nyuuzyou/wikis huggingface dataset:
Normal Vectors Form Space around Simple Complex Contour

== Theorem ==
Let $C$ be a simple contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.
Let $t \in \openint a b$ such that $\gamma$ is complex-differentiable at $t$.
Then there exists $r, R \in \R_{>0}$ such that:
:for all $s \in \openint { t-R }{ t+R }$ and for all $\epsilon \in \openint 0 r$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$, and $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$
where $\Img C$ denotes the image of $C$.
== Proof ==
Suppose there exists no $r, R \in \R_{>0}$ such that for all $s \in \openint { t-R }{ t+R }$ and for all $\epsilon \in \openint 0 r$, we have $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$.
It follows that for all $n \in \N$, there exists $t_n \in \openint { t - \dfrac 1 n }{ t + \dfrac 1 n }$ and $\epsilon \in \openint 0 {\dfrac 1 n}$ with $\tilde t_n \in \openint a b$ such that:
:$ \map \gamma {t_n} + \epsilon_n i \map {\gamma'}{t_n} = \map \gamma {\tilde t_n}$
Using the Bolzano-Weierstrass Theorem, we find a convergent subsquence $\sequence { \tilde t_{n_m} }_{m=1}^\infty$ with $\ds \lim_{m \mathop \to \infty} \tilde t_{n_m} = t_0 \in \closedint a b$.
Then:
As $C$ is a simple contour, it follows that $t_0 = t$.
Then:
Set $c := \ds \lim_{m \mathop \to \infty} \dfrac { \epsilon_{n_m} }{ \Delta t_{n_m} }$.
As $\gamma$ is differentiable at $t$, it follows that $c$ is defined as a limit.
As $\epsilon_{n_m}, \Delta t_{n_m} \in \R$ for all $m \in \N$, it follows that $c \in \R$.
As $\map {\gamma'} t \ne 0$ by definition of smooth path, it follows that $c \ne 0$.
From the equations above, we have:
As $1$ is wholly real, while $c i$ is wholly imaginary, this is a contradiction.
It follows that there exists $r_0, R_0 \in \R_{>0}$ such that:
:for all $s \in \openint { t - R_0 }{ t + R_0 }$ and for all $\epsilon \in \openint 0 {r_0}$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$
A similar argument shows that there exists $r_1, R_1 \in \R_{>0}$ such that:
:for all $s \in \openint { t - R_1 }{ t + R_1 }$ and for all $\epsilon \in \openint 0 {r_1}$: $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$
To prove the claim of the theorem, set $r = \map \min {r_0, r_1}$, and $R = \map \min {R_0, R_1}$.
Category:Orientation of Complex Contour
Normal p-Subgroup contained in All Sylow p-Subgroups

== Theorem ==
Let $G$ be a finite group.
Let $p$ be a prime number.
Let $H$ be a normal subgroup of $G$ which is a $p$-group.
Then $H$ is a subset of every Sylow $p$-subgroup of $G$.
== Proof ==
Let $P$ be a Sylow $p$-subgroup of $G$.
By Second Sylow Theorem, $H$ is a subset of a conjugate of $P$.
Then:
:$\exists g \in G: H \subseteq g P g^{-1}$.
This implies:
Since $P$ is arbitrary, $H$ is a subset of every Sylow $p$-subgroup of $G$.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \gamma$
Category:P-Groups
Category:Sylow p-Subgroups
Normal to Circle passes through Center

== Theorem ==
A normal $\NN$ to a circle $\CC$ passes through the center of $\CC$.
== Proof ==
Let $\CC$ be positioned in a Cartesian plane with its center at the origin.
Let $\NN$ pass through the point $\tuple {x_1, y_1}$.
From Equation of Normal to Circle Centered at Origin, $\NN$ has the equation:
:$y_1 x - x_1 y = 0$
or:
:$y = \dfrac {y_1} {x_1} x$
From the Equation of Straight Line in Plane: Slope-Intercept Form, this is the equation of a straight line passing through the origin.
As the geometry of a circle is unchanged by a change of coordinate axes, the result follows for a general circle in whatever frame.
== Sources ==
* : Chapter $\text {III}$. The Circle: $3$. The normal
Category:Circles
Category:Normals to Curves
Normal to Curve is Tangent to Evolute

== Theorem ==
Let $C$ be a curve defined by a real function which is twice differentiable.
Let the curvature of $C$ be non-constant.
Let $P$ be a point on $C$.
Let $Q$ be the center of curvature of $C$ at $P$.
The normal to $C$ at $P$ is tangent to the evolute $E$ of $C$ at $Q$.
== Proof ==
:
Let $P = \tuple {x, y}$ be a general point on $C$.
Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$.
From the above diagram:
:$(1): \quad \begin{cases} x - X = \pm \rho \sin \psi \\ Y - y = \pm \rho \cos \psi \end{cases}$
where:
:$\rho$ is the radius of curvature of $C$ at $P$
:$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis.
Whether the sign is plus or minus depends on whether the curve is convex or concave.
For simplicity, let it be assumed that the curvature $k$ at each point under consideration on $C$ is positive.
The case for $k < 0$ can then be treated similarly.
Thus we have $k > 0$ and so $(1)$ can be written:
:$(2): \quad \begin{cases} X = x - \rho \sin \psi \\ Y = y +\rho \cos \psi \end{cases}$
By definition of curvature:
:$k = \dfrac {\d \psi} {\d s}$
and:
:$\rho = \dfrac 1 k = \dfrac {\d s} {\d \psi}$
Hence:
and:
Differentiating $(2)$ $\psi$:
and:
By assumption, $\dfrac {\d \rho} {\d \psi} \ne 0$ on $C$.
Hence we have:
We have Slope of Normal is Minus Reciprocal of Tangent.
Thus the slope of the tangent to $E$ equals the slope of the normal to $C$.
The result follows.
Note the case when $\dfrac {\d \rho} {\d \psi} = 0$ on $C$.
In this case $C$ is a circle.
By Evolute of Circle is its Center its evolute is a single point and so has no tangent.
== Sources ==
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid
Category:Normals to Curves
Category:Tangents
Category:Evolutes
Normal to Cycloid passes through Bottom of Generating Circle

== Theorem ==
Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the normal to $C$ at a point $P$ on $C$ passes through the bottom of the generating circle of $C$.
== Proof 1 ==
== Proof 2 ==
Category:Normal to Cycloid passes through Bottom of Generating Circle
Category:Cycloids
Category:Normals to Curves
Normal to Cycloid passes through Bottom of Generating Circle/Proof 1

== Theorem ==
== Proof ==
From Normal to Cycloid, the equation for the normal to $C$ at a point $P = \tuple {x, y}$ is given by:
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $3$
Category:Normal to Cycloid passes through Bottom of Generating Circle
Normal to Cycloid passes through Bottom of Generating Circle/Proof 2

== Theorem ==
== Proof ==
From Tangent to Cycloid passes through Top of Generating Circle, the tangent to $C$ at a point $P = \tuple {x, y}$ passes through the top of the generating circle.
By definition, the normal to $C$ at $P$ is perpendicular to the tangent to $C$ at $P$.
From Thales' Theorem, the normal, the tangent and the diameter of the generating circle form a right triangle.
Thus the normal to $C$ at $P$ meets the generating circle at the opposite end of the diameter to the tangent.
Hence the result.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $3$
Category:Normal to Cycloid passes through Bottom of Generating Circle
Normality Relation is not Transitive

== Theorem ==
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $K$ be a normal subgroup of $N$.
Then it is not necessarily the case that $K$ is a normal subgroup of $G$.
== Proof 1 ==
== Proof 2 ==
Category:Normal Subgroups
Category:Normality Relation is not Transitive
Normality Relation is not Transitive/Proof 1

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
From Normality of Subgroups of Alternating Group on 4 Letters:
:$K := \set {e, t, u, v}$ is a normal subgroup of $A_4$
:$T := \set {e, t}$ is not a normal subgroup of $A_4$.
But by Subgroup of Abelian Group is Normal:
:$T$ is a normal subgroup of $K$.
Thus we have:
:$T \lhd K$, $K \lhd A_4$
but:
:$T \not \lhd A_4$
== Sources ==
* : Chapter $6$: Cosets: Exercise $11$
* : Chapter $2$: The Symmetric Groups: $\S 83 \alpha$
Category:Normality Relation is not Transitive
Category:Alternating Group on 4 Letters
Normality Relation is not Transitive/Proof 2

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $D_4$ denote the dihedral group $D_4$.
Let $D_4$ be presented in matrix representation:
Its Cayley table is given as:
Consider the subgroup $H$ whose underlying set is:
:$H = \set {\mathbf I, \mathbf A, -\mathbf I, -\mathbf A}$
From Subgroup of Index 2 is Normal, $H$ is normal in $D_4$.
Consider the subgroup $H$ whose underlying set is:
:$K = \set {\mathbf I, \mathbf A} = \gen {\mathbf A}$
From Subgroup of Index 2 is Normal, $K$ is normal in $H$.
It remains to be demonstrated that $K$ is not normal in $D_4$.
From the Cayley table:
:$\mathbf C \mathbf A \mathbf C^{-1} = \mathbf B \mathbf C = -\mathbf A \notin K$
Hence $K$ is not normal in $D_4$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.6$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $10$
* : Chapter $7$: Normal subgroups and quotient groups: Example $7.7$
Category:Normality Relation is not Transitive
Category:Dihedral Group D4
Normalized URM Program

== Theorem ==
Let $P$ be a URM program.
Let $l = \map \lambda P$ be the number of basic instructions in $P$.
Let $u = \map \rho P$ be the number of registers used by $P$.
Then $P$ can be modified as follows:
:Every Jump of the form $\map J {m, n, q}$ where $q > l$ may be replaced by $\map J {m, n, l + 1}$
:If $u > 0$, a Clear Registers Program $\map Z {2, u}$ can be appended to the end of $P$ at lines $l + 1$ to $l + u - 1$.
The new program $P'$ that results from the above modifications produces exactly the same output as $P$ for each input.
Note now though that $\map \lambda {P'} = l + u - 1$.
Such a program as $P'$ is called a normalized URM program.
The point of doing this is so as to make programs easier to concatenate. Once the above have been done, each program has a well-defined exit line which can be used as the start line of the program that immediately follows it.
== Proof ==
Each Jump of the form $\map J {m, n, q}$ where $q > l$ leads to a line which does not contain an instruction.
The line $\map J {m, n, l + 1}$ likewise contains no instructions, by definition.
Therefore, when jumping to $\map J {m, n, l + 1}$ the program behaves in exactly the same way: that is, it stops when the instruction $\map J {m, n, l + 1}$ causes the program to jump to line $l + 1$.
After the URM program has terminated, its output sits in $R_1$ by convention.
Once the URM program reaches line $l + 1$ it has by definition terminated, and because of the modifications to the Jumps as defined above, there is no other way that it can terminate.
Any further instructions that are added to a URM program that are placed at line $l + 1$ and those following will therefore be executed in order (as long as none of them are Jumps) and the program will then really terminate.
By adding the Clear Registers Program $\map Z {2, u}$ to the end of $P$, the only effect this will have on the operation of the program is to clear all the registers to $0$ except the one the output is in.
Category:URM Programs
Normalizer is Subgroup

== Theorem ==
Let $G$ be a group.
The normalizer of a subset $S \subseteq G$ is a subgroup of $G$.
:$S \subseteq G \implies \map {N_G} S \le G$
== Proof ==
Let $a, b \in \map {N_G} S$.
Then:
Therefore $a b \in \map {N_G} S$.
Now let $a \in \map {N_G} S$:
:$a \in \map {N_G} S \implies S^{a^{-1} } = \paren {S^a}^{a^{-1} } = S^{a^{-1} a} = S$
Therefore $a^{-1} \in \map {N_G} S$.
Thus, by the Two-Step Subgroup Test, $\map {N_G} S \le G$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.20$
* : $\S 1.9$: Exercise $5.15$
* : Chapter $2$: Subgroups and Cosets: $\S 35 \gamma$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$
Category:Normalizers
Normalizer of Center is Group

== Theorem ==
Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Let $x \in G$.
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Then:
:$\map Z G = \set {x \in G: \map {N_G} x = G}$
That is, the center of a group $G$ is the set of elements $x$ of $G$ such that the normalizer of $x$ is the whole of $G$.
Thus:
:$x \in \map Z G \iff \map {N_G} x = G$
and so:
:$\index G {\map {N_G} x} = 1$
where $\index G {\map {N_G} x}$ is the index of $\map {N_G} x$ in $G$.
== Proof ==
$\map {N_G} x$ is the normalizer of the set $\set x$.
Thus:
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 50$
Category:Normalizers
Category:Centers of Groups
Normalizer of Conjugate is Conjugate of Normalizer

== Theorem ==
Let $G$ be a group.
Let $a \mathop \in G$.
Let $S$ be a subset of $G$.
Let $S^a$ denote the conjugate of $S$ by $a$.
Let $\map {N_G} S$ denote the normalizer of $S$ in $G$.
Then:
:$\map {N_G} {S^a} = \paren {\map {N_G} S}^a$
That is, the normalizer of a conjugate is the conjugate of the normalizer:
== Proof ==
From the definition of conjugate:
:$S^a = \set {y \in G: \exists x \in S: y = a x a^{-1} } = a S a^{-1}$
From the definition of normalizer:
:$\map {N_G} S = \set {x \in G: S^x = S}$
Thus:
Suppose that $x \in \map {N_G} S$.
It is to be shown that:
:$a x a^{-1} \in \map {N_G} {S^a} = \map {N_G} {a S a^{-1} }$
To this end, compute:
Hence $a x a^{-1} \in \map {N_G} {S^a}$, and it follows that:
:$z \in \paren {\map {N_G} S}^a \implies z \in \map {N_G} {S^a}$
Conversely, let $x \in \map {N_G} {S^a}$.
That is, let $x \in G$ such that $x a S a^{-1} x^{-1} = a S a^{-1}$.
Now if we can show that $a^{-1} x a \in \map {N_G} S$, then:
:$x = a \left({a^{-1} x a}\right) a^{-1} \in \paren {\map {N_G} S}^a$
establishing the remaining inclusion.
Thus, we compute:
Combined with the above observation, this establishes that:
:$z \in \map {N_G} {S^a} \implies z \in \paren {\map {N_G} S}^a$
Hence the result, by definition of set equality.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48 \gamma$
Category:Normalizers
Category:Conjugacy
Normalizer of Reflection in Dihedral Group

== Theorem ==
Let $n \in \N$ be a natural number such that $n \ge 3$.
Let $D_n$ be the dihedral group of order $2 n$, given by:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Let $\map {N_{D_n} } {\set \beta}$ denote the normalizer of the singleton containing the reflection element $\beta$.
Then:
:$\map {N_{D_n} } {\set \beta} = \begin{cases} \set {e, \beta} & : n \text { odd} \\ \set {e, \beta, \alpha^{n / 2}, \alpha^{n / 2} \beta} & : n \text { even} \end{cases}$
== Proof ==
By definition, the normalizer of $\set \beta$ is:
:$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \set \beta g^{-1} = \set \beta}$
That is:
:$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \beta = \beta g}$
First let $g = \beta^k$ for $k \in \set {0, 1}$.
Then:
:$\beta \beta^k = \beta^k \beta$
Thus:
:$\forall k \in \set {0, 1}: \beta^k \in \map {N_{D_n} } {\set \beta}$
Now let $g = \alpha^j \beta^k$ for $0 < j < n, k \in \set {0, 1}$.
Suppose $g \alpha = \alpha g$.
Then:
Thus when $n$ is odd, there is no such $j$ such that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.
But when $n$ is even such that $n = 2 j$, we have that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.
Hence the result.
Category:Dihedral Groups
Category:Normalizers
Normalizer of Rotation in Dihedral Group

== Theorem ==
Let $n \in \N$ be a natural number such that $n \ge 3$.
Let $D_n$ be the dihedral group of order $2 n$, given by:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Let $\map {N_{D_n} } {\set \alpha}$ denote the normalizer of the singleton containing the rotation element $\alpha$.
Then:
:$\map {N_{D_n} } {\set \alpha} = \gen \alpha$
where $\gen \alpha$ is the subgroup generated by $\alpha$.
== Proof ==
By definition, the normalizer of $\set \alpha$ is:
:$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \set \alpha g^{-1} = \set \alpha}$
That is:
:$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \alpha = \alpha g}$
First let $g = \alpha^k$ for some $k \in \Z$.
Then:
:$\alpha \alpha^k = \alpha^k \alpha$
which includes $k = 0$, that is $e$.
Thus:
:$\forall k \in \Z: \alpha^k \in \map {N_{D_n} } {\set \alpha}$
Now let $g = \alpha^k \beta$.
Suppose $g \alpha = \alpha g$.
Then:
But $\beta \alpha \ne \alpha \beta$ in $D_n$ except for $n < 3$.
Hence the result:
:$\map {N_{D_n} } {\set \alpha} = \set {e, \alpha, \alpha^2, \ldots, \alpha^{n - 1} }$
Hence the result, by definition of generator of subgroup.
Category:Dihedral Groups
Category:Normalizers
Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then $\map {N_G} H$, the normalizer of $H$ in $G$, is the largest subgroup of $G$ containing $H$ as a normal subgroup.
== Proof ==
From Subgroup is Subgroup of Normalizer, we have that $H \le \map {N_G} H$.
Now we need to show that $H \lhd \map {N_G} H$.
For $a \in \map {N_G} H$, the conjugate of $H$ by $a$ in $\map {N_G} H$ is:
so:
:$\forall a \in \map {N_G} H: H^a = H$
and so by definition of normal subgroup:
:$H \lhd \map {N_G} H$
Now we need to show that $\map {N_G} H$ is the largest subgroup of $G$ containing $H$ such that $H \lhd \map {N_G} H$.
That is, to show that any subgroup of $G$ in which $H$ is normal is also a subset of $\map {N_G} H$.
Take any $N$ such that $H \lhd N \le G$.
In $N$, the conjugate of $H$ by $a \in N$ is $N \cap H^a = H$.
Therefore:
:$H \subseteq H^a$
Similarly, $H \subseteq H^{a^{-1} }$, so:
:$H^a \subseteq \paren {H^a}^{a^{-1} } = H$
Thus:
:$\forall a \in N: H^a = H, a \in \map {N_G} H$
That is:
:$N \subseteq \map {N_G} H$
So what we have shown is that any subgroup of $G$ in which $H$ is normal is a subset of $\map {N_G} H$, which is another way of saying that $\map {N_G} H$ is the largest such subgroup.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.20$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$
Category:Normalizers
Normalizer of Subgroup of Symmetric Group that Fixes n

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $H$ denote the subgroup of $S_n$ which consists of all $\pi \in S_n$ such that:
:$\map \pi n = n$
The normalizer of $H$ is given by:
:$\map {N_{S_n} } H = \map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$
== Proof ==
We have from Subgroup of Symmetric Group that Fixes n that $N = S_{n - 1}$.
By definition of normalizer:
:$\map {N_{S_n} } {S_{n - 1} } := \set {\rho \in S_n: \rho S_{n - 1} \rho^{-1} = S_{n - 1} }$
We have from Group is Normal in Itself that:
:$\forall \rho \in S_{n - 1}: \rho S_{n - 1} \rho^{-1} \in S_{n - 1}$
and so:
:$S_{n - 1} \subseteq \map {N_{S_n} } {S_{n - 1} }$
It remains to be shown that $\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$.
This will be done by demonstrating that:
: $S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$
where $\setminus$ denotes set difference.
Let $\rho \in S_n$ such that $\rho \notin S_{n - 1}$.
Thus:
:$\map \rho n \ne n$
and so:
:$\exists a \in S_n, a \ne n: \map \rho a = n$
for some $a \in S_{n - 1}$.
Then:
:$\map {\rho^{-1} } n = a$
Let $\pi \in S_{n - 1}$ such that:
: $\map \pi a = b$
for some $b \ne a$.
As $\rho$ is a permutation, $\rho$ is by definition a bijection.
Hence:
:$\map \rho b \ne n$
We have:
Thus $\rho \pi \rho^{-1}$ does not fix $n$.
That is:
:$\rho \pi \rho^{-1} \notin S_{n - 1}$
Since this is the case for all arbitrary $\rho \in S_n \setminus S_{n - 1}$, it follows that:
:$S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$
So from Intersection with Complement is Empty iff Subset:
:$\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$
and so by definition of set equality:
:$\map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48 \beta$
Category:Symmetric Groups
Category:Normalizers
Normalizer of Sylow p-Subgroup

== Theorem ==
Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $\map {N_G} P$ be the normalizer of $P$.
Then any $p$-subgroup of $\map {N_G} P$ is contained in $P$.
In particular, $P$ is the unique Sylow $p$-subgroup of $\map {N_G} P$.
== Proof ==
Let $Q$ be a $p$-subgroup of $N = \map {N_G} P$.
Let $\order Q = p^m, \order P = p^n$.
By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$P \lhd \map {N_G} P$
thus by Subset Product with Normal Subgroup as Generator:
:$\gen {P, Q} = P Q$
Thus by Order of Subgroup Product:
:$P Q \le G: \order {P Q} = p^{n + m - s}$
where $\order {P \cap Q} = p^s$.
Since $n$ is the highest power of $p$ dividing $\order G$, this is possible only when $m \le s$.
Since $P \cap Q \le Q, s \le m$ thus we conclude that $m = s$ and therefore $P \cap Q = Q$.
Thus from Intersection with Subset is Subset‎:
:$Q \subseteq P$
In particular, if $Q$ is a Sylow $p$-subgroup of $\map {N_G} P$, then $Q = P$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Proposition $11.9$
Category:P-Groups
Category:Subgroups
Category:Normalizers
Category:Sylow p-Subgroups
Normed Algebra Embeds into Unitization as Closed Ideal

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$ that is not unital as an algebra. 
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.
Let:
:$A_0 = \set {\tuple {x, 0} : x \in A} \subseteq A_+$
Then $A_0$ is a closed ideal of $A$.
== Proof ==
From Algebra over Field Embeds into Unitization as Ideal, $A_0$ is an ideal of $A$.
From Norm on Unitization of Normed Algebra is Equivalent to Direct Product Norm, $\norm {\, \cdot \,}_{A_+}$ is equivalent to the direct product norm $\norm {\, \cdot \,}_{A \times \GF}$.
Let $\sequence {\tuple {x_n, 0} }_{n \in \N}$ be a convergent sequence in $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ with limit $\tuple {x, \lambda} \in A_+$.
From Convergence in Direct Product Norm and Convergent Sequences in Vector Spaces with Equivalent Norms Coincide, we have $x_n \to x$ and $\lambda = 0$.
So every convergent sequence in $A_0$ has a limit in $A_0$.
So by the definition of a closed set in a normed vector space, $A_0$ is closed.
Category:Unitizations of Normed Algebras
Normed Division Algebra is Unitary Division Algebra

== Theorem ==
Let $A = \struct {A_F, \oplus}$ be a normed divison algebra over a field $F$.
Let the unit of $A$ be $1_A$, and the zero of $A$ be $0_A$.
Then $A$ is a unitary division algebra.
Also:
:$\norm {1_A} = 1$
where $\norm {1_A}$ denotes the norm of $1_A$.
== Proof ==
Let $A = \struct {A_F, \oplus}$ be a normed divison algebra as defined .
The fact that $A$ is a unitary algebra is a consequence of the definition of normed divison algebra.
From the definition of a norm, we have that:
:$\forall a \in A: \norm a = 0 \iff a = 0_A$
So, let $a, b \in A \setminus \set {0_A}$.
We have:
Thus for any arbitrary $a, b \ne 0_A$ we have shown that $a \oplus b \ne 0_A$.
Thus $A$ is a division algebra.
Next:
demonstrating that $\norm {1_A} = 1$.
== Sources ==
* : The Octonions (2002): 1.1 Preliminaries
Category:Normed Division Algebras
Category:Unitary Division Algebras