Definition:Isomorphism (Abstract Algebra)
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.
Also see
- Results about isomorphisms in the context of abstract algebra can be found here.
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures
- 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 7$: Homomorphisms and quotient algebras