# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 47/refcontrol

    Homomorphism theorems

## TheoremTheorem 47.1 change

Let $G,Q$ and $H$ be groups,MDLD/groups and let $\varphi \colon G\rightarrow H$ be a group homomorphismMDLD/group homomorphism and $\psi \colon G\rightarrow Q$ a surjectiveMDLD/surjective group homomorphism. Suppose that
    $\operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,.$
ist. Then there exists a uniquely determined group homomorphism
    ${\tilde {\varphi }}\colon Q\longrightarrow H$
such that $\varphi ={\tilde {\varphi }}\circ \psi$ holds. Put differently, the diagram
    ${\begin{matrix}G&{\stackrel {\varphi }{\longrightarrow }}&H&\\\!\!\!\!\!\psi \downarrow &\nearrow {\tilde {\varphi }}\!\!\!\!\!&\\Q&&&&\!\!\!\!\!\!\!\!\\\end{matrix}}$

### Proof

We show firstly the uniqueness. For every Element $u\in Q$, there exists some $g\in G$ with $\psi (g)=u$. The commutativity of the diagram ensures that
    ${\tilde {\varphi }}(u)=\varphi (g)\,$
holds. This means that there exists at most one ${\tilde {\varphi }}$.
We have to show that this condition yields a well-defined mapping. Hence, let $g,g'\in G$ be two preimages of $u$. Then
    $\psi {\left(g'g^{-1}\right)}=uu^{-1}=e_{Q}\,;$
therefore, we have $g'g^{-1}\in \operatorname {kern} \psi \subseteq \operatorname {kern} \varphi$. Hence, $\varphi (g)=\varphi (g')$, and the mapping is well-defined. Let $u,v\in Q$ be given, and let $g,h\in G$ be preimages. Then $gh$ is a preimage of $uv$. Therefore,
    ${\tilde {\varphi }}(uv)=\varphi (gh)=\varphi (g)\varphi (h)={\tilde {\varphi }}(u){\tilde {\varphi }}(v)\,.$
This means that ${\tilde {\varphi }}$ is a group homomorphism.
$\Box$

## Example Create referencenumber

We consider the surjective group homomorphismsMDLD/group homomorphisms
    $\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(4)$
and
    $\psi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(12).$
We have
    $\operatorname {kern} \psi =\mathbb {Z} \cdot 12\subseteq \mathbb {Z} \cdot 4=\operatorname {kern} \varphi \,.$
Due to Theorem 47.1 , there exists a uniquely determined group homomorphism
    ${\tilde {\varphi }}\colon \mathbb {Z} /(12)\longrightarrow \mathbb {Z} /(4),$
which is compatible with the remainder mappings. The morphism ${\tilde {\varphi }}$ sends the remainder of a number after division by $12$ to the remainder after division by $4$. In particular, the theorem implies that the second remainder does only depend on the first remainder, not on the number itself.
If, to the contrary, we consider
    $\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(5)$
and
    $\psi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(12),$
then
    $\operatorname {kern} \psi =\mathbb {Z} \cdot 12\not \subseteq \mathbb {Z} \cdot 5=\operatorname {kern} \varphi \,,$
and there does not exists a natural mapping
    $\mathbb {Z} /(12)\longrightarrow \mathbb {Z} /(5).$
For example, the numbers $1,13,25,37,49$ have modulo $12$ the remainder $1$ but modulo $5$ their remainders are $1,3,0,2,4$.
The mapping constructed in this theorem is called induced mapping or induced homomorphism, and the theorem is called the theorem about the induced homomorphism.

## CorollaryCorollary 47.3 change

Let $G$ and $H$ be groups,MDLD/groups and let
    $\varphi \colon G\longrightarrow H$
be a surjectiveMDLD/surjective group homomorphism.MDLD/group homomorphism Then there exists a canonical isomorphismMDLD/isomorphism (group)
    ${\tilde {\varphi }}\colon G/\operatorname {kern} \varphi \longrightarrow H.$

### Proof

We apply Theorem 47.1
to $Q=G/\operatorname {kern} \varphi$ and the canonical projectionMDLD/canonical projection (quotient set) $q\colon G\rightarrow G/\operatorname {kern} \varphi$. This induces a group homomorphism
    ${\tilde {\varphi }}\colon G/\operatorname {kern} \varphi \longrightarrow H$
fulfilling $\varphi ={\tilde {\varphi }}\circ q$, which is surjective. Let $[x]\in G/\operatorname {kern} \varphi$ and $[x]\in \operatorname {kern} {\tilde {\varphi }}$. Then
    ${\tilde {\varphi }}([x])=\varphi (x)=e_{H}\,.$
Therefore, $x\in \operatorname {kern} \varphi$. Hence, $[x]=e_{Q}$, that is, the kernel of ${\tilde {\varphi }}$ is trivial, and so, due to Fact *****, ${\tilde {\varphi }}$ is also injective.
$\Box$

## Example Create referencenumber

Let $G$ be a cyclic groupMDLD/cyclic group with a generator $g$. We consider the corresponding group homomorphismMDLD/group homomorphism
    $\varphi \colon \mathbb {Z} \longrightarrow G,n\longmapsto g^{n},$
in the sense of Fact *****. Since we have a generator, this mapping is surjective.MDLD/surjective The kernelMDLD/kernel (group) of this mapping is determined the orderMDLD/order (group element) of $g$, which we denote by $k$ (or it is $0$, in case the order is $\infty$). Due to Corollary 47.3 , there exists a canonical isomorphism
    ${\tilde {\varphi }}\colon \mathbb {Z} /(k)\longrightarrow G.$
In particular, there exists, up to isomorphism, for every $k$, exactly one cyclic group, namely $\mathbb {Z} /(k)$.

## Example Create referencenumber

The group homomorphismMDLD/group homomorphism
    $\mathbb {R} \longrightarrow S^{1},t\longmapsto {\begin{pmatrix}\cos t\\\sin t\end{pmatrix}},$
is surjective.MDLD/surjective Due to the periodicity
of the trigonometric functions,MDLD/trigonometric functions the kernelMDLD/kernel (group) equals $\mathbb {Z} 2\pi$. By the isomorphism theorem, there exists a canonical isomorphism
    $\mathbb {R} /\mathbb {Z} 2\pi \cong S^{1}\,.$

## Example Create referencenumber

The complex exponential functionMDLD/complex exponential function
    $(\mathbb {C} ,0,+)\longrightarrow (\mathbb {C} ^{\times },1,\cdot ),z\longmapsto \exp z,$
is a surjectiveMDLD/surjective group homomorphism.MDLD/group homomorphism The kernelMDLD/kernel (group) is $\mathbb {Z} 2\pi {\mathrm {i} }$. Due to the isomorphism theorem, there exists a canonical isomorphism
    $\mathbb {C} /\mathbb {Z} 2\pi {\mathrm {i} }\cong \mathbb {C} ^{\times }\,.$

## Example Create referencenumber

The determinantMDLD/determinant
    $\det \colon \operatorname {GL} _{n}\!{\left(K\right)}\longrightarrow K^{\times },M\longmapsto \det M,$
is a surjectiveMDLD/surjective group homomorphism;MDLD/group homomorphism its kernelMDLD/kernel (group) is by definition the special linear groupMDLD/special linear group $\operatorname {SL} _{n}\!{\left(K\right)}$. Due to the isomorphism theorem, there exists a canonical isomorphism
    $\operatorname {GL} _{n}\!{\left(K\right)}/\operatorname {SL} _{n}\!{\left(K\right)}\cong K^{\times }\,.$

## TheoremCreate referencenumber

Let $G$ and $H$ be groups,MDLD/groups and let
    $\varphi \colon G\longrightarrow H$
be a group homomorphism.MDLD/group homomorphism Then there exists a canonical factorization
    $G{\stackrel {q}{\longrightarrow }}G/\operatorname {kern} \varphi {\stackrel {\theta }{\longrightarrow }}\operatorname {Im} \varphi {\stackrel {\iota }{\hookrightarrow }}H,$
where $q$ is the canonical projection,MDLD/canonical projection (quotient set) $\theta$ is a group isomorphism,MDLD/group isomorphism and $\iota$ is the canonical inclusion of the

### Proof

This follows from Corollary 47.3 , applied to the image group $U=\operatorname {Im} \varphi \subseteq H$.
$\Box$
This statement is often briefly expressed by saying:
image $=$ preimage modulo kernel.

## TheoremCreate referencenumber

Let $G$ be a group,MDLD/group and let $N\subseteq G$ be a normal subgroupMDLD/normal subgroup with residue class groupMDLD/residue class group $Q=G/N$. Let $H\subseteq G$ be another normal subgroup in $G$ with $N\subseteq H$. Then the imageMDLD/image ${\overline {H}}$ of $H$ in $Q$ is a normal subgroup, and we have a canonical isomorphismMDLD/isomorphism (group)
    $G/H\cong Q/{\overline {H}}\,.$

### Proof

For the first statement, see exercise *****. Therefore, the residue class group $Q/{\overline {H}}$ is well-defined. We consider the composition
    $p\circ q:G\longrightarrow Q\longrightarrow Q/{\overline {H}}.$
Because of
    ${\begin{aligned}\operatorname {kern} {\left(p\circ q\right)}&={\left\{x\in G\mid (p\circ q)(x)=e\right\}}\\&={\left\{x\in G\mid q(x)\in \operatorname {kern} p\right\}}\\&={\left\{x\in G\mid q(x)\in {\overline {H}}\right\}}\\&=H,\end{aligned}}$
we have $\operatorname {kern} {\left(p\circ q\right)}=H$. Hence, according to Corollary 47.3 , we get the canonical isomorphism
    $G/H\longrightarrow Q/{\overline {H}}.$
$\Box$
In short, we have
    $G/H=(G/N)/(H/N)\,.$
    Residue class rings
On the residue class group for a normal subgroup in a group, there are quite often additional structures available, if the group and the normal subgroup fulfill certain properties. In the next lecture, we will look at residue class spaces for a linear subspace. Here, we discuss briefly residue class rings for an ideal in a commutative ring. We recall the definition of a ring homomorphism.

## Definition

Let $R$ and $S$ be rings.MDLD/rings A mapping
    $\varphi \colon R\longrightarrow S$
is called a ring homomorphism, if the following properties hold:
* $\varphi (a+b)=\varphi (a)+\varphi (b)$.
* $\varphi (1)=1$.
* $\varphi (a\cdot b)=\varphi (a)\cdot \varphi (b)$.

## LemmaLemma 47.11 change

Let $R$ and $S$ be commutative rings,MDLD/commutative rings and let
    $\varphi \colon R\longrightarrow S$
be a ring homomorphism.MDLD/ring homomorphism Then the kernelMDLD/kernel (ring)
    $\operatorname {kern} \varphi ={\left\{f\in R\mid \varphi (f)=0\right\}}\,$
is an idealMDLD/ideal
$R$

### Proof

$\Box$
Due to Lemma 47.11 , the kernelMDLD/kernel (ring) of a ring homomorphismMDLD/ring homomorphism is an ideal.MDLD/ideal (commutative) We will construct, for any given ideal $I\subseteq R$ in a (commutative) ring, a commutative ring $R/I$ and a surjective ring homomorphism
    $R\longrightarrow R/I,$
dessen kernel is the ideal $I$. Therefore, ideals and kernels of ring homomorphisms are essentially equivalent objects, in the same way as, in group theory, normal subgroups and kernels of group homomorphisms are the same. In fact, the same homomorphism theorems hold again, and follow easily from the group situation.

## Definition

Let $R$ be a commutative ring,MDLD/commutative ring and let $I\subseteq R$ be an idealMDLD/ideal in $R$. For $a\in R$, the subset
    $a+I={\left\{a+f\mid f\in I\right\}}\,$
$a$
$I$
$I$
These cosets are just the cosetsMDLD/cosets (group) of the (additive) subgroup $I\subseteq R$, which is a normal subgroup,MDLD/normal subgroup as $(R,+,0)$ is a commutative group. Two elements $a,b\in R$ define the same coset, that is, $a+I=b+I$, if and only if their difference $a-b$ belongs to the ideal. We also say that $a$ and $b$ represent the same coset.

## Definition

Let $R$ be a commutative ring,MDLD/commutative ring and let $I\subseteq R$ be an idealMDLD/ideal in $R$. Then the residue class ring $R/I$ (R modulo I) is a commutative ring that is determined by the following data.
* As a set, $R/I$ is the set of all cosetsMDLD/cosets (ideal) to $I$.
* An addition of cosets is defined by
    $(a+I)+(b+I):=(a+b+I)\,.$
* A multiplication of cosets is defined by
    $(a+I)\cdot (b+I):=(a\cdot b+I)\,.$
* ${\bar {0}}=0+I=I$ is the neutral element of the addition (the zero class).
* ${\bar {1}}=1+I$ is the neutral element of the multiplication (the unit class).
We have to show that these mappings (that is, addition and multiplication) are well-defined, that is, independent of the choice of representatives, and that the ring axioms are fulfilled. As $I$ is, in particular, a subgroup of the commutative group $(R,+,0)$, it is a normal subgroup, so that $R/I$ is a group and that the residue class mapping
    $R\longrightarrow R/I,a\longmapsto a+I=:{\bar {a}},$
is a group homomorphism. The only new feature compared with the group situation is that we now have also a multiplication. That the multiplication is well-defined can been seen as follows: Let two residue classes be given with different representatives, that is, ${\overline {a}}\,={\overline {a'}}\,$ and ${\overline {b}}\,={\overline {b'}}\,$. Then $a-a'\in I$ and $b-b'\in I$, and $a'=a+x$ and $b'=b+y$ with $x,y\in I$. This implies
    $a'b'=(a+x)(b+y)=ab+ay+xb+xy\,.$
The three summands on the right belong to the ideal, so that the difference $a'b'-ab\in I$. From the well-definedness, the other properties follows, in particular, that we have a ring homomorphism to the residue class ring. Again, this morphism is called the residue class mapping, or the residue class homomorphism. The image of $a\in R$ in $R/I$ is often denoted by $[a]$, ${\bar {a}}$, or just by $a$ again, and it is called the residue class of $a$. Under this mapping, exactly the elements from the ideal are sent to $0$, that is, the kernel of the residue class mapping is the given ideal.
The simplest example of this process is the mapping that sends an integer number $a$ to its remainder after division by a fixed number $d$. Every remainder is represented by one of the numbers $0,1,2,\ldots ,d-1$. In general, there does not always exist such a nice system of representatives.
    The residue class rings of $\mathbb {Z}$
We know already the residue class groups $\mathbb {Z} /(d)$; they are cyclic groups of order $d$. Moreover, these groups get now also a ring structure.

## Corollary

Let $d\geq 0$ be a natural number. Then there exists a uniquely determined ring structure on $\mathbb {Z} /(d)$ such that the residue class mappingMDLD/residue class mapping
    $\mathbb {Z} \longrightarrow \mathbb {Z} /(d),a\longmapsto {\overline {a}}\,,$
is a ring homomorphism.MDLD/ring homomorphism $\mathbb {Z} /(d)$ is a commutative ring with $d$ elements (for
$d\geq 1$

### Proof

$\Box$

## TheoremCreate referencenumber

Let $n\geq 1$ be a natural number, and let $\mathbb {Z} /(n)$ be the corresponding residue class ring.MDLD/residue class ring Then the following statements are equivalent.
* $\mathbb {Z} /(n)$ is a field.MDLD/field
* $\mathbb {Z} /(n)$ is a domain.MDLD/domain
* $n$ is a prime number.MDLD/prime number

### Proof

$\Box$
The residue class rings $S=K[X]/(P)$ are also quite easy to understand. If $P$ has degree $d$, then every residue class in $S$ is represented by a unique polynomial of degree $<d$. This polynomial is the remainder that we get by dividing through $P$.