# 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**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