# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 44/refcontrol

In the following lectures, we will enhance our methods by considering equivalence relations for algebraic structures and the formation of residue classes. For different algebraic structures (like groups, rings, vector spaces), these constructions follow the same scheme; therefore, we describe first this construction for groups.
    Groups
For an element $g\in G$ of a (multiplicatively written) group $G$ and $n\in \mathbb {N}$, we write
    $g^{n}=g\cdots g\,$
($n$ times) and
    $g^{n}={\left(g^{-1}\right)}^{-n}\,$
for $n\in \mathbb {Z} _{-}$. Due to the exponent rules, see exercise *****, this fits together well. For permutations and invertible matrices, we have encountered the order of an element several times already.

## Definition

Let $G$ be a groupMDLD/group and $g\in G$ an element. Then we call the smallest positive number $n$ with $g^{n}=e_{G}$ the order of $g$. For this, we write $\operatorname {ord} \,(g)$. If all positive powers of $g$ are different from the neutral element, then we set
$\operatorname {ord} \,(g)=\infty$

## Definition

A groupMDLD/group
$G$
This means that there exists an element $g\in G$ (a generator) such that every element in $G$ can be written as $g^{n}$ with some $n\in \mathbb {Z}$. The group $(\mathbb {Z} ,+,0)$ is cyclic, we can take $1$ or $-1$ as a generator. Also all subgroups of $\mathbb {Z}$ are cyclic themselves, as the following theorem shows.

## TheoremTheorem 44.3 change

The subgroupsMDLD/subgroups of $\mathbb {Z}$ are precisely the subsets of the form
    $\mathbb {Z} d={\left\{kd\mid k\in \mathbb {Z} \right\}}\,$
$d$

### Proof

$\Box$

## Example Create referencenumber

Let $n\in \mathbb {N} _{+}$, and consider on
    $\mathbb {Z} /(n)=\{0,1,\ldots ,n-1\}\,$
the binary operationMDLD/binary operation
    $a+b:=(a+b)\mod n={\begin{cases}a+b,{\text{ if }}a+b<n\,,\\a+b-n,{\text{ if }}a+b\geq n\,.\end{cases}}\,$
With this operation, we have a groupMDLD/group due to exercise *****. Because we can write every element as a certain sum of $1$ with itself, this is a cyclic group.MDLD/cyclic group
    Group homomorphisms

## Definition

Let $(G,\circ ,e_{G})$ and $(H,\circ ,e_{H})$ denote groups.MDLD/groups A mappingMDLD/mapping
    $\psi \colon G\longrightarrow H$
is called group homomorphism, if the equality
    $\psi (g\circ g')=\psi (g)\circ \psi (g')\,$
holds for all
$g,g'\in G$
The set of all group homomorphisms from $G$ to $H$ is denoted by
    $\operatorname {Hom} _{\left(G,H\right)}.$
Linear mappingsMDLD/Linear mappings between vector spaces are in particular group homomorphisms. The following two lemmas follow directly from the definition.

## LemmaCreate referencenumber

Let $G$ and $H$ denote groups,MDLD/groups and let $\varphi \colon G\rightarrow H$ be a group homomorphism.MDLD/group homomorphism Then $\varphi (e_{G})=e_{H}$ and $(\varphi (g))^{-1}=\varphi {\left(g^{-1}\right)}$ for every
$g\in G$

### Proof

To prove the first statement, consider
    $\varphi (e_{G})=\varphi (e_{G}e_{G})=\varphi (e_{G})\varphi (e_{G})\,.$
Multiplication with $\varphi (e_{G})^{-1}$ yields $e_{H}=\varphi (e_{G})$.
To prove the second claim, we use
    $\varphi {\left(g^{-1}\right)}\varphi (g)=\varphi {\left(g^{-1}g\right)}=\varphi (e_{G})=e_{H}\,.$
This means that $\varphi {\left(g^{-1}\right)}$ has the property that characterizes the inverse element of $\varphi (g)$. Since the inverse element in a group is, due to Lemma 3.2 , uniquely determined, we must have $\varphi {\left(g^{-1}\right)}=(\varphi (g))^{-1}$.
$\Box$

## LemmaCreate referencenumber

Let $F,G,H$ denote
* The identity is a group homomorphism.MDLD/group homomorphism
    $\operatorname {Id} \colon G\longrightarrow G$
* If $\varphi :F\rightarrow G$ and $\psi :G\rightarrow H$ are group homomorphisms, then the composition $\psi \circ \varphi \colon F\rightarrow H$ is a group homomorphism.
* For a subgroupMDLD/subgroup $F\subseteq G$, the inclusion $F\hookrightarrow G$ is a group homomorphism.
* Let $\{e\}$ be the trivial group.MDLD/trivial group Then the mapping $\{e\}\rightarrow G$ that sends $e$ to $e_{G}$ is a group homomorphism. Moreover, the (constant) mapping $G\rightarrow \{e\}$ is a group homomorphism.

### Proof

$\Box$

## Example Create referencenumber

Let $d\in \mathbb {N}$ be fixed. The mapping
    $\mathbb {Z} \longrightarrow \mathbb {Z} ,n\longmapsto dn,$
is a group homomorphism.MDLD/group homomorphism This follows immediately from the distributive law. For $d\geq 1$, this mapping is injective,MDLD/injective and the image is the subgroupMDLD/subgroup $\mathbb {Z} d\subseteq \mathbb {Z}$. For $d=0$, we have the zero mapping. For $d=1$, the mapping is the identity.MDLD/identity For $d\geq 2$, the mapping is not surjective.MDLD/surjective

## Example Create referencenumber

Let $d\in \mathbb {N} _{+}$. We consider the set
    $\mathbb {Z} /(d)=\{0,1,\ldots ,d-1\}\,,$
together with the addition described in exercise *****, which makes it a group. The mapping
    $\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(d)$
that sends an integer number $n$ to its remainder after division by $d$ is a group homomorphism.MDLD/group homomorphism For, if $m=ad+r$ and $n=bd+s$ are given with $0\leq r,s<d$, then
    $m+n=(a+b)d+r+s\,.$
Here, it may happen that $r+s\geq d$. In this case,
    $\varphi (m+n)=r+s-d\,,$
and this coincides with the addition of $r$ and $s$ in $\mathbb {Z} /(d)$. This mapping is surjective, but not injective.

## Example Create referencenumber

For a fieldMDLD/field $K$ and $n\in \mathbb {N} _{+}$, the determinantMDLD/determinant
    $\det \colon \operatorname {GL} _{n}\!{\left(K\right)}\longrightarrow K^{\times },M\longmapsto \det M,$
is a group homomorphism.MDLD/group homomorphism This follows from the multiplication theorem for the determinant and Theorem 16.11 .

## Example Create referencenumber

The assignment
    $S_{n}\longrightarrow \{1,-1\},\pi \longmapsto \operatorname {sgn} (\pi ),$
where $S_{n}$ denotes the permutation groupMDLD/permutation group for $n$ elements, is a group homomorphism,MDLD/group homomorphism due to Theorem 18.13 .

## LemmaCreate referencenumber

Let $G$ denote a group.MDLD/group Then there is a correspondence between group elements $g\in G$ and group homomorphismsMDLD/group homomorphisms $\varphi$ from $\mathbb {Z}$ to $G$, given by
    $g\longmapsto (n\mapsto g^{n}){\text{ and }}\varphi \longmapsto \varphi (1).$

### Proof

Let $g\in G$ be fixed. That the mapping
    $\varphi _{g}\colon \mathbb {Z} \longrightarrow G,n\longmapsto g^{n},$
is a
group homomorphism,MDLD/group homomorphism is just a reformulation of the exponential laws. Because of $\varphi _{g}(1)=g^{1}=g$, we obtain from the power mapping $\varphi _{g}$ the group element back. Moreover, a group homomorphism $\varphi \colon \mathbb {Z} \rightarrow G$ is uniquely determined by $\varphi (1)$, as $\varphi (n)=(\varphi (1))^{n}$ for $n$ positive, and $\varphi (n)={\left((\varphi (1))^{-1}\right)}^{-n}$ for $n$ negative must hold.
$\Box$
This lemma can be stated quickly by saying $G\cong \operatorname {Hom} _{\left(\mathbb {Z} ,G\right)}$. It is more difficult to characterize the group homomorphisms from a group $G$ to $\mathbb {Z}$. The group homomorphisms from $\mathbb {Z}$ to $\mathbb {Z}$ are just the multiplications with a fixed integer number $a$, that is,
    $\mathbb {Z} \longrightarrow \mathbb {Z} ,x\longmapsto ax.$
    Group isomorphisms

## Definition

Let $G$ and $H$ be groups.MDLD/groups A bijective group homomorphismMDLD/group homomorphism
    $\varphi \colon G\longrightarrow H$
Bijective linear mappingsMDLD/linear mappings are in particular group isomorphisms.

## Definition

The groupsMDLD/groups $H$ and $G$ are called isomorphic, if there exists a group isomorphismMDLD/group isomorphism
$\varphi \colon G\rightarrow H$

## LemmaCreate referencenumber

Let $G$ and $H$ be groups,MDLD/groups and let
    $\varphi \colon G\longrightarrow H$
be a group isomorphism.MDLD/group isomorphism Then also the inverse mappingMDLD/inverse mapping
    $\varphi ^{-1}\colon H\longrightarrow G,h\longmapsto \varphi ^{-1}(h),$

### Proof

This follows from
    ${\begin{aligned}\varphi ^{-1}(h_{1}h_{2})&=\varphi ^{-1}{\left(\varphi (\varphi ^{-1}(h_{1}))\varphi (\varphi ^{-1}(h_{2}))\right)}\\&=\varphi ^{-1}{\left(\varphi {\left(\varphi ^{-1}(h_{1})\varphi ^{-1}(h_{2})\right)}\right)}\\&=\varphi ^{-1}(h_{1})\varphi ^{-1}(h_{2}).\end{aligned}}$
$\Box$

## Example Create referencenumber

We consider the additive group of the real numbers, that is $(\mathbb {R} ,0,+)$, and the multiplicative group of the positive real numbers, thus $(\mathbb {R} _{+},1,\cdot )$. Then the exponential function
    $\exp \colon \mathbb {R} \longrightarrow \mathbb {R} _{+},x\longmapsto \exp(x),$
is a group isomorphism.MDLD/group isomorphism This rests on basic analytic properties of the exponential function. The homomorphism property is just a reformulation of the functional equation
    $\exp(x+y)=e^{x+y}=e^{x}e^{y}=\exp(x)\exp(y)\,.$
The injectivity of the mapping follows from the strict monotonicity, the surjectivity follows from the Intermediate value theorem. The inverse mapping is the natural logarithm, which is also a group isomorphism.
Isomorphic groups are equal with respect to their group-theoretic properties. An isomorphism of a group to itself is called automorphism. The set of all automorphisms on $G$ form, with the composition of mappings, a group, which is denoted by $\operatorname {Aut} \,G$ and which is called the automorphism group of $G$. Important examples of automorphisms are the so-called inner automorphisms.

## Definition

Let $G$ be a group,MDLD/group and $g\in G$ be fixed. The mapping defined by $g$,
    $\kappa _{g}\colon G\longrightarrow G,x\longmapsto gxg^{-1},$
The mapping $\kappa _{g}$ is also called the conjugation with $g$. If $G$ is a commutative Gruppe, then, because of $gxg^{-1}=xgg^{-1}=x$, the identity is the only inner automorphism. Therefore, this concept is only interesting for non-commutative groups.

## LemmaCreate referencenumber

An inner automorphismMDLD/inner automorphism is indeed an automorphism. The assignment
    $G\longrightarrow \operatorname {Aut} \,G,g\longmapsto \kappa _{g},$
is a

### Proof

We have
    $\kappa _{g}(xy)=gxyg^{-1}=gxg^{-1}gyg^{-1}=\kappa _{g}(x)\kappa _{g}(y)\,,$
so that $\kappa _{g}$ this is a group homomorphism.MDLD/group homomorphism We have
    $\kappa _{g}(\kappa _{h}(x))=\kappa _{g}(hxh^{-1})=ghxh^{-1}g^{-1}=ghx(gh)^{-1}=\kappa _{gh}\,.$
This implies, on one hand, that
    $\kappa _{g^{-1}}\circ \kappa _{g}=\kappa _{g^{-1}g}=\operatorname {Id} _{G}\,;$
therefore, $\kappa _{g}$ is bijective and an automorphism. On the other hand, this implies that the total mapping $\kappa$ is a group homomorphism.
$\Box$

## Example Create referencenumber

For a fixed invertible matrixMDLD/invertible matrix $B\in \operatorname {GL} _{n}\!{\left(K\right)}$, the conjugationMDLD/conjugation (group)
    $\kappa _{B}\colon \operatorname {GL} _{n}\!{\left(K\right)}\longrightarrow \operatorname {GL} _{n}\!{\left(K\right)},M\longmapsto BMB^{-1},$
is just the mapping that assigns, to a describing matrix $M$ of a linear mapping with respect to a basis, the describing matrix with respect to a new basis.
    The kernel of a group homomorphism

## Definition

Let $G$ and $H$ be groups,MDLD/groups and let
    $\varphi \colon G\longrightarrow H$
be a group homomorphism.MDLD/group homomorphism Then the preimageMDLD/preimage of the neutral element is called the kernel of $\varphi$, denoted by
    $\operatorname {kern} \varphi =\varphi ^{-1}(e_{H})={\left\{g\in G\mid \varphi (g)=e_{H}\right\}}\,.$

## LemmaCreate referencenumber

Let $G$ and $H$ be groups,MDLD/groups and let
    $\varphi \colon G\longrightarrow H$
be a group homomorphism.MDLD/group homomorphism Then the kernelMDLD/kernel (group) of $\varphi$ is a subgroupMDLD/subgroup
$G$

### Proof

Because of $\varphi (e_{G})=e_{H}$, we have $e_{G}\in \operatorname {kern} \varphi$. Let $g,g'\in \operatorname {kern} \varphi$. Then
    $\varphi (gg')=\varphi (g)\varphi (g')=e_{H}e_{H}=e_{H}\,;$
therefore, also $gg'\in \operatorname {kern} \varphi$. Hence, the kernel is a submonoid. Now, let $g\in \operatorname {kern} \varphi$, and consider the inverse element $g^{-1}$. Due to Fact *****, we have
    $\varphi {\left(g^{-1}\right)}=(\varphi (g))^{-1}=e_{H}^{-1}=e_{H}\,;$
Hence, $g^{-1}\in \operatorname {kern} \varphi$.
$\Box$
As for linear mappings, we have again the kernel criterion for injectivity.

## LemmaCreate referencenumber

Let $G$ and $H$ be groups.MDLD/groups A group homomorphismMDLD/group homomorphism $\varphi \colon G\rightarrow H$ is injectiveMDLD/injective if and only if the kernelMDLD/kernel
$\varphi$

### Proof

If $\varphi$ is injective, then every element $h\in H$ is hit by at most one element from $G$. As $e_{G}$ is sent to $e_{H}$, no further element can be sent to $e_{H}$. Therefore, $\operatorname {kern} \varphi =\{e_{G}\}$. Now assume that this holds. Let $g,{\tilde {g}}\in G$ be elements mapping to $h\in H$. Then
    $\varphi {\left(g{\tilde {g}}^{-1}\right)}=\varphi (g)\varphi ({\tilde {g}})^{-1}=hh^{-1}=e_{H}\,;$
hence, $g{\tilde {g}}^{-1}\in \operatorname {kern} \varphi$, and so $g{\tilde {g}}^{-1}=e_{G}$ by the condition. Therefore, $g={\tilde {g}}$.
$\Box$
    The image of a group homomorphism

## LemmaCreate referencenumber

Let $G$ and $H$ denote groups,MDLD/groups and let $\varphi \colon G\rightarrow H$ be a group homomorphism.MDLD/group homomorphism Then the imageMDLD/image of $\varphi$ is a subgroupMDLD/subgroup
$H$

### Proof

Let $B:=\operatorname {Im} \varphi$. We have $e_{H}=\varphi (e_{G})\in B$. Let $h_{1},h_{2}\in B$. Then there exist $g_{1},g_{2}\in G$ such that $\varphi (g_{1})=h_{1}$ and $\varphi (g_{2})=h_{2}$. Therefore, $h_{1}\cdot h_{2}=\varphi (g_{1})\cdot \varphi (g_{2})=\varphi (g_{1}\cdot g_{2})\in B$. Similarly, for $h\in B$ there exists a $g\in G$ fulfilling $\varphi (g)=h$. Hence, $h^{-1}=(\varphi (g))^{-1}=\varphi (g^{-1})\in B$.
$\Box$

## Example Create referencenumber

We consider the analytic mapping
    $\mathbb {R} \longrightarrow \mathbb {C} ,t\longmapsto e^{{\mathrm {i} }t}=\cos t+{\mathrm {i} }\sin t.$
Due to the exponential law (or the addition theorems for the trigonometric functions), we have $e^{{\mathrm {i} }(t+s)}=e^{{\mathrm {i} }t}e^{{\mathrm {i} }s}$. Therefore, this is a group homomorphismMDLD/group homomorphism from the additive group $(\mathbb {R} ,+,0)$ into the multiplicative group $(\mathbb {C} ^{\times },\cdot ,1)$. We determine the kernelMDLD/kernel (group) and the image of this mapping. To determine the kernel, we must identify those real numbers $t$ fulfilling
    $\cos t=1\,\,{\text{ and }}\,\,\sin t=0.$
Because of the periodicity of the trigonometric functions, this is the case if and only if $t$ is an integer multiple of $2\pi$. Hence, the kernel is the subgroupMDLD/subgroup $2\pi \mathbb {Z}$. For a point in the image, we have $\vert {e^{{\mathrm {i} }t}}\vert =\sin ^{2}t+\cos ^{2}t=1$; therefore, the image point belongs to the complex unit circle. The trigonometric functions run through the complete unit circle, so that the image group is the complex unit circle with its complex multiplication.