Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 46

An equivalence relation ${}\sim$ on a set ${}M$ defines the quotient set ${}M/\sim$ and the canonical projection ${}M\rightarrow M/\sim$. If there are further structures on ${}M$, and if the equivalence relation respects this, then it is often possible to define on ${}M/\sim$ the same structure. As the main example for this process, we consider those equivalence relations on a group that are defined by a subgroup.

Cosets

Definition

Let ${}G$ be a group, and let ${}H\subseteq G$ be a subgroup. We set ${}x\sim _{H}y$ (and say that ${}x$ and ${}y$ are equivalent) if

${}x^{-1}y\in H$.

This is indeed an equivalence relation: From ${}x^{-1}x=e_{G}\in H$ we get that this relation is reflexive. From ${}x^{-1}y\in H$ we get immediately ${}y^{-1}x=(x^{-1}y)^{-1}\in H$, and from ${}x^{-1}y\in H$ and ${}y^{-1}z\in H$ we get ${}x^{-1}z\in H$.

Two group elements ${}x$ and ${}y$ are equivalent if and only if there exists an element ${}h\in H$ of the subgroup with ${}y=xh$. In accordance with example *****, we can interpret this situation in the sense that the subgroup ${}H$ provides a set of possible moves, and two elements are equivalent if and only if they can be moved to each other by a movement from ${}H$.

Example

In an (additively written) commutative group like ${}\mathbb {Z}$ or a vector space ${}V$, and of a given subgroup ${}H$, the equivalence relation ${}x\sim _{H}y$ means ${}y-x\in H$, that is, there exists a ${}h\in H$ such that

${}y=x+h\,.$

The equivalence classes are of the form ${}x+H={\left\{x+h\mid h\in H\right\}}$. In case ${}H=\mathbb {Z} d\subseteq \mathbb {Z}$ with a fixed ${}d$, the equivalence classes have the form

$H=\mathbb {Z} d,\,1+H=\{\ldots ,1-d,1,1+d,1+2d,\ldots \},\,2+H=\{\ldots ,2-d,2,2+d,2+2d,\ldots \},\ldots .$

These classes encompass those integer numbers that have, upon division by ${}d$, the remainder ${}0$, or ${}1$, or ${}2$, etc. They form a partition of ${}\mathbb {Z}$.

The equivalence classes of a linear subspace.

If ${}H=U\subseteq V$ is a linear subspace, then the equivalence classes have the form ${}v+U={\left\{v+u\mid u\in U\right\}}$ for a vector ${}v\in V$. This is ther affine space with starting point ${}v$ and the translating space ${}U$ (in the sense of definition). The equivalence classes form a family of affine subspaces parallel to each other.

Definition

Let ${}G$ be a group, and let ${}H\subseteq G$ be a subgroup. For every ${}x\in G$, the subset

${}xH={\left\{xh\mid h\in H\right\}}\,$

is called the left coset of ${}x$ in ${}G$ with respect to ${}H$. Every subset of this form is called a left coset. Accordingly, a set of the form

${}Hx={\left\{hx\mid h\in H\right\}}\,$

is called the right coset

(of ${}x$).

The equivalence classes to the equivalence relation defined above are, because of

${}{\begin{aligned}{[}x{]}&={\left\{y\in G\mid x\sim y\right\}}\\&={\left\{y\in G\mid x^{-1}y\in H\right\}}\\&={\left\{y\in G\mid {\text{there exists }}h\in H{\text{ with }}x^{-1}y=h\right\}}\\&={\left\{y\in G\mid {\text{there exists }}h\in H{\text{ with }}y=xh\right\}}\\&=xH,\end{aligned}}$

the left cosets. The coset to the neutral element is the subgroup ${}H$ itself. Therefore, the left cosets form a disjoint decomposition (a partition) of ${}G$. This holds for the right cosets as well. In the commutative case, one does not have to distinguish between left cosets and right cosets.

Lemma

Let ${}G$ be a group, and let ${}H\subseteq G$ be a subgroup. Let ${}x,y\in G$ be elements. Then the following statements are equivalent.

1. ${}x\in yH$.
2. ${}y\in xH$.
3. ${}y^{-1}x\in H$.
4. ${}x^{-1}y\in H$.
5. ${}xH\cap yH\neq \emptyset$.
6. ${}x\sim _{H}y$.
7. ${}xH=yH$.

Proof

The equivalence between ${}(1)$ and ${}(3)$ (and between ${}(2)$ and ${}(4)$) follows by multiplication with ${}y^{-1}$ and with ${}y$. The equivalence between ${}(3)$ and ${}(4)$ follows by going to the inverse elements. From ${}(1)$ we get ${}(5)$ because of ${}1\in H$. If ${}(5)$ holds, then this means that ${}xh_{1}=yh_{2}$ holds with certain ${}h_{1},h_{2}\in H$. Therefore, ${}x=yh_{2}h_{1}^{-1}$, and ${}(1)$ is satisfied. (4) and (6) are equivalent due to the definition. Since the left cosets are the equivalence classes, the equivalence between (5) and (7) follows.

$\Box$

Lagrange's theorem

Joseph-Louis Lagrange (1736 Turin - 1813 Paris)

Definition

For a finite group ${}G$, its cardinality is called the group order, written as

${}\operatorname {ord} \,(G)={\#\left(G\right)}\,.$

With this concept, we can say that the order of a subgroup divides the order of the group.

Lemma

Let ${}G$ be a finite group. Then every element ${}g\in G$ has finite order. The powers

$g^{0}=e_{G},\,g^{1}=g,\,g^{2},\ldots ,g^{\operatorname {ord} \,(g)-1}$

are all different.

Proof

See exercise *****.

$\Box$

Theorem

Let ${}G$ be a finite group, and let ${}H\subseteq G$ denote a subgroup

of ${}G$. Then the cardinality ${}{\#\left(H\right)}$ divides the cardinality ${}{\#\left(G\right)}$.

Proof

We consider the left cosets ${}gH:={\left\{gh\mid h\in H\right\}}$ for all ${}g\in G$. The mapping

$H\longrightarrow gH,h\longmapsto gh,$

is a bijection between ${}H$ and ${}gH$, so that all cosets have the same number of elements (namely ${}{\#\left(H\right)}$). The cosets form (as they are the equivalence classes) a partition of ${}G$. Hence, ${}{\#\left(G\right)}$ is a multiple of ${}{\#\left(H\right)}$.

$\Box$

Corollary

Let ${}G$ be a finite group, and let ${}g\in G$ denote an element. Then the order of ${}g$ divides the

group order.

Proof

Let ${}H$ be the subgroup generated by ${}g$. Due to Fact *****, we have

${}\operatorname {ord} \,(g)=\operatorname {ord} \,(H)\,.$

Therefore, due to Fact *****, this number divides the group order of ${}G$.

$\Box$

Definition

For a subgroup ${}H\subseteq G$, the cardinality of the (left- or right--)cosets is called the index of ${}H$ in ${}G$, denoted as

$\operatorname {ind} _{G}H.$

In the preceding definition, the number is in general to be understood as the cardinality of a set. However, the index is mainly used if it is finite, that is, if there are only finitely many cosets. This is, for finite ${}G$, always the case but can also hold for infinite ${}G$, as already the example ${}\mathbb {Z} n\subseteq \mathbb {Z}$, ${}n\geq 1$, shows. If ${}G$ is a finite group, and ${}H\subseteq G$ is a subgroup, then Lagrange's theorem yields the simple index formula

${}{\#\left(G\right)}={\#\left(H\right)}\cdot \operatorname {ind} _{G}H\,.$

Even if ${}G$ is not finite, it is still true that the different equivalence classes are "similar“ to each other, as there is always a natural bijective mapping

$H\longrightarrow gH,h\longmapsto gh.$

Normal subgroup

Definition

Let ${}G$ be a group, and let ${}H\subseteq G$ denote a subgroup. ${}H$ is called a normal subgroup if

${}xH=Hx\,$

holds for all ${}x\in G$, that is, if every left coset

of ${}x$ coincides with the right coset of ${}x$.

For a normal subgroup, it is not necessary to distinguish between left cosets and right cosets; we just talk about cosets. Instead of ${}xH$ or ${}Hx$, we write usually ${}[x]$. The equality ${}xH=Hx$ does not mean that ${}xh=hx$ for all ${}h\in H$; it only means that for every ${}h\in H$, there exists a ${}{\tilde {h}}\in H$ fulfilling ${}xh={\tilde {h}}x$.

Lemma

Let ${}G$ be a group, and let ${}H\subseteq G$ be a subgroup. Then the following statements are equivalent.

1. ${}H$ is a normal subgroup of ${}G$.
2. We have ${}xhx^{-1}\in H$ for all ${}x\in G$ and ${}h\in H$.
3. ${}H$ is invariant under every inner automorphism of ${}G$.

Proof

(1) means for given ${}h\in H$ that we can write ${}xh={\tilde {h}}x$ with some ${}{\tilde {h}}\in H$. Multiplication by ${}x^{-1}$ from the right yields ${}xhx^{-1}={\tilde {h}}\in H$; therefore, ${}(2)$ holds. Reading this argument backwards gives the implication ${}(2)\Rightarrow (1)$. Moreover, ${}(2)$ is an explicit reformulation of ${}(3)$.

$\Box$

Example

We consider the permutation group ${}G=S_{3}$ for a set with three elements, that is, ${}S_{3}$ consists of all bijective mappings of the set ${}\{1,2,3\}$ to itself. The trivial group ${}\{\operatorname {id} \}$ and the whole group are normal subgroups. The subset ${}H=\{\operatorname {id} \,,\varphi \}$, where ${}\varphi$ is the element that swops ${}1$ and ${}2$ and fixes ${}3$, is a subgroup. However, it is not a normal subgroup. To show this, let ${}\psi$ denote the bijection that fixes ${}1$ and swops ${}2$ and ${}3$. The inverse of ${}\psi$ is ${}\psi$ itself. The conjugation ${}\psi \varphi \psi ^{-1}=\psi \varphi \psi$ is the mapping that sends ${}1$ to ${}3$, ${}2$ to ${}2$, and ${}3$ to ${}1$. This bijection does not belong to ${}H$.

Lemma

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

$\varphi \colon G\longrightarrow H$

be a group homomorphism. Then the kernel ${}\operatorname {kern} \varphi$ is a normal subgroup

in ${}G$.

Proof

By Fact *****, we know that the kernel is a subgroup. We use Fact *****. Hence, let ${}x\in G$ be arbitrary, and ${}h\in \operatorname {kern} \varphi$. Then

${}\varphi {\left(xhx^{-1}\right)}=\varphi (x)\varphi (h)\varphi {\left(x^{-1}\right)}=\varphi (x)e_{H}\varphi {\left(x^{-1}\right)}=\varphi (x)\varphi (x)^{-1}=e_{H}\,,$

therefore, ${}xhx^{-1}$ belongs to the kernel.

$\Box$

Residue class formation

The picture shows the equivalence classes for a linear subspace, together with the well-defined addition on the classes.

We show now that every normal subgroup can be realized as the kernel of a suitable surjective group homomorphism. Instead of ${}G/\sim _{H}$, we just write ${}G/H$.

The multiplication of the cosets for a normal subgroup ${}N\subseteq G$.

Theorem

Let ${}G$ be a group, and let ${}H\subseteq G$ be a normal subgroup. Let ${}G/H$ be the set of all cosets (the quotient set), and let

$q\colon G\longrightarrow G/H,g\longmapsto [g],$

denote the canonical projection. Then there exists a uniquely determined group structure on ${}G/H$ such that ${}q$ is a

group homomorphism.

Proof

Since the canonical projection shall be a group homomorphism, the operation must fulfill

${}[x][y]=[xy]\,.$

We have to show that this rule gives a well-defined operation on ${}G/H$, that is, is independent of the choice of representatives. Hence, we have to show for ${}[x]=[x']$ and ${}[y]=[y']$ that ${}[xy]=[x'y']$ holds. Due to the condition, we can write ${}x'=xh$ and ${}hy'={\tilde {h}}y=yh'$ with ${}h,{\tilde {h}},h'\in H$. Therefore,

${}x'y'=(xh)y'=x(hy')=x(yh')=xyh'\,.$

This means ${}[xy]=[x'y']$. From this, the group property, the homomorphism property of the projection and the uniqueness follows.

$\Box$

Definition

Let ${}G$ be a group, and let ${}H\subseteq G$ be a normal subgroup. The quotient set

$G/H,$

endowed with the (according to Fact *****) uniquely determined group structure, is called the factor group of ${}G$ modulo ${}H$. The elements ${}[g]\in G/H$ are called residue classes. For a residue class ${}[g]$, every element ${}g'\in G$ with ${}[g']=[g]$

is called a representative of ${}[g]$.

Example

The subgroups of the integers are of the form ${}\mathbb {Z} n$ with ${}n\geq 0$, due to Fact *****. The factor group are denoted by

$\mathbb {Z} /(n)$

("${}\mathbb {Z}$ modulo ${}n$“). For ${}n=0$, this is just ${}\mathbb {Z}$ itself; for ${}n=1$, this is the trivial group. In general, the equivalence relation on ${}\mathbb {Z}$ defined by the subgroup ${}\mathbb {Z} n$ is given in the way that ${}a$ and ${}b$ are equivalent if and only if their difference ${}a-b$ belongs to ${}\mathbb {Z} n$, that is, if it is a multiple of ${}n$. Therefore, (${}n\geq 1$), every integer number is equivalent to exactly one of the ${}n$ numbers

$0,1,2,\ldots ,n-1$

(or, as we also say, congruent modulo ${}n$), namely to the remainder upon division through ${}n$. These remainders form a system of representatives for the factor group, and contains ${}n$ elements. The fact that the quotient mapping

$\mathbb {Z} \longrightarrow \mathbb {Z} /(n),a\longmapsto [a]=a\!\!\!\mod n,$

is a homomorphism might be expressed by saying that the remainder of a sum of two integers depends only on their remainders, not on the numbers themselves. As an image of the cyclic group ${}\mathbb {Z}$, the group ${}\mathbb {Z} /(n)$ is also cyclic; ${}1$ (but also ${}-1$) is always a generator.