nyuuzyou/wikis huggingface dataset:
Group Homomorphism Preserves Subgroups

== Theorem ==
Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group homomorphism. 
Then:
:$H \le G_1 \implies \phi \sqbrk H \le G_2$
where:
:$\phi \sqbrk H$ denotes the image of $H$ under $\phi$
:$\le$ denotes subgroup.
That is, group homomorphism preserves subgroups.
== Proof ==
Let $H \le G_1$.
First note that from Null Relation is Mapping iff Domain is Empty Set:
:$H \ne \O \implies \phi \sqbrk H \ne \O$
and so $\phi \sqbrk H$ is not empty.
Next, let $x, y \in \phi \sqbrk H$.
Then:
:$\exists h_1, h_2 \in H: x = \map \phi {h_1}, y = \map \phi {h_2}$
From the definition of Group Homomorphism, we have:
:$\map \phi {h_1^{-1} \circ h_2} = x^{-1} * y$
Since $H$ is a subgroup:
:$h_1^{-1} \circ h_2 \in H$
Hence:
:$x^{-1} * y \in \phi \sqbrk H$
Thus from the One-Step Subgroup Test:
:$\phi \sqbrk H \le G_2$
== Sources ==
* : $\S 1.2$
* : $\S 47.3$ Homomorphisms and their elementary properties
Category:Subgroups
Category:Group Homomorphisms
Group Homomorphism of Product with Inverse

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
Then:
where $y^{-1}$ denotes the inverse of $y$.
== Proof ==
Let $e_G$ and $e_H$ be the identities of $\struct {G, \circ}$ and $\struct {H, *}$ respectively.
By :
:$\forall x, y \in G: x \circ y^{-1} \in G, y^{-1} \circ x \in G$
;Result $(1):$:
;Result $(2):$:
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Lemma $\text{(i)}$
* : Chapter $8$: The Homomorphism Theorem: Proposition $8.6$
Category:Group Homomorphisms
Group Identity is Right Identity for Product Inverse Operation

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\oplus: G \times G \to G$ be the product inverse of $\circ$ on $G$.
Then $e$ is also the right identity of $\struct {G, \oplus}$, in the sense that:
:$\forall x \in G: x \oplus e = x$
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(a)}: 2^\circ$
Category:Product Inverse Operation
Category:Identity Elements
Group Induced by B-Algebra Induced by Group

== Theorem ==
Let $\struct {S, \circ}$ be a group.
Let $\struct {S, *}$ be the $B$-algebra described on Group Induces $B$-Algebra.
Let $\struct {S, \circ'}$ be the group described on $B$-Algebra Induces Group.
Then $\struct {S, \circ'} = \struct {S, \circ}$.
== Proof ==
Let $a, b \in S$.
It is required to show that:
:$a \circ' b = a \circ b$
To achieve this, recall that $\circ'$ is defined on $B$-Algebra Induces Group to satisfy:
:$a \circ' b = a * \paren {e * b}$
which, by the definition of $*$ on Group Induces $B$-Algebra comes down to:
Hence the result.
== Also see ==
* $B$-Algebra Induced by Group Induced by $B$-Algebra
Category:Group Theory
Category:B-Algebras
Group Induces B-Algebra

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity element is $e$.
Let $*$ be the product inverse operation on $G$:
:$\forall a, b \in G: a * b = a \circ b^{-1}$
where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$.
Then the algebraic structure $\struct {G, *}$ is a $B$-algebra.
== Proof ==
First note that by :
:$\forall a, b \in G: a * b \in G$
and so holds.
We have that:
:$\forall x \in G: x * x = x \circ x^{-1} = e$
by definition of inverse element.
Let $0 := e$.
Then:
:: $\quad \exists 0 \in G$
:: $\quad \forall x \in G: x * x = 0$
Note that:
and so:
demonstrating:
:: $\quad \forall x \in G: x * 0 = x$
Finally, let $x, y, z \in G$:
This demonstrates:
:: $\quad \forall x, y, z \in G: \paren {x * y} * z = x * \paren {z * \paren {0 * y} }$
All axioms of the $B$-algebra have been demonstrated to hold.
Hence the result.
Category:B-Algebras
Category:Group Theory
Group Isomorphism/Examples/Congruence Modulo Initial Segment of Natural Numbers

== Example of Group Isomorphism ==
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.
Let $\phi_m: \struct {\N_m, +_m} \to \struct {\Z_m, +_\PP}$ be the mapping defined by:
:$\forall a \in \N_m: \map {\phi_m} a = \eqclass a m$
where $\struct {\N_m, +_m}$ denotes the additive group of integers modulo $m$.
Then $\phi_m$ is a (group) isomorphism.
== Proof ==
We have that:
demonstrating that $\phi_m$ is a homomorphism.
Then we have that:
Hence $\phi_m$ is injective.
We also have that $\phi_m$ is trivially surjective.
Hence $\phi_m$ is a bijection.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.2$
Category:Examples of Group Isomorphisms
Group Isomorphism/Examples/Order 2 Matrices with 1 Real Variable

== Example of Group Isomorphism ==
Let $S$ be the set defined as:
:$S := \set {\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}: t \in \R}$
Consider the algebraic structure $\struct {S, \times}$, where $\times$ is used to denote (conventional) matrix multiplication.
Then $\struct {S, \times}$ is isomorphic to the additive group of real numbers $\struct {\R, +}$.
== Proof ==
=== ===
Let $t_1, t_2 \in \R$.
We have that:
As $t_1 + t_2 \in \R$, it follows that $\struct {S, \times}$ is closed.
=== ===
By Matrix Multiplication is Associative it follows that $\struct {S, \times}$ is a semigroup.
=== ===
Then from the above:
demonstrating that $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}$.
=== ===
Then, also from above:
demonstrating that $\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}$ has an inverse $\begin{bmatrix} 1 & -t \\ 0 & 1 \end{bmatrix}$.
So all the group axioms are fulfilled, and $\struct {S, \times}$ is seen to be a group.
=== Isomorphism ===
Let $\phi: \struct {\R, +} \to \struct {S, \times}$ be the mapping defined as:
:$\forall x \in \R: \map \phi x = \begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix}$
We have that:
demonstrating that $\phi$ is a group homomorphism.
Then:
demonstrating that $\phi$ is an injection.
Finally:
demonstrating that $\phi$ is a surjection.
Thus $\phi$ is a group homomorphism which is both injective and surjective.
That is, $\phi$ is by definition a group isomorphism.
Hence the result.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $2$
Category:Examples of Group Isomorphisms
Category:Examples of Matrices
Group Isomorphism/Examples/Quotient Group of Z by 3Z with Quotient Group of A4 by K4

== Example of Group Isomorphism ==
Let $\Z / 3 \Z$ denote the quotient group of the additive group of integers by the additive group of $3 \times$ the integers.
Let $A_4 / K_4$ denote the quotient group of the alternating group on 4 letters by the Klein $4$-group.
Then $\Z / 3 \Z$ is isomorphic to $A_4 / K_4$.
== Proof ==
By Quotient Group of Integers by Multiples, $\Z / 3 \Z$ is the additive group of integers modulo $3$.
This is its Cayley table:
From Normality of Subgroups of Alternating Group on 4 Letters, $K_4$ is a normal subgroup of $A_4$.
We have that:
:$\order {A_4 / K_4} = \index {A_4} {K_4} = \dfrac {12} 4 = 3$
As $3$ is prime, it follows from Prime Group is Cyclic that $A_4 / K_4$ and $\Z / 3 \Z$ are cyclic groups of order $3$.
The result follows from Cyclic Groups of Same Order are Isomorphic.
The Cayley table of $A_4$ makes this apparent:
== Sources ==
* : $\S 7.3$. Isomorphism
Category:Examples of Group Isomorphisms
Category:Alternating Group on 4 Letters
Category:Klein Four-Group
Category:Additive Group of Integers
Category:Additive Groups of Integer Multiples
Group Isomorphism/Examples/Real Power Function

== Example of Group Isomorphism ==
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\R_{>0}, \times}$ be the multiplicative group of positive real numbers.
Let $\alpha \in \R_{>0}$ be a strictly positive real number.
Let $f: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ be the mapping:
:$\forall x \in \R: \map f x = \alpha^x$
where $\alpha^x$ denotes $\alpha$ to the power of $x$.
Then $f$ is a (group) automorphism $\alpha \ne 1$.
== Proof ==
Let $\alpha \in \R_{>0}$.
We have by definition of power to a real number that:
:$\alpha^x = e^{x \ln \alpha}$
From Natural Logarithm of 1 is 0:
:$\ln \alpha = 0 \iff \alpha = 1$
First we note that:
Thus $f$ is a (group) homomorphism from $\struct {\R, +}$ to $\struct {\R_{> 0}, \times}$ for all $\alpha \in \R_{>0}$.
=== Sufficient Condition ===
Let $\alpha \in \R_{>0}$ such that $\alpha \ne 1$.
We have that:
and so $f$ is injective by definition.
Then we have:
This demonstrates that $f$ is surjective.
So by definition $f$ is a bijection.
Thus $f$ is an (group) isomorphism from $\struct {\R, +}$ to $\struct {\R_{> 0}, \times}$.
=== Necessary Condition ===
Let $f$ be a (group) automorphism.
 $\alpha = 1$.
Then we have:
:$\forall x, y \in \R: \map f x = \map f y = 1$
and so $f$ is not injective.
Hence $f$ is not a bijection.
Therefore $f$ is not an isomorphism.
Hence by Proof by Contradiction $f$ is not an automorphism.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.4$
Category:Examples of Group Isomorphisms
Category:Exponential on Real Numbers is Group Isomorphism
Group Isomorphism Preserves Identity

== Theorem ==
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group isomorphism.
Let:
:$e_G$ be the identity of $\struct {G, \circ}$
:$e_H$ be the identity of $\struct {H, *}$.
Then:
:$\map \phi {e_G} = e_H$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 28 \gamma$
Category:Group Isomorphisms
Category:Identity Elements
Category:Group Isomorphism Preserves Identity
Group Isomorphism Preserves Identity/Proof 2

== Theorem ==
== Proof ==
It follows from Identity is only Idempotent Element in Group that $\map \phi {e_G}$ is the identity of $H$.
That is:
:$\map \phi {e_G} = e_H$
== Sources ==
* : $\S 1.5$: Theorem $4$
Category:Group Isomorphism Preserves Identity
Group Isomorphism Preserves Inverses

== Theorem ==
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group isomorphism.
Let:
:$e_G$ be the identity of $\struct {G, \circ}$
:$e_H$ be the identity of $\struct {H, *}$.
Then:
: $\forall g \in G: \map \phi {g^{-1} } = \paren {\map \phi g}^{-1}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 28 \gamma$
Category:Group Isomorphisms
Category:Inverse Elements
Category:Group Isomorphism Preserves Inverses
Group Isomorphism Preserves Inverses/Proof 2

== Theorem ==
== Proof ==
Let $g \in G$.
It follows from Inverse in Group is Unique that $\map \phi {g^{-1} }$ is the unique inverse element of $\map \phi g$ in $\struct {H, *}$.
That is:
:$\forall g \in G: \map \phi {g^{-1} } = \paren {\map \phi g}^{-1}$
== Sources ==
* : $\S 1.5$: Theorem $4$
Category:Group Isomorphism Preserves Inverses
Group Isomorphism Preserves Order of Group

== Theorem ==
Let $G$ and $H$ be groups.
Let $\phi: G \to H$ be a (group) isomorphism.
Then:
:$\order G = \order H$
where $\order {\, \cdot \,}$ denotes the order of a group.
== Proof ==
By definition, an isomorphism is a bijection.
By definition, the order of a group is the cardinality of its underlying set.
The result follows by definition of set equivalence.
== Sources ==
* : $\S 7.3$. Isomorphism
Category:Group Isomorphisms
Group Isomorphism Preserves Order of Group Element

== Theorem ==
Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$.
Let $\phi: G \to H$ be a group isomorphism.
Then:
:$a \in G \implies \order {\map \phi a} = \order a$
== Proof ==
First, suppose $a$ is of finite order.
By definition, $\phi$ is bijective, therefore injective.
The result then follows from Order of Homomorphic Image of Group Element.
Now suppose $a$ is of infinite order.
Suppose $\map \phi a$ is of finite order.
Consider the mapping $\phi^{-1}: H \to G$.
Let $b = \map \phi a$.
Let $\order b = m$.
Then:
:$\order a = \order {\map {\phi^{-1} } b} = m$
and that would mean $a$ was of finite order.
The result follows by transposition.
== Sources ==
* : $\S 7.3$. Isomorphism: Example $138$
Category:Group Isomorphisms
Group Presentation of Quaternion Group/Also presented as

== Group Presentation of Quaternion Group: Also presented as == 
The (group) presentation for the quaternion group can also be expressed in the form:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2 = \paren {a b}^2}$
== Sources ==
* 
* 
Category:Group Presentation of Quaternion Group
Group Product/Examples

== Examples of Operations on Group Product ==
=== Example: $b x a^{-1} = a^{-1} b$ ===
=== Example: $a x a^{-1} = e$ ===
=== Example: $a x a^{-1} = a$ ===
=== Example: $a x b = c$ ===
=== Example: $b a^{-1} x a b^{-1} = b a$ ===
Category:Examples of Operations on Group Product
Group Product/Examples/b a^-1 x a b^-1 = b a

== Examples of Operations on Product Elements ==
Solve for $x$ in:
:$b a^{-1} x a b^{-1} = b a$
== Solution ==
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Exercise $2 \ \text{(d)}$
Category:Examples of Operations on Group Product
Group Product/Examples/b x a^-1 = a^-1 b

== Examples of Operations on Product Elements ==
Solve for $x$ in:
:$b x a^{-1} = a^{-1} b$
== Solution ==
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.3$: Remark 1
Category:Examples of Operations on Group Product
Group Product Identity therefore Inverses

== Theorem ==
Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$
=== Part 1 ===
=== Part 2 ===
== Sources ==
* : $\S 35.3$: Elementary consequences of the group axioms
Category:Group Theory
Category:Group Product Identity therefore Inverses
Group Product Identity therefore Inverses/Part 1/Proof 1

== Theorem ==
== Proof ==
From the Division Laws for Groups:
:$g h = e \implies g = e h^{-1} = h^{-1}$
Also by the Division Laws for Groups:
:$g h = e \implies h = g^{-1} e = g^{-1}$
== Sources ==
* : $\S 35.3$: Elementary consequences of the group axioms
Category:Group Product Identity therefore Inverses
Group Product Identity therefore Inverses/Part 1/Proof 2

== Theorem ==
== Proof ==
Let $g h = e$.
Then:
and:
== Sources ==
* : Chapter $\text{II}$: Groups: The Group Property: Theorem $1 \ \text{(iii)}$
* : Chapter $2$: The Definition of Group Structure: $\S 28 \ (1)$
Category:Group Product Identity therefore Inverses
Group Product Identity therefore Inverses/Part 2

== Theorem ==
Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.
Then:
:$h g = e \implies h = g^{-1}$ and $g = h^{-1}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: The Group Property: Theorem $1 \ \text{(iv)}$
Category:Group Product Identity therefore Inverses
Group Product Identity therefore Inverses/Part 2/Proof 1

== Theorem ==
== Proof ==
From the Division Laws for Groups:
:$h g = e \implies g = e h^{-1} = h^{-1}$
Also by the Division Laws for Groups:
:$h g = e \implies h = g^{-1} e = g^{-1}$
== Sources ==
* : $\S 35.3$: Elementary consequences of the group axioms
Category:Group Product Identity therefore Inverses
Group Product Inverse Operation with Identity

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\oplus: G \times G \to G$ be the product inverse of $\circ$ on $G$.
Then:
:$\forall x, y \in G: e \oplus \paren {x \oplus y} = y \oplus x$
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(a)}: 3^\circ$
Category:Product Inverse Operation
Category:Identity Elements
Group Representation/Examples/Quaternion Group

== Example of Group Representation ==
Consider the group whose presentation can be expressed as:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2 = \paren {a b}^2}$
This has a matrix representation as:
along with $-\mathbf I$, $-\mathbf J$, $-\mathbf K$ and $-\mathbf L$.
== Proof ==
This is demonstrated in Quaternion Group/Order 4 Matrices.
== Sources ==
* 
* 
Category:Examples of Group Representations
Group Types of Order Prime Squared

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a group of order $p^2$.
Then $G$ is isomorphic either to $\Z_{p^2}$ or to $\Z_p \times \Z_p$, where $\Z_p$ denotes the additive group of integers modulo $p$.
== Proof ==
From Group of Order Prime Squared is Abelian, $G$ is an abelian group.
From Abelian Group of Prime-power Order is Product of Cyclic Groups, $G$ is either:
:the cyclic group of order $p^2$
:the direct product of the cyclic group of order $p$ with itself.
The result follows from Finite Cyclic Group is Isomorphic to Integers under Modulo Addition.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 62 \delta$
Category:Groups of Order p^2
Group does not Necessarily have Subgroup of Order of Divisor of its Order

== Theorem ==
Let $G$ be a finite group whose order is $n$.
Let $d$ be a divisor of $n$.
Then it is not necessarily the case that $G$ has a subgroup of order $d$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Lagrange's Theorem (Group Theory)
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.5$
* 
* 
Category:Group does not Necessarily have Subgroup of Order of Divisor of its Order
Category:Subgroups
Category:Order of Groups
Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 1

== Theorem ==
== Proof ==
Proof by Counterexample:
Consider $S_5$, the symmetric group on $5$ letters.
By Order of Symmetric Group, $\order {S_5} = 5! = 120$.
We have that $120 = 8 \times 15$ and so $15$ is a divisor of $120$.
However, Symmetric Group on 5 Letters has no Subgroup of Order 15.
== Sources ==
* : $\S 44$. Some consequences of Lagrange's Theorem
Category:Group does not Necessarily have Subgroup of Order of Divisor of its Order
Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 2

== Theorem ==
== Proof ==
Proof by Counterexample:
Consider the symmetric group $S_4$.
Then the order of the alternating group $A_4$ is $12$.
We list the subgroups of $A_4$:
Now $6$ divides $12$.
But there is no subgroup of $A_4$ of order $6$.
== Sources ==
* : Chapter $9$: Permutations: Exercise $3$
Category:Group does not Necessarily have Subgroup of Order of Divisor of its Order
Group has Latin Square Property

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.
Similarly, there exists a unique $h \in G$ such that $h \circ a = b$.
=== Additive Notation ===
This result can also be written in additive notation as follows:
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Group Theory
Category:Group has Latin Square Property
Group has Latin Square Property/Additive Notation

== Theorem ==
Let $\struct {G, +}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a + g = b$.
Similarly, there exists a unique $h \in G$ such that $h + a = b$.
== Proof ==
From Group has Latin Square Property, we have that:
and:
== Sources ==
* : $\S 4.6$. Elementary theorems on groups: Example $84$
Category:Group has Latin Square Property
Group has Latin Square Property/Corollary

== Corollary to Group has Latin Square Property ==
The Cayley table for any finite group is a Latin square.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 5$: The Multiplication Table
* : $\S 1.4$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(v)}$
Category:Group Theory
Category:Cayley Tables
Category:Group has Latin Square Property
Group has Latin Square Property/Corollary/Proof 2

== Corollary to Group has Latin Square Property ==
== Proof ==
Let $G$ be a finite group whose order is $n$.
Let $\tuple {x_1, x_2, \ldots, x_n}$ be the elements of the underlying set of $G$ in the order they appear in the headings of the Cayley table of $G$.
Consider the row of the Cayley table headed with $a$.
The elements of that row are:
:$\tuple {a x_1, a x_2, \ldots, a x_n}$
that is:
:$\tuple {\map {\lambda_a} {x_1}, \map {\lambda_a} {x_2}, \ldots, \map {\lambda_a} {x_n} }$
where $\lambda_a$ denotes the left regular representation of $\struct {S, \circ}$ with respect to $a$.
From Regular Representations in Group are Permutations, it follows that each of $\map {\lambda_a} {x_1}, \map {\lambda_a} {x_2}, \ldots, \map {\lambda_a} {x_n}$ appears in that row exactly once.
A similar argument based on the right regular representation proves the result for the columns.
== Sources ==
* : $\S 35.9$: Elementary consequences of the group axioms
Category:Group has Latin Square Property
Group has Latin Square Property/Proof 1

== Theorem ==
== Proof ==
Thus, such a $g$ exists.
Suppose $g, g' \in G$ where $a \circ g = b = a \circ g'$.
Then:
Thus uniqueness holds.
To prove the second part of the theorem, let $h = b \circ a^{-1}$.
The remainder of the proof follows a similar procedure to the above.
== Sources ==
* : $\S 1.4$: Theorem $1$
Category:Group has Latin Square Property
Group has Latin Square Property/Proof 2

== Theorem ==
== Proof ==
We shall prove that this is true for the first equation:
Because the statements:
:$a \circ g = b$
and
:$g = a^{-1} \circ b$
are equivalent, we may conclude that $g$ is indeed the only solution of the equation.
The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly.
== Sources ==
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups
Category:Group has Latin Square Property
Group has Latin Square Property/Proof 3

== Theorem ==
== Proof ==
Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$.
So such an element, if it exists, is unique.
Now it is demonstrated that $g = a^{-1} b$ satisfies the requirement for $a \circ g = b$
Since $a \in G$, it follows by group axiom $G3$: existence of inverses that $a^{-1} \in G$.
Then:
Thus, such a $g$ exists.
The properties of $h$ are proved similarly.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.9)$
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.3$
Category:Group has Latin Square Property
Group has Latin Square Property/Proof 4

== Theorem ==
== Proof ==
We shall prove that this is true for the first equation:
Conversely:
The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly.
== Sources ==
* : $\S 4.6$. Elementary theorems on groups: Theorem $\text{(ii)}$
* : $\S 3.2$: Groups; the axioms
Category:Group has Latin Square Property
Group has Subgroups of All Prime Power Factors

== Theorem ==
Let $p$ be a prime.
Let $G$ be a finite group of order $n$.
If $p^k \divides n$ then $G$ has at least one subgroup of order $p^k$.
== Proof ==
From Composition Series of Group of Prime Power Order, a $p$-group has subgroups corresponding to every divisor of its order.
Thus, taken with the First Sylow Theorem, a finite group has a subgroup corresponding to every prime power divisor of its order.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.6$
* 
* 
Category:Order of Groups
Category:P-Groups
Category:Subgroups