Group Magic

The magic ingredient that makes group theory versatile is that of the group homomorphism. With it, we can deduce some subsets to be groups, even if directly verifying it would take a lot more nontrivial brain power.

But first, let’s discuss a motivating example, as we should always do, when discussing abstract algebra. Fix a real number $a > 0$ . The exponential map $f : \mathbb R \to \mathbb R^{\times}$ defined by $f(x) := a^x$ satisfies the key property that for any $x, y \in \mathbb R$ ,

$f(x+y) = f(x) \cdot f(y).$

This property actually proves that $\mathbb R^+ \leq \mathbb R^{\times}$ under multiplication.

Definition 1. Let $(G, *)$ and $(H, \odot)$ be groups. We call the map $f : G \to H$ a group homomorphism from $(G, *)$ to $(H, \odot)$ if, for any $x, y \in G$ ,

$f(x * y) = f(x) \odot f(y).$

Denote the set of group homomorphisms by $\mathrm{Hom}(G, H) \subseteq \mathcal F(G, H)$ . Of course, the implicit binary operations $*$ and $\odot$ determine the group structure of $G, H$ . In the case $G=H$ as groups, we abbreviate $\mathrm{Hom}(G):= \mathrm{Hom}(G,G)$ .

When there is little ambiguity, we suppress the binary operations $(G, *) \equiv G$ and $(H, \odot) \equiv H$ .
We denote the respective group identities by $e_G$ and $e_H$ .

Example 1. For any fixed $a > 0$ , the map $f : \mathbb R \to \mathbb R^{\times}$ defined by $f(x) := a^x$ is a group homomorphism from $(\mathbb R, +)$ to $(\mathbb R^{\times}, \cdot)$ .

Homomorphisms are magical because the sets they “create” also turn out to be groups! In what follows, let $f : G \to H$ be a group homomorphism.

Lemma 1. $\{e_G\} \leq G$ .

Proof. Trivial.

Lemma 2. $f(e_G) = e_H$ and for any $x \in G$ , $f(x^{-1}) = f(x)^{-1}$ .

Proof. Since $e_G = e_G * e_G$ ,

$f(e_G) = f(e_G * e_G) = f(e_G) \odot f(e_G).$

Since $f(e_G) \in H$ , $f(e_G)^{-1}$ exists, so that

$\begin{aligned} e_H &= f(e_G)^{-1} \odot f(e_G) \\ &= f(e_G)^{-1} \odot ( f(e_G)\odot f(e_G) ) \\ &= (f(e_G)^{-1} \odot f(e_G)) \odot f(e_G) \\ &= e_H \odot f(e_G) = f(e_G). \end{aligned}$

Furthermore,

$f(x^{-1}) \odot f(x) = f(x^{-1} * x) = f(e_G) = e_H.$

Similarly, $f(x) f(x^{-1}) = e_H$ . Therefore, $f(x^{-1})$ is an inverse of $f(x)$ . Since inverses are unique, $f(x)^{-1} = f(x^{-1})$ .

Lemma 3. If $K \leq H$ , then $f^{-1}(K) \leq G$ .

Proof. Fix $x, y \in f^{-1}(K)$ so that $f(x), f(y) \in K$ . Since $f$ is a group homomorphism,

$f(x * y^{-1}) = f(x) \odot f(y^{-1}) = f(x) \odot f(y)^{-1} \in K.$

Hence, $x * y^{-1} \in f^{-1}(K)$ . Since $x,y \in G$ are arbitrary, $f^{-1}(K) \leq G$ .

Theorem 1. Define the kernel of $f$ by

$\ker(f) := f^{-1}(\{e_H\}).$

Then $\ker(f) \leq G$ and $f(G) \leq H$ .

Proof. For the first claim, Lemma 1 asserts that $\{e_H\} \leq H$ . By Lemma 3,

$\ker(f) = f^{-1}(\{e_H\}) \leq G.$

For the second claim, fix $x, y \in f(G)$ . Find $x_0, y_0 \in G$ such that $x = f(x_0)$ and $y = f(y_0)$ . By Lemma 2,

$y^{-1} = f(y_0)^{-1} = f(y_0^{-1}).$

Since $G$ is a group, $x_0 * y_0^{-1} \in G$ . Since $f$ is a group homomorphism,

$x \odot y^{-1} = f(x_0) \odot f(y_0^{-1}) = f(x_0 * y_0^{-1}) \in f(G).$

Example 2. Fix $a > 0$ . Then

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

under multiplication. Furthermore, all groups here are Abelian.

Proof. By the definition of $f$ as per Example 1, $f(\mathbb R) = \mathbb R^+$ . By Theorem 1,

$\mathbb R^+ = f(\mathbb R) \leq \mathbb R^{\times}.$

Furthermore, define $g : \mathbb Z \to \mathbb R^+$ by $g := f|_{\mathbb Z}$ , which can be checked to be a group homomorphism (Exercise!). Since $\mathbb Z$ forms a group under $+$ , by Theorem 1 again,

$\{a^n : n \in \mathbb Z\} = f(\mathbb Z) \leq \mathbb R^+.$

Example 3. Define $S^1 := \{z \in \mathbb C : |z| = 1\}$ . Then $S^1 \leq \mathbb C^{\times}$ under multiplication.

Proof. Recall that $z \in S^1$ if and only if $z = e^{it}$ for some $t \in \mathbb R$ . Define $f : \mathbb R \to \mathbb C^{\times}$ by $f(t) = e^{it}$ . Using the complex exponential,

$f(r+s)=e^{i(r+s)} = e^{ir} \cdot e^{is} = f(r) \cdot f(s).$

Thus, $f$ is a group homomorphism from $(\mathbb R, +)$ to $(\mathbb C^{\times}, \cdot)$ . By Theorem 1,

$S^1 = f(\mathbb R) \leq \mathbb C^{\times}.$

Example 4. Define the special linear group by

$\mathrm{SL}_n(\mathbb F) := \{\mathbf A \in \mathrm{GL}_n(\mathbb F) : \det(\mathbf A) = 1\}.$

Then $\mathrm{SL}_n(\mathbb F) \leq \mathrm{GL}_n(\mathbb F)$ under matrix multiplication.

Proof. By properties of the determinant, for any $\mathbf A, \mathbf B \in \mathrm{GL}_n(\mathbb F)$ ,

$\det(\mathbf A \mathbf B) = \det(\mathbf A) \det(\mathbf B).$

Hence, $\det|_{\mathrm{GL}_n(\mathbb F)} : \mathrm{GL}_n(\mathbb F) \to \mathbb F^{\times}$ is a group homomorphism. By Theorem 1,

$\mathrm{SL}_n(\mathbb F) = \ker(\det|_{\mathrm{GL}_n(\mathbb F)}) \leq \mathrm{GL}_n(\mathbb F).$

Lemma 4. Given groups $G,H,K$ ,

$f \in \mathrm{Hom}(G,H) \quad \text{and}\quad g \in \mathrm{Hom}(H,K)$

implies that $g \circ f \in \mathrm{Hom}(G, K)$ .

Proof. Exercise.

Example 5. Define the automorphisms on $G$ via

$\mathrm{Aut}(G) := \mathrm{Sym}(G) \cap \mathrm{Hom}(G).$

Furthermore, if $K \leq G$ , define

$\mathrm{Aut}_K(G) := \mathrm{Sym}_K(G) \cap \mathrm{Hom}(G).$

Then $\mathrm{Aut}_K(G) \leq \mathrm{Aut}(G) \leq \mathrm{Sym}(G)$ .

Proof. Use Lemma 4.

Just these first two topics alone establish all of the following useful groups and subgroups under the relevant group operations:

$k \mathbb Z \leq \mathbb Z \leq \mathbb Q \leq \mathbb R \leq \mathbb C$ under addition,
$\mathbb Q^{\times} \leq \mathbb R^{\times} \leq \mathbb C^{\times}$ under multiplication,
$\{a^n : n \in \mathbb Z\} \leq \mathbb R^+ \leq \mathbb R^{\times}$ under multiplication,
$S^1 \leq \mathbb C^{\times}$ under multiplication,
$\mathcal F(K, \mathbb F)$ under addition,
$\mathrm{SL}_n(\mathbb F) \leq \mathrm{GL}_n(\mathbb F)$ under matrix multiplication,
$\mathrm{Aut}_K(G) \leq \mathrm{Aut}(G) \leq \mathrm{Sym}(G)$ under function composition.

Therefore, all of the group-theoretic results that we eventually develop could, in principle, apply to all of these groups. That’s a pretty versatile list!

The group $\mathrm{Sym}(K)$ turns out to be rather interesting when we particularise to finite $K$ . In particular, if $K$ is a group, we can discuss finite groups, and a key example of such arises from returning to the integers. We will explore these ideas in future posts.

—Joel Kindiak, 15 Dec 25, 1348H

KindiakMath

Group Magic

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Group Magic

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