org/wiki/Center_of_Dihedral_Group
https://proofwiki.org/wiki/Center_of_Dihedral_Group
[ "Dihedral Groups", "Centers of Groups" ]
[ "Definition:Natural Numbers", "Definition:Dihedral Group", "Definition:Order of Structure", "Definition:Center (Abstract Algebra)/Group" ]
[ "Definition:Center (Abstract Algebra)/Group", "Group of Order less than 6 is Abelian", "Definition:Abelian Group", "Definition:Abelian Group", "Dihedral Group/Group Presentation", "Product of Generating Elements of Dihedral Group", "Definition:Generator of Group", "Definition:Order of Group Element", ...
proofwiki-15074
Intersection of Additive Groups of Integer Multiples
Let $m, n \in \Z_{> 0}$ be (strictly) positive integers. Let $\struct {m \Z, +}$ and $\struct {n \Z, +}$ be the corresponding additive groups of integer multiples. Then: :$\struct {m \Z, +} \cap \struct {n \Z, +} = \struct {\lcm \set {m, n} \Z, +}$
By definition: :$m \Z = \set {x \in \Z: m \divides x}$ Thus: {{begin-eqn}} {{eqn | l = m \Z \cap n \Z | r = \set {x \in \Z: n \divides x} \cap \set {x \in \Z: m \divides x} | c = }} {{eqn | r = \set {x \in \Z: n \divides x \land m \divides x} | c = }} {{eqn | r = \set {x \in \Z: \lcm \set {m, n} \di...
Let $m, n \in \Z_{> 0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]]. Let $\struct {m \Z, +}$ and $\struct {n \Z, +}$ be the corresponding [[Definition:Additive Group of Integer Multiples|additive groups of integer multiples]]. Then: :$\struct {m \Z, +} \cap \struct {n \Z, +} = \struct {\...
By definition: :$m \Z = \set {x \in \Z: m \divides x}$ Thus: {{begin-eqn}} {{eqn | l = m \Z \cap n \Z | r = \set {x \in \Z: n \divides x} \cap \set {x \in \Z: m \divides x} | c = }} {{eqn | r = \set {x \in \Z: n \divides x \land m \divides x} | c = }} {{eqn | r = \set {x \in \Z: \lcm \set {m, n} \...
Intersection of Additive Groups of Integer Multiples
https://proofwiki.org/wiki/Intersection_of_Additive_Groups_of_Integer_Multiples
https://proofwiki.org/wiki/Intersection_of_Additive_Groups_of_Integer_Multiples
[ "Additive Groups of Integer Multiples" ]
[ "Definition:Strictly Positive/Integer", "Definition:Additive Group of Integer Multiples" ]
[]
proofwiki-15075
Subgroups of Additive Group of Integers Modulo m
Let $n \in \Z_{> 0}$ be a (strictly) positive integer. Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$. The subgroups of $\struct {\Z_m, +_m}$ are the additive groups of integers modulo $k$ where: :$k \divides m$
From Integers Modulo m under Addition form Cyclic Group, $\struct {\Z_m, +_m}$ is cyclic. Let $H$ be a subgroup of $\struct {\Z_m, +_m}$ From Subgroup of Cyclic Group is Cyclic, $H$ is of the form $\struct {\Z_k, +_k}$ for some $k \in \Z$. From Lagrange's Theorem, it follows that $k \divides m$. Hence the result. {{qed...
Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]]. Let $\struct {\Z_m, +_m}$ denote the [[Definition:Additive Group of Integers|additive group of integers modulo $m$]]. The [[Definition:Subgroup|subgroups]] of $\struct {\Z_m, +_m}$ are the [[Definition:Additive Group of In...
From [[Integers Modulo m under Addition form Cyclic Group]], $\struct {\Z_m, +_m}$ is [[Definition:Cyclic Group|cyclic]]. Let $H$ be a [[Definition:Subgroup|subgroup]] of $\struct {\Z_m, +_m}$ From [[Subgroup of Cyclic Group is Cyclic]], $H$ is of the form $\struct {\Z_k, +_k}$ for some $k \in \Z$. From [[Lagrange's...
Subgroups of Additive Group of Integers Modulo m
https://proofwiki.org/wiki/Subgroups_of_Additive_Group_of_Integers_Modulo_m
https://proofwiki.org/wiki/Subgroups_of_Additive_Group_of_Integers_Modulo_m
[ "Additive Groups of Integers Modulo m" ]
[ "Definition:Strictly Positive/Integer", "Definition:Additive Group of Integers", "Definition:Subgroup", "Definition:Additive Group of Integers Modulo m" ]
[ "Integers Modulo m under Addition form Cyclic Group", "Definition:Cyclic Group", "Definition:Subgroup", "Subgroup of Cyclic Group is Cyclic", "Lagrange's Theorem (Group Theory)" ]
proofwiki-15076
Subgroup of Additive Group of Integers Generated by Two Integers
Let $m, n \in \Z_{> 0}$ be (strictly) positive integers. Let $\struct {\Z, +}$ denote the additive group of integers. Let $\gen {m, n}$ be the subgroup of $\struct {\Z, +}$ generated by $m$ and $n$. Then: :$\gen {m, n} = \struct {\gcd \set {m, n} \Z, +}$ That is, the additive groups of integer multiples of $\gcd \set {...
By definition: :$\gen {m, n} = \set {x \in \Z: \gcd \set {m, n} \divides x}$ {{Handwaving|Sorry, I would make the effort, but it's tedious.}} Hence the result. {{qed}}
Let $m, n \in \Z_{> 0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]]. Let $\struct {\Z, +}$ denote the [[Definition:Additive Group of Integers|additive group of integers]]. Let $\gen {m, n}$ be the [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$ [[Definition:Generator of Subgroup|gen...
By definition: :$\gen {m, n} = \set {x \in \Z: \gcd \set {m, n} \divides x}$ {{Handwaving|Sorry, I would make the effort, but it's tedious.}} Hence the result. {{qed}}
Subgroup of Additive Group of Integers Generated by Two Integers
https://proofwiki.org/wiki/Subgroup_of_Additive_Group_of_Integers_Generated_by_Two_Integers
https://proofwiki.org/wiki/Subgroup_of_Additive_Group_of_Integers_Generated_by_Two_Integers
[ "Additive Groups of Integer Multiples" ]
[ "Definition:Strictly Positive/Integer", "Definition:Additive Group of Integers", "Definition:Subgroup", "Definition:Generator of Subgroup", "Definition:Additive Group of Integer Multiples", "Definition:Greatest Common Divisor/Integers" ]
[]
proofwiki-15077
Subgroups of Cartesian Product of Additive Group of Integers
Let $\struct {\Z, +}$ denote the additive group of integers. Let $m, n \in \Z_{> 0}$ be (strictly) positive integers. Let $\struct {\Z \times \Z, +}$ denote the Cartesian product of $\struct {\Z, +}$ with itself. The subgroups of $\struct {\Z \times \Z, +}$ are not all of the form: :$\struct {m \Z, +} \times \struct {n...
Consider the map $\phi: \struct {m \Z, +} \times \struct {n \Z, +} \mapsto \struct {\Z, +} \times \struct {\Z, +}$ defined by: :$\forall c, d \in \Z: \map \phi {m c, n d} = \tuple {c, d}$ which is a group isomorphism. {{explain|Prove the above statement}} Hence, $\struct {m \Z, +} \times \struct {n \Z, +}$ is a free ab...
Let $\struct {\Z, +}$ denote the [[Definition:Additive Group of Integers|additive group of integers]]. Let $m, n \in \Z_{> 0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]]. Let $\struct {\Z \times \Z, +}$ denote the [[Definition:Cartesian Product|Cartesian product]] of $\struct {\Z, +}$ wi...
Consider the map $\phi: \struct {m \Z, +} \times \struct {n \Z, +} \mapsto \struct {\Z, +} \times \struct {\Z, +}$ defined by: :$\forall c, d \in \Z: \map \phi {m c, n d} = \tuple {c, d}$ which is a [[Definition:Group Isomorphism|group isomorphism]]. {{explain|Prove the above statement}} Hence, $\struct {m \Z, +} \t...
Subgroups of Cartesian Product of Additive Group of Integers
https://proofwiki.org/wiki/Subgroups_of_Cartesian_Product_of_Additive_Group_of_Integers
https://proofwiki.org/wiki/Subgroups_of_Cartesian_Product_of_Additive_Group_of_Integers
[ "Additive Group of Integers", "Additive Groups of Integer Multiples" ]
[ "Definition:Additive Group of Integers", "Definition:Strictly Positive/Integer", "Definition:Cartesian Product", "Definition:Subgroup", "Definition:Additive Group of Integer Multiples" ]
[ "Definition:Isomorphism (Abstract Algebra)/Group Isomorphism", "Definition:Free Abelian Group on Set", "Definition:Rank (Free Abelian Group)", "Definition:Subgroup", "Definition:Generator of Subgroup", "Definition:Singleton", "Definition:Subgroup" ]
proofwiki-15078
Equivalence Relation on Symmetric Group by Image of n is Congruence Modulo Subgroup
Let $S_n$ denote the symmetric group on $n$ letters $\set {1, \dots, n}$. Let $\sim$ be the relation on $S_n$ defined as: :$\forall \pi, \tau \in S_n: \pi \sim \tau \iff \map \pi n = \map \tau n$ Then $\sim$ is an equivalence relation which is congruence modulo a subgroup. {{explain|Work needed to be done to explain ex...
We claim that $\sim$ is left congruence modulo $S_{n - 1}$, the symmetric group on $n - 1$ letters $\set {1, \dots, n - 1}$. Notice that every element of $S_{n - 1}$ fixes $n$. For all $\pi, \tau \in S_n$ such that $\pi \sim \tau$: {{begin-eqn}} {{eqn | l = \map {\paren {\pi^{-1} \circ \tau} } n | r = \map {\pi^{...
Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]] $\set {1, \dots, n}$. Let $\sim$ be the [[Definition:Relation|relation on $S_n$]] defined as: :$\forall \pi, \tau \in S_n: \pi \sim \tau \iff \map \pi n = \map \tau n$ Then $\sim$ is an [[Definition:Equivalence Relation|eq...
We claim that $\sim$ is [[Definition:Left Congruence Modulo Subgroup|left congruence modulo $S_{n - 1}$]], the [[Definition:Symmetric Group on n Letters|symmetric group on $n - 1$ letters]] $\set {1, \dots, n - 1}$. Notice that every element of $S_{n - 1}$ fixes $n$. For all $\pi, \tau \in S_n$ such that $\pi \sim \...
Equivalence Relation on Symmetric Group by Image of n is Congruence Modulo Subgroup
https://proofwiki.org/wiki/Equivalence_Relation_on_Symmetric_Group_by_Image_of_n_is_Congruence_Modulo_Subgroup
https://proofwiki.org/wiki/Equivalence_Relation_on_Symmetric_Group_by_Image_of_n_is_Congruence_Modulo_Subgroup
[ "Symmetric Groups" ]
[ "Definition:Symmetric Group/n Letters", "Definition:Relation", "Definition:Equivalence Relation", "Definition:Congruence Modulo Subgroup" ]
[ "Definition:Congruence Modulo Subgroup/Left Congruence", "Definition:Symmetric Group/n Letters", "Definition:Congruence Modulo Subgroup/Left Congruence", "Definition:Congruence Modulo Subgroup/Left Congruence", "Definition:Equivalence Relation", "Left Congruence Modulo Subgroup is Equivalence Relation" ]
proofwiki-15079
Order of Additive Group of Integers Modulo m
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$. The order of $\struct {\Z_m, +_m}$ is $m$.
By definition, the order of a group is the cardinality of its underlying set. By definition, the underlying set of $\struct {\Z_m, +_m}$ is the set of residue classes $\Z_m$: :$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$ From Cardinality of Set of Residue Classes, $\Z_m$ has $m$ elements. Henc...
Let $\struct {\Z_m, +_m}$ denote the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]]. The [[Definition:Order of Group|order]] of $\struct {\Z_m, +_m}$ is $m$.
By definition, the [[Definition:Order of Group|order]] of a [[Definition:Group|group]] is the [[Definition:Cardinality|cardinality]] of its [[Definition:Underlying Set of Structure|underlying set]]. By definition, the [[Definition:Underlying Set of Structure|underlying set]] of $\struct {\Z_m, +_m}$ is the [[Definitio...
Order of Additive Group of Integers Modulo m
https://proofwiki.org/wiki/Order_of_Additive_Group_of_Integers_Modulo_m
https://proofwiki.org/wiki/Order_of_Additive_Group_of_Integers_Modulo_m
[ "Additive Groups of Integers Modulo m" ]
[ "Definition:Additive Group of Integers Modulo m", "Definition:Order of Structure" ]
[ "Definition:Order of Structure", "Definition:Group", "Definition:Cardinality", "Definition:Underlying Set/Abstract Algebra", "Definition:Underlying Set/Abstract Algebra", "Definition:Set of Residue Classes", "Cardinality of Set of Residue Classes", "Definition:Element" ]
proofwiki-15080
Order of Multiplicative Group of Reduced Residues
Let $\struct {\Z'_m, \times_m}$ denote the multiplicative group of reduced residues modulo $m$. The order of $\struct {\Z'_m, \times_m}$ is $\map \phi m$, where $\phi$ denotes the Euler $\phi$ function.
By definition, the order of a group is the cardinality of its underlying set. By definition, the underlying set of $\struct {\Z'_m, \times_m}$ is the reduced residue system $\Z'_m$: :$\Z'_m = \set {\eqclass {a_1} m, \eqclass {a_2} m, \ldots, \eqclass {a_{\map \phi m} } m}$ where: :$\forall k: a_k \perp m$ From Cardina...
Let $\struct {\Z'_m, \times_m}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $m$]]. The [[Definition:Order of Group|order]] of $\struct {\Z'_m, \times_m}$ is $\map \phi m$, where $\phi$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function...
By definition, the [[Definition:Order of Group|order]] of a [[Definition:Group|group]] is the [[Definition:Cardinality|cardinality]] of its [[Definition:Underlying Set of Structure|underlying set]]. By definition, the [[Definition:Underlying Set of Structure|underlying set]] of $\struct {\Z'_m, \times_m}$ is the [[Def...
Order of Multiplicative Group of Reduced Residues
https://proofwiki.org/wiki/Order_of_Multiplicative_Group_of_Reduced_Residues
https://proofwiki.org/wiki/Order_of_Multiplicative_Group_of_Reduced_Residues
[ "Multiplicative Groups of Reduced Residues" ]
[ "Definition:Multiplicative Group of Reduced Residues", "Definition:Order of Structure", "Definition:Euler Phi Function" ]
[ "Definition:Order of Structure", "Definition:Group", "Definition:Cardinality", "Definition:Underlying Set/Abstract Algebra", "Definition:Underlying Set/Abstract Algebra", "Definition:Reduced Residue System", "Cardinality of Reduced Residue System", "Definition:Element" ]
proofwiki-15081
Circle Group is Uncountably Infinite
The circle group $\struct {K, \times}$ is an uncountably infinite group.
From Quotient Group of Reals by Integers is Circle Group, $\struct {K, \times}$ is isomorphic to the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$. But $\dfrac {\struct {\R, +} } {\struct {\Z, +} }$ is the half-open interval $\hointr 0 1$. A real interval is uncountable by Real Numbers are Uncountable. {{exp...
The [[Definition:Circle Group|circle group]] $\struct {K, \times}$ is an [[Definition:Uncountable Set|uncountably]] [[Definition:Infinite Group|infinite group]].
From [[Quotient Group of Reals by Integers is Circle Group]], $\struct {K, \times}$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Quotient Group|quotient group]] of $\struct {\R, +}$ by $\struct {\Z, +}$. But $\dfrac {\struct {\R, +} } {\struct {\Z, +} }$ is the [[Definition:Half-Open Real Interva...
Circle Group is Uncountably Infinite
https://proofwiki.org/wiki/Circle_Group_is_Uncountably_Infinite
https://proofwiki.org/wiki/Circle_Group_is_Uncountably_Infinite
[ "Circle Group", "Uncountable Sets" ]
[ "Definition:Circle Group", "Definition:Uncountable/Set", "Definition:Infinite Group" ]
[ "Quotient Group of Reals by Integers is Circle Group", "Definition:Isomorphism (Abstract Algebra)/Group Isomorphism", "Definition:Quotient Group", "Definition:Real Interval/Half-Open", "Definition:Real Interval", "Definition:Uncountable/Set", "Real Numbers are Uncountably Infinite", "Real Numbers are ...
proofwiki-15082
Subgroup of Symmetric Group that Fixes n
Let $S_n$ denote the symmetric group on $n$ letters. Let $H$ denote the subgroup of $S_n$ which consists of all $\pi \in S_n$ such that: :$\map \pi n = n$ Then: :$H = S_{n - 1}$ and the index of $H$ in $S_n$ is given by: :$\index {S_n} H = n$
We have that $S_{n - 1}$ is the symmetric group on $n - 1$ letters. Let $\pi \in S_{n - 1}$. Then $\pi$ is a permutation on $n - 1$ letters. Hence $\pi$ is also a permutation on $n$ letters which fixes $n$. So $S_{n - 1} \subseteq H$. Now let $\pi \in H$. Then $\pi$ is a permutation on $n$ letters which fixes $n$. That...
Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]]. Let $H$ denote the [[Definition:Subgroup|subgroup]] of $S_n$ which consists of all $\pi \in S_n$ such that: :$\map \pi n = n$ Then: :$H = S_{n - 1}$ and the [[Definition:Index of Subgroup|index]] of $H$ in $S_n$ is give...
We have that $S_{n - 1}$ is the [[Definition:Symmetric Group on n Letters|symmetric group on $n - 1$ letters]]. Let $\pi \in S_{n - 1}$. Then $\pi$ is a [[Definition:Permutation on n Letters|permutation on $n - 1$ letters]]. Hence $\pi$ is also a [[Definition:Permutation on n Letters|permutation on $n$ letters]] wh...
Subgroup of Symmetric Group that Fixes n
https://proofwiki.org/wiki/Subgroup_of_Symmetric_Group_that_Fixes_n
https://proofwiki.org/wiki/Subgroup_of_Symmetric_Group_that_Fixes_n
[ "Symmetric Groups" ]
[ "Definition:Symmetric Group/n Letters", "Definition:Subgroup", "Definition:Index of Subgroup" ]
[ "Definition:Symmetric Group/n Letters", "Definition:Permutation on n Letters", "Definition:Permutation on n Letters", "Definition:Fixed Element under Permutation", "Definition:Permutation on n Letters", "Definition:Fixed Element under Permutation", "Definition:Permutation on n Letters", "Definition:Gr...
proofwiki-15083
Order of Product of Abelian Group Elements Divides LCM of Orders of Elements
Let $G$ be an abelian group. Let $a, b \in G$. Then: :$\order {a b} \divides \lcm \set {\order a, \order b}$ where: :$\order a$ denotes the order of $a$ :$\divides$ denotes divisibility :$\lcm$ denotes the lowest common multiple.
Let $\order a = m, \order b = n$. Let $c = \lcm \set {m, n}$. Then: {{begin-eqn}} {{eqn | l = c | r = r m | c = for some $r \in \Z$ }} {{eqn | r = s n | c = for some $s \in \Z$ }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \paren {a b}^c | r = a^c b^c | c = Power of Product of Commuting Ele...
Let $G$ be an [[Definition:Abelian Group|abelian group]]. Let $a, b \in G$. Then: :$\order {a b} \divides \lcm \set {\order a, \order b}$ where: :$\order a$ denotes the [[Definition:Order of Group Element|order]] of $a$ :$\divides$ denotes [[Definition:Divisor of Integer|divisibility]] :$\lcm$ denotes the [[Definitio...
Let $\order a = m, \order b = n$. Let $c = \lcm \set {m, n}$. Then: {{begin-eqn}} {{eqn | l = c | r = r m | c = for some $r \in \Z$ }} {{eqn | r = s n | c = for some $s \in \Z$ }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \paren {a b}^c | r = a^c b^c | c = [[Power of Product of Commu...
Order of Product of Abelian Group Elements Divides LCM of Orders of Elements
https://proofwiki.org/wiki/Order_of_Product_of_Abelian_Group_Elements_Divides_LCM_of_Orders_of_Elements
https://proofwiki.org/wiki/Order_of_Product_of_Abelian_Group_Elements_Divides_LCM_of_Orders_of_Elements
[ "Order of Group Elements", "Abelian Groups" ]
[ "Definition:Abelian Group", "Definition:Order of Group Element", "Definition:Divisor (Algebra)/Integer", "Definition:Lowest Common Multiple/Integers" ]
[ "Power of Product of Commuting Elements in Semigroup equals Product of Powers", "Element to Power of Multiple of Order is Identity" ]
proofwiki-15084
Elements of Abelian Group whose Order Divides n is Subgroup
Let $G$ be an abelian group whose identity element is $e$. Let $n \in \Z_{>0}$ be a (strictly) positive integer . Let $G_n$ be the subset of $G$ defined as: :$G_n = \set {x \in G: \order x \divides n}$ where: :$\order x$ denotes the order of $x$ :$\divides$ denotes divisibility. Then $G_n$ is a subgroup of $G$.
From Identity is Only Group Element of Order 1: :$\order e = 1$ and so from One Divides all Integers: :$\order e \divides n$ Thus $G_n \ne \O$. Then: {{begin-eqn}} {{eqn | l = x | o = \in | r = G_n | c = }} {{eqn | ll= \leadsto | l = \order x | o = \divides | r = n | c = }} {...
Let $G$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity element]] is $e$. Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]] . Let $G_n$ be the [[Definition:Subset|subset]] of $G$ defined as: :$G_n = \set {x \in G: \order x \div...
From [[Identity is Only Group Element of Order 1]]: :$\order e = 1$ and so from [[One Divides all Integers]]: :$\order e \divides n$ Thus $G_n \ne \O$. Then: {{begin-eqn}} {{eqn | l = x | o = \in | r = G_n | c = }} {{eqn | ll= \leadsto | l = \order x | o = \divides | r = n ...
Elements of Abelian Group whose Order Divides n is Subgroup
https://proofwiki.org/wiki/Elements_of_Abelian_Group_whose_Order_Divides_n_is_Subgroup
https://proofwiki.org/wiki/Elements_of_Abelian_Group_whose_Order_Divides_n_is_Subgroup
[ "Order of Group Elements", "Subgroups", "Abelian Groups" ]
[ "Definition:Abelian Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Strictly Positive/Integer", "Definition:Subset", "Definition:Order of Group Element", "Definition:Divisor (Algebra)/Integer", "Definition:Subgroup" ]
[ "Identity is Only Group Element of Order 1", "Integer Divisor Results/One Divides all Integers", "Order of Group Element equals Order of Inverse", "Order of Product of Abelian Group Elements Divides LCM of Orders of Elements", "Definition:Lowest Common Multiple/Integers", "Two-Step Subgroup Test" ]
proofwiki-15085
Groups of Order 4
There exist exactly $2$ groups of order $4$, up to isomorphism: :$C_4$, the cyclic group of order $4$ :$K_4$, the Klein $4$-group.
From Existence of Cyclic Group of Order n we have that one such group of order $4$ is the cyclic group of order $4$: This is exemplified by the additive group of integers modulo $4$, whose Cayley table can be presented as: {{:Modulo Addition/Cayley Table/Modulo 4}} From Group whose Order equals Order of Element is Cycl...
There exist exactly $2$ [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $4$, up to [[Definition:Group Isomorphism|isomorphism]]: :$C_4$, the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $4$ :$K_4$, the [[Definition:Klein Four-Group|Klein $4$-group]].
From [[Existence of Cyclic Group of Order n]] we have that one such [[Definition:Group|group]] of [[Definition:Order of Group|order]] $4$ is the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $4$: This is exemplified by the [[Definition:Additive Group of Integers Modulo m|additive grou...
Groups of Order 4
https://proofwiki.org/wiki/Groups_of_Order_4
https://proofwiki.org/wiki/Groups_of_Order_4
[ "Groups of Order 4" ]
[ "Definition:Group", "Definition:Order of Structure", "Definition:Isomorphism (Abstract Algebra)/Group Isomorphism", "Definition:Cyclic Group", "Definition:Order of Structure", "Definition:Klein Four-Group" ]
[ "Existence of Cyclic Group of Order n", "Definition:Group", "Definition:Order of Structure", "Definition:Cyclic Group", "Definition:Order of Structure", "Definition:Additive Group of Integers Modulo m", "Modulo Addition/Cayley Table/Modulo 4", "Group whose Order equals Order of Element is Cyclic", "...
proofwiki-15086
Groups of Order 6
There exist exactly $2$ groups of order $6$, up to isomorphism: :$C_6$, the cyclic group of order $6$ :$S_3$, the symmetric group on $3$ letters.
From Existence of Cyclic Group of Order n we have that one such group of order $6$ is $C_6$ the cyclic group of order $6$: This is exemplified by the additive group of integers modulo $6$, whose Cayley table can be presented as: {{:Modulo Addition/Cayley Table/Modulo 6}} Then we have the symmetric group on $3$ letters....
There exist exactly $2$ [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $6$, up to [[Definition:Group Isomorphism|isomorphism]]: :$C_6$, the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $6$ :$S_3$, the [[Definition:Symmetric Group on n Letters|symmetric group on $...
From [[Existence of Cyclic Group of Order n]] we have that one such [[Definition:Group|group]] of [[Definition:Order of Group|order]] $6$ is $C_6$ the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $6$: This is exemplified by the [[Definition:Additive Group of Integers Modulo m|additiv...
Groups of Order 6
https://proofwiki.org/wiki/Groups_of_Order_6
https://proofwiki.org/wiki/Groups_of_Order_6
[ "Groups of Order 6" ]
[ "Definition:Group", "Definition:Order of Structure", "Definition:Isomorphism (Abstract Algebra)/Group Isomorphism", "Definition:Cyclic Group", "Definition:Order of Structure", "Definition:Symmetric Group/n Letters" ]
[ "Existence of Cyclic Group of Order n", "Definition:Group", "Definition:Order of Structure", "Definition:Cyclic Group", "Definition:Order of Structure", "Definition:Additive Group of Integers Modulo m", "Modulo Addition/Cayley Table/Modulo 6", "Definition:Symmetric Group/n Letters", "Order of Symmet...
proofwiki-15087
Finite Group whose Subsets form Chain is Cyclic P-Group
Let $G$ be a group. Let $G$ be such that its subgroups form a chain. Then $G$ is a cyclic $p$-group.
Suppose $G$ is not a $p$-group. Then there exist two distinct primes $p_1, p_2$. By Cauchy's Group Theorem, there exist subgroups $H, K$ such that: :$\order H = p_1$ :$\order K = p_2$ That is: :$H = \gen a$ :$k = \gen b$ for some $a, b \in G: a \ne b$, where: :$\order a = p_1$ :$\order b = p_2$ and so both $H \nsubsete...
Let $G$ be a [[Definition:Group|group]]. Let $G$ be such that its [[Definition:Subgroup|subgroups]] form a [[Definition:Chain of Sets|chain]]. Then $G$ is a [[Definition:Cyclic Group|cyclic]] [[Definition:P-Group|$p$-group]].
Suppose $G$ is not a [[Definition:P-Group|$p$-group]]. Then there exist two [[Definition:Distinct Elements|distinct]] [[Definition:Prime Number|primes]] $p_1, p_2$. By [[Cauchy's Group Theorem]], there exist [[Definition:Subgroup|subgroups]] $H, K$ such that: :$\order H = p_1$ :$\order K = p_2$ That is: :$H = \gen a...
Finite Group whose Subsets form Chain is Cyclic P-Group
https://proofwiki.org/wiki/Finite_Group_whose_Subsets_form_Chain_is_Cyclic_P-Group
https://proofwiki.org/wiki/Finite_Group_whose_Subsets_form_Chain_is_Cyclic_P-Group
[ "Cyclic Groups", "P-Groups" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Chain (Order Theory)/Subset Relation", "Definition:Cyclic Group", "Definition:P-Group" ]
[ "Definition:P-Group", "Definition:Distinct/Plural", "Definition:Prime Number", "Cauchy's Group Theorem", "Definition:Subgroup", "Definition:Subgroup", "Definition:Chain (Order Theory)/Subset Relation", "Rule of Transposition", "Definition:P-Group", "Definition:Cyclic Group", "Definition:Cyclic G...
proofwiki-15088
Subgroup of Circle Group Generated by Distinct Roots of Unity
Let $K$ be the circle group. Let $m, n \in \Z_{>0}$ be (strictly) positive integers. Let $d = \lcm \set {m, n}$ be the least common multiple of $m$ and $n$. Let $\alpha$ be a primitive $n$th root of unity. Let $\beta$ be a primitive $m$th root of unity. Let $\gamma$ be a primitive $d$th root of unity. Let $H = \gen {\a...
{{ProofWanted|This is (probably) a specialisation of a more general result on cyclic groups.}}
Let $K$ be the [[Definition:Circle Group|circle group]]. Let $m, n \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]]. Let $d = \lcm \set {m, n}$ be the [[Definition:Least Common Multiple|least common multiple]] of $m$ and $n$. Let $\alpha$ be a [[Definition:Primitive Complex Root...
{{ProofWanted|This is (probably) a specialisation of a more general result on cyclic groups.}}
Subgroup of Circle Group Generated by Distinct Roots of Unity
https://proofwiki.org/wiki/Subgroup_of_Circle_Group_Generated_by_Distinct_Roots_of_Unity
https://proofwiki.org/wiki/Subgroup_of_Circle_Group_Generated_by_Distinct_Roots_of_Unity
[ "Multiplicative Groups of Complex Roots of Unity", "Circle Group" ]
[ "Definition:Circle Group", "Definition:Strictly Positive/Integer", "Definition:Lowest Common Multiple", "Definition:Root of Unity/Complex/Primitive", "Definition:Root of Unity/Complex/Primitive", "Definition:Root of Unity/Complex/Primitive", "Definition:Subgroup", "Definition:Generator of Subgroup" ]
[]
proofwiki-15089
Multiplicative Group of Complex Roots of Unity is Subgroup of Circle Group
Let $n \in \Z$ be an integer such that $n > 0$. Let $\struct {U_n, \times}$ denote the multiplicative group of complex $n$th roots of unity. Let $\struct {K, \times}$ denote the circle group. Then $\struct {U_n, \times}$ is a subgroup of $\struct {K, \times}$.
By definition of the multiplicative group of complex $n$th roots of unity: :$U_n := \set {z \in \C: z^n = 1}$ By definition of the circle group: :$K = \set {z \in \C: \cmod z = 1}$ By Modulus of Complex Root of Unity equals 1: :$\forall z \in U_n: \cmod z = 1$ Thus: :$U_n \subseteq K$ We further have that the operation...
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$. Let $\struct {U_n, \times}$ denote the [[Definition:Multiplicative Group of Complex Roots of Unity|multiplicative group of complex $n$th roots of unity]]. Let $\struct {K, \times}$ denote the [[Definition:Circle Group|circle group]]. Then $\stru...
By definition of the [[Definition:Multiplicative Group of Complex Roots of Unity|multiplicative group of complex $n$th roots of unity]]: :$U_n := \set {z \in \C: z^n = 1}$ By definition of the [[Definition:Circle Group|circle group]]: :$K = \set {z \in \C: \cmod z = 1}$ By [[Modulus of Complex Root of Unity equals 1]...
Multiplicative Group of Complex Roots of Unity is Subgroup of Circle Group
https://proofwiki.org/wiki/Multiplicative_Group_of_Complex_Roots_of_Unity_is_Subgroup_of_Circle_Group
https://proofwiki.org/wiki/Multiplicative_Group_of_Complex_Roots_of_Unity_is_Subgroup_of_Circle_Group
[ "Circle Group", "Multiplicative Groups of Complex Roots of Unity" ]
[ "Definition:Integer", "Definition:Multiplicative Group of Complex Roots of Unity", "Definition:Circle Group", "Definition:Subgroup" ]
[ "Definition:Multiplicative Group of Complex Roots of Unity", "Definition:Circle Group", "Modulus of Complex Root of Unity equals 1", "Definition:Operation/Binary Operation", "Definition:Multiplication/Complex Numbers", "Roots of Unity under Multiplication form Cyclic Group", "Definition:Group", "Defin...
proofwiki-15090
Intersection of Multiplicative Groups of Complex Roots of Unity
Let $\struct {K, \times}$ denote the circle group. Let $m, n \in \Z_{>0}$ be (strictly) positive integers. Let $c = \lcm \set {m, n}$ be the lowest common multiple of $m$ and $n$. Let $\struct {U_n, \times}$ denote the multiplicative group of complex $n$th roots of unity. Let $\struct {U_m, \times}$ denote the multipli...
{{ProofWanted|This is (probably) a specialisation of a more general result on cyclic groups.}}
Let $\struct {K, \times}$ denote the [[Definition:Circle Group|circle group]]. Let $m, n \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]]. Let $c = \lcm \set {m, n}$ be the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $m$ and $n$. Let $\struct {U_n...
{{ProofWanted|This is (probably) a specialisation of a more general result on cyclic groups.}}
Intersection of Multiplicative Groups of Complex Roots of Unity
https://proofwiki.org/wiki/Intersection_of_Multiplicative_Groups_of_Complex_Roots_of_Unity
https://proofwiki.org/wiki/Intersection_of_Multiplicative_Groups_of_Complex_Roots_of_Unity
[ "Multiplicative Groups of Complex Roots of Unity", "Circle Group" ]
[ "Definition:Circle Group", "Definition:Strictly Positive/Integer", "Definition:Lowest Common Multiple/Integers", "Definition:Multiplicative Group of Complex Roots of Unity", "Definition:Multiplicative Group of Complex Roots of Unity" ]
[]
proofwiki-15091
Direct Product of Normal Subgroups is Normal
Let $G$ and $G'$ be groups. Let: :$H \lhd G$ :$H' \lhd G'$ where $\lhd$ denotes the relation of being a normal subgroup. Then: :$\paren {H \times H'} \lhd \paren {G \times G'}$ where $H \times H'$ denotes the group direct product of $H$ and $H'$
Let $\tuple {x, x'} \in G \times G'$ and $\tuple {y, y'} \in H \times H'$. Then: {{begin-eqn}} {{eqn | l = \tuple {x, x'} \tuple {y, y'} \tuple {x, x'}^{-1} | r = \tuple {x, x'} \tuple {y, y'} \tuple {x^{-1}, x'^{-1} } | c = }} {{eqn | r = \tuple {x y x^{-1}, x' y' x'^{-1} } | c = }} {{eqn | o = \in...
Let $G$ and $G'$ be [[Definition:Group|groups]]. Let: :$H \lhd G$ :$H' \lhd G'$ where $\lhd$ denotes the relation of being a [[Definition:Normal Subgroup|normal subgroup]]. Then: :$\paren {H \times H'} \lhd \paren {G \times G'}$ where $H \times H'$ denotes the [[Definition:Group Direct Product|group direct product]...
Let $\tuple {x, x'} \in G \times G'$ and $\tuple {y, y'} \in H \times H'$. Then: {{begin-eqn}} {{eqn | l = \tuple {x, x'} \tuple {y, y'} \tuple {x, x'}^{-1} | r = \tuple {x, x'} \tuple {y, y'} \tuple {x^{-1}, x'^{-1} } | c = }} {{eqn | r = \tuple {x y x^{-1}, x' y' x'^{-1} } | c = }} {{eqn | o = \i...
Direct Product of Normal Subgroups is Normal
https://proofwiki.org/wiki/Direct_Product_of_Normal_Subgroups_is_Normal
https://proofwiki.org/wiki/Direct_Product_of_Normal_Subgroups_is_Normal
[ "Group Direct Products", "Normal Subgroups" ]
[ "Definition:Group", "Definition:Normal Subgroup", "Definition:Group Direct Product" ]
[]
proofwiki-15092
Normalizer of Rotation in Dihedral Group
Let $n \in \N$ be a natural number such that $n \ge 3$. Let $D_n$ be the dihedral group of order $2 n$, given by: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Let $\map {N_{D_n} } {\set \alpha}$ denote the normalizer of the singleton containing the rotation element $\alpha$. ...
By definition, the normalizer of $\set \alpha$ is: :$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \set \alpha g^{-1} = \set \alpha}$ That is: :$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \alpha = \alpha g}$ First let $g = \alpha^k$ for some $k \in \Z$. Then: :$\alpha \alpha^k = \alpha^k \alpha$ which in...
Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n \ge 3$. Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$, given by: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Let $\map {N_{D_n} ...
By definition, the [[Definition:Normalizer|normalizer]] of $\set \alpha$ is: :$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \set \alpha g^{-1} = \set \alpha}$ That is: :$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \alpha = \alpha g}$ First let $g = \alpha^k$ for some $k \in \Z$. Then: :$\alpha \alph...
Normalizer of Rotation in Dihedral Group
https://proofwiki.org/wiki/Normalizer_of_Rotation_in_Dihedral_Group
https://proofwiki.org/wiki/Normalizer_of_Rotation_in_Dihedral_Group
[ "Dihedral Groups", "Normalizers" ]
[ "Definition:Natural Numbers", "Definition:Dihedral Group", "Definition:Order of Structure", "Definition:Normalizer", "Definition:Singleton", "Definition:Rotation (Geometry)/Plane", "Definition:Generated Subgroup" ]
[ "Definition:Normalizer", "Cancellation Laws", "Definition:Generator of Subgroup", "Category:Dihedral Groups", "Category:Normalizers" ]
proofwiki-15093
Normalizer of Reflection in Dihedral Group
Let $n \in \N$ be a natural number such that $n \ge 3$. Let $D_n$ be the dihedral group of order $2 n$, given by: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Let $\map {N_{D_n} } {\set \beta}$ denote the normalizer of the singleton containing the reflection element $\beta$. ...
By definition, the normalizer of $\set \beta$ is: :$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \set \beta g^{-1} = \set \beta}$ That is: :$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \beta = \beta g}$ First let $g = \beta^k$ for $k \in \set {0, 1}$. Then: :$\beta \beta^k = \beta^k \beta$ Thus: :$\forall ...
Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n \ge 3$. Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$, given by: :$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$ Let $\map {N_{D_n} ...
By definition, the [[Definition:Normalizer|normalizer]] of $\set \beta$ is: :$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \set \beta g^{-1} = \set \beta}$ That is: :$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \beta = \beta g}$ First let $g = \beta^k$ for $k \in \set {0, 1}$. Then: :$\beta \beta^k = ...
Normalizer of Reflection in Dihedral Group
https://proofwiki.org/wiki/Normalizer_of_Reflection_in_Dihedral_Group
https://proofwiki.org/wiki/Normalizer_of_Reflection_in_Dihedral_Group
[ "Dihedral Groups", "Normalizers" ]
[ "Definition:Natural Numbers", "Definition:Dihedral Group", "Definition:Order of Structure", "Definition:Normalizer", "Definition:Singleton", "Definition:Reflection (Geometry)/Plane" ]
[ "Definition:Normalizer", "Cancellation Laws", "Product of Generating Elements of Dihedral Group", "Definition:Odd Integer", "Definition:Even Integer", "Category:Dihedral Groups", "Category:Normalizers" ]
proofwiki-15094
Normalizer of Subgroup of Symmetric Group that Fixes n
Let $S_n$ denote the symmetric group on $n$ letters. Let $H$ denote the subgroup of $S_n$ which consists of all $\pi \in S_n$ such that: :$\map \pi n = n$ The normalizer of $H$ is given by: :$\map {N_{S_n} } H = \map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$
We have from Subgroup of Symmetric Group that Fixes n that $N = S_{n - 1}$. By definition of normalizer: :$\map {N_{S_n} } {S_{n - 1} } := \set {\rho \in S_n: \rho S_{n - 1} \rho^{-1} = S_{n - 1} }$ We have from Group is Normal in Itself that: :$\forall \rho \in S_{n - 1}: \rho S_{n - 1} \rho^{-1} \in S_{n - 1}$ and so...
Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]]. Let $H$ denote the [[Definition:Subgroup|subgroup]] of $S_n$ which consists of all $\pi \in S_n$ such that: :$\map \pi n = n$ The [[Definition:Normalizer|normalizer]] of $H$ is given by: :$\map {N_{S_n} } H = \map {N_{S...
We have from [[Subgroup of Symmetric Group that Fixes n]] that $N = S_{n - 1}$. By definition of [[Definition:Normalizer|normalizer]]: :$\map {N_{S_n} } {S_{n - 1} } := \set {\rho \in S_n: \rho S_{n - 1} \rho^{-1} = S_{n - 1} }$ We have from [[Group is Normal in Itself]] that: :$\forall \rho \in S_{n - 1}: \rho S_...
Normalizer of Subgroup of Symmetric Group that Fixes n
https://proofwiki.org/wiki/Normalizer_of_Subgroup_of_Symmetric_Group_that_Fixes_n
https://proofwiki.org/wiki/Normalizer_of_Subgroup_of_Symmetric_Group_that_Fixes_n
[ "Symmetric Groups", "Normalizers" ]
[ "Definition:Symmetric Group/n Letters", "Definition:Subgroup", "Definition:Normalizer" ]
[ "Subgroup of Symmetric Group that Fixes n", "Definition:Normalizer", "Group is Normal in Itself", "Definition:Set Difference", "Definition:Permutation on n Letters", "Definition:Bijection", "Definition:Fixed Element under Permutation", "Intersection with Complement is Empty iff Subset", "Definition:...
proofwiki-15095
Center of Quaternion Group
Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the quaternion group. Let $\map Z {\Dic 2}$ denote the center of $\Dic 2$. Then: :$\map Z {\Dic 2} = \set {e, a^2}$
By definition, the center of $\Dic 2$ is: :$\map Z {\Dic 2} = \set {g \in \Dic 2: g x = x g, \forall x \in \Dic 2}$ We are given that: :$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ We have that $\Dic 2$ is generated by $\alpha$ and $\beta$. Thus: :$x \in \map Z {\Dic 2} \iff x a = a x \land x b = b x$ Let $x \...
Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the [[Definition:Quaternion Group|quaternion group]]. Let $\map Z {\Dic 2}$ denote the [[Definition:Center of Group|center of $\Dic 2$]]. Then: :$\map Z {\Dic 2} = \set {e, a^2}$
By definition, the [[Definition:Center of Group|center]] of $\Dic 2$ is: :$\map Z {\Dic 2} = \set {g \in \Dic 2: g x = x g, \forall x \in \Dic 2}$ We are given that: :$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ We have that $\Dic 2$ is [[Definition:Generator of Group|generated]] by $\alpha$ and $\beta$. ...
Center of Quaternion Group
https://proofwiki.org/wiki/Center_of_Quaternion_Group
https://proofwiki.org/wiki/Center_of_Quaternion_Group
[ "Quaternion Group", "Centers of Groups" ]
[ "Definition:Dicyclic Group/Quaternion Group", "Definition:Center (Abstract Algebra)/Group" ]
[ "Definition:Center (Abstract Algebra)/Group", "Definition:Generator of Group", "Definition:Order of Group Element", "Product of Generating Elements of Quaternion Group" ]
proofwiki-15096
Product of Generating Elements of Quaternion Group
Let $Q = \Dic 2$ be the quaternion group: :$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ Then for all $k \in \Z_{\ge 0}$: :$b a^k = a^{-k} b$
The proof proceeds by induction. For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition: :$b a^k = a^{-k} b$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = b a^0 | r = b e | c = }} {{eqn | r = e b | c = }} {{eqn | r = a^{-0} b | c = }} {{end-eqn}} Thus $\map P 0$ is seen to hold.
Let $Q = \Dic 2$ be the [[Definition:Quaternion Group|quaternion group]]: :$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ Then for all $k \in \Z_{\ge 0}$: :$b a^k = a^{-k} b$
The proof proceeds by [[Principle of Mathematical Induction|induction]]. For all $k \in \Z_{\ge 0}$, let $\map P k$ be the [[Definition:Proposition|proposition]]: :$b a^k = a^{-k} b$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = b a^0 | r = b e | c = }} {{eqn | r = e b | c = }} {{eqn | r = a...
Product of Generating Elements of Quaternion Group
https://proofwiki.org/wiki/Product_of_Generating_Elements_of_Quaternion_Group
https://proofwiki.org/wiki/Product_of_Generating_Elements_of_Quaternion_Group
[ "Quaternion Group" ]
[ "Definition:Dicyclic Group/Quaternion Group" ]
[ "Principle of Mathematical Induction", "Definition:Proposition", "Principle of Mathematical Induction" ]
proofwiki-15097
Conjugacy Classes of Quaternion Group
Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the quaternion group. The conjugacy classes of $\Dic 2$ are: :$\set e, \set {a^2}, \set {a, a^3}, \set {b, a^2 b}, \set {a b, a^3 b}$
From Center of Quaternion Group, we have: :$\map Z {\Dic 2} = \set {e, a^2}$ Thus from Conjugacy Class of Element of Center is Singleton, $\set e$ and $\set {a^2}$ are two of those conjugacy classes. By inspection of the Cayley table: {{:Quaternion Group/Cayley Table}} we investigate the remaining $6$ elements in turn,...
Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the [[Definition:Quaternion Group|quaternion group]]. The [[Definition:Conjugacy Class|conjugacy classes]] of $\Dic 2$ are: :$\set e, \set {a^2}, \set {a, a^3}, \set {b, a^2 b}, \set {a b, a^3 b}$
From [[Center of Quaternion Group]], we have: :$\map Z {\Dic 2} = \set {e, a^2}$ Thus from [[Conjugacy Class of Element of Center is Singleton]], $\set e$ and $\set {a^2}$ are two of those [[Definition:Conjugacy Class|conjugacy classes]]. By inspection of the [[Quaternion Group/Cayley Table|Cayley table]]: {{:Quater...
Conjugacy Classes of Quaternion Group
https://proofwiki.org/wiki/Conjugacy_Classes_of_Quaternion_Group
https://proofwiki.org/wiki/Conjugacy_Classes_of_Quaternion_Group
[ "Quaternion Group", "Examples of Conjugacy Classes" ]
[ "Definition:Dicyclic Group/Quaternion Group", "Definition:Conjugacy Class" ]
[ "Center of Quaternion Group", "Conjugacy Class of Element of Center is Singleton", "Definition:Conjugacy Class", "Quaternion Group/Cayley Table", "Definition:Conjugacy Class", "Definition:Conjugacy Class", "Definition:Conjugacy Class" ]
proofwiki-15098
Conjugacy Action on Subsets is Group Action
Let $\powerset G$ be the set of all subgroups of $G$. For any $S \in \powerset G$ and for any $g \in G$, the conjugacy action: :$g * S := g \circ S \circ g^{-1}$ is a group action.
By {{Group-axiom|2}}, $e \in G$, thus: {{begin-eqn}} {{eqn | l = e * S | r = e \circ S \circ e^{-1} | c = Definition of $*$ }} {{eqn | r = S | c = {{Defof|Identity Element}} }} {{end-eqn}} Thus {{GroupActionAxiom|2}} is seen to be fulfilled. Then: {{begin-eqn}} {{eqn | l = \paren {g_1 \circ g_2} * S ...
Let $\powerset G$ be the set of all [[Definition:Subgroup|subgroups]] of $G$. For any $S \in \powerset G$ and for any $g \in G$, the [[Definition:Conjugacy Action on Subsets|conjugacy action]]: :$g * S := g \circ S \circ g^{-1}$ is a [[Definition:Group Action|group action]].
By {{Group-axiom|2}}, $e \in G$, thus: {{begin-eqn}} {{eqn | l = e * S | r = e \circ S \circ e^{-1} | c = Definition of $*$ }} {{eqn | r = S | c = {{Defof|Identity Element}} }} {{end-eqn}} Thus {{GroupActionAxiom|2}} is seen to be fulfilled. Then: {{begin-eqn}} {{eqn | l = \paren {g_1 \circ g_2}...
Conjugacy Action on Subsets is Group Action
https://proofwiki.org/wiki/Conjugacy_Action_on_Subsets_is_Group_Action
https://proofwiki.org/wiki/Conjugacy_Action_on_Subsets_is_Group_Action
[ "Conjugacy Action" ]
[ "Definition:Subgroup", "Definition:Conjugacy Action/Subsets", "Definition:Group Action" ]
[ "Inverse of Group Product" ]
proofwiki-15099
Normed Vector Space Requires Multiplicative Norm on Division Ring
Let $R$ be a normed division ring with a submultiplicative norm $\norm {\, \cdot \,}_R$. Let $V$ be a vector space that is not a trivial vector space. Let $\norm {\, \cdot \,}: V \to \R_{\ge 0}$ be a mapping from $V$ to the positive real numbers satisfying the vector space norm axioms. Then $\norm {\, \cdot \,}_R$ is a...
Since $V$ is not a trivial vector space: :$\exists \mathbf v \in V: \mathbf v \ne 0$ By {{NormAxiomVector|1}}: :$\norm {\mathbf v} > 0$ Let $r, s \in R$: {{begin-eqn}} {{eqn | l = \norm {r s}_R \norm {\mathbf v} | r = \norm {\paren {r s} \mathbf v} | c = {{NormAxiomVector|2}} }} {{eqn | r = \norm {r \paren ...
Let $R$ be a [[Definition:Normed Division Ring|normed division ring]] with a [[Definition:Submultiplicative Norm on Ring|submultiplicative norm]] $\norm {\, \cdot \,}_R$. Let $V$ be a [[Definition:Vector Space|vector space]] that is not a [[Definition:Trivial Vector Space|trivial vector space]]. Let $\norm {\, \cdot ...
Since $V$ is not a [[Definition:Trivial Vector Space|trivial vector space]]: :$\exists \mathbf v \in V: \mathbf v \ne 0$ By {{NormAxiomVector|1}}: :$\norm {\mathbf v} > 0$ Let $r, s \in R$: {{begin-eqn}} {{eqn | l = \norm {r s}_R \norm {\mathbf v} | r = \norm {\paren {r s} \mathbf v} | c = {{NormAxiomV...
Normed Vector Space Requires Multiplicative Norm on Division Ring
https://proofwiki.org/wiki/Normed_Vector_Space_Requires_Multiplicative_Norm_on_Division_Ring
https://proofwiki.org/wiki/Normed_Vector_Space_Requires_Multiplicative_Norm_on_Division_Ring
[ "Normed Division Rings", "Norm Theory" ]
[ "Definition:Normed Division Ring", "Definition:Norm/Ring/Submultiplicative", "Definition:Vector Space", "Definition:Trivial Vector Space", "Definition:Positive/Real Number", "Axiom:Vector Space Norm Axioms", "Definition:Norm/Ring/Multiplicative" ]
[ "Definition:Trivial Vector Space", "Category:Normed Division Rings", "Category:Norm Theory" ]