KindiakMath

Baby Group Theory

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

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