nyuuzyou/wikis huggingface dataset:
Surjection iff Right Cancellable

== Theorem ==
Let $f$ be a mapping.
Then $f$ is a surjection $f$ is right cancellable.
== Necessary Condition ==
=== Proof 1 ===
=== Proof 2 ===
== Sufficient Condition ==
=== Proof 1 ===
=== Proof 2 ===
== Also see ==
* Injection iff Left Cancellable
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.10 \ \text{(a)}$
* : Chapter $\text{I}$: Sets and Functions: Problem $\text{BB}$: Categorical Matters
* : Appendix: Elementary set and number theory
* : Appendix $\text{A}.5$: Identity, One-one, and Onto Functions: Proposition $\text{A}.5.1: 2 \ \text{(e)}$
Category:Surjections
Category:Cancellability
Category:Surjection iff Right Cancellable
Surjection iff Right Cancellable/Necessary Condition/Proof 1

== Theorem ==
== Proof ==
Let $f: X \to Y$ be surjective.
Let $h_1: Y \to Z, h_2: Y \to Z: h_1 \circ f = h_2 \circ f$.
As $f$ is a surjection:
:$\Img f = Y$
by definition.
But in order for $h_1 \circ f$ to be defined, it is necessary that $Y = \Dom {h_1}$.
Similarly, for $h_2 \circ f$ to be defined, it is necessary that $Y = \Dom {h_2}$.
So it follows that the domains of $h_1$ and $h_2$ are the same.
Also:
:The codomain of $h_1$ equals the codomain of $h_1 \circ f$
:The codomain of $h_2$ equals the codomain of $h_2 \circ f$
again by definition of composition of mappings.
Now, we have shown that the domains and codomains of $h_1$ and $h_2$ are the same.
All we need to do now to prove that $h_1 = h_2$, and therefore that $f$ is right cancellable, is to show that:
:$\forall y \in Y: h_1 \paren y = h_2 \paren y$.
So, let $y \in Y$.
As $f$ is surjective:
: $\exists x \in X: y = f \paren x$
Thus:
Thus $h_1 \paren y = h_2 \paren y$ and thus $f$ is right cancellable.
== Sources ==
* : $\S 25$: Some further results and examples on mappings: Worked Example $1$
Category:Surjection iff Right Cancellable
Surjection iff Right Cancellable/Necessary Condition/Proof 2

== Theorem ==
== Proof ==
Let $f: X \to Y$ be surjective.
Then from Surjection iff Right Inverse:
:$\exists g: Y \to X: f \circ g = I_Y$
Suppose $h \circ f = k \circ f$ for two mappings $h: Y \to Z$ and $k: Y \to Z$.
Then:
Thus $f$ is right cancellable.
So surjectivity implies right cancellability.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.6$
Category:Surjection iff Right Cancellable
Surjection iff Right Cancellable/Sufficient Condition/Proof 1

== Theorem ==
== Proof ==
Suppose $f$ is a mapping which is not surjective.
Then:
:$\exists y_1 \in Y: \neg \exists x \in X: \map f x = y_1$
Let $Z = \set {a, b}$.
Let $h_1$ and $h_2$ be defined as follows. 
:$\map {h_1} y = a: y \in Y$
:$\map {h_2} y = \begin {cases}
a & : y \ne y_1 \\
b & : y = y_1
\end {cases}$
Thus we have $h_1 \ne h_2$ such that $h_1 \circ f = h_2 \circ f$.
Therefore $f$ is not right cancellable.
It follows from the Rule of Transposition that if $f$ is right cancellable, then $f$ must be surjective.
== Sources ==
* : Chapter $4$: Mappings: Exercise $16$
Category:Surjection iff Right Cancellable
Surjection iff Right Cancellable/Sufficient Condition/Proof 2

== Theorem ==
== Proof ==
Let $f: X \to Y$ be a right cancellable mapping.
Let $Y$ contain exactly one element.
Then by definition $Y$ is a singleton.
From Mapping to Singleton is Surjection it follows that $f$ is a surjection.
So let $Y$ contain at least two elements.
Call those two elements $a$ and $b$, and we note that $a \ne b$.
We define the two mappings $h, k$ as follows:
:$h: Y \to Y: \forall x \in Y: \map h x = \begin{cases}
x & : x \in \Img f \\
a & : x \notin \Img f
\end{cases}$
:$k: Y \to Y: \forall x \in Y: \map k x = \begin{cases}
x & : x \in \Img f \\
b & : x \notin \Img f
\end{cases}$
It is clear that:
:$\forall y \in X: \map h {\map f y} = \map f y = \map k {\map f y}$
and so:
:$h \circ f = k \circ f$
But by hypothesis, $f$ is right cancellable.
Thus $h = k$.
 $Y \ne \Img f$.
Then:
:$\Img f \subsetneq Y$
That is:
:$\exists x \in Y: x \notin \Img f$
It follows that:
:$a = \map h x = \map k x = b$
But we posited that $a \ne b$.
From this contradiction we conclude that:
:$Y = \Img f$
So, by definition, $f$ must be a surjection.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.6$
Category:Surjection iff Right Cancellable
Surjection iff Right Inverse

== Theorem ==
A mapping $f: S \to T, S \ne \O$ is a surjection :
:$\exists g: T \to S: f \circ g = I_T$
where:
:$g$ is a mapping
:$I_T$ is the identity mapping on $T$.
That is, $f$ has a right inverse.
=== Non-Uniqueness ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Injection iff Left Inverse
== Sources ==
* : Chapter $3$. Mappings: Exercise $6$
* : $\S 1.3$: Exercise $5$
* : Appendix: Elementary set and number theory
* : $0$: Some Conventions and some Basic Facts
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $5$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.5 \ \text{(a)}$
Category:Surjections
Category:Right Inverse Mappings
Category:Surjection iff Right Inverse
Surjection iff Right Inverse/Non-Uniqueness

== Theorem ==
Let $S$ and $T$ be sets such that $S \ne \O$.
Let $f: S \to T$ be a surjection.
A right inverse of $f$ is in general not unique.
Uniqueness occurs $f$ is a bijection.
== Proof ==
If $f$ is not an injection then:
:$\exists y \in T: \exists x_1, x_2 \in S: \map f {x_1} = y = \map f {x_2}$
Hence we have more than one choice in $\map {f^{-1} } {\set y}$ for how to map $\map g y$.
That is, $\map g y$ is not unique.
This does not happen $f$ is an injection.
Hence the result.
== Examples ==
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $6$
Category:Surjection iff Right Inverse
Surjection iff Right Inverse/Non-Uniqueness/Examples/Arbitrary Example

== Example of Surjection iff Right Inverse: Non-Uniqueness ==
Let $S = \set {0, 1}$.
Let $T = \set a$.
Let $f: S \to T$ be defined as:
:$\forall x \in S: \map f x = a$
Then $f$ has $2$ distinct right inverses.
== Proof ==
Let $g_0: T \to S$ and $g_1: T \to S$ be the mappings defined as:
We have that:
:$\Dom {g_0} = \Cdm f = \Dom {g_1}$
and that $f$ is a surjection.
Hence we can construct:
and:
Hence both $g_0$ and $g_1$ are distinct right inverses of $f$.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{G}$
Category:Surjection iff Right Inverse
Surjection iff Right Inverse/Proof 1

== Theorem ==
== Proof ==
Assume $\exists g: T \to S: f \circ g = I_T$.
From Identity Mapping is Surjection, $I_T$ is surjective, so $f \circ g$ is surjective.
So from Surjection if Composite is Surjection, $f$ is a surjection.
Note that the existence of such a $g$ requires that $S$ is non-empty.
Now, assume $f$ is a surjection.
Consider the indexed family of non-empty sets $\set {\map {f^{-1} } y}_{y \mathop \in T}$ where $\map {f^{-1} } y$ denotes the preimage of $y$ under $f$.
Using the axiom of choice, there exists a mapping $g: T \to S$ such that $\map g y \in \set {\map {f^{-1} } y}$ for all $y \in T$.
:
That is, $\map {\paren {f \circ g} } y = y$, as desired.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions: Theorem $8$
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.2$
Category:Surjection iff Right Inverse
Surjection iff Right Inverse/Proof 2

== Theorem ==
== Proof ==
Take the result Condition for Composite Mapping on Right:
Let $A, B, C$ be sets.
Let $f: B \to A$ and $g: C \to A$ be mappings.
Then:
:$\Img g \subseteq \Img f$
:
:$\exists h: C \to B$ such that $h$ is a mapping and $f \circ h = g$.
Let $C = A = T$, let $B = S$ and let $g = I_T$.
Then the above translates into:
:$\Img {I_T} \subseteq \Img f$
:
:$\exists g: T \to S$ such that $g$ is a mapping and $f \circ g = I_T$.
But we know that:
:$\Img f \subseteq T = \Img {I_T}$
So by definition of set equality, the result follows.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.6$
Category:Surjection iff Right Inverse
Surjection on Total Ordering reflects Preordering

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $f: S \to T$ be a mapping to an arbitrary set $T$.
Let $\RR$ be a relation on $T$ defined such that:
:$\RR: = \set {\tuple {\map f x, \map f y}: x \preccurlyeq y}$
That is, $a$ is related to $b$ in $T$ they have preimages $x$ and $y$ under $f$ such that $x$ precedes $y$.
Then:
:$f$ is a surjection onto set $T$
:
:$\RR$ is a preordering on $T$
== Proof ==
=== Sufficient Condition ===
Let $f: S \to T$ be a surjection.
By definition, a preordering is a relation which is both transitive and reflexive.
From Mapping on Total Ordering reflects Transitivity, $\RR$ is a transitive relation.
It remains to be shown that $\RR$ is a reflexive relation.
Let $a \in T$.
Because $f$ is a surjection, it follows that:
:$\exists x \in S: a = \map f x$
Because $\preccurlyeq$ is a total ordering on $S$, it follows that:
:$x \preccurlyeq x$
Hence by definition of $\RR$:
:$\map f x \mathrel \RR \map f x$
That is:
:$a \mathrel \RR a$
As $a$ was arbitrary, it follows that $\RR$ is reflexive.
Hence it follows that $\RR$ is a preordering.
=== Necessary Condition ===
Let $\RR$ be a preordering on $T$.
Then $\RR$ is a relation which is both transitive and reflexive.
Let $a \in T$.
As $\RR$ is reflexive, it follows that:
:$\forall a \in T: a \mathrel \RR a$
That is:
:$\forall a \in T: \exists x \in S: \map f x \mathrel \RR \map f x$
where $a = \map f x$
That means:
:$\forall a \in T: \exists x \in S: a = \map f x$
That is:
:$f$ is a surjection.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $41 \ \text {(b)}$
Category:Total Orderings
Category:Preorderings
Surjection that Preserves Inner Product is Linear

== Theorem ==
Let $H, K$ be Hilbert spaces, and denote by ${\innerprod \cdot \cdot}_H$ and ${\innerprod \cdot \cdot}_K$ their respective inner products.
Let $U: H \to K$ be a surjection such that:
:$\forall g, h \in H: {\innerprod g h}_H = {\innerprod {Ug} {Uh} }_K$
Then $U$ is a linear map, and hence an isomorphism.
== Proof ==
Let $x, y \in H$.
Let $\alpha \in \GF$.
By surjectivity of $U$, choose $z \in H$ such that $Uz = \map U {\alpha x + y} - \paren { \alpha Ux + Uy }$.
Then:
By positivity, $Uz = {\bf 0}_K$.
Hence:
:$\map U {\alpha x + y} = \alpha U x + U y$
Thus, $U$ is linear.
== Sources ==
* : $\text I.5$: Exercise $9$
Category:Hilbert Spaces
Category:Linear Transformations on Hilbert Spaces
Surjective Embedding is Homeomorphism

== Theorem ==
Let $T_1 = \struct{S_1, \tau_1}$ and $T_2 = \struct{S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a surjective embedding.
Then $f$ is a homeomorphism. 
== Proof ==
Let $f {\restriction_{S_1 \times f\sqbrk {S_1} }}$ be the restriction of $f$ to its image.
By definition of surjection, $f \sqbrk {S_1} = S_2$.
Therefore, $f {\restriction_{S_1 \times f\sqbrk {S_1} }} = f$
By definition of embedding, $f {\restriction_{S_1 \times f\sqbrk {S_1} }}$ of $f$ to its image is a homeomorphism.
Therefore, $f$ is a homeomorphism.
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces. Topological Embeddings
Category:Embeddings (Topology)
Category:Homeomorphisms (Topological Spaces)
Surjective Field Homomorphism is Field Isomorphism

== Theorem ==
Let $E$ and $F$ be fields.
Let $\phi: E \to F$ be a (field) homomorphism.
Let $\phi$ be a surjection.
Then $\phi$ is an isomorphism.
== Proof ==
As asserted, let $\phi$ be a surjection.
From Field Homomorphism is either Trivial or Injection, $\phi$ is either an injection or the trivial homomorphism.
If $\phi$ is an injection, then, by definition, $\phi$ is a bijection.
Hence, again by definition, $\phi$ is an isomorphism.
If $\phi$ is not an injection, then $\phi$ is the trivial homomorphism.
But from Field Contains at least 2 Elements, $\Img \phi$ cannot in that case be a field.
Hence if $\phi$ is not an injection, then $\phi$ cannot be a surjection.
Hence the result.
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \eta$
Category:Field Homomorphisms
Category:Field Isomorphisms
Surjective Monotone Function is Continuous

== Theorem ==
Let $X$ be an open set of $\R$.
Let $Y$ be a real interval.
Let $f: X \to Y$ be a surjective monotone real function.
Then $f$ is continuous on $X$.
== Proof ==
, let $f$ be increasing.
Let $c \in X$.
From Limit of Monotone Real Function: Corollary, the one sided limits of monotone functions exist:
and satisfy:
:$L^-_c, L^+_c \in Y$
:$L^-_c \le \map f c \le L^+_c$
Suppose that $\ds L = \lim_{x \mathop \to c} \map f x$ exists.
From Limit iff Limits from Left and Right:
:$L = L^-_c$
This leads to:
:$L \le \map f c$
Similarly:
:$L = L^+_c$
which leads to:
:$L \ge \map f c$
 
Hence:
:$\ds \lim_{x \mathop \to c} \map f x = \map f c$
proving continuity at $c$.
By assumption, $f$ is increasing.
Suppose $\ds \lim_{x \mathop \to c} \map f x$ does not exist.
Then from Discontinuity of Monotonic Function is Jump Discontinuity, there is a jump discontinuity at $c$.
 $f$ has a jump discontinuity at $c$.
From Real Numbers are Densely Ordered:
:$L^-_c < y < L^+_c$
for some $y \in Y$.
By surjectivity, $y = \map f a$ for some $a \in X$.
Hence:
:$L^-_c < \map f a < L^+_c$
If $a < c$ then $\map f a \le L^-_c$.
This contradicts the previous inequality.
There is a similar contradiction if $a \ge c$.
Category:Monotone Real Functions
Category:Real Analysis
Category:Continuity
Surjective Restriction of Real Exponential Function

== Theorem ==
Let $\exp: \R \to \R$ be the exponential function:
:$\map \exp x = e^x$
Then the restriction of the codomain of $\exp$ to the strictly positive real numbers:
:$\exp: \R \to \R_{>0}$
is a surjective restriction.
Hence:
:$\exp: \R \to \R_{>0}$
is a bijection.
== Proof ==
We have Exponential on Real Numbers is Injection.
Let $y \in \R_{> 0}$.
Then $\exists x \in \R: x = \map \ln y$
That is:
:$\exp x = y$
and so $\exp: \R \to \R_{>0}$ is a surjection.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: Topological Spaces: $\S 9$. Functions: Example $2$
* : $\S 3.3$. Injective, surjective, bijective; inverse mappings: Example $48$
Category:Exponential Function
Category:Surjections
Survival Analysis/Examples/Effect of Censored Observations

== Example of Survival Analysis ==
However, sometimes censored observations contain some useful information.
Consider a person who withdraws from a medical study after $100$ days after treatment, but before showing the response of interest.
Clearly that person has a longer survival time than somebody who survives only $90$ days.
However, it is unknown as to whether such a person might or might not have a survival time of somebody who survives, for example, for $120$ days.
== Sources ==
* 
Category:Examples of Survival Analysis
Survival Function is Decreasing

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $F_f$ be the survival function of $f$.
Then $F_f$ is a decreasing function.
== Proof ==
Let $\alpha, \beta \in \hointr 0 \infty$ with $\alpha \le \beta$.
We show: 
:$\map {F_f} \beta \le \map {F_f} \alpha$
We first show: 
:$\set {x \in X : \size {\map f x} \ge \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha}$
Let $x \in X$ have $\size {\map f x} \ge \beta$.
Then, since $\beta \ge \alpha$ we have $\size {\map f x} \ge \alpha$.
So by the definition of set inclusion, we have: 
:$\set {x \in X : \size {\map f x} \ge \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha}$
So, by Measure is Monotone we have: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge \beta} } \le \map \mu {\set {x \in X : \size {\map f x} \ge \alpha} }$
So by the definition of the survival function we have: 
:$\map {F_f} \beta \le \map {F_f} \alpha$
So $F_f$ is a decreasing function.
== Sources ==
* : $1.1.1$: The Distribution Function
Category:Survival Functions
Survival Function is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable function.
Then the survival function $F_f$ is well-defined.
== Proof ==
From Absolute Value of Measurable Function is Measurable:
:$\size f$ is $\Sigma$-measurable.
Then from Characterization of Measurable Functions, we have: 
:$\set {x \in X : \size {\map f x} \ge \alpha} \in \Sigma$ for all $\alpha \in \hointr 0 \infty$.
So:
:$\map \mu {\set {x \in X : \size {\map f x} \ge \alpha} }$
is well-understood, and the survival function $F_f$ is well-defined.
Category:Survival Functions
Category:Examples of Well-Defined Mappings‎
Survival Function of Pointwise Scalar Multiple

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $F_{c f}$ be the survival function of the pointwise scalar multiple $c f$. 
Let $F_f$ be the survival function of $f$. 
Let $c \in \R \setminus \set 0$.
Then: 
:$\ds \map {F_{c f} } \alpha = \map {F_f} {\frac \alpha {\size c} }$ for all $\alpha \in \hointr 0 \infty$.
== Proof ==
Let $\alpha \in \hointr 0 \infty$.
Then, we have: 
== Sources ==
* : $1.1.1$: The Distribution Function
Category:Survival Functions
Survival Function preserves Inequality

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be $\Sigma$-measurable functions such that: 
:$\size f \le \size g$ $\mu$-almost everywhere.
Let $F_f$ and $F_g$ be the survival functions of $f$ and $g$ respectively.
Then: 
:$F_f \le F_g$ 
== Proof ==
We aim to show that: 
:$\map {F_f} \alpha \le \map {F_g} \alpha$ for all $\alpha \in \hointr 0 \infty$.
That is: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge \alpha} } \le \map \mu {\set {x \in X : \size {\map g x} \ge \alpha} }$ for all $\alpha \in \hointr 0 \infty$.
Since $f \le g$ $\mu$-almost everywhere, there exists a $\mu$-null set such that: 
:if $x \in X$ has $\size {\map f x} > \size {\map g x}$ then $x \in N$.
From Measurable Functions Determine Measurable Sets, in particular we have: 
:$\set {x \in X : \size {\map f x} > \size {\map g x} }$ is $\mu$-null
from Null Sets Closed under Subset.
Let $\alpha \in \hointr 0 \infty$.
We have: 
Swapping $f$ for $g$ in this computation also gives: 
:$\ds \map \mu {\set {x \in X : \size {\map g x} \ge \alpha} } = \map \mu {\set {x \in X \setminus N : \size {\map g x} \ge \alpha} }$
So we aim to show that: 
:$\ds \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} } \le \map \mu {\set {x \in X \setminus N : \size {\map g x} \ge \alpha} }$
From Measure is Monotone, it suffices to show:
:$\set {x \in X \setminus N : \size {\map f x} \ge \alpha} \subseteq \set {x \in X \setminus N : \size {\map g x} \ge \alpha}$
Let $x \in X \setminus N$ have: 
:$\size {\map f x} \ge \alpha$
Then, we have: 
:$\alpha \le \size {\map f x} \le \size {\map g x}$
So we have: 
:$\size {\map g x} \ge \alpha$
So, from the definition of set inclusion, we have: 
:$\set {x \in X \setminus N : \size {\map f x} \ge \alpha} \subseteq \set {x \in X \setminus N : \size {\map g x} \ge \alpha}$ for all $\alpha \in \hointr 0 \infty$
and hence the demand.
== Sources ==
* : $1.1.1$: The Distribution Function
Category:Survival Functions
Swelled Class contains Empty Set

== Theorem ==
Let $A$ be a swelled class.
Then the empty set is an element of $A$.
== Proof ==
By definition of swelled class, every subclass of every element of $A$ is also an element of $A$.
Let $x \in A$.
Then by Empty Class is Subclass of All Classes, the empty class is an element of $A$.
By the , the empty class is a set.
Hence the result.
Category:Empty Set
Category:Swelled Classes
Swelled Set which is Closed under Chain Unions with Choice Function is Type M

== Theorem ==
Let $S$ be a set of sets which:
:is closed under chain unions
:has a choice function $C$ for its union $\ds \bigcup S$.
Then:
:$S$ is swelled
:
:$S$ is of type $M$.
== Proof ==
=== Sufficient Condition ===
Let $S$ be swelled.
Let $b \in S$ be arbitrary.
From Closure under Chain Unions with Choice Function implies Elements with no Immediate Extension:
:$b$ is the subset of an element of $S$ which has no immediate extension in $S$.
Let $x \in S$ have no immediate extension in $S$.
Then from Element of Swelled Set with no Immediate Extension is Maximal, $x$ is a maximal element under the subset relation on $S$.
That is:
:$b$ is the subset of a maximal element of $S$ under the subset relation.
As $b$ is arbitrary:
:every element of $S$ is a subset of a maximal element of $S$ under the subset relation.
Hence, by definition, $S$ is of type $M$.
== Necessary Condition ==
Let $S$ be of type $M$.
== Historical Note ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles: Corollary $6.3$
Category:Closure under Chain Unions
Category:Choice Functions
Category:Swelled Classes
Category:Type M Sets
Swelled Set which is Closed under Chain Unions with Choice Function is Type M/Historical Note

== Historical Note on Swelled Set which is Closed under Chain Unions with Choice Function is Type $M$ ==
 and , in their of $2010$, demonstrate only the sufficient condition: that a swelled set with the given criteria is of type $M$.
In raising the question as to whether such a set of sets which is not swelled would also be of type $M$, they say:
:We posed this question to Professor , who informed us that it would not.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles
Category:Historical Notes
Sylow Subgroup is Hall Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a Sylow $p$-subgroup of $G$.
Then $H$ is a Hall subgroup of $G$.
== Proof ==
Let $p$ be prime.
Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$.
By definition, a Sylow $p$-subgroup $H$ of $G$ is a subgroup of $G$ of order $p^n$.
By Lagrange's Theorem, the index of $H$ in $G$ is given by:
:$\index G H = \dfrac {\order G} {\order H}$
So in this case:
:$\index G H = \dfrac {k p^n} {p^n} = k$
As $p \nmid k$ it follows from Prime not Divisor implies Coprime that $k \perp p$.
The result follows from the definition of Hall subgroup.
Category:Sylow p-Subgroups
Category:Hall Subgroups
Sylow Theorems

== Theorem ==
=== First Sylow Theorem ===
=== Second Sylow Theorem ===
=== Third Sylow Theorem ===
=== Fourth Sylow Theorem ===
=== Fifth Sylow Theorem ===
== Examples ==
== Historical Note ==
Category:P-Groups
Category:Sylow p-Subgroups
Category:Subgroups
Category:Sylow Theorems
Sylow Theorems/Examples

== Examples of Use of Sylow Theorems ==
=== Sylow $2$-Subgroups in Group of Order $12$ ===
=== Sylow $3$-Subgroups in Group of Order $12$ ===
=== Sylow $7$-Subgroups in Group of Order $28$ ===
=== Sylow $2$-Subgroups in Group of Order $32$ ===
=== Sylow $2$-Subgroups in Group of Order $48$ ===
Category:Sylow Theorems
Sylow Theorems/Examples/Sylow 2-Subgroups in Group of Order 12

== Example of Use of Sylow Theorems ==
In a group of order $12$, there are either $1$ or $3$ Sylow $2$-subgroups.
== Proof ==
Let $G$ be a group of order $12$.
Let $n_2$ be the number of Sylow $2$-subgroups in $G$.
From the Fourth Sylow Theorem, $n_2$ is congruent to $1$ modulo $2$, that is, odd.
Let $H$ be a Sylow $2$-subgroup of $G$.
We have that:
:$12 = 3 \times 2^2$
and so the order of $H$ is $4$.
Thus:
From the Fifth Sylow Theorem:
:$n_2 \divides 3$
where $\divides$ denotes divisibility.
Thus there may be $1$ or $3$ Sylow $2$-subgroups of $G$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Exercise $1 \ \text{(d)}$
Category:Sylow Theorems
Category:Groups of Order 12
Sylow Theorems/Examples/Sylow 2-Subgroups in Group of Order 32

== Example of Use of Sylow Theorems ==
The Sylow Theorems tell us nothing about the Sylow $2$-subgroups in a group of order $32$.
== Proof ==
We have that $32 = 2^5$.
Hence the only Sylow $2$-subgroup in $G$ is $G$ itself.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Exercise $1 \ \text{(c)}$
Category:Sylow Theorems
Category:Groups of Order 32