hcju/proofwikips huggingface dataset:
Well-Founded Recursion
Let $A$ be a class. Let $\mathcal R$ be a relation that is foundational on $A$. Let every $\mathcal R$-initial segment of any element $x$ of $A$ be a small class. Let $K$ be a class of mappings $f$ that satisfy: :$(1): \quad$ The domain of $f$ is some subset $y \subseteq A$ such that $y$ is transitive with respect to $\mathcal R$. :$(2): \quad \forall x \in y: f \left({x}\right) = G \left({F \restriction \mathcal R^{-1} x}\right)$ where $F \restriction R^{-1} x$ denotes the restriction of $F$ to the $\mathcal R$-initial segment of $x$. {{explain|What is $G$? And should $F$ not be introduced before it is used, instead of after?}} Let $F = \bigcup K$, the union of $K$. Then: : $F$ is a mapping with domain $A$ : $\forall x \in A: F \left({x}\right) = G \left({F \restriction \mathcal R^{-1} x}\right)$ : $F$ is unique, in the sense that if another mapping $A$ has the above two properties, then $A = F$.
Von Neumann Hierarchy is Supertransitive
Let $V$ denote the Von Neumann Hierarchy. Let $x$ be an ordinal. Then $\map V x$ is supertransitive.
Von Neumann Hierarchy Comparison
Let $x$ and $y$ be ordinals such that $x < y$. Then: :$\map V x \in \map V y$ :$\map V x \subset \map V y$ {{explain|$\map V x$ etc.}}
Every Set in Von Neumann Universe
Let $S$ be a small class. Then $S$ is well-founded. {{explain|The connection between "well-founded" and "in von Neumann universe".}}
Ordinal Equal to Rank
Let $x$ be an ordinal. Let $S$ be a small class. Let $V \left({ x }\right)$ denote the von Neumann hierarchy on the ordinal $x$. Then $x$ equals the rank of $S$ iff $S \in V \left({x+1}\right) \land S \notin V \left({x}\right)$.
Tau of Prime Number
Let $p \in \Z_{> 0}$. Then $p$ is a prime number : :$\map \tau p = 2$ where $\map \tau p$ denotes the tau function of $p$.
Join is Commutative
Let $\left({S, \vee, \preceq}\right)$ be a join semilattice. Then $\vee$ is commutative.
Meet is Commutative
Let $\left({S, \wedge, \preceq}\right)$ be a meet semilattice. Then $\wedge$ is commutative.
Join Succeeds Operands
Let $\left({S, \preceq}\right)$ be an ordered set. Let $a, b \in S$ admit a join $a \vee b \in S$. Then: :$a \preceq a \vee b$ :$b \preceq a \vee b$ i.e., $a \vee b$ succeeds its operands $a$ and $b$.
Meet Precedes Operands
Let $\left({S, \preceq}\right)$ be an ordered set. Let $a, b \in S$ admit a meet $a \wedge b \in S$. Then: :$a \wedge b \preceq a$ :$a \wedge b \preceq b$ i.e., $a \wedge b$ precedes its operands $a$ and $b$.
Meet is Associative
Let $\left({S, \wedge, \preceq}\right)$ be a meet semilattice. Then $\wedge$ is associative.
Poset Elements Equal iff Equal Weak Upper Closure
Let $\left({S, \preccurlyeq}\right)$ be an ordered set. Let $s, t \in S$. Then $s = t$ : :$s^\succcurlyeq = t^\succcurlyeq$ where $s^\succcurlyeq$ denotes weak upper closure of $s$. That is, , for all $r \in S$: :$s \preccurlyeq r \iff t \preccurlyeq r$
Join is Associative
Let $\struct {S, \vee, \preceq}$ be a join semilattice. Then $\vee$ is associative.
Prime iff Equal to Product
Let $p \in \Z$ be an integer such that $p \ne 0$ and $p \ne \pm 1$. Then $p$ is prime : :$\forall a, b \in \Z: p = ab \implies p = \pm a \lor p = \pm b$
Multiple of Divisor Divides Multiple
Let $a, b, c \in \Z$. Let: :$a \divides b$ where $\divides$ denotes divisibility. Then: :$a c \divides b c$
Divisor Relation is Transitive
The divisibility relation is a transitive relation on $\Z$, the set of integers. That is: :$\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$
Polynomial Forms over Field form Integral Domain/Formulation 1
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $X$ be transcendental in $F$. Let $F \sqbrk X$ be the ring of polynomial forms in $X$ over $F$. Then $F \sqbrk X$ is an integral domain.
Sum of Indices of Real Number/Positive Integers
Let $n, m \in \Z_{\ge 0}$ be positive integers. Let $r^n$ be defined as $r$ to the power of $n$. Then: : $r^{n + m} = r^n \times r^m$
Product of Indices of Real Number/Positive Integers
Let $n, m \in \Z_{\ge 0}$ be positive integers. Let $r^n$ be defined as $r$ to the power of $n$. Then: :$\paren {r^n}^m = r^{n m}$
Smallest Element is Initial Object
Let $\mathbf P$ be an order category. Suppose the objects $\mathbf P_0$ of $\mathbf P$, considered as an ordered set, have a smallest element $p$. Then $p$ is an initial object of $\mathbf P$.
Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy
Let $x$ be an ordinal. Let $S$ be a small class. Let $V \left({ x }\right)$ denote the von Neumann hierarchy on the ordinal $x$. Then $x$ is a subset of the rank of $S$  $S \notin V \left({x}\right)$.
Membership Rank Inequality
Let $S$ and $T$ be sets. Let $\operatorname{rank} \left({ S }\right)$ denote the rank of $S$. Then: :$S \in T \implies \operatorname{rank} \left({S}\right) < \operatorname{rank} \left({ T }\right)$
Greatest Element is Terminal Object
Let $\mathbf P$ be an order category. Let $p$ be the greatest element of the objects $\mathbf P_0$ of $\mathbf P$, considered as a ordered set. Then $p$ is a terminal object of $\mathbf P$.
Rank of Set Determined by Members
Let $S$ be a set. Let $\operatorname{rank} \left({ S }\right)$ denote the rank of $S$. Then: :$\operatorname{rank} \left({ S }\right) = \bigcap \left\{ x \in \operatorname{On} : \forall y \in S: \operatorname{rank} \left({ y }\right) < x \right\}$
Bounded Rank implies Small Class
Let $S$ be a class. Suppose the rank, denoted $\operatorname{rank} \left({x}\right)$, of each $x \in S$ is bounded above by some ordinal $y$. {{MissingLinks|rank}} Then $S$ is a small class.
Count of Subsets with Even Cardinality
Let $S$ be a set whose cardinality is $n$. Then the number of subsets of $S$ whose cardinality is even is $2^{n-1}$.
Count of Subsets with Odd Cardinality
Let $S$ be a set whose cardinality is $n$. Then the number of subsets of $S$ whose cardinality is odd is $2^{n-1}$.
Well-Founded Relation is Strongly Well-Founded
Let $A$ be a class. Let $\prec$ be a foundational relation on $A$. Let $B$ be a nonempty class such that $B \subseteq A$. Then $B$ has a $\prec$-minimal element.
Category of Ordered Sets has Enough Constants
Let $\mathbf{OrdSet}$ be the category of ordered sets. Then $\mathbf{OrdSet}$ has enough constants.
Cardinal Number is Ordinal
Let $S$ be a set such that $S \sim x$ for some ordinal $x$. Let $\left|{S}\right|$ denote the cardinal number of $S$. Then: :$\left|{S}\right| \in \operatorname{On}$ where $\operatorname{On}$ denotes the class of ordinals.
Condition for Set Equivalent to Cardinal Number
Let $S$ be a set. Let $\card S$ denote the cardinal number of $S$. That is, let $\card S$ be the intersection of all ordinals equivalent to $S$. {{refactor|This following statement needs to go somewhere else as it's a non-sequitur in this context.}} Note that in the absence of the Axiom of Choice, $\card S$ may be the class of all sets. Then the following are equivalent: : $(1): \quad S \sim \card S$ : $(2): \quad \exists x \in \On: S \sim x$ : $(3): \quad \exists x \in \On: \exists y: \paren {y \subseteq x \land S \sim y}$ where $A \sim B$ means that there is a bijection from $A$ onto $B$, and the quantification over $y$ is unbounded (so $y$ may be any set).
Cardinal Number Equivalence or Equal to Universe
Let $S$ be a set. Let $\left\vert{S}\right\vert$ denote the cardinal number of $S$. Let $\mathbb U$ denote the universal class. Then: :$S \sim \left\vert{S}\right\vert \lor \left\vert{S}\right\vert = \mathbb U$
Ordinal Number Equivalent to Cardinal Number
Let $x$ be an ordinal. Let $\card x$ denote the cardinal number of $x$. Then: :$x \sim \card x$ where $\sim$ denotes set equivalence.
Cardinal Number Less than Ordinal
Let $S$ be a set. Let $\left|{S}\right|$ denote the cardinal number of $S$. Let $x$ be an ordinal such that $S \sim x$. Then: :$\left|{S}\right| \le x$
Equivalent Sets have Equal Cardinal Numbers
Let $S$ and $T$ be sets. Let $\card S$ denote the cardinal number of $S$. Then: :$S \sim T \implies \card S = \card T$
Condition for Set Union Equivalent to Associated Cardinal Number
Let $S$ and $T$ be sets. Let $\left|{S}\right|$ denote the cardinal number of $S$. Let $\sim$ denote set equivalence. Then: :$S \cup T \sim \left|{S \cup T}\right| \iff S \sim \left|{S}\right| \land T \sim \left|{T}\right|$
Cardinal of Cardinal Equal to Cardinal
Let $S$ be a set such that $S$ is equivalent to its cardinal. If the axiom of choice holds, then this condition holds for any set. Then: :$\left|{ \left({ \left|{S}\right| }\right) }\right| = \left|{S}\right|$ where $\left|{S}\right|$ denotes the cardinal number of $S$.
Cardinal of Finite Ordinal
Let $n$ be a finite ordinal. Let $\left|{n}\right|$ denote the cardinal number of $n$. Then: :$\left|{n}\right| = n$
Finite Ordinal is equal to Natural Number
Let $n$ be an element of the minimal infinite successor set. Let $x$ be an ordinal. Then: :$n \sim x \implies n = x$
Product (Category Theory) is Unique
Let $\mathbf C$ be a metacategory. Let $A$ and $B$ be objects of $\mathbf C$. Let $A \times B$ and $A \times' B$ both be products of $A$ and $B$. Then there is a unique isomorphism $u: A \times B \to A \times' B$. That is, products are unique up to unique isomorphism. {{expand|cover extended theorem (for general def)}}
Subset implies Cardinal Inequality
Let $S$ and $T$ be sets such that $S \subseteq T$. Furthermore, let: :$T \sim \card T$ where $\card T$ denotes the cardinality of $T$. Then: :$\card S \le \card T$
Subset of Ordinal implies Cardinal Inequality
Let $S$ be a set. Let $x$ be an ordinal such that $S \subseteq x$. Then: :$\left|{S}\right| \le \left|{x}\right|$ where $\left|{S}\right|$ denotes the cardinality of $S$.
Set Less than Cardinal Product
Let $S$ and $T$ be sets. Let $T$ be nonempty. Suppose that $S \times T \sim \left|{ S \times T }\right|$. Then: :$\left|{ S }\right| \le \left|{ S \times T }\right|$
Injection implies Cardinal Inequality
Let $S$ and $T$ be sets. Let $f: S \to T$ be an injection. Let $\card T$ denote the cardinal number of $T$. Let: :$T \sim \card T$ where $\sim$ denotes set equivalence Then: :$\card S \le \card T$
Set is Equivalent to Image under Injection
Let $S$ and $T$ be sets. Let $f: S \to T$ be an injection. Then the image of $S$ under $f$ is equivalent to $S$.
Cardinality of Image of Mapping not greater than Cardinality of Domain
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $\card S$ denote the cardinal number of $S$. Let $S \sim \card S$. Then: :$\card {\Img f} \le \card S$
Injection iff Cardinal Inequality
Let $\card T$ denote the cardinal number of $T$. Let $S$ and $T$ be sets such that $S \sim \card S$ and $T \sim \card T$. Then: :$\card S \le \card T $  there exists an injection $f: S \to T$.
Surjection iff Cardinal Inequality
Let $S$ and $T$ be sets such that $S \sim \card S$ and $T \sim \card T$. Furthermore, let $S$ be non-empty. Then: :$0 < \card T \le \card S$  there exists a surjection $f: S \to T$.
Right Identity while exists Right Inverse for All is Identity
Let $\struct  {S, \circ}$ be a semigroup with a right identity $e_R$ such that: :$\forall x \in S: \exists x_R: x \circ x_R = e_R$ That is, every element of $S$ has a right inverse with respect to the right identity. Then $e_R$ is also a left identity, that is, is an identity.
Left Identity while exists Left Inverse for All is Identity
Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that: :$\forall x \in S: \exists x_L: x_L \circ x = e_L$ That is, every element of $S$ has a left inverse with respect to the left identity. Then $e_L$ is also a right identity, that is, is an identity.
Right Inverse for All is Left Inverse
Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$ such that: :$\forall x \in S: \exists x_R: x \circ x_R = e_R$ That is, every element of $S$ has a right inverse with respect to the right identity. Then $x_R \circ x = e_R$, that is, $x_R$ is also a left inverse with respect to the right identity.
Exponential on Complex Plane is Group Homomorphism
Let $\struct {\C, +}$ be the additive group of complex numbers. Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers. Let $\exp: \struct {\C, +} \to \struct {\C_{\ne 0}, \times}$ be the mapping: :$x \mapsto \map \exp x$ where $\exp$ is the complex exponential function. Then $\exp$ is a group homomorphism.
Group Direct Product is Product in Category of Groups
Let $\mathbf{Grp}$ be the category of groups. Let $G$ and $H$ be groups, and let $G \times H$ be their direct product. Then $G \times H$ is a binary product of $G$ and $H$ in $\mathbf{Grp}$.
Product Category is Product in Category of Categories
Let $\mathbf{Cat}$ be the category of categories. Let $\mathbf C$ and $\mathbf D$ be small categories, and let $\mathbf C \times \mathbf D$ be their product category. Then $\mathbf C \times \mathbf D$ is a binary product of $\mathbf C$ and $\mathbf D$ in $\mathbf{Cat}$.
Infimum is Product in Order Category
Let $\mathbf P$ be an order category whose ordering is $\preceq$. Let $p,q \in \mathbf P_0$, and suppose that they have some infimum $r = \inf \left\{{p, q}\right\}$. Then $r$ is a binary product of $p$ and $q$ in $\mathbf P$.
Cardinal of Union Less than Cardinal of Cartesian Product
Let $S$ and $T$ be sets that are equivalent to their cardinal numbers. Let $\card S$ denote the cardinal number of $S$. Let $\card S > 1$ and $\card T > 1$. Then: :$\card {S \cup T} \le \card {S \times T}$
Cardinal Product Equinumerous to Ordinal Product
Let $S$ and $T$ be sets that are equivalent to their cardinal numbers. Let $\left|{ S }\right|$ denote the cardinal number of $S$. Let $\cdot$ denote ordinal multiplication and let $\times$ denote the Cartesian product. Then: :$S \times T \sim \left|{ S }\right| \cdot \left|{ T }\right|$
Product of Finite Sets is Finite
Let $S$ and $T$ be finite sets. Then $S \times T$ is a finite set.
Union of Finite Sets is Finite
Let $S$ and $T$ be finite sets. Then $S \cup T$ is a finite set.
Ordinal is Finite iff Natural Number
Let $x$ be an ordinal. Then $x$ is a finite set  $x$ is an element of the minimal infinite successor set.
Cardinal Inequality implies Ordinal Inequality
Let $T$ be a set. Let $\left|{T}\right|$ denote the cardinal number of $T$. Let $x$ be an ordinal. Then: :$x < \left|{T}\right| \iff \left|{x}\right| < \left|{T}\right|$
Cardinal Number Plus One Less than Cardinal Product
Let $x$ be an ordinal such that $x > 1$. Then: :$\left|{ x + 1 }\right| \le \left|{ x \times x }\right|$ Where $\times$ denotes the Cartesian product.
Non-Finite Cardinal is equal to Cardinal Product
Let $\omega$ denote the minimal infinite successor set. Let $x$ be an ordinal such that $x \ge \omega$. Then: :$\left|{x}\right| = \left|{ x\times x}\right|$ where $\times$ denotes the Cartesian product.
Cartesian Product Preserves Cardinality
Let $R$, $S$, and $T$ be sets. Suppose that $S$ is equivalent to $T$. Then: :$R \times S \sim R \times T$ :$S \times R \sim T \times R$
Product of Composite Morphisms
Let $\mathbf C$ be a metacategory. Let $f \times f': A \times A' \to B \times B'$ and $g \times g': B \times B' \to C \times C'$ be two composable products of morphisms in $\mathbf C$. Then: :$\left({g \circ f}\right) \times \left({g' \circ f'}\right) = \left({g \times g'}\right) \circ \left({f \times f'}\right)$ where $\times$ signifies product of morphisms.
Cardinal Product Equal to Maximum
Let $S$ and $T$ be sets that are equinumerous to their cardinal number. Let $\left|{S}\right|$ denote the cardinal number of $S$. Suppose $S$ is infinite. Suppose $T > 0$. Then: :$\left|{S \times T}\right| = \max\left({\left|{S}\right|, \left|{T}\right| }\right)$
Cardinal of Union Equal to Maximum
Let $S$ and $T$ be sets that are equinumerous to their cardinal number. Let $\left|{ S }\right|$ denote the cardinal number of $S$. Suppose $S$ is infinite. Then: :$\left|{ S \cup T }\right| = \max\left({ \left|{ S }\right| , \left|{ T }\right| }\right)$
Cardinal Class is Subset of Ordinal Class
Let $\mathcal N$ denote the class of all cardinal numbers. Let $\operatorname{On}$ denote the class of all ordinals. Then: :$\mathcal N \subseteq \operatorname{On}$
Cardinal of Cardinal Equal to Cardinal/Corollary
Let $\mathcal N$ denote the class of all cardinal numbers. Let $x$ be an ordinal. Then: :$x \in \mathcal N \iff x = \left|{ x }\right|$
Cardinal Equal to Collection of All Dominated Ordinals
Let $S$ be a set. Let $\preccurlyeq$ denote the dominance relation. Let $\On$ denote the class of all ordinals. Let $x = \set {y \in \On: y \preccurlyeq S}$ Then: :$(1): \quad x$ is an element of the class of cardinals. :$(2): \quad$ There is no injection $f$ such that $f : x \to S$
Cardinal Class is Proper Class
The class $\mathcal N$ of all cardinal numbers is a proper class.
Infinite Cardinal Class is Proper Class
The infinite cardinal class $\mathcal N’$ is a proper class.
Ordinal in Aleph iff Cardinal in Aleph
Let $x$ and $y$ be ordinals. Then: :$x \in \aleph_y \iff \left|{ x }\right| \in \aleph_y$ where $\aleph$ denotes the aleph mapping.
Aleph Product is Aleph
Let $x$ be an ordinal. Then: :$\left|{\aleph_x \times \aleph_x}\right| = \aleph_x$ where $\aleph$ denotes the aleph mapping.
Aleph is Infinite Cardinal
Let $x$ be an ordinal. Then $\aleph_x$ is an infinite cardinal where $\aleph$ denotes the aleph mapping.
Surjection from Aleph to Ordinal
Let $x$ and $y$ be ordinals. Suppose that: :$0 < y < \aleph_{x+1}$ Then there is a surjection: :$f : \aleph_x \to y$
Ordinal Less than Successor Aleph
Let $x$ and $y$ be ordinals. Then: :$y < \aleph_{x+1} \iff y < \aleph_x \lor y \sim \aleph_x$
Aleph is Infinite
Let $x$ be an ordinal. :$\aleph_x \ge \omega$ where: :$\aleph$ denotes the aleph mapping :$\omega$ denotes the minimal infinite successor set.
Aleph-Null
Let $\omega$ denote the minimal infinite successor set. :$\omega = \aleph_0$ where $\aleph$ denotes the aleph mapping.
Minimal Infinite Successor Set is Infinite Cardinal
$\omega$, the minimal infinite successor set, is an element of the infinite cardinal class $\mathcal N’$.
Set of All Mappings is Small Class
Let $S$ and $T$ be small classes. It follows that the set of all mappings $S^T$ is a small class.
Covariant Hom Functor is Functor
Let $\mathbf{Set}$ be the category of sets. Let $\mathbf C$ be a locally small category. Let $C \in \mathbf C_0$ be an object of $\mathbf C$. Let $\hom_\mathbf C \paren {C, \cdot}: \mathbf C \to \mathbf{Set}$ be the covariant representable functor based at $C$. Then $\hom_\mathbf C \paren {C, \cdot}$ is a functor.
Inverse Morphism is Unique
Let $\mathbf C$ be a metacategory. Let $f: C \to D$ be an isomorphism of $\mathbf C$. Then $f$ admits a unique inverse morphism $g: D \to C$.
Cofinal Ordinal Relation is Reflexive
Let $x$ be an ordinal. Then $x$ is cofinal to itself. That is: :$\operatorname{cof} \left({x, x}\right)$
Cofinal Ordinal Relation is Transitive
Let $x$, $y$, and $z$ be ordinals. Let $\operatorname{cof}$ denote the cofinal relation. Then: :$\operatorname{cof} \left({ x,y }\right) \land \operatorname{cof} \left({ y,z }\right) \implies \operatorname{cof} \left({ x,z }\right)$
Cofinal to Zero iff Ordinal is Zero
Let $x$ be an ordinal. Let $\operatorname{cof}$ denote the cofinal relation. Let $0$ denote the zero ordinal. Then the following are equivalent: :$\operatorname{cof} \left({ x,0 }\right)$ :$\operatorname{cof} \left({ 0,x }\right)$ :$x = 0$
Condition for Cofinal Nonlimit Ordinals
Let $x$ and $y$ be nonlimit ordinals. Let $\operatorname{cof}$ denote the cofinal relation. Let $\le$ denote the subset relation. {{explain|This statement could be worded a little more carefully. It is, from examining the link, clear that the subset relation and the ordering relation are the same thing, but it grates to see $\le$ being defined as the subset relation. I'm not saying it ''is'' wrong, because it's not wrong, but it ''looks'' uncomfortable.}} Furthermore, let $x$ and $y$ satisfy the condition: :$0 < x \le y$ Then: :$\map {\operatorname{cof}} { y,x }$
Cofinal Limit Ordinals
Let $x$ and $y$ be ordinals. Let $\operatorname{cof}$ denote the cofinal relation. Let $K_{II}$ denote the class of all limit ordinals. Then: :$\operatorname{cof} \left({ x,y }\right) \implies \left({ x \in K_{II} \iff y \in K_{II} }\right)$
Matrix is Invertible iff Determinant has Multiplicative Inverse
Let $R$ be a commutative ring with unity. Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$. Then $\mathbf A$ is invertible  its determinant is invertible in $R$. If $R$ is one of the standard number fields $\Q$, $\R$ or $\C$, this translates into: :$\mathbf A$ is invertible  its determinant is non-zero.
Prime Groups of Same Order are Isomorphic
Two prime groups of the same order are isomorphic to each other.
Infinite Cyclic Group is Unique up to Isomorphism
All infinite cyclic groups are isomorphic. That is, up to isomorphism, there is only one infinite cyclic group.
Intersection of Subgroups is Subgroup/General Result
Let $\mathbb S$ be a set of subgroups of $\struct {G, \circ}$, where $\mathbb S \ne \O$. Then the intersection $\displaystyle \bigcap \mathbb S$ of the elements of $\mathbb S$ is itself a subgroup of $G$. Also, $\displaystyle \bigcap \mathbb S$ is the largest subgroup of $\struct {G, \circ}$ contained in each element of $\mathbb S$.
Union of Subgroups/Corollary 2
Let $H \vee K$ be the join of $H$ and $K$. Then $H \vee K = H \cup K$  $H \subseteq K$ or $K \subseteq H$.
Existence of Unique Subgroup Generated by Subset/Singleton Generator
Let $a \in G$. Then $H = \gen a = \set {a^n: n \in \Z}$ is the unique smallest subgroup of $G$ such that $a \in H$. That is: : $K \le G: a \in K \implies H \subseteq K$
Join of Subgroups is Group Generated by Union
Let $G$ be a group. Let $H$ and $K$ be subgroups of $G$. Let $S$ be the set of words of $H \cup K$. Then $S$ is a subgroup of $K$ such that: :$S = \gen {H \cup K} = H \vee K$ where $H \vee K$ denotes the join of $H$ and $K$.
Order of Subgroup Product/Corollary
:$\size {H \vee K} \ge \dfrac {\order H \order K} {\order {H \cap K} }$ or :$\dfrac {\size {H \vee K} } {\order H} \ge \dfrac {\order K} {\order {H \cap K} }$ where $H \vee K$ denotes join and $\order H$ denotes the order of $H$.
Homomorphic Image of Group Element is Coset
Let $\phi: \left({G, \circ}\right) \to \left({H, *}\right)$ be a group homomorphism. Let $\ker \left({\phi}\right)$ be the kernel of $\phi$. Let $h \in H$. Then $\operatorname{Im}^{-1} \left({h}\right)$ is either the empty set or a coset of $\ker \left({\phi}\right)$.