nyuuzyou/wikis huggingface dataset:
Center of Conic is Center of Symmetry

== Theorem ==
Let $\KK$ be a central conic.
Let $C$ be the center of $\KK$.
Then $C$ is a center of symmetry for $\KK$.
== Proof ==
First we note from Parabola has No Center that we do not need to consider the parabola.
Then we note from Circle is Ellipse with Equal Major and Minor Axes that a circle is a special case of the ellipse.
Hence there is no need to investigate the circle separately.
It remains to demonstrate the result for the ellipse and the hyperbola.
Let the central conic be expressed in reduced form.
By definition of reduced form, the center of $\KK$ lies at the origin.
From Equation of Ellipse in Reduced Form and Equation of Hyperbola in Reduced Form, $\KK$ can be expressed as:
:$\dfrac {x^2} {a^2} \pm \dfrac {y^2} {b^2} = 1$
Let $\tuple {p, q}$ be a point on $\KK$.
Then:
and so $\tuple {-p, -q}$ is also a point on $\KK$.
Hence the result by definition of center of symmetry.
Category:Centers of Conic Sections
Center of 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 $\map Z {D_n}$ denote the center of $D_n$.
Then:
:$\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, \alpha^{n / 2} } & : n \text { even} \end{cases}$
== Proof ==
By definition, the center of $D_n$ is:
:$\map Z {D_n} = \set {g \in D_n: g x = x g, \forall x \in D_n}$
For $n \le 2$ we have that $\order {D_n} \le 4$ and so by Group of Order less than 6 is Abelian $D_n$ is abelian for $n < 3$.
Hence by definition of abelian group:
:$\map Z {D_n} = D_n$
for $n < 3$.
So, let $n \ge 3$.
By Group Presentation of Dihedral Group:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
From Product of Generating Elements of Dihedral Group:
:$\beta \alpha^k = \alpha^{n - k} \beta$
for all $k \in \Z_{\ge 0}$.
We have that $D_n$ is generated by $\alpha$ and $\beta$.
Thus:
:$x \in \map Z {D_n} \iff x \alpha = \alpha x \land x \beta = \beta x$
Let $x \in \map Z {D_n}$.
We have that $x$ can be expressed in the form:
:$x = \alpha^i \beta^j$
As $x \in \map Z {D_n}$, we have:
But for $j = 1$ this means:
But the order of $\alpha$ is $n$ and $n > 2$, and hence:
:$\alpha^2 \ne e$
So if $x \in \map Z {D_n}$ it follows that $x$ has to be in the form:
:$x = \alpha^i$
for some $i \in \Z_{\ge 0}$.
Again, as $x \in \map Z {D_n}$, we have:
So either $i = 0$ or $n = 2 i$, as $0 \le i \le n$.
If $i = 0$ then $x = \alpha^0 = e$.
If $2 i = n$ then $n$ is even and so:
:$x = \alpha^{n / 2}$
Hence the result.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 50 \beta$
Category:Dihedral Groups
Category:Centers of Groups
Center of Division Ring is Subfield

== Theorem ==
Let $\struct {K, +, \circ}$ be an division ring.
Let $\map Z K$ be the center of $K$.
Then $\map Z K$ is a subfield of $K$.
== Proof ==
For $\map Z K$ to be a subfield of $K$, it needs to be a division ring that is commutative.
Thus the result follows directly from Center of Ring is Commutative Subring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.9$
Category:Division Rings
Category:Subfields
Center of Gravity equals Center of Mass if it exists

== Theorem ==
Let $B$ be a body in a gravitational field $\mathbf G$.
Let $B$ have a center of gravity $P$.
Then the center of mass of $B$ is also $P$.
== Proof ==
Let $Q$ denote the center of mass of $B$
First suppose that $\mathbf G$ is uniform.
From Center of Gravity in Uniform Gravitational Field is Center of Mass:
:$P = Q$
Now suppose that $\mathbf G$ is non-uniform.
There are two possibilities:
:$(1): \quad$ $B$ is barycentric
:$(2): \quad$ $B$ is not barycentric.
If $(1)$, then from Center of Gravity of Barycentric Body is Center of Mass:
:$P = Q$
If $(2)$, then from Center of Gravity in Non-Uniform Gravitational Field, the forces on $B$ induced by $\mathbf G$ consist of:
:a single force $\mathbf F$
:a couple $C$ whose plane is perpendicular to the line of action of $\mathbf F$.
The line of action of $\mathbf F$ does not necessarily pass through some fixed point as $B$ rotates in $\mathbf G$.
Hence $B$ has no center of gravity.
All cases are covered, and the result follows.
== Sources ==
* 
* 
Category:Centers of Mass
Category:Centers of Gravity
Center of Gravity in Non-Uniform Gravitational Field

== Theorem ==
Let $B$ be a body in a gravitational field $\mathbf G$ which is non-uniform.
The forces on $B$ are reducible to:
:$(1): \quad$ a single force $\mathbf F$
:$(2): \quad$ a couple $C$ whose plane is perpendicular to the line of action of $\mathbf F$.
The line of action of $\mathbf F$ does not pass through some fixed point as $B$ rotates in $\mathbf G$ unless $C$ is zero.
== Proof ==
== Sources ==
* 
* 
Category:Centers of Gravity
Center of Gravity of Barycentric Body is Center of Mass

== Theorem ==
Let $B$ be a body in a gravitational field $\mathbf G$ which is non-uniform.
Let $B$ be barycentric.
Then the couple on $B$ induced by $\mathbf G$ is reduced to zero
As a consequence, the center of gravity of $B$ is coincident with the center of mass of $B$.
== Proof ==
== Sources ==
* 
* 
Category:Centers of Gravity
Center of Gravity of Cycloid/Historical Note

== Historical Note on Center of Gravity of Cycloid ==
The question of the location of the Center of Gravity of Cycloid was posed by in $1658$.
Some sources suggest that it was solved by , but this is not easy to corroborate.
== Sources ==
* : Chapter $\text{V}$: "Greatness and Misery of Man"
Category:Historical Notes
Center of Group is Abelian Subgroup

== Theorem ==
Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Then $\map Z G$ is an abelian subgroup of $G$.
== Proof ==
By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.
The definition of the center $\map Z G$ grants that all elements of $\map Z G)$ commute with all elements of $G$.
In particular, all elements of $\map Z G$ commute with all elements of $\map Z G$ as $\map Z G \subseteq G$.
Therefore $\map Z G$ is abelian.
== Sources ==
* : $\S 1.9$: Exercise $5.14$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(vii)}$
Category:Subgroups
Category:Abelian Groups
Category:Centers of Groups
Center of Group is Normal Subgroup

== Theorem ==
Let $G$ be a group
The center $\map Z G$ of $G$ is a normal subgroup of $G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $10$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.11 \ \text{(a)}$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \theta$
* 
* : Chapter $7$: Normal subgroups and quotient groups: Exercise $5$
Category:Normal Subgroups
Category:Centers of Groups
Category:Center of Group is Normal Subgroup
Center of Group is Normal Subgroup/Proof 1

== Theorem ==
== Proof ==
Recall that Center of Group is Abelian Subgroup.
Since $g x = x g$ for each $g \in G$ and $x \in \map Z G$:
:$g \map Z G = \map Z G g$
Thus:
:$\map Z G \lhd G$
Category:Center of Group is Normal Subgroup
Center of Group is Normal Subgroup/Proof 2

== Theorem ==
== Proof ==
We have:
:$\forall a \in G: x \in \map Z G^a \iff a x a^{-1} = x a a^{-1} = x \in \map Z G$
Therefore:
:$\forall a \in G: \map Z G^a = \map Z G$
and $\map Z G$ is a normal subgroup of $G$.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 50$
Category:Center of Group is Normal Subgroup
Center of Group is Subgroup

== Theorem ==
Let $G$ be a group.
The center $\map Z G$ of $G$ is a subgroup of $G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \delta$
* : $1$: Introduction to Finite Group Theory: $1.13$
* : $\S 37.2$ Some important general examples of subgroups
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(vii)}$
Category:Subgroups
Category:Direct Proofs
Category:Centers of Groups
Category:Center of Group is Subgroup
Center of Group is Subgroup/Proof 1

== Theorem ==
== Proof ==
For brevity, suppress the symbol for the group operation (which may be $\circ$, or $+$).
Apply the Two-Step Subgroup Test:
=== Condition $(1)$ ===
By the definition of identity, $e g = g e = g$ for all $g \in G$.
So, $e \in \map Z G$, meaning $\map Z G$ is non-empty.
=== Condition $(2)$ ===
Suppose $a, b \in \map Z G$.
Using the associative property and the definition of center, we have:
Thus, $a b \in \map Z G$.
=== Condition $(3)$ ===
Suppose $c \in \map Z G$. Then:
Therefore:
: $\map Z G \le G$
== Sources ==
* : $\S 37.2$ Some important general examples of subgroups
Category:Center of Group is Subgroup
Center of Group is Subgroup/Proof 2

== Theorem ==
== Proof ==
We have the result Center is Intersection of Centralizers.
That is, $\map Z G$ is the intersection of all the centralizers of $G$.
All of these are subgroups of $G$ by Centralizer of Group Element is Subgroup.
Thus from Intersection of Subgroups is Subgroup, $\map Z G$ is also a subgroup of $G$.
== Sources ==
* : $\S 37.2$ Some important general examples of subgroups
Category:Center of Group is Subgroup
Center of Group of Order Prime Cubed

== Theorem ==
Let $G$ be a group of order $p^3$, where $p$ is a prime.
Let $\map Z G$ be the center of $G$.
Then $\order {\map Z G} \ne p^2$.
== Proof ==
 $\order {\map Z G} = p^2$.
Then $\order {G / \map Z G} = p$.
From Prime Group is Cyclic it follows that $G / \map Z G$ is cyclic.
From Quotient of Group by Center Cyclic implies Abelian it follows that $G$ is abelian.
by definition of abelian group it follows that $\order {\map Z G} = p^3$.
The result follows by Proof by Contradiction.
Category:Normal Subgroups
Category:Centers of Groups
Category:P-Groups
Center of Group of Prime Power Order is Non-Trivial

== Theorem ==
Let $G$ be a group whose order is the power of a prime.
Then the center of $G$ is non-trivial:
:$\forall G: \order G = p^r: p \in \mathbb P, r \in \N_{>0}: \map Z G \ne \set e$
== Proof ==
Suppose $G$ is abelian.
By definition of abelian group:
:$\map Z G = G$
and the result is seen to be true as $G$ is itself non-trivial.
From Prime Group is Cyclic and Cyclic Group is Abelian, this will always be the case for $r = 1$.
So, suppose $G$ is non-abelian.
Thus $\map Z G \ne G$ and therefore $G \setminus \map Z G \ne \O$.
Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the conjugacy classes into which $G \setminus \map Z G$ is partitioned.
From Conjugacy Class of Element of Center is Singleton, all of these will have more than one element.
From the Conjugacy Class Equation:
:$\ds \order {\map Z G} = \order G - \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
From Number of Conjugates is Number of Cosets of Centralizer:
:$\order {\conjclass {x_j} } \divides \order G$
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Then:
:$\forall j: 1 \le j \le m: \index G {\map {N_G} {x_j} } > 1 \implies p \divides \index G {\map {N_G} {x_j} }$
Since $p \divides \order G$, it follows that:
:$p \divides \order {\map Z G}$
and the result follows.
== Sources ==
* : $\S 1.9$: Exercise $5.17$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 52$ Lemma $1$
* : $\S 48.4$ Conjugacy
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.20$
Category:Group Theory
Category:P-Groups
Category:Centers of Groups
Center of Mass/Examples/Uniform Lamina

== Theorem ==
Let $\LL$ be a uniform lamina embedded in a cartesian plane in the shape of the area between the curve $\map f x$, the straight lines $x = a$ and $x = b$, and the $x$-axis.
Let the area of $\LL$ be $A$.
Then the coordinates $\tuple {\bar x, \bar y}$ of the center of mass of $B$ are given by:
== Proof ==
== Sources ==
* 
* 
Category:Centers of Mass
Center of Mass Operation is Associative

== Theorem ==
Let $S$ denote the set of massy particles in ordinary space.
Let $\circ$ denote the binary operation defined as:
:$\forall x, y \in S: x \circ y =$ the center of mass of $x$ and $y$
Then $\circ$ is an associative operation.
== Proof ==
== Sources ==
* 
* 
Category:Centers of Mass
Category:Examples of Associative Operations‎
Center of Mass in Barycentric Coordinates

== Theorem ==
Let $p_0, p_1, p_2, p_3$ be fixed non-coplanar points, such that $p_i = \tuple {x_i, y_1, z_i}$.
Let $P$ be a point in ordinary space expressed in barycentric coordinates $\set {p_0, p_1, p_2, p_3}$:
:$p = \lambda_0 p_0 + \lambda_1 p_1 + \lambda_2 p_2 + \lambda_3 p_3$
such that:
:$\lambda_0 + \lambda_1 + \lambda_2 + \lambda_3 = 0$
Let point masses of mass $\lambda_0, \lambda_1, \lambda_2, \lambda_3$ be placed at $p_0, p_1, p_2, p_3$ respectively.
Then $p$ is the center of mass of $\set {p_0, p_1, p_2, p_3}$.
== Proof ==
== Sources ==
* 
* 
Category:Centers of Mass
Category:Barycentric Coordinates
Center of Mass of System of Particles in Cartesian Plane

== Theorem ==
Let $B$ be a system of $n$ discrete particles embedded in a cartesian plane, each with:
:mass $m_i$
:position $\tuple {x_i, y_i}$
where $i \in \set {1, 2, \ldots, n}$.
Then the coordinates $\tuple {\bar x, \bar y}$ of the center of mass of $B$ are given by:
where:
:$\ds M = \sum_{i \mathop = 1}^n m_i$
== Proof ==
== Sources ==
* 
* 
Category:Centers of Mass
Center of Mass of Uniform Circular Arc

== Theorem ==
Let $\WW$ be a wire of uniform density.
Let $\WW$ be bent into the shape of the arc of a circle $\CC$ of radius $r$ subtending an angle of $2 \alpha$ from the center of $\CC$.
Then the center of mass of $\PP$ is the point $\dfrac {r \sin \alpha} \alpha$ from the center of $\CC$.
== Proof ==
== Sources ==
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
Category:Centers of Mass
Category:Circles
Category:Wires
Category:Uniform Density
Center of Mass of Uniform Circular Sector

== Theorem ==
Let $\WW$ be a uniform lamina in the shape of the sector of a circle $\CC$ of radius $r$ of an angle of $2 \alpha$.
Then the center of mass of $\PP$ is the point $\dfrac {2 r \sin \alpha} {3 \alpha}$ from the center of $\CC$.
== Proof ==
== Sources ==
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
Category:Centers of Mass
Category:Sectors of Circles
Category:Uniform Laminae
Center of Mass of Uniform Hemispherical Shell

== Theorem ==
Let $\HH$ be a uniform lamina in the shape of a hemisphere of radius $r$.
Then the center of mass of $\HH$ is the point $\dfrac r 2$ from the center of $\HH$ along the radius of $\HH$ perpendicular to the base of $\HH$.
== Proof ==
== Sources ==
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
Category:Centers of Mass
Category:Hemispheres
Category:Uniform Density
Center of Mass of Uniform Solid Cone

== Theorem ==
Let $\CC$ be a solid cone of uniform density.
Then the center of mass of $\CC$ is the point $\dfrac 3 4$ of the distance from the vertex of $\CC$ to the center of mass of the base of $\CC$.
== Proof ==
== Sources ==
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
Category:Centers of Mass
Category:Cones
Category:Uniform Density
Center of Mass of Uniform Solid Hemisphere

== Theorem ==
Let $\BB$ be a solid hemisphere of radius $r$ of uniform density.
Then the center of mass of $\BB$ is the point $\dfrac {3 r} 8$ from the center of $\BB$ along the radius of $\BB$ perpendicular to the base of $\BB$.
== Proof ==
== Sources ==
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
Category:Centers of Mass
Category:Hemispheres
Category:Uniform Density
Center of Mass of Uniform Solid Pyramid

== Theorem ==
Let $\PP$ be a solid pyramid of uniform density.
Then the center of mass of $\PP$ is the point $\dfrac 3 4$ of the distance from the vertex of $\PP$ to the center of mass of the base of $\PP$.
== Proof ==
== Sources ==
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
Category:Centers of Mass
Category:Pyramids
Category:Uniform Density
Center of Mass of Uniform Triangular Lamina

== Theorem ==
Let $\LL$ be a uniform lamina in the shape of a triangle.
Then the center of mass of $\LL$ is at the centroid of $\LL$.
That is, it lies $\dfrac 2 3$ of the way along an arbitrary median of $\LL$ from its vertex.
== Proof ==
== Sources ==
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
* : Appendix: Table $4$ Centres of mass The position of the centre of mass of certain uniform bodies.
Category:Uniform Laminae
Category:Centers of Mass
Category:Triangles
Center of Non-Abelian Group of Order pq is Trivial

== Theorem ==
Let $p$ and $q$ be distinct prime numbers.
Let $G$ be a non-abelian group of order $p q$ whose identity is $e$.
Then the center of $G$ is trivial:
:$\map Z G = \set e$
== Proof ==
From Center of Group is Normal Subgroup, $\map Z G$ is a normal subgroup of $G$.
By Lagrange's Theorem, the order of $\map Z G$ is either $1$, $p$, $q$ or $p q$.
Because $G$ is not abelian, $G \ne \map Z G$.
Hence $\order {\map Z G} \ne p q$.
From Quotient of Group by Center Cyclic implies Abelian:
:$G / \map Z G$ cannot be a cyclic group which is non-trivial.
Then we have:
:$\map C x \subset G$
where $\map C x$ is the centralizer of $x$.
Hence by Prime Group is Cyclic, $G / \map Z G$ cannot be cyclic of order $p$ or $q$.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $19 \ \text {(i)}$
Category:Centers of Groups
Category:Groups of Order p q
Center of Opposite Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be the opposite group to $G$.
Let $\map Z {G, \circ}$ and $\map Z {G, *}$ be the centers of $\struct {G, \circ}$ and $\struct {G, *}$, respectively.
Then:
:$\map Z {G, \circ} = \map Z {G, *}$
== Proof ==
We have, for $g \in G$:
Hence the result, by definition of set equality.
Category:Opposite Groups
Category:Centers of Groups
Center of Quaternion Group

== Theorem ==
Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the quaternion group.
Let $\map Z {\Dic 2}$ denote the center of $\Dic 2$.
Then:
:$\map Z {\Dic 2} = \set {e, a^2}$
== Proof ==
By definition, the center of $\Dic 2$ is:
:$\map Z {\Dic 2} = \set {g \in \Dic 2: g x = x g, \forall x \in \Dic 2}$
We are given that:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
We have that $\Dic 2$ is generated by $\alpha$ and $\beta$.
Thus:
:$x \in \map Z {\Dic 2} \iff x a = a x \land x b = b x$
Let $x \in \map Z {\Dic 2}$.
We have that $x$ can be expressed in the form:
:$x = a^i b^j$
for $i \in \set {0, 1, 2, 3}$ and $j \in \set {0, 1}$.
As $x \in \map Z {\Dic 2}$, we have:
For $j = 1$ this means:
But the order of $a$ is $4$, not $2$, and hence:
:$a^2 \ne e$
So if $x \in \map Z {\Dic 2}$ it follows that $x$ has to be in the form:
:$x = a^i$
for some $i \in \Z_{\ge 0}$.
Again, as $x \in \map Z {\Dic 2}$, we have:
So either $i = 0$ or $2 i = 4$, as $0 \le i \le 4$.
If $i = 0$ then $x = a^0 = e$.
If $2 i = 4$ then:
:$x = a^{4 / 2} = a^2$
Hence the result.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 50 \gamma$
Category:Quaternion Group
Category:Centers of Groups
Center of Ring is Commutative Subring

== Theorem ==
The center $\map Z R$ of a ring $R$ is a commutative subring of $R$.
== Proof ==
Follows directly from the definition of center and Centralizer of Ring Subset is Subring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.5$: Corollary
* : Chapter $9$: Rings: Exercise $2$
Category:Ring Theory
Category:Subrings
Center of Square is Axis of Rotational Symmetry of Order 4

== Theorem ==
Let $\SS$ be a square.
Let $\AA$ be a straight line passing through the center of $\SS$.
Then $\AA$ is an axis of rotational symmetry of $\SS$ of order $4$.
== Proof ==
== Sources ==
* 
* 
Category:Axes of Rotational Symmetry
Category:Order of Rotational Symmetry
Category:Squares
Center of Symmetric Group is Trivial

== Theorem ==
Let $n \in \N$ be a natural number.
Let $S_n$ denote the symmetric group of order $n$.
Let $n \ge 3$.
Then the center $\map Z {S_n}$ of $S_n$ is trivial.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Symmetric Group is not Abelian
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $11$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 50 \alpha$
* : Chapter $6$: An Introduction to Groups: Exercise $6 \ \text{(ii)}$
Category:Symmetric Groups
Category:Center of Symmetric Group is Trivial
Category:Centers of Groups
Center of Symmetric Group is Trivial/Proof 1

== Theorem ==
== Proof ==
From its definition, the identity (here denoted by $e$) of a group $G$ commutes with all elements of $G$.
So by definition of center:
:$e \in \map Z {S_n}$
By definition of center:
:$\map Z {S_n} = \set {\tau \in S_n: \forall \sigma \in S_n: \tau \sigma = \sigma \tau}$
Let $\pi, \rho \in S_n$ be permutations of $\N_n$.
Let us choose an arbitrary $\pi \in S_n: \pi \ne e, \map \pi i = j, i \ne j$.
Since $n \ge 3$, we can find $\rho \in S_n$ which interchanges $j$ and $k$ (where $k \ne i, j$) and fixes everything else.
It follows that $\rho^{-1}$ does the same thing, and in particular both $\rho$ and $\rho^{-1}$ fix $i$.
So:
So:
:$\map {\rho \pi \rho^{-1} } i = k \ne j = \map \pi i$
From Conjugate of Commuting Elements, if $\rho$ and $\pi$ were to commute, $\rho \pi \rho^{-1} = \pi$.
But they don't.
Whatever $\pi \in S_n$ is, you can always find a $\rho$ such that $\rho \pi \rho^{-1} \ne \pi$.
So no non-identity elements of $S_n$ commute with all elements of $S_n$.
Hence:
:$\map Z {S_n} = \set e$
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 79 \beta$
Category:Center of Symmetric Group is Trivial
Center of Symmetric Group is Trivial/Proof 2

== Theorem ==
== Proof ==
Let $\pi \in S_n$ such that $\pi \ne e$ be arbitrary.
Let $i, j \in \set {1, 2, \ldots, n}$ such that $\map \pi i = j \ne i$.
Let $m = \map \pi j$.
Then $m \ne j$.
Since $n \ge 3$, there exists $k \in \N$ such that $k \ne j, k \ne m$.
Let $\rho \in S_n$ interchange $j, k$ and fix everything else.
Then $\rho$ fixes $m$.
Then:
By definition of $\rho$:
:$k = \map \rho j$
So:
:$\map \pi k = \map {\pi \rho} j$
But because permutations are injective:
:$\map \pi j \ne \map \pi k$
Thus:
:$\map {\rho \pi} j \ne \map {\pi \rho} j$
That is, there exists a $\rho$ with which $\pi$ does not commute.
So $\pi \notin \map Z {S_n}$.
As $\pi \ne e$ is arbitrary, it follows that only $e$ is in $\map Z {S_n}$.
That is:
:$\map Z {S_n} = \set e$
Hence, by definition, $\map Z {S_n}$ is trivial.
Category:Center of Symmetric Group is Trivial
Centered Hexagonal Number as Sum of Triangular Numbers

== Theorem ==
Let $C_n$ be the $n$th centered hexagonal number.
Then:
:$C_n = 6 T_{n - 1} + 1$
where $T_{n - 1}$ denotes the $n - 1$th triangular number.
== Proof ==
== Visual Demonstration ==
== Sources ==
* : $37$
* : $37$
Category:Centered Hexagonal Number as Sum of Triangular Numbers
Category:Centered Hexagonal Numbers
Category:Triangular Numbers