nyuuzyou/wikis huggingface dataset:
Norm satisfying Parallelogram Law induced by Inner Product/Real Case/Lemma

== Lemma ==
Let $V$ be a vector space over $\R$.
Let $\norm \cdot : V \to \R$ be a norm on $V$ such that: 
:$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2}$
for each $x, y \in V$.
Define the function $\innerprod \cdot \cdot : V \times V \to \R$ by: 
:$\ds \innerprod x y = \frac {\norm {x + y}^2 - \norm {x - y}^2} 4$
for each $x, y \in V$.
Then: 
:$\innerprod {n x} y = n \innerprod x y$
for each integer $n \ge 0$ and $x, y \in V$.
== Proof ==
Let $x, y \in V$.
In the case $n = 0$, we have: 
:$\innerprod {n x} y = \innerprod 0 y$
From Inner Product with Zero Vector, we then have: 
:$\innerprod 0 y = 0$
so:
:$\innerprod {n x} y = n \innerprod x y$
in this case. 
Now we prove the statement for $n \ge 1$. 
We proceed by induction.
For each $n \in \N$ let $\map P n$ be the proposition:
:$\innerprod {n x} y = n \innerprod x y$ 
=== Basis for the Induction ===
We have: 
:$\innerprod {1 \times x} y = \innerprod x y$
So $\map P 1$ holds.
=== Induction Hypothesis ===
Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.
So this is our induction hypothesis:
:$\ds \innerprod {n x} y = n \innerprod x y$ 
=== Induction Step ===
Now we need to show:
:$\innerprod {\paren {n + 1} x} y = \paren {n + 1} \innerprod x y$ 
We have: 
So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.
Category:Norm satisfying Parallelogram Law induced by Inner Product
Normal Bundle Theorem

== Theorem ==
Let $\tilde M$ be an $m$-dimensional Riemannian manifold.
Let $M \subseteq \tilde M$ be an immersed or embedded $n$-dimensional submanifold with or without boundary.
Let $\valueat {T \tilde M} M$ be the ambient tangent bundle.
Let $NM$ be the normal bundle of $M$.
Then $NM$ is a rank-$\paren {m - n}$ smooth vector subbundle of $\valueat {T \tilde M} M$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Riemannian Geometry
Normal Distribution CDF in terms of Error Function

== Theorem ==
Let $\map \Phi x$ denote the cumulative distribution function of the standard normal distribution.
Then:
:$\ds \map \Phi x = \dfrac 1 2 \paren {1 - \map \erf {\dfrac x {\sqrt 2} } }$
where $\erf$ denotes the error function.
== Proof ==
== Sources ==
* 
Category:Error Function
Category:Normal Distribution
Category:Examples of Cumulative Distribution Functions‎
Normal Distribution is Pearson Distribution

== Theorem ==
The normal distribution is an example of a Pearson distribution.
== Proof ==
Recall the definition of the Pearson distribution:
Setting $a = -1$, $b = c = d = 0$ gives:
:$\map {f'} x = -x \map f x$
== Sources ==
* 
* 
Category:Normal Distribution
Category:Pearson Distributions
Normal Element of C*-Algebra is Hermitian iff Spectrum is Real

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Then $x$ is Hermitian :
:$\map {\sigma_A} x \subseteq \R$ 
== Proof ==
We first take $A$ unital.
=== Necessary Condition: Unital Case ===
Let $x \in A$ be Hermitian.
From Spectrum of Hermitian Element in Unital C*-Algebra is Real, we have:
:$\map {\sigma_A} x \subseteq \R$
=== Sufficient Condition: Unital Case ===
Suppose that $x \in A$ has:
:$\map {\sigma_A} x \subseteq \R$ 
Let $\Theta_x$ be the continuous functional calculus for $x$.
Let $\iota : \map {\sigma_A} x \to \C$ be the inclusion mapping.
Since $\map {\sigma_A} x \subseteq \R$, we have:
:$z = \overline z$ for each $z \in \map {\sigma_A} x$
from Complex Number equals Conjugate iff Wholly Real. 
Hence we have $\iota = \overline \iota$.
We then have: 
=== General Case ===
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $A$ as a $\text C^\ast$-algebra given by Existence of Unitization of C*-Algebra.
By the definition of the spectrum, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} } \cup \set 0$
and hence:
:$\map {\sigma_A} x \subseteq \R$ $\map {\sigma_{A_+} } {\tuple {x, 0} } \subseteq \R$
By the unital case, we have:
:$\map {\sigma_{A_+} } {\tuple {x, 0} } \subseteq \R$ $\tuple {x, 0}$ is Hermitian.
That is, $\tuple {x, 0}^\ast = \tuple {x, 0}$.
We have $\tuple {x, 0}^\ast = \tuple {x^\ast, 0}$, hence this is equivalent to $x = x^\ast$.
So $\map {\sigma_A} x \subseteq \R$ $x$ is Hermitian.
Category:Hermitian Elements of *-Algebras
Category:C*-Algebras
Normal Element of C*-Algebra is Projection iff Spectrum contains only Zero and One

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $p \in A$ be normal. 
Let $\map {\sigma_A} p$ denote the spectrum of $p$ in $A$.
Then $p$ is a projection :
:$\map {\sigma_A} p \subseteq \set {0, 1}$
== Proof ==
We first take $A$ unital.
=== Necessary Condition: Unital Case ===
Let $p \in A$ be a projection.
Then from Spectrum of Projection in *-Algebra:
:$\map {\sigma_A} p \subseteq \set {0, 1}$
=== Sufficient Condition: Unital Case ===
Suppose that:
:$\map {\sigma_A} p \subseteq \set {0, 1}$
From Normal Element of C*-Algebra is Hermitian iff Spectrum is Real, $p$ is Hermitian. 
Let $\Theta_p$ be the continuous functional calculus for $x$.
Let $\iota : \map {\sigma_A} p \to \C$ be the inclusion mapping.
Since $\map {\sigma_A} p \subseteq \set {0, 1}$, we have:
:$z^2 = z$ for each $z \in \map {\sigma_A} p$
We then have:
=== General Case ===
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $A$ as a $\text C^\ast$-algebra given by Existence of Unitization of C*-Algebra.
By the definition of the spectrum, we have:
:$\map {\sigma_A} p = \map {\sigma_{A_+} } {\tuple {p, 0} } \cup \set 0$
Note that $\tuple {p, 0}^2 = \tuple {p^2, 0}$ and $\tuple {p, 0}^\ast = \tuple {p^\ast, 0}$.
Hence $\tuple {p, 0}$ is a projection on $A_+$ $p$ is a projection on $A$.
By the unital case, we have that $\tuple {p, 0}$ is a projection on $A_+$ $\map {\sigma_{A_+} } {\tuple {p, 0} } \subseteq \set {0, 1}$. 
That is, $\map {\sigma_A} p \subseteq \set {0, 1}$.
So $p$ is a projection on $A$ $\map {\sigma_A} p \subseteq \set {0, 1}$. 
Category:C*-Algebras
Normal Element of Unital C*-Algebra is Unitary iff Spectrum is Subset of Unit Circle

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Then $x$ is unitary :
:$\map {\sigma_A} x \subseteq \set {z \in \C : \cmod z = 1}$ 
== Proof ==
=== Necessary Condition ===
Let $A' \subseteq A$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x}$. 
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $A'$ is commutative.
Let $\Phi_{A'}$ be the spectrum of $A'$.
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra, we have:
:$\map {\sigma_A} x = \map {\sigma_{A'} } x$
From Spectrum of Element of Unital Commutative Banach Algebra, we have $\map {\sigma_A} x = \map {\sigma_{A'} } x = \set {\map \phi x : \phi \in \Phi_{A'} }$. 
From Character on Unital C*-Algebra has Modulus One at Unitary Elements, we have $\cmod {\map \phi x} = 1$ for each $\phi \in \Phi_{A'}$.
Hence $\map {\sigma_A} x \subseteq \set {z \in \C : \cmod z = 1}$. 
=== Sufficient Condition ===
Suppose that $x \in A$ has:
:$\map {\sigma_A} x \subseteq \set {z \in \C : \cmod z = 1}$ 
Let $\Theta_x$ be the continuous functional calculus for $x$.
Let $\iota : \map {\sigma_A} x \to \C$ be the inclusion mapping.
Since $\map {\sigma_A} x \subseteq \set {z \in \C : \cmod z = 1}$, we have:
:$z \overline z = \cmod z^2 = 1$ for each $z \in \map {\sigma_A} x$
from Product of Complex Number with Conjugate. 
Hence we have $\iota \overline \iota = 1$. 
We therefore have: 
Since $x$ is normal, we also have:
:$x x^\ast = {\mathbf 1}_A$
Hence $x$ is unitary.
Category:Unitary Elements of Unital *-Algebras
Category:C*-Algebras
Normal First Order Ordinary Differential Equation/Examples/y' = -x / y

== Example of Normal First Order Ordinary Differential Equation ==
Consider the first order ordinary differential equation:
:$x + y y' = 0$
This can be expressed in normal form as:
:$y' = -x / y$
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $1$ Introduction
Category:Examples of Normal Differential Equations
Normal Space is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a normal space, then so is $T_B$.
== Proof ==
We have that $\struct {S, \tau}$ is a normal space :
:$\struct {S, \tau}$ is a $T_4$ space
:$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
From $T_4$ Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_4$ space, then so is $T_B$.
From $T_1$ Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Normal Spaces
Normal Space is Regular Space

== Theorem ==
Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a regular space.
== Proof ==
Let $T = \struct {S, \tau}$ be a normal space.
From Normal Space is $T_3$ Space, we have that $T$ is a $T_3$ space.
We also have by definition of normal space that $T$ is a $T_1$ (Fréchet) space.
From $T_1$ Space is $T_0$ Space we have that $T$ is a $T_0$ (Kolmogorov) space
So $T$ is both a $T_3$ space and a $T_0$ (Kolmogorov) space.
Hence $T$ is a regular space by definition.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Regular and Normal Spaces
Category:Regular Spaces
Category:Normal Spaces
Normal Space is T3 Space

== Theorem ==
Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a $T_3$ space.
=== Corollary ===
== Proof ==
Let $T = \struct {S, \tau}$ be a normal space.
From the definition of normal space:
:$\struct {S, \tau}$ is a $T_4$ space
:$\struct {S, \tau}$ is a Fréchet ($T_1$) space.
Let $F$ be any closed set in $T$.
Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.
As $T$ is a Fréchet ($T_1$) space it follows from Equivalence of Definitions of $T_1$ Space that $\set y$ is closed.
As $T = \struct {S, \tau}$ is a normal space, we have that:
:$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ 
That is, for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.
But $F$ and $\set y$ are disjoint closed sets.
So:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
which is precisely the definition of a $T_3$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Regular and Normal Spaces
Category:T3 Spaces
Category:Normal Spaces
Normal Space is Tychonoff Space

== Theorem ==
Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a Tychonoff (completely regular) space.
== Proof ==
Let $T = \struct {S, \tau}$ be a normal space.
From the definition of normal space:
:$\struct {S, \tau}$ is a $T_4$ space
:$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
Let $F$ be any closed set in $T$.
Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.
As $T$ is a $T_1$ (Fréchet) space it follows from Equivalence of Definitions of $T_1$ Space that $\set y$ is closed.
As $T = \struct {S, \tau}$ is a $T_4$ space, we have that for any two disjoint closed sets $A, B \subseteq S$ there exists an Urysohn function for $A$ and $B$.
But $F$ and $\set y$ are disjoint closed sets.
So there exists an Urysohn function for $F$ and $\set y$.
This is the definition of a $T_{3 \frac 1 2}$ space.
Next we note that as $T$ is a $T_1$ (Fréchet) space, from $T_1$ Space is $T_0$ Space it follows that $T$ is a $T_0$ (Kolmogorov) space.
So $T$ is both a $T_{3 \frac 1 2}$ space and $T_0$ (Kolmogorov) space.
So, by definition, $T$ is a Tychonoff space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:Tychonoff Spaces
Category:Normal Spaces
Normal Subgroup Test

== Theorem ==
Let $G$ be a group and $H \le G$.
Then $H$ is a normal subgroup (by definition 1) of $G$ :
:$\forall x \in G: x H x^{-1} \subseteq H$.
== Proof ==
Let $H$ be a subgroup of $G$.
Suppose $H$ is normal in $G$.
Then $\forall x \in G, a \in H: \exists b \in H: x a = b x$.
Thus, $x a x^{-1} = b \in H$ implying $x H x^{-1} \subseteq H$.
Conversely, suppose $\forall x \in G: x H x^{-1} \subseteq H$.
Then for $g \in G$, we have $g H g^{-1} \subseteq H$, which implies $g H \subseteq H g$.
Also, for $g^{-1} \in G$, we have $g^{-1} H (g^{-1})^{-1} = g^{-1} H g \subseteq H$ which implies $H g \subseteq g H$.
Therefore, $g H = H g$ meaning $H \lhd G$.
Category:Normal Subgroups
Normal Subgroup iff Normalizer is Group

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then $H$ is normal in $G$ the normalizer of $H$ is equal to $G$:
:$H \lhd G \iff \map {N_G} H = G$
== Proof ==
=== Sufficient Condition ===
Let $H$ be normal in $G$.
Then $G$ is trivially the largest subgroup of $G$ in which $H$ is normal.
Thus from Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$\map {N_G} H = G$
=== Necessary Condition ===
Let $\map {N_G} H = G$.
From Subgroup is Normal Subgroup of Normalizer, $H$ is normal in $\map {N_G} H$.
Hence $H$ is normal in $G$.
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Normal Subgroups
Category:Normalizers
Normal Subgroup induced by Congruence Relation defines that Congruence

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\RR$ be a congruence relation for $\circ$.
Let $\eqclass e \RR$ be the equivalence class of $e$ under $\RR$.
Let $N = \eqclass e \RR$ be the normal subgroup induced by $\RR$.
Then $\RR$ is the equivalence relation $\RR_N$ defined by $N$.
== Proof ==
Let $\RR_N$ be the equivalence defined by $N$.
Then:
But from Congruence Class Modulo Subgroup is Coset:
:$x \mathrel {\RR_N} y \iff x^{-1} \circ y \in N$
Thus:
:$\RR = \RR_N$
== Also see ==
* Congruence Relation induces Normal Subgroup
* Congruence Relation on Group induces Normal Subgroup
* Quotient Structure on Group defined by Congruence equals Quotient Group
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.5$
Category:Normal Subgroups
Category:Congruence Relations
Normal Subgroup is Kernel of Group Homomorphism

== Theorem ==
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Then there exists a group homomorphism of which $N$ is the kernel.
== Proof ==
Let $G / N$ be the quotient group of $G$ by $N$.
Let $q_N: G \to G / N$ be the quotient epimorphism from $G$ to $G / N$:
:$\forall x \in G: \map {q_N} x = x N$
Then from Quotient Group Epimorphism is Epimorphism, $N$ is the kernel of $q_n$
Thus $q_N$ is that group homomorphism of which $N$ is the kernel.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $6$
Category:Normal Subgroups
Category:Group Homomorphisms
Category:Kernels of Group Homomorphisms
Normal Subgroup of Group of Order 24

== Theorem ==
Let $G$ be a group of order $24$.
Then $G$ has either:
:a normal subgroup of order $8$
or:
:a normal subgroup of order $4$.
== Proof ==
We note that:
:$24 = 3 \times 2^3$
Hence a Sylow $2$-subgroup of $G$ is of order $8$.
Let $n_2$ denote the number of Sylow $2$-subgroups of $G$.
By the Fourth Sylow Theorem:
:$n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is odd
and from the Fifth Sylow Theorem:
:$n_2 \divides 24$
where $\divides$ denotes divisibility.
It follows that $n_3 \in \set {1, 3}$.
Suppose $n_2 = 1$.
Let $P$ denote the unique Sylow $2$-subgroup of $G$.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
This is therefore the required normal subgroup of order $8$.
Suppose $n_2 = 3$.
Let $S_1$, $S_2$ and $S_3$ denote the $3$ Sylow $2$-subgroups of $G$.
Each has order $8$, as noted earlier.
Consider the product $S_1 S_2$.
By Order of Subgroup Product:
:$\order {S_1 S_2} = \dfrac {2^3 2^3} {2^r}$
where:
:$\order {S_1 S_2}$ denotes the order of $S_1 S_2$
:$\order {S_1 \cap S_2} = 2^r$
As $S_1 S_2 \subseteq G$ and $\order G = 24$:
:$\order {S_1 S_2} \le 24$
Hence:
As $S_1 \cap S_2$ is a proper subgroup of $S_1$, it can have no more than $2^2 = 4$ elements.
So if $G$ has $3$ Sylow $2$-subgroups, the intersection of any $2$ of them is of order $4$.
Let $T = S_1 \cap S_2$, so that $\order T = 4$.
We have that:
:$\index T {S_1} = 2$
where $\index T {S_1}$ denotes the index of $T$ in $S_1$.
From Subgroup of Index 2 is Normal, $T$ is normal in $S_1$.
Similarly, $T$ is normal in $S_2$.
Thus $S_1$ and $S_2$ are both subgroups of the normalizer $\map {N_G} T$ of $T$ in $G$.
Thus $H = \gen {S_1, S_2}$ is a subgroup of $\map {N_G} T$.
Hence $T$ is a normal subgroup of $H$.
As $H$ is a subgroup of $\map {N_G} T$, it contains $S_1 S_2$ as a subset.
But $S_1 S_2$ contains $\dfrac {2^6} {2^2} = 16$ elements.
The only subgroup of $G$ containing $16$ elements has to be $G$ itself.
So $H = G$ and so $\map {N_G} T = G$.
That is, $T$ is a normal subgroup of $G$.
This is therefore the required normal subgroup of order $4$.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(6)$ Groups of order $24$: Proposition $12.7$
Category:Groups of Order 24
Normal Subgroup of Group of Order 24/Mistake

== Source Work ==
:
:Chapter $12$: Applications of Sylow Theory: $(6)$ Groups of order $24$:
::Proposition $12.7$
== Mistake ==
:Thus $S_1$ and $S_2$ are both subgroups of $\map {N_G} T$, so $H = \gen {S_1, S_2}$ is a subgroup of $\map {N_T} G$ and hence $T$ is a normal subgroup of $H$.
== Correction ==
The notation:
:$\map {N_G} T$
denotes the normalizer of $T$ in $G$: the largest subgroup of $G$ in which $T$ is a normal subgroup.
Hence the notation:
:$\map {N_T} G$
makes no sense in this context, and it should be another instance of $\map {N_G} T$.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(6)$ Groups of order $24$: Proposition $12.7$
Category:Mistakes/A Course in Group Theory
Normal Subgroup of Order 25 in Group of Order 100

== Theorem ==
Let $G$ be a group of order $100$.
Then $G$ has a normal subgroup of order $25$.
== Proof ==
Let $r$ be the number of Sylow $p$-subgroup of order $5^2 = 25$
The First Sylow Theorem guarantees existence, so $r \ge 1$.
From the Fourth Sylow Theorem:
:$r \equiv 1 \pmod 5$
That is:
:$r \in \set {1, 6, 11, 16, \ldots}$
From the Fifth Sylow Theorem:
:$r \divides \dfrac {100} {25} = 4$
from which it follows that $r = 1$.
From Sylow p-Subgroup is Unique iff Normal, this unique Sylow $5$-subgroup must be normal.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59$
Category:Groups of Order 100
Normal Subgroup of Subset Product of Subgroups

== 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:
: $N \lhd N H$
where:
: $\lhd$ denotes normal subgroup
: $N H$ denotes subset product.
== Proof ==
From Subset Product with Normal Subgroup is Subgroup:
:$N H = H N$ is a subgroup of $G$.
By definition of subset product all elements of $H N$ can be written in the form:
:$h n \in H N$
where $h \in H, n \in N$.
Let $h n \in H N$.
Let $n_1 \in N$.
From Inverse of Group Product:
:$\paren {h n} n_1 \paren {h n}^{-1} = h n n_1 n^{-1} h^{-1}$
We have that:
:$n n_1 n \in N$
:$h, h^{-1} \in G$.
Then, since $N$ is a normal subgroup of $G$:
:$\paren {h n} n_1 \paren {n^{-1} h^{-1} } \in N$
Thus:
: $N \lhd N H$
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \epsilon$
Category:Normal Subgroups
Category:Subset Products
Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group

== Theorem ==
Let $n \in \N$ be a natural number such that $n > 4$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ denote the alternating group on $n$ letters.
$A_n$ is the only proper non-trivial normal subgroup of $S_n$.
== Proof ==
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is seen to be normal in $S_n$.
It remains to be shown that $A_n$ is the only such normal subgroup of $S_n$.
 $N$ is a proper non-trivial normal subgroup of $S_n$ such that $N$ is a proper subset of $A_n$.
From Intersection with Normal Subgroup is Normal, we have that:
:$A_n \cap N$ is a normal subgroup of $A_n$.
As $N \subseteq A_n$, from Intersection with Subset is Subset‎ we have that:
:$A_n \cap N = N$
That is:
:$N$ is a normal subgroup of $A_n$.
But from Alternating Group is Simple except on 4 Letters, $A_n$ has no proper non-trivial normal subgroup.
So $N$ cannot be a normal subgroup of $A_n$.
Hence by Proof by Contradiction it follows that $N$ cannot be a normal subgroup of $S_n$.
So $S_n$ has no proper non-trivial normal subgroup apart from $A_n$.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 84 \beta$
* 
* 
Category:Alternating Groups
Category:Symmetric Groups
Category:Examples of Normal Subgroups
Normal Subgroup of p-Group of Order p is Subset of Center

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a $p$-group.
Let $N$ be a normal subgroup of $G$ of order $p$.
Then:
:$N \subseteq \map Z G$
where $\map Z G$ denotes the center of $G$.
== Proof ==
From Intersection of Normal Subgroup with Center in p-Group:
:$\order {N \cap \map Z G} > 1$
From Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $N$.
It follows from Lagrange's Theorem that:
:$\order {N \cap \map Z G} = p$
and so:
:$N \cap \map Z G = N$
But from Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $\map Z G$
That is:
:$N$ is a subgroup of $\map Z G$
and the result follows.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $21$
Category:Normal Subgroups
Category:Centers of Groups
Category:P-Groups
Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ and $K$ be normal subgroups of $G$.
Let $H \subseteq K$.
Then $H$ is a normal subgroup of $K$.
== Proof ==
Hence the result by definition of normal subgroup.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.15$
Category:Normal Subgroups
Normal Sylow p-Subgroups in Group of Order 12

== Theorem ==
Let $G$ be of order $12$.
Then $G$ has either:
:a normal Sylow $2$-subgroup
or:
:a normal Sylow $3$-subgroup.
== Proof ==
Note that a Sylow $2$-subgroup of $G$ is of order $4$.
From Sylow $3$-Subgroups in Group of Order 12, there are either $1$ or $4$ Sylow $3$-subgroups.
Suppose there is exactly $1$ Sylow $3$-subgroup $P$.
Then from Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
Suppose there are $4$ Sylow $3$-subgroups $P_1$, $P_2$, $P_3$ and $P_4$.
Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, 3, 4}, i \ne j$ is the trivial subgroup of $G$:
:$P_i \cap P_j = \set e$
Thus $G$ contains:
:The identity element $e$
:$8$ elements of order $3$, of which $2$ each are in $P_1$, $P_2$, $P_3$ and $P_4$
:$3$ more elements, which (along with $e$) must form the Sylow $2$-subgroup of order $4$.
This Sylow $2$-subgroup $Q$ must be unique.
Hence by Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal.
Hence the result.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(3)$ Groups of order $12$: Proposition $12.3$
Category:Groups of Order 12
Category:Sylow p-Subgroups