nyuuzyou/wikis huggingface dataset:
Sylow Theorems/Examples/Sylow 2-Subgroups in Group of Order 48

== Example of Use of Sylow Theorems ==
In a group of order $48$, there are either $1$ or $3$ Sylow $2$-subgroups.
== Proof ==
Let $G$ be a group of order $48$.
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:
:$48 = 3 \times 2^4$
and so the order of $H$ is $16$.
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 $1 \ \text{(b)}$
Category:Sylow Theorems
Category:Groups of Order 48
Sylow Theorems/Examples/Sylow 3-Subgroups in Group of Order 12

== Example of Use of Sylow Theorems ==
In a group of order $12$, there are either $1$ or $4$ Sylow $3$-subgroups.
== Proof ==
Let $G$ be a group of order $12$.
Let $n_3$ be the number of Sylow $3$-subgroups in $G$.
From the Fourth Sylow Theorem, $n_3$ is congruent to $1$ modulo $3$, that is, in $\set {1, 4, 7, \ldots}$
Let $H$ be a Sylow $3$-subgroup of $G$.
We have that:
:$12 = 4 \times 3$
and so the order of $H$ is $3$.
Thus:
From the Fifth Sylow Theorem:
:$n_3 \divides 4$
where $\divides$ denotes divisibility.
Thus there may be $1$ or $4$ Sylow $3$-subgroups of $G$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Exercise $1 \ \text{(e)}$
Category:Sylow Theorems
Category:Groups of Order 12
Sylow Theorems/Examples/Sylow 7-Subgroups in Group of Order 28

== Example of Use of Sylow Theorems ==
In a group of order $28$, there exists exactly $1$ Sylow $7$-subgroup.
This Sylow $7$-subgroup is normal in $G$.
== Proof ==
Let $G$ be a group of order $28$.
From the Fourth Sylow Theorem, the number of Sylow $7$-subgroups is congruent to $1$ modulo $7$.
From the Fifth Sylow Theorem, the number of Sylow $7$-subgroups is a divisor of $\dfrac {28} 7 = 4$.
Thus there is exactly $1$ Sylow $7$-subgroup of $G$.
From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $7$-subgroup of $G$ is normal in $G$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Exercise $1 \ \text{(a)}$
Category:Sylow Theorems
Category:Groups of Order 28
Sylow Theorems/Historical Note

== Historical Note on Sylow Theorems ==
When cracking open the structure of a group, it is a useful plan to start with investigating the prime subgroups.
The Sylow Theorems are a set of results which provide us with just the sort of information we need.
 was a Norwegian mathematician who established some important facts on this subject.
He published what are now referred to as the Sylow Theorems in $1872$.
The name is pronounced something like Soolof.
There is no standard numbering for the Sylow Theorems.
Different authors use different labellings.
Therefore, the nomenclature as defined on is to a greater or lesser extent arbitrary.
=== First Sylow Theorem ===
== Sources ==
* 
* : $\S 44$. Some consequences of Lagrange's Theorem
* : Chapter $11$: The Sylow Theorems: Summary for Chapter $11$
Category:Historical Notes
Sylow Theorems/Historical Note/First Sylow Theorem

== Historical Note on First Sylow Theorem ==
's original work in $1872$ demonstrated the existence of what is now known as a Sylow $p$-subgroup.
The corollary, that there exists a subgroup of order $p^n$ for all $p^n \divides \order G$, was deduced later, but is frequently itself referred to as the First Sylow Theorem.
The proof using the Orbit-Stabilizer Theorem is based on one published by in $1959$.
== Sources ==
* : $\S 6.5$. Orbits: Example $121$
* : Chapter $2$: The Sylow Theorems: $\S 56$. First Sylow Theorem
* : Chapter $11$: The Sylow Theorems: Summary for Chapter $11$
Category:Historical Notes
Sylow p-Subgroup is Unique iff Normal

== Theorem ==
A group $G$ has exactly one Sylow $p$-subgroup $P$ $P$ is normal.
== Proof ==
If $G$ has precisely one Sylow $p$-subgroup, it must be normal from Unique Subgroup of a Given Order is Normal.
Suppose a Sylow $p$-subgroup $P$ is normal.
Then it equals its conjugates.
Thus, by the Third Sylow Theorem, there can be only one such Sylow $p$-subgroup.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Example $11.12$: Remark
Category:Normal Subgroups
Category:Sylow p-Subgroups
Sylow p-Subgroups of Group of Order 2p

== Theorem ==
Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Then $G$ has exactly one Sylow $p$-subgroup.
This Sylow $p$-subgroup is normal.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Sylow p-Subgroups
Category:Groups of Order 2 p
Category:Sylow p-Subgroups of Group of Order 2p
Sylow p-Subgroups of Group of Order 2p/Proof 1

== Theorem ==
== Proof ==
Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
:$n_p \divides 2 p$
that is:
:$n_p \in \set {1, 2, p, 2 p}$
But $p$ and $2 p$ are congruent to $0$ modulo $p$
So:
:$n_p \notin \set {p, 2 p}$
Also we have that $p > 2$.
Hence:
:$2 \not \equiv 1 \pmod p$
and so it must be that $n_p = 1$.
It follows from Sylow $p$-Subgroup is Unique iff Normal that this Sylow $p$-subgroup is normal.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(1)$ Groups of order $2p$
Category:Sylow p-Subgroups of Group of Order 2p
Sylow p-Subgroups of Group of Order 2p/Proof 2

== Theorem ==
== Proof ==
Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.
From the First Sylow Theorem, there exists at least $1$ Sylow $p$-subgroup of $G$.
Let $P$ be such a Sylow $p$-subgroup of $G$.
The index of $P$ is $2$.
From Subgroup of Index 2 is Normal, $P$ is normal in $G$.
From Sylow $p$-Subgroup is Unique iff Normal it follows that there is only $1$ Sylow $p$-subgroup of $G$.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(1)$ Groups of order $2p$
Category:Sylow p-Subgroups of Group of Order 2p
Sylvester's Law of Inertia

== Theorem ==
Let $\struct {V, q}$ be the scalar product space.
Let $V^*$ be the vector space dual to $V$.
Suppose there exists a basis $\tuple {\beta^i}$ for $V^*$ such that $q$ is expressible as:
:$q = \paren {\beta^1}^2 + \ldots + \paren {\beta^r}^2 - \paren {\beta^{r + 1}}^2 - \ldots - \paren {\beta^{r + s}}^2$
where:
:$r, s \in \N : r + s = n$
Then $r$ is the maximum dimension among all subspaces on which the restriction of $q$ is positive definite.
Furthermore, $r$ and $s$ are independent of the choice of basis.
== Also presented as ==
== Proof ==
== Historical Note ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Sylvester's Law of Inertia
Category:Scalar Product Spaces
Symbol Count by Finite State Machine

== Theorem ==
Let $\Sigma$ be a finite set.
Let $\Pi \subseteq \Sigma$.
For a finite sequence $w = \sequence {a_1, a_2, \dotsc, a_n}$ of elements of $\Sigma$:
:Let $\map c w = \size {\set {i : a_i \in \Pi} }$
Let $k \in \N$.
Then, for each of the following, there exists a finite state machine whose accepted languages is all $w \in \Sigma^*$ such that the condition holds:
* $\map c w < k$
* $\map c w \le k$
* $\map c w > k$
* $\map c w \ge k$
* $\map c w = k$
* $\map c w \ne k$
== Proof ==
=== $\map c w < k$ ===
Define $F = \tuple {S, A, I, \Sigma, T}$ where:
:$S = \set {s_0, s_1, \dotsc, s_k}$
:$A = \set {s_0, s_1, \dotsc, s_{k - 1} }$
:$I = s_0$
:$\map T {s_i, \sigma} = \begin{cases}
s_{i + 1} & : \sigma \in \Pi \land i < k \\
s_i & : \text {otherwise}
\end{cases}$
The machine being in the state $s_i$ indicates that $i$ symbols in $\Pi$ have been read so far, up to a maximum of $k$.
So long as the word ends before the $k$-th symbol in $\Pi$ is read, the machine accepts.
If a $k$-th symbol is read, it is no longer possible for the machine to transition out of $s_k$, and the machine rejects.
=== $\map c w \le k$ ===
This is equivalent to $\map c w < k + 1$, which has already been shown to exist.
=== $\map c w > k$ ===
This is equivalent to $\neg \paren {\map c w \le k}$, which exists by Negation of Finite State Machine in addition to the above.
=== $\map c w \ge k$ ===
This is equivalent to $\neg \paren {\map c w < k}$, which exists by Negation of Finite State Machine in addition to the above.
=== $\map c w = k$ ===
This is equivalent to $\map c w \le k \land \map c w \ge k$, which exists by Conjunction of Finite State Machines in addition to the above.
=== $\map c w \ne k$ ===
This is equivalent to $\map c w < k \lor \map c w > k$, which exists by Disjunction of Finite State Machines in addition to the above.
Category:Abstract Machines
Symbol Reduction of Turing Machine

== Theorem ==
Let $T$ be a Turing machine with input symbols $\Sigma$ and blank symbol $B$.
Then there exists a Turing machine $T'$ such that:
* The input symbols of $T'$ are $\Sigma$
* The tape symbols of $T'$ are $\Sigma \cup \set B$
* The language accepted by $T'$ is exactly the language accepted by $T$
* The inputs on which $T'$ halts are exactly the inputs on which $T$ halts
== Proof ==
== Sources ==
*
Symmetric Bilinear Form can be Diagonalized

== Theorem ==
Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$ of finite dimension $n>0$.
Let $f$ be a symmetric bilinear form on $V$.
Then there exists an ordered basis for which the relative matrix of $f$ is diagonal.
== Proof ==
Category:Bilinear Forms (Linear Algebra)
Symmetric Bilinear Form is Reflexive

== Theorem ==
Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$.
Let $b$ be a bilinear form on $V$.
Let $b$ be symmetric.
Then $b$ is reflexive.
== Proof ==
Let $\tuple {v, w} \in V \times V$ with $\map b {v, w} = 0$.
Because $b$ is symmetric, $\map b {w, v} = 0$.
Because $\tuple {v, w}$ was arbitrary, $b$ is reflexive.
== Also see ==
* Bilinear Form is Reflexive iff Symmetric or Alternating
Category:Bilinear Forms (Linear Algebra)
Symmetric Closure/Examples/Arbitrary Example 1

== Examples of Symmetric Closures ==
Let $S = \set {1, 2, 3, 4, 5}$ be a set.
Let $\RR$ be the relation on $S$ defined as:
:$\RR = \set {\tuple {1, 2}, \tuple {2, 3}, \tuple {3, 4}, \tuple {5, 4} }$
The symmetric closure $\RR^\leftrightarrow$ of $\RR$ is given by:
:$\RR^* = \set {\tuple {1, 2}, \tuple {2, 3}, \tuple {3, 4}, \tuple {5, 4}, \tuple {2, 1}, \tuple {3, 2}, \tuple {4, 3}, \tuple {4, 5} }$
== Proof ==
By definition of symmetric closure:
:$\RR^\leftrightarrow = \RR \cup \RR^{-1}$
where $\RR^{-1}$ denotes the inverse of $\RR$.
By definition of inverse relation:
:$\RR^{-1} = \set {\tuple {2, 1}, \tuple {3, 2}, \tuple {4, 3}, \tuple {4, 5} }$
The result then follows by definition of set union.
</onlyinclude>
== Sources ==
* : Chapter $1$: Preliminaries: Exercises: $1.7$
Category:Examples of Symmetric Closures
Symmetric Closure of Inverse Relation

== Definition ==
Let $\RR$ be a relation on a set $S$.
Let $\RR^{-1}$ be the inverse of $\RR$.
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$.
Then: 
:$\RR^\leftrightarrow = \paren {\RR^{-1} }^\leftrightarrow$
== Proof ==
Category:Symmetric Closures
Category:Inverse Relations
Symmetric Closure of Ordering may not be Transitive

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\preceq^\leftrightarrow$ be the symmetric closure of $\preceq$.
Then it is not necessarily the case that $\preceq^\leftrightarrow$ is transitive.
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b, c}$ where $a$, $b$, and $c$ are distinct.
Let:
:${\preceq} = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {b, c} }$:
Then $\preceq$ is an ordering, but $\preceq^\leftrightarrow$ is not transitive, as follows:
$\preceq$ is reflexive because it contains the diagonal relation on $S$.
That $\preceq$ is transitive and antisymmetric can be verified by inspecting all ordered pairs of its elements.
Thus $\preceq$ is an ordering.
Now consider $\preceq^\leftrightarrow$, the symmetric closure of $\preceq$:
:${\preceq^\leftrightarrow} = {\preceq} \cup {\preceq}^{-1} = \set{\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {c, a}, \tuple {b, c}, \tuple {c, b} }$
by inspection.
Now $\tuple {a, c} \in {\preceq^\leftrightarrow}$ and $\tuple {c, b} \in {\preceq^\leftrightarrow}$, but $\tuple {a, b} \notin {\preceq^\leftrightarrow}$.
Thus $\preceq^\leftrightarrow$ is not transitive.
Category:Orderings
Category:Symmetric Closures
Category:Transitive Relations
Symmetric Closure of Relation Compatible with Operation is Compatible

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\RR$ be a relation compatible with $\circ$.
Let $\RR^\leftrightarrow$ denote the symmetric closure of $\RR$.
Then $\RR^\leftrightarrow$ is compatible with $\circ$.
== Proof ==
By the definition of symmetric closure:
:$\RR^\leftrightarrow = \RR \cup \RR^{-1}$.
Here $\RR^{-1}$ is the inverse of $\RR$.
By Inverse of Relation Compatible with Operation is Compatible, $\RR^{-1}$ is compatible with $\circ$.
Thus by Union of Relations Compatible with Operation is Compatible:
:$\RR^\leftrightarrow = \RR \cup \RR^{-1}$ is compatible with $\circ$.
Category:Compatible Relations
Category:Symmetric Closures
Symmetric Densely-Defined Linear Operator is Closable

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a symmetric densely-defined linear operator.
Then $\tuple {\map D T, T}$ is closable.
== Proof ==
Let $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoint of $\tuple {\map D T, T}$.
From Adjoint of Densely-Defined Linear Operator is Closed, $\tuple {\map D {T^\ast}, T^\ast}$ is closed.
From Adjoint of Symmetric Densely-Defined Linear Operator Extends Operator, $\tuple {\map D {T^\ast}, T^\ast}$ extends $\tuple {\map D T, T}$.
Hence $\tuple {\map D T, T}$ is closable.
Category:Symmetric Densely-Defined Linear Operators
Category:Closable Densely-Defined Linear Operators
Symmetric Difference/Examples/Arbitrary Example 1

== Example of Symmetric Difference ==
Let:
:$A = \set {1, 2, 3}$
:$B = \set {2, 3, 4}$
Then their symmetric difference is given by:
:$A \symdif B = \set {1, 4}$
== Sources ==
* 
Category:Examples of Symmetric Difference
Symmetric Difference/Examples/Arbitrary Example 2

== Example of Symmetric Difference ==
Let:
:$A = \set {1, 2, 3, 4}$
:$B = \set {3, 4, 5, 6}$
Then their symmetric difference is given by:
:$A \symdif B = \set {1, 2, 5, 6}$
== Sources ==
* 
* 
Category:Examples of Symmetric Difference
Symmetric Difference is Associative

== Theorem ==
Symmetric difference is associative:
:$R \symdif \paren {S \symdif T} = \paren {R \symdif S} \symdif T$
== Proof 1 ==
== Proof 2 ==
There is another way of doing this.
== Comment ==
This illustrates that you can express the symmetric difference of three sets as the union of four intersections (which seems more intuitively obvious) as well as the intersection of four unions (which is not quite so obvious).
== Also see ==
* Intersection is Associative
* Union is Associative
* Set Difference is not Associative
== Sources ==
* : $\S 5$: Complements and Powers
* : Chapter $1$. Sets: Exercise $7 \ \text{(iii)}$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \alpha$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $14$
* : Chapter $1$: Sets and Logic: Exercise $9 \ \text{(ii)}$
* : $\S 2$: Problem $6 \ \text{(ii)}$
* : $\text {(iv)}$
Category:Symmetric Difference
Category:Examples of Associative Operations‎
Category:Symmetric Difference is Associative
Symmetric Difference is Associative/Proof 1

== Theorem ==
== Proof ==
We can directly expand the expressions for $R \symdif \paren {S \symdif T}$ and $\paren {R \symdif S} \symdif T$, and see that they come to the same thing.
Expanding the :
Expanding the :
From Union is Commutative it is seen that the and are the same, and the result is proved.
Category:Symmetric Difference is Associative
Symmetric Difference is Commutative

== Theorem ==
Symmetric difference is commutative:
:$S \symdif T = T \symdif S$
== Proof ==
== Also see ==
* Union is Commutative
* Intersection is Commutative
* Set Difference is Anticommutative
== Sources ==
* : $\S 5$: Complements and Powers
* : Chapter $1$. Sets: Exercise $7 \ \text{(i)}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $14$
* : $\text {(iii)}$
Category:Symmetric Difference
Category:Examples of Commutative Operations
Symmetric Difference is Subset of Union of Symmetric Differences

== Theorem ==
Let $R, S, T$ be sets.
Then:
:$R \symdif S \subseteq \paren {R \symdif T} \cup \paren {S \symdif T}$
where $R \symdif S$ denotes the symmetric difference between $R$ and $S$.
== Proof ==
From the definition of symmetric difference, we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Then from Set Difference is Subset of Union of Differences, we have:
:$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$
:$S \setminus R \subseteq \paren {S \setminus T} \cup \paren {T \setminus R}$
Thus:
Category:Symmetric Difference
Category:Set Union
Symmetric Difference of Complements

== Theorem ==
The symmetric difference of two sets equals the symmetric difference of their complements:
:$\map \complement S \symdif \map \complement T = S \symdif T$
== Proof ==
== Sources ==
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \beta$
Category:Symmetric Difference
Category:Set Complement
Symmetric Difference of Events is Event

== Theorem ==
Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
The event space $\Sigma$ of $\EE$ has the property that:
:$A, B \in \Sigma \implies A \ast B \in \Sigma$
That is, the symmetric difference of two events is also an event in the event space.
== Proof ==
== Also see ==
* Elementary Properties of Event Space
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Exercise $3$
Category:Event Spaces
Symmetric Difference of Unions

== Theorem ==
Let $R$, $S$ and $T$ be sets.
Then:
:$\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
where:
:$\symdif$ denotes the symmetric difference
:$\setminus$ denotes set difference
:$\cup$ denotes set union
== Proof ==
Category:Symmetric Difference
Category:Set Union
Symmetric Difference of Unions is Subset of Union of Symmetric Differences

== Theorem ==
Let $I$ be an indexing set.
Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.
Then:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}$
where $S \symdif T$ is the symmetric difference between $S$ and $T$.
== Proof ==
From Difference of Unions is Subset of Union of Differences, we have:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
:$\ds \bigcup_{\alpha \mathop \in I} T_i \setminus \bigcup_{\alpha \mathop \in I} S_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}$
where $\setminus$ denotes set difference.
Thus we have:
Category:Set Union
Category:Symmetric Difference
Symmetric Difference on Power Set forms Abelian Group

== Theorem ==
Let $S$ be a set such that $\O \subset S$ (that is, $S$ is non-empty).
Let $A \symdif B$ be defined as the symmetric difference between $A$ and $B$.
Let $\powerset S$ denote the power set of $S$.
Then the algebraic structure $\struct {\powerset S, \symdif}$ is an abelian group.
== Proof ==
From Power Set is Closed under Symmetric Difference, we have that $\struct {\powerset S, \symdif}$ is closed.
The result follows directly from Set System Closed under Symmetric Difference is Abelian Group.
== Sources ==
* : Chapter $4$. Groups: Exercise $12$
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.4$
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.1 \ \text{(b)}$
* : Chapter $\text{II}$: Groups: Exercise $\text{T}$
* : Chapter $2$: The Definition of Group Structure: $\S 26 \kappa$
Category:Abelian Groups
Category:Symmetric Difference
Category:Power Set
Symmetric Difference with Complement

== Theorem ==
The symmetric difference of a set with its complement is the universal set:
:$S \symdif \relcomp {} S = \mathbb U$
== Proof ==
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $9 \ \text {(i)}$
* : $\text {(i)}$
Category:Symmetric Difference
Category:Set Complement
Category:Universal Set
Symmetric Difference with Empty Set

== Theorem ==
:$S \symdif \O = S$
where $\symdif$ denotes the symmetric difference.
== Proof ==
== Sources ==
* : $\S 5$: Complements and Powers
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $6$
* : Chapter $1$: Sets and Logic: Exercise $9 \ \text{(i)}$
* : $\text {(i)}$
Category:Symmetric Difference
Category:Empty Set
Symmetric Difference with Intersection forms Boolean Ring

== Theorem ==
Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cap$ denote the set intersection operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a Boolean ring.
== Proof ==
From Symmetric Difference with Intersection forms Ring:
:$\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity.
From Set Intersection is Idempotent, $\cap$ is an idempotent operation on $S$.
Hence the result by definition of Boolean ring.
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
Category:Idempotent Rings
Category:Set Intersection
Category:Symmetric Difference
Category:Power Set
Symmetric Difference with Intersection forms Ring

== Theorem ==
Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cap$ denote the set intersection operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity, in which the unity is $S$.
This ring is not an integral domain.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Symmetric Difference with Union does not form Ring
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.3$
* : $\S 2$: Problem $6 \ \text{(iii)}$
Category:Commutative Algebra
Category:Set Intersection
Category:Symmetric Difference
Category:Power Set
Category:Symmetric Difference with Intersection forms Ring
Category:Examples of Commutative and Unitary Rings
Symmetric Difference with Intersection forms Ring/Proof 1

== Theorem ==
== Proof ==
From Symmetric Difference on Power Set forms Abelian Group, $\struct {\powerset S, \symdif}$ is an abelian group, where $\O$ is the identity and each element is self-inverse.
From Power Set with Intersection is Monoid, $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.
Also Intersection Distributes over Symmetric Difference.
Thus $\struct {\powerset S, \cap}$ is a commutative ring with a unity which is $S$.
From Intersection with Empty Set:
:$\forall A \in \powerset S: A \cap \O = \O = \O \cap A$
Thus $\O$ is indeed the zero.
However, from Set Intersection Not Cancellable, it follows that $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $6$
Category:Symmetric Difference with Intersection forms Ring
Symmetric Difference with Intersection forms Ring/Proof 2

== Theorem ==
== Proof ==
From Power Set is Closed under Symmetric Difference and Power Set is Closed under Intersection, we have that both $\struct {\powerset S, \symdif}$ and $\struct {\powerset S, \cap}$ are closed.
Hence $\powerset S$ is a ring of sets, and hence a commutative ring.
From Intersection with Subset is Subset‎, we have $A \subseteq S \iff A \cap S = A$.
Thus we see that $S$ is the unity.
Also during the proof of Power Set with Intersection is Monoid, it was established that $S$ is the identity of $\struct {\powerset S, \cap}$.
We also note that set intersection is not cancellable, so $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.
The result follows.
Category:Symmetric Difference with Intersection forms Ring