Theory – KindiakMath

I’ll be honest—much of my blogposts so far have only dealt with the analysis-side of mathematics and not so much the algebra-side. Let’s fix that.

To begin, let’s talk about the integers, commonly denoted $\mathbb Z$ . The integers are an algebraically rich structure whose calculations we take for granted:

Given integers $x, y$ , the result $x + y$ is also an integer.
Given integers $x, y, z$ , addition is associative: $(x+y) + z = x + (y+z)$ .
Given any integer $x$ , $x + 0 = x = 0 + x$ .
Given any integer $x$ , there exists a unique integer $-x$ such that $x+ (-x) = 0 = (-x)+ x$ .

Furthermore, given integers $x, y$ , addition is commutative: $x+y = y+x$ . In this case, we say that the integers $\mathbb Z$ form an (Abelian) group under addition.

Definition 1. Let $G$ be a set. A map $* : G \times G \to G$ is called a binary map on $G$ . We call $(G, *)$ a group if it satisfies the following properties:

For any $x,y,z \in G$ , $(x * y) * z = x * (y * z)$ .
There exists some $e \in G$ such that for any $x \in G$ , $x * e = x = e* x$ .
For any $x \in G$ , there exists some $y \in G$ such that $x * y = e = y* x$ .

Equivalently, we say that $G$ forms a group under $*$ . Additionally, if for any $x, y \in G$ we have $x * y = y * x$ , we say that $(G, *)$ is Abelian.

Example 1. $\mathbb Z$ forms an Abelian group under $+$ .

Lemma 1. Let $G$ be a group under $*$ .

There exists a unique $e \in G$ such that for any $x \in G$ , $x * e = x = e* x$ .
For any $x \in G$ , there exists a unique $y \in G$ such that $x * y = e = y* x$ .

We call $e$ the identity of $G$ under $*$ , and denote $x^{-1} := y$ as the inverse of $G$ under $*$ . In the case that the operation is addition (i.e. $+$ ), we denote the additive inverse of $x$ by $-x$ .

Proof. Suppose $e_1, e_2$ are identities for $G$ . Then

$e_1 = e_1 * e_2 = e_2.$

Similarly, suppose $y_1, y_2$ are inverses for $x$ . Then

$x * y_1 = e = y_1 * x,\quad x * y_2 = e = y_2 * x.$

In particular,

$\begin{aligned} y_1 &= y_1 * e \\ &= y_1 * (x * y_2) \\ &= (y_1 * x) * y_2 \\ &= e * y_2 \\ &= y_2. \end{aligned}$

Group theory pervades all commonly-used mathematical structures.

Example 2. For any vector space $V$ with addition $+$ , $V$ forms an Abelian group under $+$ , almost by definition. In particular:

Since the number systems $\mathbb Q, \mathbb R, \mathbb C$ are fields, they are vector spaces too, and thus all form Abelian groups under $+$ . The additive identity is, not surprisingly, $0$ .
Given any set $K$ and any field $\mathbb F$ , the vector space of functions $\mathcal F(K, \mathbb F)$ also forms an Abelian group under $+$ , with additive identity $0 \equiv 0(\cdot)$ .
In particular, the commonly-used vector spaces $\mathbb F^n \cong \mathcal F(\{1,\dots, n\}, \mathbb F)$ , $\mathcal F(\mathbb N, \mathbb F)$ , $\mathcal F(\mathbb R, \mathbb F)$ , and $\mathcal F(\mathbb C, \mathbb F)$ all form Abelian groups under $+$ .

Example 3. Let $K$ be any set, and $\mathcal F(K, K)$ denote the collection of functions from $K$ to $K$ . Define the symmetric group on $K$ by

$\mathrm{Sym}(K) := \{ f \in \mathcal F(K, K) : f\ \text{is bijective}\}.$

Then $\mathrm{Sym}(K)$ forms a group under function composition $\circ$ :

In particular, by viewing matrices as linear transformations, the set of invertible $n \times n$ $\mathbb F$ -valued matrices $\mathrm{GL}_{n}(\mathbb F)$ forms a group under matrix multiplication.
Since the latter is not commutative in general, we also conclude that $\mathrm{Sym}(K)$ would not be Abelian in general.

Proof. Firstly, we check that $\circ$ is a binary operation on $\mathrm{Sym}(K)$ . Given $f, g \in \mathrm{Sym}(K)$ , both functions are bijective. Hence, their composition $g \circ f : K \to K$ is bijective as well, yielding $g \circ f \in \mathrm{Sym}(K)$ .

Associativity of function composition works in $\mathcal F(K, K)$ , and thus works in $\mathrm{Sym}(K)$ . The identity function $\mathrm{id}_K$ is clearly bijective. Finally for any $f \in \mathrm{Sym}(K)$ , $f^{-1} : K \to K$ is also bijective, and thus $f^{-1} \in \mathrm{Sym}(K)$ . Furthermore,

$f \circ f^{-1} = \mathrm{id}_K = f^{-1} \circ f.$

Therefore, every $f \in \mathrm{Sym}(K)$ has a unique inverse $f^{-1} \in \mathrm{Sym}(K)$ .

Furthermore, there are many smaller groups derived from these other groups. For example, the set of even numbers

$2 \mathbb Z := \{2 n : n \in \mathbb Z\} \subseteq \mathbb Z$

satisfies the properties of an Abelian group as laid out in Definition 1. We call them sub-groups.

In what follows, let $(G, *)$ be a group.

Definition 2. We call $H \subseteq G$ a sub-group of $G$ , denoted $H \leq G$ , if the following properties hold:

For any $x, y \in H$ , $x * y \in H$ .
$H$ forms a group under $*$ .

Theorem 1. $H \leq G$ if and only if for any $x, y \in H$ , $x * y^{-1} \in H$ .

Proof. $(\Rightarrow)$ Suppose $H \leq G$ . Fix $x, y \in H$ . Since $H$ forms a group under $*$ , $y^{-1} \in H$ and hence $x * y^{-1} \in H$ .

$(\Leftarrow)$ Suppose for any $x, y \in H$ , $x * y^{-1} \in H$ . In particular, $e = x * x^{-1} \in H$ . This result implies

$x^{-1} = e * x^{-1} \in H,$

which in turn implies that

$x * y = x * (y^{-1})^{-1} \in H,$

since the uniqueness of inverses yields

$y^{-1} * y = e \quad \Rightarrow\quad (y^{-1})^{-1} = y.$

Finally, associativity is inherited from the associativity of $*$ in $G$ .

Example 4. $\mathbb Z \leq \mathbb Q \leq \mathbb R \leq \mathbb C$ as subgroups under addition. Furthermore, defining

$K^{\times} := K \backslash \{0\}$

for $K \subseteq \mathbb C$ , we have $\mathbb Q^{\times} \leq \mathbb R^{\times} \leq \mathbb C^{\times}$ as subgroups under multiplication.

Example 5. Given $K \subseteq L$ , define the set of bijections from $L$ to $L$ that fixes $K$ by

$\mathrm{Sym}_K(L):= \{f \in \mathrm{Sym}(L) : f|_K = \mathrm{id}|_K\}.$

Then $\mathrm{Sym}_K(L) \leq \mathrm{Sym}(L)$ under function composition.

Proof. Fix $f, g \in \mathrm{Sym}_K(L)$ . Then

$\begin{aligned} (g^{-1} \circ f)|_K(x) &= (g^{-1} \circ f)(x) \\ &= g^{-1}(f(x)) \\ &= g^{-1}(x) \\ &= x, \end{aligned}$

where the last equality follows from $g$ having an inverse:

$\begin{aligned} g \in \mathrm{Sym}_K(L) \quad &\Rightarrow \quad g(x) = x \\ & \Rightarrow \quad x = g^{-1}(x). \end{aligned}$

Therefore, $(g^{-1} \circ f)|_K = \mathrm{id}_K$ . Since $g^{-1} \circ f \in \mathrm{Sym}(L)$ , we have $g^{-1} \circ f \in \mathrm{Sym}_K(L)$ . By Theorem 1, $\mathrm{Sym}_K(L) \leq \mathrm{Sym}(L)$ .

Example 6. Given $k > 0$ , define the set of multiples of $k$ by

$k\mathbb Z:= \{kn : n \in \mathbb Z\} \subseteq \mathbb R.$

In particular, suppose $k$ is an integer:

$k\mathbb Z \leq \mathbb Z$ under addition.
If $\ell > 0$ is an integer, then $k \mathbb Z \leq \ell \mathbb Z$ if and only if $\ell \mid k$ .

Finally, $2\mathbb Z \leq \mathbb Z$ under addition.

Proof. For the first claim, we remark that the (additive) inverse of $kn$ is $k(-n)$ , so that

$\begin{aligned} km + (-kn) &= km + k(-n) \\ &= k(m-n) \\ &= k(m+(-n)) \in k \mathbb Z.\end{aligned}$

For the second claim, we first use the division algorithm to find unique non-negative integers $q, r$ such that

$k = q\ell + r,\quad 0 \leq r < \ell.$

$(\Leftarrow)$ If $\ell \mid k$ , then $r = 0$ , so that for any $kn \in k \mathbb Z$ ,

$kn = (q\ell)n = \ell(qn) \in \ell \mathbb Z,$

yielding $k \mathbb Z \subseteq \ell \mathbb Z$ . Since both sets are groups, we have $k \mathbb Z \leq \ell \mathbb Z$ by Definition 2.

$(\Rightarrow)$ . It is clear that $k = k \cdot 1 \in k \mathbb Z \subseteq \ell \mathbb Z$ . Therefore,

$r = k + \ell(-q) \in \ell \mathbb Z.$

Therefore, there exists $s \in \mathbb Z$ such that $r = s \ell$ . Hence,

$0 \leq s\ell < \ell \quad \Rightarrow \quad 0 \leq s < 1.$

Therefore, $s = 0$ implies $r = 0\cdot \ell = 0$ . Hence, $k =q\ell \Rightarrow \ell \mid k$ .

More groups arise from transforming these groups. For example, by judiciously choosing $k$ in Example 6, given $a > 0$ , we can show that under multiplication,

$\{a^n : n \in \mathbb Z\} \leq \mathbb R^+ \leq \mathbb R^{\times}.$

We could verify this fact directly, but transformations help elucidate its relationship with pre-existing groups far more vividly. We explore this idea next time via homomorphisms.

—Joel Kindiak, 15 Dec 25, 1533H

Category: Theory

Baby Group Theory