Definition:Isomorphism (Abstract Algebra)

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This page is about isomorphism in the context of abstract algebra. For other uses, see isomorphism.

Definition

An isomorphism is a homomorphism which is a bijection.

That is, it is a mapping which is both a monomorphism and an epimorphism.


An algebraic structure $\struct {S, \circ}$ is isomorphic to another algebraic structure $\struct {T, *}$ if and only if there exists an isomorphism from $\struct {S, \circ}$ to $\struct {T, *}$, and we can write $S \cong T$ (although notation may vary).


Semigroup Isomorphism

Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups.

Let $\phi: S \to T$ be a (semigroup) homomorphism.


Then $\phi$ is a semigroup isomorphism if and only if $\phi$ is a bijection.


Monoid Isomorphism

Let $\struct {S, \circ}$ and $\struct {T, *}$ be monoids.

Let $\phi: S \to T$ be a (monoid) homomorphism.


Then $\phi$ is a monoid isomorphism if and only if $\phi$ is a bijection.


Group Isomorphism

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

Let $\phi: G \to H$ be a (group) homomorphism.


Then $\phi$ is a group isomorphism if and only if $\phi$ is a bijection.


Ring Isomorphism

Let $\struct {R, +, \circ}$ and $\struct {S, \oplus, *}$ be rings.

Let $\phi: R \to S$ be a (ring) homomorphism.


Then $\phi$ is a ring isomorphism if and only if $\phi$ is a bijection.


$F$-Isomorphism

Let $R, S$ be rings with unity.

Let $F$ be a subfield of both $R$ and $S$.


Let $\phi: R \to S$ be an $F$-homomorphism such that $\phi$ is bijective.


Then $\phi$ is an $F$-isomorphism.


The relationship between $R$ and $S$ is denoted $R \cong_F S$.


Field Isomorphism

Let $\struct {F, +, \circ}$ and $\struct {K, \oplus, *}$ be fields.

Let $\phi: F \to K$ be a (field) homomorphism.


Then $\phi$ is a field isomorphism if and only if $\phi$ is a bijection.


$R$-Algebraic Structure Isomorphism

Let $\struct {S, \ast_1, \ast_2, \ldots, \ast_n, \circ}_R$ and $\struct {T, \odot_1, \odot_2, \ldots, \odot_n, \otimes}_R$ be $R$-algebraic structures.

Let $\phi: S \to T$ be an $R$-algebraic structure homomorphism.


Then $\phi$ is an $R$-algebraic structure isomorphism if and only if $\phi$ is a bijection.


Ordered Structure Isomorphism

An ordered structure isomorphism from an ordered structure $\struct {S, \circ, \preceq}$ to another $\struct {T, *, \preccurlyeq}$ is a mapping $\phi: S \to T$ that is both:

$(1): \quad$ An isomorphism, that is a bijective homomorphism, from the structure $\struct {S, \circ}$ to the structure $\struct {T, *}$
$(2): \quad$ An order isomorphism from the ordered set $\struct {S, \preceq}$ to the ordered set $\struct {T, \preccurlyeq}$.


Isomorphic Copy

Let $\phi: S \to T$ be an isomorphism.

Let $x \in S$.


Then $\map \phi x \in T$ is known as the isomorphic copy of $x$ (under $\phi$).


Examples

$\struct {\Z \sqbrk {\sqrt 3}, +}$ with Numbers of Form $2^m 3^n$ under $\times$

Let $\Z \sqbrk {\sqrt 3}$ denote the set of quadratic integers over $3$:

$\Z \sqbrk {\sqrt 3} = \set {a + b \sqrt 3: a, b \in \Z}$


Let $S$ be the set defined as:

$S := \set {2^m 3^n: m, n \in \Z}$


Let $\struct {\Z \sqbrk {\sqrt 3}, +}$ and $\struct {S, \times}$ be the algebraic structures formed from the above with addition and multiplication respectively.


Then $\struct {\Z \sqbrk {\sqrt 3}, +}$ and $\struct {S, \times}$ are isomorphic.


$\struct {\N, +}$ under Doubling

Let $\N$ denote the set of natural numbers.

Let $2 \N$ denote the set of even non-negative integers:

$2 \N := \set {0, 2, 4, 6, \ldots}$

Let $\struct {\N, +}$ and $\struct {2 \N, +}$ be the algebraic structures formed from the above with addition.

Let $f: \N \to 2 \N$ be the mapping defined as:

$\forall n \in \N: \map f n = 2 n$


Then $f$ is an isomorphism.


Also see


Linguistic Note

The word isomorphism derives from the Greek morphe (μορφή) meaning form or structure, with the prefix iso- meaning equal.

Thus isomorphism means equal structure.


Sources

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