nyuuzyou/wikis huggingface dataset:
Subset/Examples

== Examples of Subsets ==
=== British People are Subset of People ===
=== Subset of Alphabet ===
=== Integers are Subset of Real Numbers ===
=== Initial Segment is Subset of Integers ===
=== Even Numbers form Subset of Integers ===
=== Even Numbers form Subset of Real Numbers ===
=== Abstract Example $1$ ===
=== Abstract Example $2$ ===
=== Abstract Example $3$ ===
=== Abstract Example $4$ ===
=== Romeo and Juliet ===
Category:Examples of Subsets
Subset/Examples/Abstract Example 1

== Example of Subset ==
Let $P$ denote the set:
:$P = \set {1, 2, 3, 4}$
Let $Q$ denote the set:
:$Q = \set {2, 4}$
Then:
:$Q \subseteq P$
but note that:
:$Q \notin P$
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.1$: Set Notation
Category:Examples of Subsets
Subset/Examples/Abstract Example 2

== Example of Subset ==
Let $A$ denote the set:
:$A := \set {\dfrac 1 2, 1, \sqrt 2, e, \pi}$
Let $C$ denote the set:
:$C := \set {x \in \R: x > 0}$
Then:
:$A \subseteq C$
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.1$: Set Notation
Category:Examples of Subsets
Subset/Examples/Abstract Example 3

== Example of Subset ==
Let $A$ denote the set:
:$A := \set {1, 2, 3}$
Let $B$ denote the set:
:$B := \set {\&, 3, +, 1, 2}$
Then:
:$A \subseteq B$
:$A \subseteq A$
:$B \subseteq B$
== Sources ==
* : $2$. Graphs: Sets: Definition $3$
Category:Examples of Subsets
Subset/Examples/British People are Subset of People

== Example of Subset ==
Let $B$ denote the set of British people.
Let $P$ denote the set of people.
Then $B$ is a proper subset of $P$:
:$B \subsetneq P$
== Sources ==
* : $\S 1.2$. Subsets: Example $9$
Category:Examples of Subsets
Subset/Examples/Initial Segment is Subset of Integers

== Example of Subset ==
The (one-based) initial segment of the natural numbers $\N^*_{\le n}$ of integers $\Z$ is a proper subset of the set of integers $\Z$:
:$\N^*_{\le n} \subsetneq \Z$
== Sources ==
* : $\S 1.2$. Subsets: Example $9$
Category:Examples of Subsets
Subset/Examples/Romeo and Juliet

== Example of Subset ==
Let $B$ denote the set:
:$B := \set {\textrm {Romeo}, \textrm {Juliet} }$
Let $D$ denote the set:
:$D := \set {y: \text {$y$ loves Romeo} }$
Then:
:$B \subseteq D$
presuming, of course, that Romeo actually loves himself, which may be doubted considering he took his own life.
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.1$: Set Notation
Category:Examples of Subsets
Subset/Examples/Subset of Alphabet

== Example of Subset ==
Let $S$ denote the capital letters of the (English) alphabet:
:$S = \set {A, B, C, D, \dotsc, Z}$
Then $\set {A, B, C}$ is a subset of $S$:
:$\set {A, B, C} \subseteq S$
== Sources ==
* : Chapter $1$: Numbers: Sets
Category:Examples of Subsets
Subset Equivalences

== Definitions ==
In the following:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$S \cup T$ denotes the union of $S$ and $T$
:$S \cap T$ denotes the intersection of $S$ and $T$
:$S \setminus T$ denotes the set difference between $S$ and $T$
:$\O$ denotes the empty set
:$\mathbb U$ denotes the universal set
:$\complement$ denotes set complement.
=== Union with Superset is Superset‎ ===
=== Intersection with Subset is Subset‎ ===
=== Set Difference with Superset is Empty Set‎ ===
=== Intersection with Complement is Empty iff Subset ===
=== Complement Union with Superset is Universe ===
=== Set Complement inverts Subsets ===
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.3 \ \text{(a)}$
Category:Subsets
Subset Product/Examples

== Examples of Subset Product ==
Let $G$ be a group.
=== Example 1 ===
=== Example 2 ===
=== Example 3 ===
=== Example 4 ===
=== Subset Product with Empty Set ===
=== Subsets of $\R$ under Multiplication ===
=== Congruence Modulo $m$ in $\N_{<m}$ ===
Category:Examples of Subset Product
Subset Product/Examples/Congruence Modulo Initial Segment of Natural Numbers

== Example of Subset Product ==
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\N_{<m}$ denote the initial segment of the natural numbers $\N$:
:$\N_{<m} = \set {0, 1, \ldots, m - 1}$
Let $\RR_m$ denote the equivalence relation:
:$\forall x, y \in \Z: x \mathrel {\RR_m} y \iff \exists k \in \Z: x = y + k m$
For each $a \in \N_{<m}$, let $\eqclass a m$ be the equivalence class of $a \in \N_{<m}$ under $\RR_m$:
:$\eqclass a m := \set {a + z m: z \in \Z}$
Let $\Z_m$ be the set defined as:
:$\Z_m := \set {\eqclass a m: a \in \N_{<m} }$
Let $+_\PP$ denote the operation induced on $\powerset \Z$ by integer addition.
Then the algebraic structure $\struct {\Z_m, +_\PP}$ is closed in the sense:
:$\forall \eqclass a m, \eqclass b m \in \Z_m: \eqclass a m +_\PP \eqclass b m \in \Z_m$
== Proof ==
From Equivalence Class Example: Congruence Modulo $\N_{<m}$:
:$\RR_m$ is an equivalence relation
:$\eqclass a m$ is the equivalence class of $a \in \N_{<m}$ under $\RR_m$.
Let $\eqclass a m, \eqclass b m \in \RR_m$ be arbitrary.
Then we have:
It also follows that:
:$\forall a, b \in \N_{<m}: \eqclass a m = \eqclass b m \iff a = b$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets
Category:Examples of Subset Product
Subset Product/Examples/Example 1

== Examples of Subset Product ==
Let $G$ be a group.
Let $a \in G$ be an element of $G$.
Let:
Let $\order a = 4$.
Then:
:$\card {X Y} = 4$
where $\card {\, \cdot \,}$ denotes cardinality.
== Proof ==
Calculating the elements of $X Y$
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $1 \ \text{(i)}$
Category:Examples of Subset Product
Subset Product/Examples/Example 2

== Examples of Subset Product ==
Let $G$ be a group.
Let $a \in G$ be an element of $G$.
Let:
Let $\order a = 6$.
Then:
:$\card {X Y} = 5$
where $\card {\, \cdot \,}$ denotes cardinality.
== Proof ==
Calculating the elements of $X Y$
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $1 \ \text{(ii)}$
Category:Examples of Subset Product
Subset Product/Examples/Example 3

== Example of Subset Product ==
Let $G$ be a group.
Let the order of $G$ be $n \in \Z_{>0}$.
Let $X \subseteq G$ be a subset of $G$.
Let $\card X > \dfrac n 2$.
Then:
:$X X = G$
where $X X$ denotes subset product.
== Proof ==
Let $g \in G$ be arbitrary.
Let $Y := \set {x^{-1} g: x \in X}$.
Then:
:$\card Y = \card X$
As $\card X > \dfrac {\order g} 2$ we have:
:$X \cap Y \ne \O$
So:
:$\exists x_1, x_2 \in X: x_1^{-1} g = x_2$
That is:
:$g = x_1 x_2$
and so:
:$g \in X X$
As $g$ is arbitrary, it follows that:
:$X X = G$
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $3$
Category:Examples of Subset Product
Subset Product/Examples/Example 4

== Example of Subset Product ==
Let $S$ be the initial segment of the natural numbers $\N_{<3}$:
:$\N_{<3} = \set {0, 1, 2}$
Let $\circ$ be the operation defined on $S$ by the Cayley table:
:$\begin {array} {c|cccc}
\circ & 0 & 1 & 2 \\
\hline
0 & 0 & 1 & 2 \\
1 & 1 & 0 & 0 \\
2 & 2 & 0 & 0 \\
\end {array}$
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Then every non-empty subset of $S$ which does not contain $0$ is invertible in $\struct {\powerset S, \circ_\PP}$.
== Proof ==
By inspection of the Cayley table for $\circ$, it is apparent that $0$ is an identity element of $\struct {S, \circ}$.
From Identity Element for Power Structure, $\set 0$ is an identity element of $\struct {\powerset S, \circ_\PP}$.
Let $X \in \powerset S$ such that $0 \notin X$.
Let $Y \in \powerset S$ be an inverse of $X$ such that $0 \notin Y$.
Then we have:
That is, for this particular algebraic structure, every non-empty subset of $S$ that does not contain $0$ is an inverse of every other non-empty subset of $S$ that does not contain $0$.
It follows that every non-empty subset of $S$ that does not contain $0$ is invertible in $\struct {\powerset S, \circ_\PP}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.7 \ \text {(b)}$
Category:Examples of Subset Product
Subset Product/Examples/Subset Product with Empty Set

== Example of Subset Product ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $A, B \in \powerset S$.
If $A = \O$ or $B = \O$, then $A \circ B = \O$.
== Proof ==
We show the contrapositive:
Suppose that $A \circ B \ne \O$.
Then:
:$\exists x: x \in A \circ B$
From definition of subset product:
:$A \circ B = \set {a \circ b: a \in A, b \in B}$
Therefore:
:$\exists a \in A, b \in B: x = a \circ b$
In particular, both $A$ and $B$ are non-empty.
By De Morgan's Laws (Logic)/Conjunction of Negations, it is not the case that $A = \O$ or $B = \O$.
Hence the result by Proof by Contraposition.
== Sources ==
* : $\S 41$: Multiplication of subsets of a group
Category:Examples of Subset Product
Subset Product/Examples/Subsets of Reals under Multiplication

== Example of Subset Product ==
Let $\struct {\R, \times}$ be the multiplicative group of (non-zero) real numbers.
Let $S = \set {-1, 2}$.
Let $T = \set {1, 2, 3}$.
Then the subset product $S T$ is:
:$ST = \set {-1, -2, -3, 2, 4, 6}$
== Proof ==
Taking each ordered pair $\tuple {s, t}$ from $S \times T$:
== Sources ==
* : Chapter $\text{II}$: Groups: Subgroups
Category:Examples of Subset Product
Subset Product/Singleton/Examples/Subsets of Reals under Multiplication

== Example of Subset Product ==
Let $\struct {\R, \times}$ be the multiplicative group of (non-zero) real numbers.
Let $S = \set 3$.
Let $T = \set {-1, 2}$.
Then the subset product $S T$ is:
:$ST = \set {-3, 6}$
== Proof ==
Taking each ordered pair $\tuple {s, t}$ from $S \times T$:
== Sources ==
* : Chapter $\text{II}$: Groups: Subgroups
Category:Examples of Subset Product
Subset Product Action is Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ be the power set of $\struct {G, \circ}$.
For any $S \in \powerset G$ and for any $g \in G$, the subset product action:
:$\forall g \in G: \forall S \in \powerset G: g * S = g \circ S$
is a group action.
== Proof ==
Let $g \in G$.
First we note that since $G$ is closed, and $g \circ S$ consists of products of elements of $G$, it follows that:
:$g * S \subseteq G$
Next we note:
:$e * S = e \circ S = \set {e \circ s: s \in S} = \set {s: s \in S} = S$
and so is satisfied.
Now let $g, h \in G$.
We have:
and so is satisfied.
Hence the result.
== Also see ==
* Stabilizer of Subset Product Action on Power Set
* Stabilizer of Coset Action on Set of Subgroups
* Orbit of Subgroup under Coset Action is Coset Space
== Sources ==
* : $\S 5.5$. Groups acting on sets: Example $104$
* : $\S 3.3$: Group actions and coset decompositions
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.6$
Category:Subset Product Action
Category:Subset Products
Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$S \circ' C^{-1}$ is a commutative semigroup
where $S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$ under $\circ'$ in $T$.
== Proof ==
Note that by definition of inverse completion, $\struct {T, \circ'}$ is a semigroup.
Thus $\circ'$ is associative.
First it is demonstrated that $S \circ' C^{-1}$ is a semigroup.
Let $x, z \in S$.
Let $y, w \in C$.
Then:
Thus:
:$\paren {x \circ z} \circ' \paren {w \circ y}^{-1} \in S \circ' C^{-1}$
proving that $S \circ' C^{-1}$ is closed.
Therefore by Subsemigroup Closure Test:
:$S \circ' C^{-1}$ is a subsemigroup of $\struct {T, \circ'}$
and thus a semigroup.
It remains to be shown that $\circ'$ is a commutative operation.
Let $\paren {x \circ' y^{-1} }$ and $\paren {z \circ' w^{-1} }$ be two arbitrary elements of $S \circ' C^{-1}$.
By Element Commutes with Product of Commuting Elements, $x, y, z, w$ all commute with each other under $\circ$.
As $\circ'$ is an extension of $\circ$, it follows that $x, y, z, w$ also all commute with each other under $\circ'$.
Then:
So $x \circ' y^{-1}$ commutes with $z \circ' w^{-1}$.
It follows by definition that $S \circ' C^{-1}$ is a commutative semigroup.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 2^\circ$
Category:Inverse Completions
Subset Product is Subset of Generator

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq \struct {G, \circ}$.
Then $X \circ Y \subseteq \gen {X, Y}$ where:
:$X \circ Y$ is the Subset Product of $X$ and $Y$ in $G$.
:$\gen {X, Y}$ is the subgroup of $G$ generated by $X$ and $Y$.
== Proof ==
It is clear from Set of Words Generates Group that $\map W {\hat X \cup \hat Y} = \gen {X, Y}$.
It is equally clear that $X \circ Y \subseteq \map W {\hat X \cup \hat Y}$.
Category:Group Theory
Category:Subset Products
Subset Product of Abelian Subgroups

== Theorem ==
Let $\left({G, \circ}\right)$ be an abelian group.
Let $H_1$ and $H_2$ be subgroups of $G$.
Then $H_1 \circ H_2$ is a subgroup of $G$.
== Proof ==
From Subgroup of Abelian Group is Normal, $H_1$ and $H_2$ are normal.
The result follows from Subset Product with Normal Subgroup is Subgroup‎.
Category:Abelian Groups
Category:Subset Products
Subset Product of Normal Subgroups is Normal

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $N$ and $N'$ be normal subgroups of $G$.
Then $N N'$ is also a normal subgroup of $G$.
== Proof ==
From Subset Product with Normal Subgroup is Subgroup‎, we already have that $N N'$ is a subgroup of $G$.
Let $n n' \in N N'$, so that $n \in N, n' \in N'$.
Let $g \in G$.
From Subgroup is Normal iff Contains Conjugate Elements:
:$g n g^{-1}\in N, g n' g^{-1}\in N'$
So:
:$\paren {g n g^{-1} } \paren {g n' g^{-1} } = g n n' g^{-1} \in N N'$
So $N N'$ is normal.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \zeta$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $9$
Category:Normal Subgroups
Category:Subset Products
Subset Product of Normal Subgroups with Trivial Intersection

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H, K$ be normal subgroups of $G$.
Let $H \cap K = e$.
Then $H K$ is isomorphic to $H \times K$ where:
: $H K$ denotes the subset product of $H$ and $K$
: $H \times K$ denotes the direct product of $H$ and $K$.
== Proof ==
Let $G' = H K$.
From Subset Product of Normal Subgroups is Normal, $G'$ is a normal subgroup of $G$.
That is $G'$ is itself a group.
So by the Internal Direct Product Theorem, $G'$ is the internal group direct product of $H$ and $K$.
The result follows by definition of the internal group direct product.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \mu$
Category:Normal Subgroups
Category:Group Direct Products
Subset Product of Subgroups

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Then $H \circ K$ is a subgroup of $G$ $H$ and $K$ are permutable.
That is:
$H \circ K$ is a subgroup of $G$ :
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product.
=== Necessary Condition ===
=== Sufficient Condition ===
== Examples ==
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 38 \alpha$
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6$
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.17$
Category:Subgroups
Category:Subset Products
Category:Subset Product of Subgroups
Subset Product of Subgroups/Examples/Subgroups Generated by b and a b in D3

== Examples of Use of Subset Product of Subgroups ==
Consider the dihedral group $D_3$, given as the group presentation:
Consider the generated subgroups $H := \gen b$ and $K := \gen {a b}$:
Then $H$ and $K$ are not permutable, and neither $H K$ nor $K H$ is a subgroup of $D_3$.
== Proof ==
Consider the subset product $H K$:
But $\set {e, b, a b, a^2}$ has $4$ elements.
Thus by Lagrange's Theorem (Group Theory), $H K$ is not a subgroup of $D_3$.
Then we see:
So $H K \ne K H$ and so $H$ and $K$ are not permutable.
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Exercise $5$
Category:Subset Product of Subgroups
Subset Product of Subgroups/Necessary Condition

== Theorem ==
Let $\left({G, \circ}\right)$ be a group.
Let $H, K$ be subgroups of $G$.
Let $H \circ K$ be a subgroup of $G$.
Then $H$ and $K$ are permutable.
That is:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product.
== Proof 1 ==
== Proof 2 ==
Category:Subset Product of Subgroups
Subset Product of Subgroups/Necessary Condition/Proof 1

== Theorem ==
== Proof ==
Suppose $H \circ K$ is a subgroup of $G$.
Let $h \circ k \in H \circ K$.
Then $h \circ k$ is the inverse of some element $g$ of $H \circ K$.
Thus we can write $g = h' \circ k'$ for some $h' \in H$ and $k' \in K$.
So:
So by definition of subset:
: $H \circ K \subseteq K \circ H$
Now suppose $x \in K \circ H$.
Then:
So by definition of subset:
:$K \circ H \subseteq H \circ K$
The result follows by definition of set equality.
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.17$
Category:Subset Product of Subgroups
Subset Product of Subgroups/Sufficient Condition

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Let $H$ and $K$ be permutable subgroups of $G$.
That is, suppose:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product.
Then $H \circ K$ is a subgroup of $G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 5$: Proposition $5.17$
Category:Subset Product of Subgroups
Subset Product of Subgroups/Sufficient Condition/Proof 1

== Theorem ==
== Proof ==
Suppose $H \circ K = K \circ H$.
First note that $H \circ K \ne \O$, as $e_G = e_G \circ e_G \in H \circ K$, from Identity of Subgroup.
Suppose $a_1, a_2 \in H, b_1, b_2 \in K$.
Then:
:$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = a_1 \circ \paren {b_1 \circ a_2} \circ b_2$.
Since $H \circ K = K \circ H$, we see that, for some $a \in H, b \in K$:
:$b_1 \circ a_2 = a \circ b$
Thus:
:$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = \paren {a_1 \circ a} \circ \paren {b \circ b_2}$
As $H, K \le G$, we have $a_1 \circ a \in H$ and $b \circ b_2 \in K$, hence:
:$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} \in H \circ K$
thus demonstrating closure of $H \circ K$ under $\circ$.
Finally, if $a \circ b \in H \circ K$, then by Inverse of Group Product:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$
Since $b^{-1} \in K$ and $a^{-1} \in H$, we have:
:$\paren {a \circ b}^{-1} \in K \circ H$
and hence $H \circ K$ is shown to be closed under inverses.
Thus, from the Two-Step Subgroup Test, $H \circ K$ is a subgroup of $G$.
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.17$
Category:Subset Product of Subgroups
Subset Product of Subgroups/Sufficient Condition/Proof 2

== Theorem ==
== Proof ==
Suppose $H \circ K = K \circ H$.
Then:
That is:
:$\paren {H \circ K} \circ \paren {H \circ K}^{-1} = H \circ K$
Thus by definition of set equality:
:$\paren {H \circ K} \circ \paren {H \circ K}^{-1} \subseteq H \circ K$
So from One-Step Subgroup Test using Subset Product:
:$H \circ K \le G$
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6$
Category:Subset Product of Subgroups
Subset Product with Identity

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\struct {S, \circ}$ have an identity element $e$.
Then $e \circ S = S \circ e = S$, where $\circ$ is understood to be the subset product with singleton.
== Proof ==
Thus:
:$e \circ S = S$
A similar argument shows that:
:$S \circ e = S$
Category:Subset Products
Subset Product with Normal Subgroup as Generator

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let:
:$H$ be a subgroup of $G$
:$N$ be a normal subgroup of $G$.
Then:
:$N \lhd \gen {N, H} = N H = H N \le G$
where:
:$\le$ denotes subgroup
:$\lhd$ denotes normal subgroup
:$\gen {N, H}$ denotes a subgroup generator
:$N H$ denotes subset product.
== Proof ==
From Subset Product is Subset of Generator:
:$N H \subseteq \gen {N, H}$
From Subset Product with Normal Subgroup is Subgroup:
:$N H = H N \le G$
Then by the definition of a subgroup generator, $\gen {N, H}$ is the smallest subgroup containing $N H$ and so:
:$\gen {N, H} = N H = H N \le G$
From Normal Subgroup of Subset Product of Subgroups we have that:
:$N \lhd N H$
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.16$
* : Chapter $\text{II}$: Groups: Problem $\text{HH}$
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.10$
Category:Normal Subgroups
Category:Subset Products
Subset Product with Normal Subgroup is Subgroup

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let:
:$(1): \quad H$ be a subgroup of $G$
:$(2): \quad N$ be a normal subgroup of $G$.
Let $H N$ denote subset product.
Then $H N$ and $N H$ are both subgroups of $G$.
== Proof ==
It is clear that $e \in N H$, so $N H \ne \O$.
Suppose $n_1, n_2 \in N$ and $h_1, h_2 \in H$.
Then:
Since $N$ is normal in $G$:
:$\exists n \in N: n = h_1 n_2 h_1^{-1}$
Thus:
Also:
so from the Two-Step Subgroup Test, $N H$ is a subgroup of $G$.
The fact that $N H = H N$ follows from Subset Product of Subgroups.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.16$
* : Chapter $\text{II}$: Groups: Problem $\text{HH}$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \epsilon$
* : $\S 5$: Groups $\text{I}$: Subgroups
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $9$
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.10$
Category:Normal Subgroups
Category:Subset Products
Subset Product within Commutative Structure is Commutative

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
If $\circ$ is commutative, then the operation $\circ_\PP$ induced on the power set of $S$ is also commutative.
== Proof ==
Let $\struct {S, \circ}$ be a magma in which $\circ$ is commutative.
Let $X, Y \in \powerset S$.
Then:
Hence $\circ_\PP$ is commutative on $\powerset S$.
== Also see ==
* Subset Product within Semigroup is Associative
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets
Category:Subset Products
Subset Product within Semigroup is Associative

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Then the operation $\circ_\PP$ induced on the power set of $S$ is also associative.
=== Corollary ===
== Proof ==
Let $X, Y, Z \in \powerset S$.
Then:
demonstrating that $\circ_\PP$ is associative on $\powerset S$.
== Also see ==
* Subset Product within Commutative Structure is Commutative
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets
* : Chapter $\text{II}$: Groups: Exercise $\text{G}$
* : $\S 41.1$ Multiplication of subsets of a group
Category:Subset Products
Category:Associativity
Category:Semigroups
Category:Subset Product within Semigroup is Associative
Subset Product within Semigroup is Associative/Corollary

== Corollary to Subset Product within Semigroup is Associative ==
Let $\struct {S, \circ}$ be a semigroup.
Then:
== Proof ==
From the definition of Subset Product with Singleton:
The result then follows directly from Subset Product within Semigroup is Associative.
== Also see ==
* Subset Product within Commutative Structure is Commutative
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup
Category:Subset Product within Semigroup is Associative