Isomorphism theorems#first isomorphism theorem

The isomorphism theorems were formulated in some generality for homomorphisms of modules by Emmy Noether in her paper Abstrakter Aufbau der Idealtheorie in algebraischen Zahl- und Funktionenkörpern, which was published in 1927 in Mathematische Annalen. Less general versions of these theorems can be found in work of Richard Dedekind and previous papers by Noether.

Three years later, B.L. van der Waerden published his influential Moderne Algebra, the first abstract algebra textbook that took the groups-rings-fields approach to the subject. Van der Waerden credited lectures by Noether on group theory and Emil Artin on algebra, as well as a seminar conducted by Artin, Wilhelm Blaschke, Otto Schreier, and van der Waerden himself on ideals as the main references. The three isomorphism theorems, called homomorphism theorem, and two laws of isomorphism when applied to groups, appear explicitly.

We first present the isomorphism theorems of the groups.

{{see also | Fundamental theorem on homomorphisms}}

Image:First-isomorphism-theorem.svg

Let $G$ and $H$ be groups, and let $f\; :\; G\; \backslash rightarrow\; H$ be a homomorphism. Then:

1. The kernel of $f$ is a normal subgroup of $G$,
2. The image of $f$ is a subgroup of $H$, and
3. The image of $f$ is isomorphic to the quotient group $G\; /\; \backslash ker\; f$.

In particular, if $f$ is surjective then $H$ is isomorphic to $G\; /\; \backslash ker\; f$.

This theorem is usually called the {{anchor|first isomorphism theorem}}first isomorphism theorem.

File:Diagram for the First Isomorphism Theorem.png

Let $G$ be a group. Let $S$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$. Then the following hold:

1. The product $SN$ is a subgroup of $G$,
2. The subgroup $N$ is a normal subgroup of $SN$,
3. The intersection $S\; \backslash cap\; N$ is a normal subgroup of $S$, and
4. The quotient groups $(SN)/N$ and $S/(S\backslash cap\; N)$ are isomorphic.

Technically, it is not necessary for $N$ to be a normal subgroup, as long as $S$ is a subgroup of the normalizer of $N$ in $G$. In this case, $N$ is not a normal subgroup of $G$, but $N$ is still a normal subgroup of the product $SN$.

This theorem is sometimes called the second isomorphism theorem, diamond theorem{{cite book|author=I. Martin Isaacs|author-link=Martin Isaacs|title=Algebra: A Graduate Course|url=https://archive.org/details/algebragraduatec00isaa|url-access=limited|year=1994|publisher=American Mathematical Soc.|isbn=978-0-8218-4799-2|page=[https://archive.org/details/algebragraduatec00isaa/page/n45 33]}} or the parallelogram theorem.{{cite book|author=Paul Moritz Cohn|author-link=Paul Moritz Cohn|title=Classic Algebra|url=https://archive.org/details/classicalgebra00cohn_300|url-access=limited|year=2000|publisher=Wiley|isbn=978-0-471-87731-8|page=[https://archive.org/details/classicalgebra00cohn_300/page/n256 245]}}

An application of the second isomorphism theorem identifies projective linear groups: for example, the group on the complex projective line starts with setting $G\; =\; \backslash operatorname\{GL\}\_2(\backslash mathbb\{C\})$, the group of invertible 2 × 2 complex matrices, $S\; =\; \backslash operatorname\{SL\}\_2(\backslash mathbb\{C\})$, the subgroup of determinant 1 matrices, and $N$ the normal subgroup of scalar matrices , we have $S\; \backslash cap\; N\; =\; \backslash \{\backslash pm\; I\backslash \}$, where $I$ is the identity matrix, and $SN\; =\; \backslash operatorname\{GL\}\_2(\backslash mathbb\{C\})$. Then the second isomorphism theorem states that:

: $\backslash operatorname\{PGL\}\_2(\backslash mathbb\{C\})\; :=\; \backslash operatorname\{GL\}\_2\; \backslash left(\backslash mathbb\{C\})/(\backslash mathbb\{C\}^\{\backslash times\}\backslash !I\backslash right)\; \backslash cong\; \backslash operatorname\{SL\}\_2(\backslash mathbb\{C\})/\backslash \{\backslash pm\; I\backslash \}\; =:\; \backslash operatorname\{PSL\}\_2(\backslash mathbb\{C\})$

Let $G$ be a group, and $N$ a normal subgroup of $G$.

Then

1. If $K$ is a subgroup of $G$ such that $N\; \backslash subseteq\; K\; \backslash subseteq\; G$, then $G/N$ has a subgroup isomorphic to $K/N$.
2. Every subgroup of $G/N$ is of the form $K/N$ for some subgroup $K$ of $G$ such that $N\; \backslash subseteq\; K\; \backslash subseteq\; G$.
3. If $K$ is a normal subgroup of $G$ such that $N\; \backslash subseteq\; K\; \backslash subseteq\; G$, then $G/N$ has a normal subgroup isomorphic to $K/N$.
4. Every normal subgroup of $G/N$ is of the form $K/N$ for some normal subgroup $K$ of $G$ such that $N\; \backslash subseteq\; K\; \backslash subseteq\; G$.
5. If $K$ is a normal subgroup of $G$ such that $N\; \backslash subseteq\; K\; \backslash subseteq\; G$, then the quotient group $(G/N)/(K/N)$ is isomorphic to $G/K$.

The last statement is sometimes referred to as the third isomorphism theorem. The first four statements are often subsumed under Theorem D below, and referred to as the lattice theorem, correspondence theorem, or fourth isomorphism theorem.

{{main | Lattice theorem}}

Let $G$ be a group, and $N$ a normal subgroup of $G$.

The canonical projection homomorphism $G\backslash rightarrow\; G/N$ defines a bijective correspondence

between the set of subgroups of $G$ containing $N$ and the set of (all) subgroups of $G/N$. Under this correspondence normal subgroups correspond to normal subgroups.

This theorem is sometimes called the correspondence theorem, the lattice theorem, and the fourth isomorphism theorem.

The Zassenhaus lemma (also known as the butterfly lemma) is sometimes called the fourth isomorphism theorem.{{cite book |last=Wilson |first=Robert A. |author-link=Robert A. Wilson (mathematician) |title=The Finite Simple Groups |date=2009 |doi=10.1007/978-1-84800-988-2 |at=p. 7 |publisher=Springer-Verlag London |isbn=978-1-4471-2527-3 |series=Graduate Texts in Mathematics 251|volume=251 }}

The first isomorphism theorem can be expressed in category theoretical language by saying that the category of groups is (normal epi, mono)-factorizable; in other words, the normal epimorphisms and the monomorphisms form a factorization system for the category. This is captured in the commutative diagram in the margin, which shows the objects and morphisms whose existence can be deduced from the morphism $f\; :\; G\; \backslash rightarrow\; H$. The diagram shows that every morphism in the category of groups has a kernel in the category theoretical sense; the arbitrary morphism f factors into $\backslash iota\; \backslash circ\; \backslash pi$, where ι is a monomorphism and π is an epimorphism (in a conormal category, all epimorphisms are normal). This is represented in the diagram by an object $\backslash ker\; f$ and a monomorphism $\backslash kappa:\; \backslash ker\; f\; \backslash rightarrow\; G$ (kernels are always monomorphisms), which complete the short exact sequence running from the lower left to the upper right of the diagram. The use of the exact sequence convention saves us from having to draw the zero morphisms from $\backslash ker\; f$ to $H$ and $G\; /\; \backslash ker\; f$.

If the sequence is right split (i.e., there is a morphism σ that maps $G\; /\; \backslash operatorname\{ker\}\; f$ to a {{pi}}-preimage of itself), then G is the semidirect product of the normal subgroup $\backslash operatorname\{im\}\; \backslash kappa$ and the subgroup $\backslash operatorname\{im\}\; \backslash sigma$. If it is left split (i.e., there exists some $\backslash rho:\; G\; \backslash rightarrow\; \backslash operatorname\{ker\}\; f$ such that $\backslash rho\; \backslash circ\; \backslash kappa\; =\; \backslash operatorname\{id\}\_\{\backslash text\{ker\}\; f\}$), then it must also be right split, and $\backslash operatorname\{im\}\; \backslash kappa\; \backslash times\; \backslash operatorname\{im\}\; \backslash sigma$ is a direct product decomposition of G. In general, the existence of a right split does not imply the existence of a left split; but in an abelian category (such as that of abelian groups), left splits and right splits are equivalent by the splitting lemma, and a right split is sufficient to produce a direct sum decomposition $\backslash operatorname\{im\}\; \backslash kappa\; \backslash oplus\; \backslash operatorname\{im\}\; \backslash sigma$. In an abelian category, all monomorphisms are also normal, and the diagram may be extended by a second short exact sequence $0\; \backslash rightarrow\; G\; /\; \backslash operatorname\{ker\}\; f\; \backslash rightarrow\; H\; \backslash rightarrow\; \backslash operatorname\{coker\}\; f\; \backslash rightarrow\; 0$.

In the second isomorphism theorem, the product SN is the join of S and N in the lattice of subgroups of G, while the intersection S ∩ N is the meet.

The third isomorphism theorem is generalized by the nine lemma to abelian categories and more general maps between objects.

Below we present four theorems, labelled A, B, C and D. They are often numbered as "First isomorphism theorem", "Second..." and so on; however, there is no universal agreement on the numbering. Here we give some examples of the group isomorphism theorems in the literature. Notice that these theorems have analogs for rings and modules.

It is less common to include the Theorem D, usually known as the lattice theorem or the correspondence theorem, as one of isomorphism theorems, but when included, it is the last one.

The statements of the theorems for rings are similar, with the notion of a normal subgroup replaced by the notion of an ideal.

Let $R$ and $S$ be rings, and let $\backslash varphi:R\backslash rightarrow\; S$ be a ring homomorphism. Then:

1. The kernel of $\backslash varphi$ is an ideal of $R$,
2. The image of $\backslash varphi$ is a subring of $S$, and
3. The image of $\backslash varphi$ is isomorphic to the quotient ring $R/\backslash ker\backslash varphi$.

In particular, if $\backslash varphi$ is surjective then $S$ is isomorphic to $R/\backslash ker\backslash varphi$.{{Cite web |last=Moy |first=Samuel |date=2022 |title=An Introduction to the Theory of Field Extensions |url=https://math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Moy.pdf |access-date=Dec 20, 2022 |website=UChicago Department of Math}}

Let $R$ be a ring. Let $S$ be a subring of $R$, and let $I$ be an ideal of $R$. Then:

1. The sum $S+I=\backslash \{s+i\backslash mid\; s\backslash in\; S,i\backslash in\; I\backslash \}$ is a subring of $R$,
2. The intersection $S\backslash cap\; I$ is an ideal of $S$, and
3. The quotient rings $(S+I)/I$ and $S/(S\backslash cap\; I)$ are isomorphic.

Let R be a ring, and I an ideal of R. Then

1. If $A$ is a subring of $R$ such that $I\; \backslash subseteq\; A\; \backslash subseteq\; R$, then $A/I$ is a subring of $R/I$.
2. Every subring of $R/I$ is of the form $A/I$ for some subring $A$ of $R$ such that $I\; \backslash subseteq\; A\; \backslash subseteq\; R$.
3. If $J$ is an ideal of $R$ such that $I\; \backslash subseteq\; J\; \backslash subseteq\; R$, then $J/I$ is an ideal of $R/I$.
4. Every ideal of $R/I$ is of the form $J/I$ for some ideal $J$ of $R$ such that $I\; \backslash subseteq\; J\; \backslash subseteq\; R$.
5. If $J$ is an ideal of $R$ such that $I\; \backslash subseteq\; J\; \backslash subseteq\; R$, then the quotient ring $(R/I)/(J/I)$ is isomorphic to $R/J$.

Let $I$ be an ideal of $R$. The correspondence $A\backslash leftrightarrow\; A/I$ is an inclusion-preserving bijection between the set of subrings $A$ of $R$ that contain $I$ and the set of subrings of $R/I$. Furthermore, $A$ (a subring containing $I$) is an ideal of $R$ if and only if $A/I$ is an ideal of $R/I$.{{cite book | last1=Dummit | first1=David S. | first2=Richard M. | last2=Foote | title=Abstract algebra | url=https://archive.org/details/abstractalgebra00dumm_304 | url-access=limited | location=Hoboken, NJ | publisher=Wiley | date=2004 | page=[https://archive.org/details/abstractalgebra00dumm_304/page/n259 246] | isbn=978-0-471-43334-7}}

The statements of the isomorphism theorems for modules are particularly simple, since it is possible to form a quotient module from any submodule. The isomorphism theorems for vector spaces (modules over a field) and abelian groups (modules over $\backslash mathbb\{Z\}$) are special cases of these. For finite-dimensional vector spaces, all of these theorems follow from the rank–nullity theorem.

In the following, "module" will mean "R-module" for some fixed ring R.

Let $M$ and $N$ be modules, and let $\backslash varphi:M\backslash rightarrow\; N$ be a module homomorphism. Then:

1. The kernel of $\backslash varphi$ is a submodule of $M$,
2. The image of $\backslash varphi$ is a submodule of $N$, and
3. The image of $\backslash varphi$ is isomorphic to the quotient module $M/\backslash ker\backslash varphi$.

In particular, if $\backslash varphi$ is surjective then $N$ is isomorphic to $M/\backslash ker\backslash varphi$.

Let $M$ be a module, and let $S$ and $T$ be submodules of $M$. Then:

1. The sum $S+T=\backslash \{s+t\backslash mid\; s\backslash in\; S,t\backslash in\; T\backslash \}$ is a submodule of $M$,
2. The intersection $S\backslash cap\; T$ is a submodule of $M$, and
3. The quotient modules $(S+T)/T$ and $S/(S\backslash cap\; T)$ are isomorphic.

Let M be a module, T a submodule of M.

1. If $S$ is a submodule of $M$ such that $T\; \backslash subseteq\; S\; \backslash subseteq\; M$, then $S/T$ is a submodule of $M/T$.
2. Every submodule of $M/T$ is of the form $S/T$ for some submodule $S$ of $M$ such that $T\; \backslash subseteq\; S\; \backslash subseteq\; M$.
3. If $S$ is a submodule of $M$ such that $T\; \backslash subseteq\; S\; \backslash subseteq\; M$, then the quotient module $(M/T)/(S/T)$ is isomorphic to $M/S$.

Let $M$ be a module, $N$ a submodule of $M$. There is a bijection between the submodules of $M$ that contain $N$ and the submodules of $M/N$. The correspondence is given by $A\backslash leftrightarrow\; A/N$ for all $A\backslash supseteq\; N$. This correspondence commutes with the processes of taking sums and intersections (i.e., is a lattice isomorphism between the lattice of submodules of $M/N$ and the lattice of submodules of $M$ that contain $N$).Dummit and Foote (2004), p. 349

To generalise this to universal algebra, normal subgroups need to be replaced by congruence relations.

A congruence on an algebra $A$ is an equivalence relation $\backslash Phi\backslash subseteq\; A\; \backslash times\; A$ that forms a subalgebra of $A\; \backslash times\; A$ considered as an algebra with componentwise operations. One can make the set of equivalence classes $A/\backslash Phi$ into an algebra of the same type by defining the operations via representatives; this will be well-defined since $\backslash Phi$ is a subalgebra of $A\; \backslash times\; A$. The resulting structure is the quotient algebra.

Let $f:A\; \backslash rightarrow\; B$ be an algebra homomorphism. Then the image of $f$ is a subalgebra of $B$, the relation given by $\backslash Phi:f(x)=f(y)$ (i.e. the kernel of $f$) is a congruence on $A$, and the algebras $A/\backslash Phi$ and $\backslash operatorname\{im\}\; f$ are isomorphic. (Note that in the case of a group, $f(x)=f(y)$ iff $f(xy^\{-1\})\; =\; 1$, so one recovers the notion of kernel used in group theory in this case.)

Given an algebra $A$, a subalgebra $B$ of $A$, and a congruence $\backslash Phi$ on $A$, let $\backslash Phi\_B\; =\; \backslash Phi\; \backslash cap\; (B\; \backslash times\; B)$ be the trace of $\backslash Phi$ in $B$ and $[B]^\backslash Phi=\backslash \{K\; \backslash in\; A/\backslash Phi:\; K\; \backslash cap\; B\; \backslash neq\backslash emptyset\backslash \}$ the collection of equivalence classes that intersect $B$. Then

1. $\backslash Phi\_B$ is a congruence on $B$,
2. $\backslash \; [B]^\backslash Phi$ is a subalgebra of $A/\backslash Phi$, and
3. the algebra $[B]^\backslash Phi$ is isomorphic to the algebra $B/\backslash Phi\_B$.

Let $A$ be an algebra and $\backslash Phi,\; \backslash Psi$ two congruence relations on $A$ such that $\backslash Psi\; \backslash subseteq\; \backslash Phi$. Then $\backslash Phi/\backslash Psi\; =\; \backslash \{\; ([a\text{'}]\_\backslash Psi,[a$]_\Psi): (a',a)\in \Phi\} = [\ ]_\Psi \circ \Phi \circ [\ ]_\Psi^{-1} is a congruence on $A/\backslash Psi$, and $A/\backslash Phi$ is isomorphic to $(A/\backslash Psi)/(\backslash Phi/\backslash Psi).$

Let $A$ be an algebra and denote $\backslash operatorname\{Con\}A$ the set of all congruences on $A$. The set

$\backslash operatorname\{Con\}A$ is a complete lattice ordered by inclusion.Burris and Sankappanavar (2012), p. 37

If $\backslash Phi\backslash in\backslash operatorname\{Con\}A$ is a congruence and we denote by $\backslash left[\backslash Phi,A\backslash times\; A\backslash right]\backslash subseteq\backslash operatorname\{Con\}A$ the set of all congruences that contain $\backslash Phi$ (i.e. $\backslash left[\backslash Phi,A\backslash times\; A\backslash right]$ is a principal filter in $\backslash operatorname\{Con\}A$, moreover it is a sublattice), then

the map $\backslash alpha:\backslash left[\backslash Phi,A\backslash times\; A\backslash right]\backslash to\backslash operatorname\{Con\}(A/\backslash Phi),\backslash Psi\backslash mapsto\backslash Psi/\backslash Phi$ is a lattice isomorphism.Burris and Sankappanavar (2012), p. 49{{cite web |first1=William |last1=Sun |title=Is there a general form of the correspondence theorem? |url=https://math.stackexchange.com/q/2850331 |website=Mathematics StackExchange |access-date=20 July 2019}}

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