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

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 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 group

${}G$ is called cyclic if it is generated by one element.

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.

Theorem

The subgroups of ${}\mathbb {Z}$ are precisely the subsets of the form

${}\mathbb {Z} d={\left\{kd\mid k\in \mathbb {Z} \right\}}\,$

with a uniquely determined nonnegative number ${}d$.

Proof

See exercise *****.

$\Box$

Example

Let ${}n\in \mathbb {N} _{+}$, and consider on

${}\mathbb {Z} /(n)=\{0,1,\ldots ,n-1\}\,$

the 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 group due to exercise *****. Because we can write every element as a certain sum of ${}1$ with itself, this is a cyclic group.

Group homomorphisms

We have mentioned group homomorphisms already in the 18th lecture in the context of the sign of a permutation.

Definition

Let ${}(G,\circ ,e_{G})$ and ${}(H,\circ ,e_{H})$ denote groups. A 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 mappings between vector spaces are in particular group homomorphisms. The following two lemmas follow directly from the definition.

Lemma

Let ${}G$ and ${}H$ denote groups, and let ${}\varphi \colon G\rightarrow H$ be a 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 ${}\varphi {\left(g^{-1}\right)}=(\varphi (g))^{-1}$.

$\Box$

Lemma

Let ${}F,G,H$ denote

groups. Then the following properties hold.

1. The identity

   $\operatorname {Id} \colon G\longrightarrow G$

   is a group homomorphism.

2. 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.
3. For a subgroup ${}F\subseteq G$, the inclusion ${}F\hookrightarrow G$ is a group homomorphism.
4. Let ${}\{e\}$ be the 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

This is trivial.

$\Box$

Example

Let ${}d\in \mathbb {N}$ be fixed. The mapping

$\mathbb {Z} \longrightarrow \mathbb {Z} ,n\longmapsto dn,$

is a group homomorphism. This follows immediately from the distributive law. For ${}d\geq 1$, this mapping is injective, and the image is the subgroup ${}\mathbb {Z} d\subseteq \mathbb {Z}$. For ${}d=0$, we have the zero mapping. For ${}d=1$, the mapping is the identity. For ${}d\geq 2$, the mapping is not surjective.

Example

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. 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

For a field ${}K$ and ${}n\in \mathbb {N} _{+}$, the determinant

$\det \colon \operatorname {GL} _{n}\!{\left(K\right)}\longrightarrow K^{\times },M\longmapsto \det M,$

is a group homomorphism. This follows from the multiplication theorem for the determinant and Theorem 16.11 .

Example

The assignment

$S_{n}\longrightarrow \{1,-1\},\pi \longmapsto \operatorname {sgn} (\pi ),$

where ${}S_{n}$ denotes the permutation group for ${}n$ elements, is a group homomorphism, due to Theorem 18.13 .

Lemma

Let ${}G$ denote a group. Then there is a correspondence between group elements ${}g\in G$ and 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, 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. A bijective group homomorphism

$\varphi \colon G\longrightarrow H$

is called an isomorphism.

Bijective linear mappings are in particular group isomorphisms.

Definition

The groups ${}H$ and ${}G$ are called isomorphic, if there exists a group isomorphism

${}\varphi \colon G\rightarrow H$.

Lemma

Let ${}G$ and ${}H$ be groups, and let

$\varphi \colon G\longrightarrow H$

be a group isomorphism. Then also the inverse mapping

$\varphi ^{-1}\colon H\longrightarrow G,h\longmapsto \varphi ^{-1}(h),$

is a group isomorphism.

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

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. 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, and ${}g\in G$ be fixed. The mapping defined by ${}g$,

$\kappa _{g}\colon G\longrightarrow G,x\longmapsto gxg^{-1},$

is called an inner automorphism.

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.

Lemma

An inner automorphism is indeed an automorphism. The assignment

$G\longrightarrow \operatorname {Aut} \,G,g\longmapsto \kappa _{g},$

is a

group homomorphism.

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. 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

For a fixed invertible matrix ${}B\in \operatorname {GL} _{n}\!{\left(K\right)}$, the conjugation

$\kappa _{B}\colon \operatorname {GL} _{n}\!{\left(K\right)}\longrightarrow \operatorname {GL} _{n}\!{\left(K\right)},M\longmapsto BMB^{-1},$

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, and let

$\varphi \colon G\longrightarrow H$

be a group homomorphism. Then the preimage of the neutral element is called the kernel of ${}\varphi$, denoted by

${}\operatorname {kern} \varphi =\varphi ^{-1}(e_{H})={\left\{g\in G\mid \varphi (g)=e_{H}\right\}}\,.$

Lemma

Let ${}G$ and ${}H$ be groups, and let

$\varphi \colon G\longrightarrow H$

be a group homomorphism. Then the kernel of ${}\varphi$ is a subgroup

of ${}G$.

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.

Lemma

Let ${}G$ and ${}H$ be groups. A group homomorphism ${}\varphi \colon G\rightarrow H$ is injective if and only if the kernel

of ${}\varphi$ is trivial.

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

Lemma

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

of ${}H$.

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

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 homomorphism from the additive group ${}(\mathbb {R} ,+,0)$ into the multiplicative group ${}(\mathbb {C} ^{\times },\cdot ,1)$. We determine the kernel 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 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.