nyuuzyou/wikis huggingface dataset:
Isomorphism Preserves Associativity/Proof 1

== Theorem ==
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
As an isomorphism is surjective, it follows that:
:$\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \map \phi y = v, \map \phi z = w$
So:
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also an isomorphism.
Thus the result for $\phi$ can be applied to $\phi^{-1}$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.2: \ 1^\circ$
Category:Isomorphism Preserves Associativity
Isomorphism Preserves Cancellability

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then:
:$a \in S$ is cancellable in $\struct {S, \circ}$ $\map \phi a \in T$ is cancellable in $\struct {T, *}$.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $a$ is cancellable.
From Isomorphism Preserves Left Cancellability and Isomorphism Preserves Right Cancellability:
:$a \in S$ is left cancellable in $\struct {S, \circ}$ $\map \phi a \in T$ is left cancellable in $\struct {T, *}$
and
:$a \in S$ is right cancellable in $\struct {S, \circ}$ $\map \phi a \in T$ is right cancellable in $\struct {T, *}$.
It follows that:
:$a \in S$ is cancellable in $\struct {S, \circ}$
:
:$\map \phi a \in T$ is cancellable in $\struct {T, *}$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$
Category:Isomorphism Preserves Cancellability
Category:Isomorphisms (Abstract Algebra)‎
Category:Cancellability
Isomorphism Preserves Commutativity

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is commutative $*$ is commutative.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.2: \ 2^\circ$
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.3$
Category:Isomorphism Preserves Commutativity
Category:Isomorphisms (Abstract Algebra)
Category:Commutativity
Isomorphism Preserves Commutativity/Proof 1

== Theorem ==
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
As an isomorphism is surjective, it follows that:
:$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$
So:
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.
Thus the result for $\phi$ can be applied to $\phi^{-1}$.
Category:Isomorphism Preserves Commutativity
Isomorphism Preserves Commutativity/Proof 2

== Theorem ==
== Proof ==
We have that an isomorphism is a homomorphism which is also a bijection.
By definition, an epimorphism is a homomorphism which is also a surjection.
That is, an isomorphism is an epimorphism which is also an injection.
Thus Epimorphism Preserves Commutativity can be applied.
Category:Isomorphism Preserves Commutativity
Isomorphism Preserves Groups

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
If $\struct {S, \circ}$ is a group, then so is $\struct {T, *}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.5$: Exercise $3.4$
Category:Isomorphisms (Abstract Algebra)
Category:Group Theory
Category:Isomorphism Preserves Groups
Isomorphism Preserves Groups/Proof 1

== Theorem ==
== Proof ==
From Isomorphism Preserves Semigroups, if $\struct {S, \circ}$ is a semigroup then so is $\struct {T, *}$.
From Isomorphism Preserves Identity, if $\struct {S, \circ}$ has an identity $e_S$, then $\map \phi {e_S}$ is the identity for $*$.
From Isomorphism Preserves Inverses, if $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
The result follows from the definition of group.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups
Category:Isomorphism Preserves Groups
Isomorphism Preserves Identity

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ $\map \phi {e_S}$ is the identity for $*$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.2: \ 3^\circ$
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.3$
Category:Isomorphism Preserves Identity
Category:Isomorphisms (Abstract Algebra)
Category:Identity Elements
Isomorphism Preserves Identity/Proof 1

== Theorem ==
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Then $\forall x \in S: x \circ e_S = x = e_S \circ x$.
The result follows directly from the morphism property of $\circ$ under $\phi$:
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.
Thus the result for $\map \phi {e_S}$ can be applied to $\map {\phi^{-1} } {\map \phi {e_S} }$.
Category:Isomorphism Preserves Identity
Isomorphism Preserves Identity/Proof 2

== Theorem ==
== Proof ==
We have that an isomorphism is a homomorphism which is also a bijection.
By definition, an epimorphism is a homomorphism which is also a surjection.
That is, an isomorphism is an epimorphism which is also an injection.
Thus Epimorphism Preserves Identity can be applied.
Category:Isomorphism Preserves Identity
Isomorphism Preserves Inverses

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Then $x^{-1}$ is an inverse of $x$ for $\circ$ $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
That is, :
:$\map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$
== Proof 1 ==
== Proof 2 ==
Category:Isomorphism Preserves Inverses
Category:Isomorphisms (Abstract Algebra)
Category:Inverse Elements
Isomorphism Preserves Inverses/Proof 1

== Theorem ==
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
From Epimorphism Preserves Identity, it follows that $\struct {T, *}$ also has an identity, which is $\map \phi {e_S}$.
Let $y$ be an inverse of $x$ in $\struct {S, \circ}$. 
Then:
So $\map \phi y$ is an inverse of $\map \phi x$ in $\struct {T, *}$.
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.
Thus the result for $\map \phi x$ can be applied to $\map {\phi^{-1} } {\map \phi x}$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.2: \ 4^\circ$
Category:Isomorphism Preserves Inverses
Isomorphism Preserves Left Cancellability

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then:
:$a \in S$ is left cancellable in $\struct {S, \circ}$
:
:$\map \phi a \in T$ is left cancellable in $\struct {T, *}$.
== Proof ==
Let $a$ be left cancellable in $\struct {S, \circ}$.
Let $x \in S$ and $y \in S$ be arbitrary.
Then:
That is, $\map \phi a$ is left cancellable in $\struct {T, *}$.
As $\phi$ is an isomorphism, then so is $\phi^{-1}$.
So the same proof works in reverse in exactly the same way.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$
Category:Isomorphism Preserves Cancellability
Isomorphism Preserves Right Cancellability

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then:
:$a \in S$ is right cancellable in $\struct {S, \circ}$ $\map \phi a \in T$ is right cancellable in $\struct {T, *}$.
== Proof ==
Let $a$ be right cancellable in $\struct {S, \circ}$.
Let $x \in S$ and $y \in S$ be arbitrary.
Then:
That is, $\map \phi a$ is right cancellable in $T$.
As $\phi$ is an isomorphism, then so is $\phi^{-1}$.
So the same proof works in reverse in exactly the same way.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$
Category:Isomorphism Preserves Cancellability
Isomorphism Preserves Semigroups

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: S \to T$ be an isomorphism.
If $\struct {S, \circ}$ is a semigroup, then so is $\struct {T, *}$.
== Proof 1 ==
== Proof 2 ==
Category:Isomorphism Preserves Semigroups
Category:Isomorphisms (Abstract Algebra)
Category:Semigroups
Isomorphism Preserves Semigroups/Proof 1

== Theorem ==
== Proof ==
If $\struct {S, \circ}$ is a semigroup, then by definition it is closed.
From Morphism Property Preserves Closure, $\struct {T, *}$ is therefore also closed.
If $\struct {S, \circ}$ is a semigroup, then by definition $\circ$ is associative.
From Isomorphism Preserves Associativity, $*$ is therefore also associative.
So $\struct {T, *}$ is closed, and $*$ is associative, and therefore by definition, $\struct {T, *}$ is a semigroup.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups
Category:Isomorphism Preserves Semigroups
Isomorphism Theorems

== Preface ==
There is no standard numbering for the Isomorphism Theorems. Different authors use different labellings.
Therefore, the following nomenclature is to a greater or lesser extent arbitrary, and not necessarily the most widely used or standard. Please take care.
== First Isomorphism Theorem ==
== Second Isomorphism Theorem ==
== Third Isomorphism Theorem ==
== Fourth Isomorphism Theorem ==
Category:Isomorphism Theorems
Category:Quotient Groups
Category:Quotient Rings
Category:Isomorphisms (Abstract Algebra)
Category:Homomorphisms (Abstract Algebra)
Category:Named Theorems
Isomorphism between Additive Group Modulo 16 and Multiplicative Group Modulo 17

== Theorem ==
Let $\struct {\Z_{16}, +}$ denote the additive group of integers modulo $16$.
Let $\struct {\Z'_{17}, \times}$ denote the multiplicative group of reduced residues modulo $17$.
Let $\phi: \struct {\Z_{16}, +} \to \struct {\Z'_{17}, \times}$ be the mapping defined as:
:$\forall \eqclass k {16} \in \struct {\Z_{16}, +}: \map \phi {\eqclass k {16} } = \eqclass {3^k} {17}$
Then $\phi$ is a group isomorphism.
== Proof ==
Let $\eqclass x {16}, \eqclass y {16} \in \struct {\Z_{16}, +}$.
Then:
Thus it is seen that $\phi$ is a group homomorphism.
It remains to be seen that $\phi$ is a bijection.
Because $17$ is prime:
$\forall x \in \Z, 1 \le x < 17: x \perp 17$
where $\perp$ denotes coprimality.
Thus by definition of multiplicative group of reduced residues modulo $17$:
:$\order {\struct {\Z'_{17}, \times} } = 16$
where $\order {\, \cdot \,}$ denotes the order of a group.
Similarly, by definition of additive group of integers modulo $16$:
:$\order {\struct {\Z_{16}, +} } = 16$
So:
:$\order {\struct {\Z'_{17}, \times} } = \order {\struct {\Z_{16}, +} }$
which is a necessary condition for group isomorphism.
Now we have:
Now let $\eqclass x {16}, \eqclass y {16} \in \Z_{16}$ such that $\map \phi {\eqclass x {16} } = \map \phi {\eqclass y {16} }$.
We have:
Thus $\phi$ is an injection.
From Equivalence of Mappings between Finite Sets of Same Cardinality it follows that $\phi$ is a bijection.
Thus $\phi$ is a bijective group homomorphism.
Hence the result by definition of group isomorphism.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 62 \beta$
Category:Additive Groups of Integers Modulo m
Category:Multiplicative Groups of Reduced Residues
Category:Groups of Order 16
Category:Examples of Group Isomorphisms
Isomorphism between Algebraic Structures induces Isomorphism between Induced Structures

== Theorem ==
Let $A$ be a set.
Let $\struct {S, \odot}$ and $\struct {T, \otimes}$ be algebraic structures.
Let:
:$S^A$ denote the set of mappings from $A$ to $S$
:$T^A$ denote the set of mappings from $A$ to $T$.
Let:
:$\struct {S^A, \odot}$ denote the algebraic structure on $S^A$ induced by $\odot$
:$\struct {T^A, \otimes}$ denote the algebraic structure on $T^A$ induced by $\otimes$.
Let $\phi$ be an isomorphism from $S$ to $T$.
Let $\chi: S^A \to T^A$ be the mapping defined as:
:$\forall f \in S^A: \map \chi f = \phi \circ f$
where:
:$\phi \circ f$ denotes the composition of $\phi$ with $f$.
Then:
:$\chi$ is an isomorphism from $\struct {S^A, \odot}$ to $\struct {T^A, \otimes}$
== Proof ==
Let $f: A \to S$ and $g: A \to S$ be arbitrary elements of $S^A$.
We have that $\phi$ is an isomorphism.
Hence $\phi$ is a homomorphism which is a bijection.
Hence, from Bijection iff exists Mapping which is Left and Right Inverse, $\phi$ has an inverse mapping $\phi^{-1}: T \to S$ such that:
:$\phi \circ \phi^{-1} = I_T$
:$\phi^{-1} \circ \phi = I_S$
where $I_S$ and $I_T$ denote the identity mappings on $S$ and $T$ respectively.
From Inverse of Bijection is Bijection, $\phi^{-1}$ is also a bijection.
We have that:
demonstrating that $\chi$ is a homomorphism.
Then:
Hence $\chi$ is an injection.
Let $g: A \to T$ be an arbitrary element of $T^A$.
As $\phi^{-1}: T \to S$ is a bijection:
:$\exists h \in S^A: h = \map {\phi^{-1} } g = \phi^{-1} \circ g$
That is:
:$\exists h \in S^A: \map \phi h = g$
As $g$ is arbitrary:
:$\forall g \in T^A: \exists h \in S^A: g = \phi \circ h = \map \chi h$
That is, $\chi$ is a surjection.
We have shown that $\chi$ is:
:an injection
:a surjection
:a homomorphism
and so by definition $\chi$ is an isomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.11 \ \text{(c)}$
Category:Composite Mappings
Category:Isomorphisms (Abstract Algebra)
Category:Pointwise Operations
Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition

== Theorem ==
Let $\struct {U_\C, \times}$ be the group of Gaussian integer units under complex multiplication.
Let $\struct {\Z_n, +_4}$ be the integers modulo $4$ under modulo addition.
Then $\struct {U_\C, \times}$ and $\struct {\Z_4, +_4}$ are isomorphic algebraic structures.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Group Isomorphisms/Order 4
Category:Additive Groups of Integers Modulo m
Category:Group of Gaussian Integer Units
Category:Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition
Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition/Proof 1

== Theorem ==
== Proof ==
From Gaussian Integer Units are 4th Roots of Unity:
:$U_\C$ is the set consisting of the (complex) $4$th roots of $1$.
The result follows from Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.2$
Category:Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition
Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition/Proof 2

== Theorem ==
== Proof ==
Let the mapping $f: \Z_4 \to U_\C$ be defined as:
From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $f$ is an isomorphism:
=== Cayley Table of Integers Modulo $4$ ===
The Cayley table for $\struct {\Z_4, +_4}$ is as follows:
=== Cayley Table of Gaussian Integer Units ===
The Cayley table for $\struct {U_\C, \times}$ is as follows:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.2$
Category:Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition
Isomorphism between Gaussian Integer Units and Reduced Residue System Modulo 5 under Multiplication

== Theorem ==
Let $\struct {U_\C, \times}$ be the group of Gaussian integer units under complex multiplication.
Let $\struct {\Z'_5, \times_5}$ be the multiplicative group of reduced residues modulo $5$.
Then $\struct {U_\C, \times}$ and $\struct {\Z'_5, \times_5}$ are isomorphic algebraic structures.
== Proof ==
Establish the mapping $f: U_C \to \Z'_5$ as follows:
From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $f$ is an isomorphism:
=== Cayley Table of Gaussian Integer Units ===
The Cayley table for $\struct {U_\C, \times}$ is as follows:
</onlyinclude>
=== Multiplicative Group of Reduced Residues Modulo $5$ ===
The Cayley table for $\struct {\Z'_5, \times_5}$ is as follows:
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $1$
Category:Examples of Group Isomorphisms/Order 4
Category:Multiplicative Group of Reduced Residues Modulo 5
Category:Group of Gaussian Integer Units
Isomorphism between Gaussian Integer Units and Rotation Matrices Order 4

== Theorem ==
Let $\struct {U_\C, \times}$ be the group of Gaussian integer units under complex multiplication.
Let $\struct {R_4, \times}$ be the group of rotation matrices of order $4$ under modulo addition.
Then $\struct {U_\C, \times}$ and $\struct {R_4, \times}$ are isomorphic algebraic structures.
== Proof ==
Establish the mapping $f: U_C \to R_4$ as follows:
From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $f$ is an isomorphism:
=== Cayley Table of Gaussian Integer Units ===
The Cayley table for $\struct {U_\C, \times}$ is as follows:
</onlyinclude>
=== Group of Rotation Matrices Order $4$ ===
The Cayley table for $\struct {R_4, \times}$ is as follows:
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $1$
Category:Examples of Group Isomorphisms/Order 4
Category:Group of Rotation Matrices Order 4
Category:Group of Gaussian Integer Units
Isomorphism between Group Generated by Reciprocal of z and 1 minus z and Symmetric Group on 3 Letters

== Theorem ==
Let $S_3$ denote the symmetric group on $3$ letters.
Let $G$ be the group generated by $1 / z$ and $1 - z$.
Then $S_3$ and $G$ are isomorphic algebraic structures.
== Proof ==
Establish the mapping $\phi: S_3 \to G$ as follows:
From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $\phi$ is an isomorphism:
=== Cayley Table of Symmetric Group on $3$ Letters ===
The Cayley table for $S_3$ is as follows:
</onlyinclude>
=== Group Generated by $1 / z$ and $1 - z$ ===
The Cayley table for $S$ is as follows:
== Sources ==
* : Introduction
Category:Examples of Group Isomorphisms/Order 6
Category:Symmetric Group on 3 Letters
Category:Group Generated by Reciprocal of z and 1 minus z
Isomorphism between Group of Units Ring of Integers Modulo 2^n and C2 x C2^(n-2)

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer and $n \ge 2$.
Let $R = \struct {\Z / 2^n \Z, +, \times}$ be the ring of integers modulo $2^n$.
Let $U = \struct {\paren {\Z / 2^n \Z}^\times, \times}$ denote the group of units of $R$.
Let $C_2$ be cyclic group of order $2$.
Let $C_{2^{n - 2} }$ be the cyclic group of order $2^{n - 2}$.
Then $U$ is isomorphic to the group direct product of $C_2$ and $C_{2^{n - 2} }$.
== Proof ==
=== Lemma ===
Let $\gen 5$ be the subgroup of $U$ generated by $5$.
Then $-1 \notin \gen 5$.
=== Proof of lemma ===
 that $-1 \in \gen 5$, then
:$-1 \equiv 5^k \pmod {2^n}$ for some $k$
then
:$1 \equiv 5^{2 k} \pmod {2^n}$
so $2 k$ divides the order of $5$ in $U$ which is $2^{n - 2}$ by Order of $5$ in Units of Ring of Integers Modulo $2^n$.
hence 
:$k \divides 2^{n - 3}$
hence
:$\paren {-1}^{\frac {2^{n - 3}} k} \equiv 5^{2^{n - 3}} \pmod {2^n}$
contradicting with $5^{2^{n-3}} \equiv 1 + 2^{n-1} \pmod {2^n}$ which is proved in Order of $5$ in Units of Ring of Integers Modulo $2^n$. 
From the lemma, the elements
:$\set {\paren {-1}^e 5^j: e=0,1; j=0,1, \ldots, 2^{n - 2}-1}$
are all distinct.
Since there are $2^{n - 1}$ of these elements and $\order U = 2^{n - 1}$ we have
:$U \simeq \set{ \pm 1} \times \gen 5$
By Isomorphism between Ring of Integers Modulo 2 and Parity Ring, $\set{\pm 1} \simeq C_2$
By Order of $5$ in Units of Ring of Integers Modulo $2^n$, $\gen 5 \simeq C_{2^{n - 2} }$
:$U \simeq C_2 \times C_{2^{n - 2} }$
== Source ==
* , arts. $90$–$91$
Category:Ring of Integers Modulo m
Isomorphism between Group of Units of Ring of Integers Modulo p^n and C((p-1)p^(n-1))

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $p$ be an odd prime.
Let $R = \struct {\Z / p^n \Z, +, \times}$ be the ring of integers modulo $p^n$.
Let $U = \struct {\paren {\Z / p^n \Z}^\times, \times}$ denote the group of units of $R$.
Let $C_{(p - 1)p^{n - 1}}$ be be cyclic group of order $(p - 1)p^{n - 1}$.
Then $U$ is isomorphic to $C_{(p - 1)p^{n - 1}}$.
== Proof ==
The case $n = 1$ is proved in Ring of Integers Modulo Prime is Field.
Suppose $n \ge 2$.
== Source ==
* , arts. 84–89
Isomorphism between Ring of Integers Modulo 2 and Parity Ring

== Theorem ==
The ring of integers modulo $2$ and the parity ring are isomorphic.
== Proof ==
To simplify the notation, let the elements of $\Z_2$ be identified as $0$ for $\eqclass 0 2$ and $1$ for $\eqclass 1 2$.
Let $f$ be the mapping from the parity ring $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ and the ring of integers modulo $2$ $\struct {\Z_2, +_2, \times_2}$:
:$f: \struct {\set {\text{even}, \text{odd} }, +, \times} \to \struct {\Z_2, +_2, \times_2}$:
::$\forall x \in R: \map f x = \begin{cases}
0 & : x = \text{even} \\
1 & : x = \text{odd} \end{cases}$
The bijective nature of $f$ is apparent:
:$f^{-1}: \struct {\Z_2, +_2, \times_2} \to \struct {\set {\text{even}, \text{odd} }, +, \times}$:
::$\forall x \in \Z_2: \map {f^{-1} } x = \begin{cases}
\text{even} & : x = 0 \\
\text{odd} & : x = 1 \end{cases}$
Thus the following equations can be checked:
and:
These results can be determined from their Cayley tables:
=== Cayley Tables for Parity Ring ===
=== Cayley Tables for $\Z_2$ ===
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.1$
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.1$
Category:Ring of Integers Modulo m
Category:Parity Ring
Category:Ring Isomorphisms
Category:Field Isomorphisms
Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {R_n, \times}$ be the complex $n$th roots of unity under complex multiplication.
Let $\struct {\Z_n, +_n}$ be the integers modulo $n$ under modulo addition.
Then $\struct {R_n, \times}$ and $\struct {\Z_n, +_n}$ are isomorphic algebraic structures.
== Proof ==
The set of integers modulo $n$ is the set exemplified by the integers:
:$\Z_n = \set {0, 1, \ldots, n - 1}$
The complex $n$th roots of unity is the set:
:$R_n = \set {z \in \C: z^n = 1}$
From Complex Roots of Unity in Exponential Form:
:$R_n = \set {1, e^{\theta / n}, e^{2 \theta / n}, \ldots, e^{\left({n - 1}\right) \theta / n} }$
where $\theta = 2 i \pi$.
Let $z, w, \in R_n$.
Then:
:$\paren {z w}^n = z^n w^n = 1$
and so $z w \in R_n$.
Thus $\struct {R_n, \times}$ is a closed algebraic structure.
Consider the mapping $f: \Z_n \to R_n$ defined as:
:$\forall r \in \Z_n: \map f r = e^{r \theta / n}$
which can be seen to be a bijection by inspection.
Let $j, k \in \Z_n$.
Then:
Thus $f$ is an isomorphism.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.2$
Category:Examples of Group Isomorphisms
Category:Roots of Unity
Category:Additive Groups of Integers Modulo m