Chapter 02

Medium

Groups — Definition and Examples

00 · Symbol Glossary

$G$A group

The standard letter for a group — a set together with an operation, usually written multiplicatively as $ab$ even when the actual operation is something else (composition, addition, matrix multiplication).

$|G|$Order of the group

The number of elements in $G$ . If $G$ is infinite, $|G| = \infty$ . Every finite group has a well-defined, finite order — one of the first questions asked about any new group.

$|g|$Order of an element

The smallest positive integer $k$ with $g^k = e$ , if one exists; otherwise $g$ has infinite order. Not to be confused with $|G|$ — context (a group vs. an element) always disambiguates.

$S_n$Symmetric group on n letters

The group of all bijections (rearrangements) of $\{1,2,\ldots,n\}$ to itself, under composition. $|S_n| = n!$ . The prototypical example of a noncommutative group, and — as later chapters show — secretly the only kind of finite group there is, up to relabeling.

$(a_1\, a_2\, \cdots\, a_k)$Cycle notation

Shorthand for the permutation sending $a_1 \mapsto a_2 \mapsto \cdots \mapsto a_k \mapsto a_1$ , and fixing everything else. Read left to right as "where does each element go."

$D_n$Dihedral group of order 2n

The group of symmetries (rotations and reflections) of a regular $n$ -gon. Contains $n$ rotations and $n$ reflections, so $|D_n| = 2n$ .

$r, s$Standard dihedral generators

Conventional names for a single rotation ( $r$ , by $360°/n$ ) and a single reflection ( $s$ ) in $D_n$ . Every element of $D_n$ can be written using only these two, combined repeatedly — the seed of the "generators and relations" idea developed fully in Chapter 10.

$GL_n(\mathbb{R})$General linear group

The group of all invertible $n\times n$ real matrices, under matrix multiplication. "General linear" refers to invertible linear maps on $\mathbb{R}^n$ ; " $GL$ " is standard notation across all of mathematics.

$\mathbb{Z}_n^\times$Group of units mod n

The elements of $\mathbb{Z}_n$ that have a multiplicative inverse, under multiplication. As Exercise 1.5 showed for $n=5$ : when $n$ is prime, $\mathbb{Z}_n^\times = \{1,\ldots,n-1\}$ , everything nonzero.

$ab$Multiplicative notation for the group operation

Convention: juxtaposition $ab$ means $a \ast b$ for whatever the group's actual operation is. Used for any group, even ones whose operation is really addition or composition — a notational simplification, not a claim about numbers.

01 · The Definition, at Last

Chapter 01 built a hierarchy — magma, semigroup, monoid, group — one property at a time. A group is simply the top row of that table, promoted to a definition worth naming on its own, because it is the single most useful and most common structure in all of algebra.

Definition — Group

A group is a set $G$ together with a binary operation $\ast$ satisfying:

G1. Closure. $a \ast b \in G$ for all $a, b \in G$ .

G2. Associativity. $(a\ast b)\ast c = a\ast(b\ast c)$ for all $a,b,c\in G$ .

G3. Identity. There exists $e \in G$ with $e\ast a = a\ast e = a$ for all $a\in G$ .

G4. Inverses. For every $a \in G$ , there exists $a^{-1} \in G$ with $a\ast a^{-1} = a^{-1}\ast a = e$ .

If additionally $a\ast b = b\ast a$ for all $a,b\in G$ , the group is called abelian (commutative); otherwise it is nonabelian.

By Theorem 1.1 and Theorem 1.2 from Chapter 01, $e$ and each $a^{-1}$ are automatically unique — there's no need to specify "the" identity or "the" inverse as extra data; the axioms pin them down.

Multiplicative notation by convention

From here on, group operations default to multiplicative notation: $ab$ for $a\ast b$ , $a^2$ for $a\ast a$ , $a^n$ for $a$ combined with itself $n$ times, $a^{-1}$ for the inverse, and $e$ (or sometimes $1$ ) for the identity. When the group happens to be $(\mathbb{Z},+)$ or similar, this is purely notational — $a^2$ there really means $a+a$ , and $a^{-1}$ means $-a$ . Additive notation ( $0$ , $-a$ , $na$ ) is used instead whenever the group is explicitly abelian and additive, like $(\mathbb{Z}_n,+)$ .

Example — Groups already in hand

$(\mathbb{Z}, +)$ : identity $0$ , inverse of $a$ is $-a$ . Abelian. Infinite order.

$(\mathbb{Q}\setminus\{0\}, \times)$ : identity $1$ , inverse of $a$ is $1/a$ . Abelian. Infinite order.

$(\mathbb{Z}_n, +)$ : identity $[0]$ , inverse of $[a]$ is $[-a] = [n-a]$ . Abelian. Order $n$ .

Each of these was verified, axiom by axiom, in Chapter 01 — nothing new is being claimed, only relabeled under the single word "group."

02 · First Consequences of the Axioms

A handful of facts follow from G1–G4 alone, before a single specific group is chosen. Because they hold in every group, they can be used freely from here on without re-derivation.

Theorem 2.1 — Cancellation Laws

Let $G$ be a group and $a,b,c \in G$ .

Left cancellation. If $ab = ac$ , then $b=c$ .

Right cancellation. If $ba = ca$ , then $b=c$ .

Proof

Suppose $ab=ac$ . Multiply both sides on the left by $a^{-1}$ (which exists by G4):

a^{-1}(ab) = a^{-1}(ac)

By associativity (G2), regroup each side: $(a^{-1}a)b = (a^{-1}a)c$ . By G4, $a^{-1}a=e$ , so this becomes $eb=ec$ , and by G3, $b=c$ .

Right cancellation is the mirror argument: multiply both sides of $ba=ca$ on the right by $a^{-1}$ , regroup, and use $aa^{-1}=e$ .

Why cancellation is not automatic in general

Cancellation uses all four axioms: inverses to introduce $a^{-1}$ , associativity to regroup, and the identity law to simplify. In a structure lacking inverses — like $(\mathbb{Z}_4, \times)$ from Chapter 01 — cancellation can fail entirely: $2\times1 = 2\times3 = 2$ in $\mathbb{Z}_4$ , yet $1 \neq 3$ . Groups are exactly strong enough to rule this out.

Theorem 2.2 — Unique Solvability

For any $a,b\in G$ , the equations $ax=b$ and $ya=b$ each have exactly one solution in $G$ : $x = a^{-1}b$ and $y = ba^{-1}$ .

Proof

Existence. Check directly: $a(a^{-1}b) = (aa^{-1})b = eb = b$ , so $x=a^{-1}b$ is a solution to $ax=b$ .

Uniqueness. Suppose $x'$ is also a solution: $ax' = b = ax$ . By left cancellation (Theorem 2.1), $x'=x$ . The argument for $ya=b$ is symmetric, using right cancellation.

Example — Solving inside a concrete group

In $(\mathbb{Z}_7,+)$ , solve $3+x=5$ (writing additively, so $a^{-1}$ is $-a$ ). By Theorem 2.2, $x = -3+5 = 2$ (computed mod 7). Check: $3+2=5$ . ✓ Exactly one solution, guaranteed before even computing it.

Theorem 2.3 — Inverse of a Product

For all $a,b\in G$ : $(ab)^{-1} = b^{-1}a^{-1}$ — note the order reverses.

Proof

By the uniqueness of inverses (Theorem 1.2), it suffices to check that $b^{-1}a^{-1}$ satisfies the defining property of $(ab)^{-1}$ , namely that it multiplies with $ab$ (on both sides) to give $e$ .

(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = a\, e\, a^{-1} = aa^{-1} = e

(b^{-1}a^{-1})(ab) = b^{-1}(a^{-1}a)b = b^{-1}\, e\, b = b^{-1}b = e

Both products give $e$ , in both orders, so $b^{-1}a^{-1}$ is an inverse of $ab$ ; by uniqueness, it is the inverse.

Common mistake — Forgetting to reverse the order

It is tempting to guess $(ab)^{-1} = a^{-1}b^{-1}$ . This is only correct when $G$ is abelian! Concretely: think of $a$ ="put on socks," $b$ ="put on shoes." Then $ab$ = "socks, then shoes." Undoing this correctly means "take off shoes, then take off socks" — that's $b^{-1}a^{-1}$ , not $a^{-1}b^{-1}$ ("take off socks first" would trap the shoes). The everyday intuition and the algebra agree exactly.

03 · Order of a Group and of an Element

Definition — Order of a Group

The order of $G$ , written $|G|$ , is the number of elements in $G$ (its cardinality). If this is finite, $G$ is a finite group.

Definition — Order of an Element

For $g\in G$ , define $g^0 = e$ , $g^n = g\cdot g\cdots g$ ( $n$ times) for $n>0$ , and $g^{-n} = (g^{-1})^n$ for $n>0$ . The order of $g$ , written $|g|$ , is the smallest positive integer $k$ such that $g^k = e$ . If no such $k$ exists, $g$ has infinite order.

Step-by-step — Order of $2 \in \mathbb{Z}_6$ (additive group)

Translate to additive notation: in

(\mathbb{Z}_6,+)

, "

g^k

" means

g+g+\cdots+g

(

k

times), i.e.

kg

, and "

=e

" means "

\equiv 0 \pmod 6

Try $k=1$ :

1\cdot 2 = 2 \neq 0

Try $k=2$ :

2\cdot 2 = 4 \neq 0

Try $k=3$ :

3\cdot 2 = 6 \equiv 0 \pmod 6

. Stop — this is the smallest such

k

Conclusion:

|2| = 3

(\mathbb{Z}_6,+)

Example — Every nonidentity element of $(\mathbb{Z}, +)$ has infinite order

For any nonzero $a \in \mathbb{Z}$ , $ka = a+a+\cdots+a$ ( $k$ times) is never $0$ for $k \geq 1$ — it's a nonzero multiple of $a$ . So no finite $k$ works; $|a| = \infty$ . This matches intuition: $(\mathbb{Z},+)$ is an infinite group precisely because repeatedly adding any nonzero integer never returns to $0$ .

Groups of numbers are only half the story. The symmetric group is built from rearrangements — and it is the first genuinely noncommutative group in this course built from first principles, rather than borrowed from arithmetic.

Definition — Symmetric Group

$S_n$ is the set of all bijections $\sigma : \{1,\ldots,n\} \to \{1,\ldots,n\}$ , called permutations, with the operation being composition $\sigma\tau$ (apply $\tau$ first, then $\sigma$ ). The identity is the permutation fixing every point.

Why S_n is a group

Closure: composing two bijections gives a bijection. Associativity: function composition is always associative — $(fg)h$ and $f(gh)$ both mean "apply $h$ , then $g$ , then $f$ ." Identity: the "do nothing" permutation. Inverses: every bijection has an inverse bijection (undo the rearrangement). All four axioms hold for free, just from the definition of "bijection."

Definition — Cycle Notation

The cycle $(a_1\, a_2\, \cdots\, a_k)$ denotes the permutation sending $a_1\mapsto a_2, a_2\mapsto a_3, \ldots, a_{k-1}\mapsto a_k, a_k\mapsto a_1$ , and fixing every element not listed. A 1-cycle like $(4)$ (a fixed point) is usually omitted entirely.

Example — Reading a permutation of $\{1,2,3,4\}$

The cycle $(1\,3\,4)$ on $\{1,2,3,4\}$ sends $1\mapsto3$ , $3\mapsto4$ , $4\mapsto1$ , and fixes $2$ (since $2$ isn't listed). As a full table:

$x$	1	2	3	4
$\sigma(x)$	3	2	4	1

Step-by-step — Composing $\sigma=(1\,2\,3)$ and $\tau=(1\,2)$ in $S_3$, computing $\sigma\tau$

Convention:

\sigma\tau

means "apply

\tau

first, then

\sigma

" — read right to left, same convention as function composition

f\circ g

Track where 1 goes:

\tau

sends

1\mapsto 2

; then

\sigma

sends

2\mapsto 3

. So

(\sigma\tau)(1)=3

Track where 2 goes:

\tau

sends

2\mapsto 1

; then

\sigma

sends

1\mapsto 2

. So

(\sigma\tau)(2)=2

— a fixed point.

Track where 3 goes:

\tau

fixes

3

(

3

isn't in

\tau

's cycle); then

\sigma

sends

3\mapsto1

. So

(\sigma\tau)(3)=1

Assemble:

1\mapsto3

3\mapsto1

, and

2

fixed — this is the transposition

\sigma\tau = (1\,3)

Common mistake — $S_n$ is not abelian for $n\geq 3$

Compute $\tau\sigma$ from the same example (apply $\sigma$ first, then $\tau$ ): $1\mapsto2\mapsto1$ (fixed), $2\mapsto3\mapsto3$ , $3\mapsto1\mapsto2$ . So $\tau\sigma = (2\,3)$ — completely different from $\sigma\tau=(1\,3)$ computed above. Order of composition matters. $S_1$ and $S_2$ happen to be abelian (too small to have room for disagreement), but $S_n$ for $n\geq3$ never is.

Order of S_n

$|S_n| = n!$ , since a bijection of $n$ objects is exactly a choice of ordering: $n$ choices for where $1$ goes, then $n-1$ remaining choices for $2$ , and so on. $|S_3|=6$ , $|S_4|=24$ — small numbers that grow explosively, foreshadowing why Chapter 09's Sylow theorems are needed to say anything structural about groups of specific orders.

05 · The Dihedral Group $D_n$

The dihedral group $D_n$ collects all rigid motions (rotations and flips, no stretching) that carry a regular $n$ -gon back onto itself.

Definition — Dihedral Group

Let $r$ denote rotation by $360°/n$ and $s$ denote a reflection across a fixed axis of symmetry. $D_n$ consists of the $n$ rotations $e, r, r^2,\ldots, r^{n-1}$ and the $n$ reflections $s, rs, r^2s,\ldots, r^{n-1}s$ , under composition. $|D_n| = 2n$ .

Example — $D_3$, the symmetries of an equilateral triangle

Label the triangle's vertices $1,2,3$ . Rotations: $e$ (identity), $r$ (rotate $120°$ , sends $1\to2\to3\to1$ ), $r^2$ (rotate $240°$ ). Reflections: three, each flipping across an axis through one vertex and the midpoint of the opposite side.

Concretely, as permutations of $\{1,2,3\}$ : $r = (1\,2\,3)$ , and a reflection fixing vertex $1$ is $s=(2\,3)$ . This identifies $D_3$ with all of $S_3$ — both have order $6$ , and in fact $D_3 \cong S_3$ (the $\cong$ notation for "structurally identical" is made precise in Chapter 06).

Relation — The Defining Rule of $D_n$

Every element of $D_n$ has a unique expression $r^i s^j$ with $0\le i<n$ , $j\in\{0,1\}$ , governed by two rules:

r^n = e, \qquad s^2 = e, \qquad srs = r^{-1}

Step-by-step — Verifying $srs=r^{-1}$ makes sense geometrically in $D_4$ (square)

Setup:

r

= rotate

90°

counterclockwise.

s

= reflect across a fixed axis.

Apply $r$ first (rightmost): rotate

90°

CCW.

Then apply $s$ : reflect the rotated square.

Then apply $s$ again (leftmost): reflecting twice across the same axis undoes the first reflection — but the square has already been rotated in between, so the net effect is a rotation, not the identity.

Result: the net motion is rotation by

90°

clockwise — exactly

r^{-1}

. A reflection "flips the direction" of any rotation sandwiched inside it; this is the geometric content of

srs=r^{-1}

Common mistake — $D_n$ is never abelian for $n \geq 3$

If $D_n$ were abelian, $sr$ would equal $rs$ . But the relation gives $srs=r^{-1}$ , i.e. $sr = r^{-1}s^{-1}\cdot s \cdot s= r^{-1}s$ (using $s^2=e$ , so $s^{-1}=s$ ), which is $r^{-1}s$ — not $rs$ , unless $r=r^{-1}$ , i.e. $r^2=e$ . That only happens when $n\le 2$ . For $n\geq3$ , rotating then reflecting is genuinely different from reflecting then rotating.

06 · Matrix Groups: $GL_n(\mathbb{R})$

Definition — General Linear Group

$GL_n(\mathbb{R})$ is the set of all invertible $n\times n$ matrices with real entries, under matrix multiplication.

Example — Verifying the group axioms for $GL_n(\mathbb{R})$

Closure: if $A,B$ are invertible, $(AB)^{-1}=B^{-1}A^{-1}$ exists (a matrix analog of Theorem 2.3), so $AB$ is invertible too — $AB \in GL_n(\mathbb{R})$ .

Associativity: matrix multiplication is associative for all matrices, invertible or not — inherited automatically.

Identity: the identity matrix $I$ satisfies $IA=AI=A$ .

Inverses: by definition of "invertible," every $A\in GL_n(\mathbb{R})$ has a matrix inverse $A^{-1}$ , and $A^{-1}$ is itself invertible (its inverse is $A$ ), so $A^{-1}\in GL_n(\mathbb{R})$ .

All four axioms hold — $GL_n(\mathbb{R})$ is a group, of infinite order for every $n\geq1$ .

Common mistake — All $n\times n$ matrices do not form a group under multiplication

Let $M_n(\mathbb{R})$ be all real $n\times n$ matrices (dropping the invertibility requirement). The zero matrix $0$ is in $M_n(\mathbb{R})$ , but it has no inverse — there's no matrix $B$ with $0B=I$ , since $0B=0\neq I$ . G4 fails. Restricting to invertible matrices — $GL_n(\mathbb{R})$ specifically — is exactly what repairs this.

Noncommutative for n ≥ 2

Section 03's Chapter 01 example already showed two $2\times2$ matrices with $AB\neq BA$ . So $GL_n(\mathbb{R})$ is nonabelian for $n\geq2$ — another concrete, infinite example alongside the finite examples $S_n$ and $D_n$ above. Nonabelian groups are not rare exceptions; they are the norm once you leave pure arithmetic.

07 · Recognizing Structure in a Cayley Table

For small finite groups, the entire multiplication table (Cayley table) can be inspected directly for structural facts — a preview of techniques that scale up, via theory instead of brute force, starting in Chapter 03.

Fact — Symmetry Detects Commutativity

A finite group's Cayley table is symmetric across its main diagonal if and only if the group is abelian.

Example — Reading abelian-ness off a table

$(\mathbb{Z}_4,+)$ 's table from Chapter 01 was symmetric across the diagonal (e.g. entry at row $1$ , column $3$ equals entry at row $3$ , column $1$ : both give $0$ ) — confirming abelian, as expected for any $(\mathbb{Z}_n,+)$ .

By contrast, $S_3$ 's table is not symmetric: the entry for $\sigma\tau$ (row $\sigma$ , column $\tau$ ) differs from $\tau\sigma$ (row $\tau$ , column $\sigma$ ) whenever $\sigma\tau\neq\tau\sigma$ — exactly the failure demonstrated in Section 04's FailBlock.

Fact — Every Row and Column Is a Rearrangement of G

In any group's Cayley table, each row and each column contains every element of $G$ exactly once.

Proof

Fix a row, labeled by $a\in G$ . The entries in that row are $ab$ as $b$ ranges over $G$ . By Theorem 2.2, for any target $c\in G$ , the equation $ax=c$ has exactly one solution $x=a^{-1}c \in G$ — so $c$ appears in row $a$ exactly once (at column $x=a^{-1}c$ ). Since $c$ was arbitrary, every element of $G$ appears exactly once in row $a$ . The column argument is symmetric, using $ya=c$ .

A fast sanity check

This gives an instant way to spot a fake group table: if any row or column repeats an entry, or omits one, G1–G4 cannot all hold. It is not a proof of correctness by itself (a table can pass this check yet fail associativity), but it is the cheapest first filter, and it is exactly why the $\mathbb{Z}_4$ addition table in Chapter 01 "looked right" at a glance.

08 · Exercises

EXERCISE 2.1

$G4$ requires every element to have an inverse. Test the specific element $2$ : is there a real number $x$ with $2\cdot x$ equal to the identity of this operation?

The operation is ordinary multiplication, so the identity would need to be $1$ (since $a\times1=a$ ). Does $2$ have an inverse? We'd need $x$ with $2x=1$ , i.e. $x=1/2$ — which is a real number. In fact every nonzero real has a reciprocal in $\mathbb{R}$ , so $(\mathbb{R}\setminus\{0\},\times)$ is a group. The subtlety is $0$ : if $0$ were included, $0$ would need an inverse $x$ with $0\cdot x=1$ , impossible. That is exactly why the set excludes $0$ .

Is $(\mathbb{R}\setminus\{0\}, \times)$ a group? If you claimed $(\mathbb{R}, \times)$ (including $0$ ) instead, which axiom fails, and for which element?

EXERCISE 2.2

Use Theorem 2.3 directly, remembering that the order reverses.

By Theorem 2.3, $(abc)^{-1}$ can be computed by treating $abc$ as $(ab)c$ : $\big((ab)c\big)^{-1} = c^{-1}(ab)^{-1} = c^{-1}(b^{-1}a^{-1}) = c^{-1}b^{-1}a^{-1}$ .

$(abc)^{-1} = c^{-1}b^{-1}a^{-1}$ — the three inverses appear in exactly the reverse order of the original product.

Using Theorem 2.3, find a formula for $(abc)^{-1}$ in terms of $a^{-1}, b^{-1}, c^{-1}$ , for $a,b,c$ in a (possibly nonabelian) group $G$ .

EXERCISE 2.3

Compute $3\cdot k \pmod 8$ for $k=1,2,3,\ldots$ until you hit $0$ .

$1\cdot3=3$ , $2\cdot3=6$ , $3\cdot3=9\equiv1$ , $4\cdot3=12\equiv4$ , $5\cdot3=15\equiv7$ , $6\cdot3=18\equiv2$ , $7\cdot3=21\equiv5$ , $8\cdot3=24\equiv0$ .

The smallest positive $k$ with $3k\equiv0\pmod8$ is $k=8$ . So $|3|=8$ in $(\mathbb{Z}_8,+)$ — element $3$ generates the entire group by repeated addition, since $8=|\mathbb{Z}_8|$ .

Find the order of the element $3$ in $(\mathbb{Z}_8, +)$ .

EXERCISE 2.4

Track where each of $1,2,3,4$ goes under $\tau$ first, then $\sigma$ .

$\tau=(1\,2)(3\,4)$ sends $1\to2$ , $2\to1$ , $3\to4$ , $4\to3$ . Then $\sigma=(1\,3)$ sends $1\to3$ , $3\to1$ , fixes $2,4$ .

Compose ( $\sigma\tau$ , apply $\tau$ then $\sigma$ ): $1\xrightarrow{\tau}2\xrightarrow{\sigma}2$ . $2\xrightarrow{\tau}1\xrightarrow{\sigma}3$ . $3\xrightarrow{\tau}4\xrightarrow{\sigma}4$ . $4\xrightarrow{\tau}3\xrightarrow{\sigma}1$ .

So $1\to2$ , $2\to3$ , $3\to4$ , $4\to1$ : this is the single 4-cycle $\sigma\tau = (1\,2\,3\,4)$ .

In $S_4$ , let $\sigma = (1\,3)$ and $\tau = (1\,2)(3\,4)$ . Compute $\sigma\tau$ in cycle notation.

EXERCISE 2.5

Recall from Section 05 that $D_n$ has order $2n$ , split evenly between rotations and reflections.

$D_6$ (symmetries of a regular hexagon) has order $2(6)=12$ : rotations $e,r,r^2,r^3,r^4,r^5$ (rotations by multiples of $60°$ ) and six reflections $s,rs,r^2s,r^3s,r^4s,r^5s$ .

$|D_6| = 12$ .

What is $|D_6|$ , the order of the symmetry group of a regular hexagon? List how many rotations and how many reflections it contains.

EXERCISE 2.6

Use the row/column rearrangement fact from Section 07: what must be true about repeated entries in a valid Cayley table?

By the Fact in Section 07, every row of a genuine group table must contain each element of $G=\{e,a,b,c\}$ exactly once. In the given (broken) table, if row $a$ reads $e, e, b, c$ (with $e$ appearing twice and $a$ missing), this violates the rearrangement property — $a$ never appears in its own row, and $e$ appears twice.

This cannot be a valid Cayley table for a group. At minimum, G4 (inverses) or the underlying binary operation itself must be inconsistent — a genuine group operation forces every row to be a full rearrangement of $G$ , with no repeats and no omissions.

A student writes down a $4\times4$ table for a claimed group $G=\{e,a,b,c\}$ , but row $a$ reads $e, e, b, c$ (as $a$ combines with $e,a,b,c$ respectively). Without knowing anything else about the table, explain why this cannot be a valid Cayley table for a group.

09 · Chapter Summary

Concept	Statement
Group	Set + operation satisfying G1–G4 (closure, associativity, identity, inverses)
Abelian	$ab=ba$ for all $a,b$ ; nonabelian otherwise
Cancellation	$ab=ac \implies b=c$ (Thm 2.1)
Unique solvability	$ax=b$ has exactly one solution $x=a^{-1}b$ (Thm 2.2)
Inverse of a product	$(ab)^{-1}=b^{-1}a^{-1}$ — order reverses (Thm 2.3)
Order of $G$	$
Order of $g$	Smallest $k>0$ with $g^k=e$ ; possibly infinite
$S_n$	All bijections of $\{1,\ldots,n\}$ ; order $n!$ ; nonabelian for $n\geq3$
$D_n$	Symmetries of a regular $n$ -gon; order $2n$ ; relations $r^n=e,\,s^2=e,\,srs=r^{-1}$
$GL_n(\mathbb{R})$	Invertible $n\times n$ matrices; nonabelian for $n\geq2$
Cayley table	Symmetric $\iff$ abelian; every row/column is a full rearrangement of $G$

Next: Chapter 03 — Subgroups & Cyclic Groups zooms in on substructures: which subsets of a group are themselves groups, and what happens when a single element is used to generate an entire group by repeated combination.

00 · Symbol Glossary

01 · The Definition, at Last

02 · First Consequences of the Axioms

03 · Order of a Group and of an Element

04 · The Symmetric Group SnS_nSn​

05 · The Dihedral Group DnD_nDn​

06 · Matrix Groups: GLn(R)GL_n(\mathbb{R})GLn​(R)

07 · Recognizing Structure in a Cayley Table

08 · Exercises

09 · Chapter Summary

04 · The Symmetric Group $S_n$

05 · The Dihedral Group $D_n$

06 · Matrix Groups: $GL_n(\mathbb{R})$