Algebra, at its core, is the study of structure: the hidden patterns that make collections of objects behave in regular, predictable ways. Where analysis asks *how fast does this change?*, algebra asks *what is preserved?* Three layers of structure are the subject of these notes. Groups capture the pure essence of symmetry — the transformations of an object that leave it unchanged. Rings generalize the integers, equipping a set with two compatible operations and asking when the arithmetic of $\mathbb{Z}$ — divisibility, primality, factorization — extends to this broader setting. Modules generalize vector spaces by allowing scalars to come from a ring rather than a field; the reward is a unified theory that simultaneously classifies finite abelian groups and explains why every complex matrix is similar to a block-diagonal one.

The common thread through all three is the same methodology: identify the correct notion of *subobject*, form the *quotient* by identifying elements related by that subobject, and use the *isomorphism theorems* to read off the relationship between original, subobject, and quotient. Learning this methodology for groups — where it first appears in its cleanest form — is what makes the transition to rings and modules almost automati

.

These notes follow the Cambridge Part IB course in Groups, Rings, and Modules. Each chapter is self-contained, but the progression is deliberate: groups provide the language, rings provide the arithmetic, and modules provide the classification machinery that ties everything toget

These notes follow the Cambridge Part IB course in Groups, Rings, and Modules. Each chapter is self-contained, but the progression is deliberate: groups provide the language, rings provide the arithmetic, and modules provide the classification machinery that ties everything toget

roups

A group is the mathematical distillation of the concept of symmetry. This chapter begins with the basic definitions — groups, subgroups, cosets — and the first deep result, Lagrange's theorem. It then builds the structural theory: the notion of a normal subgroup and quotient group, the three isomorphism theorems, group actions and the orbit-stabilizer theorem, conjugacy classes and the class equation, Sylow's theorems, and finally the classification of finite abelian groups and the simplicity of the alternating groups. Each of these topics answers a progressively sharper version of the same question: how much can we determine about a group's structure from limit

ro

ps

A g

d data?

## Symmetry and the Gr

up Axioms

### What

oup is the mathematical distillation of the concept of symmetry. This chapter begins with the basic definitions — groups, subgroups, cosets — and the first deep result, Lagrange's theorem. It then builds the structural theory: the notion of a normal subgroup and quotient group, the three isomorphism theorems, group actions and the orbit-stabilizer theorem, conjugacy classes and the class equation, Sylow's theorems, and finally the classification of finite abelian groups and the simplicity of the alternating groups. Each of these topics answers a progressively sharper version of the same question: how much can we determine about a group's structure from limit

d data?

##

Is a Group?

Before writing down an axiom, consider what we want to capture. The symmetries of an equilateral triangle are the six transformations that map the triangle to itself: three rotations ($0°$, $120°$, $240°$) and three reflections. These transformations can be *composed* — "rotate, then reflect" is itself a transformation of the triangle. This composition is associative, there is a "do nothing" transformation, and every transformation can be undone. The six symmetries, together with composition, form the dihedral group $D_6$. What makes this interesting is not the triangle, but the algebraic structure of the six-element set itself — and a group is precisely the abstraction of t

Symmetry and the Gr

up Axioms

#

efinition:Group]
A **group** is a triple $(G, \cdot, e)$ where $G$ is a set, $\cdot : G \times G \to G$ is a binary operation, and $e \in G$ is an ele

ment, satisfyin

# What

Is a Group?

g:
\begin{align*}
&\text{(Associativity)} \quad (a \cdot b) \cdot c = a \cdot (b \cdot c) \quad \text{for all 

} a, b, c \in G, \\
&\text{(Identity)} \quad a \cdot e = e \cdot a = a \quad \text{f

efinition:Group]

A **group** is a t

= a^{-1} \cdo

t a = e.
\end

riple $(G, \cdot, e)$ where $G$ is a set, $\cdot : G \times G \to G$ is a binary operation, and $e \in G$ is an ele

ment, satisfyin

align*}
[/definition]

We usually suppress the operation and write $ab$ for $a \cdot b$. Observe that neither the identity nor inverses are assumed to be unique in the definition — but both turn out to be. If $b$ is also an identity, then $e = eb = b$; if $b$ is also an inverse of $a$, then $b = be = b(aa^{-1}) = (ba)a^{-1} = ea^{-1} = a^{-1}$. So the structure is more rig

d than it first appears.

g:
\begin{align*}
&

\text{(Associat

ivity)} \quad (a \cdot b) \cdot c = a \cdot (b \cdot c) \quad \text{for all 

 = ba$ for all

} a, b, c \in G, \\
&\text{(Identit

$a, b \in G$.
[/definition]

The distinction between abelian and non-abelian groups is fundamental: in an abelian group, the order of composition is irrelevant, while in a non-abelian group it is not. The symmetry group of the equilateral triangle is non-abelian: rotating by $120°$ and then reflecting across a fixed axis gives a different result than refl

cting first and then rotating.

[example

y)} \quad a \cdot e = e \cdot a = a \quad \text{f

:Foundational Examples of Groups]
The following are the core examples that

t a = e.
\end

will recur throughout the course.

*Additive number groups.* The sets $(\mathbb{Z}, +, 0)$, $(\mathbb{Q}, +, 0)$, $(\mathbb{R}, +, 0)$, $(\mathbb{C}, +, 0)$ are all abelian groups. The key point is that subtraction

is always possible: $a + (-a) = 0$.

*Symmetric group.* The **symmetric group** $S_n$ is the group of all bijections $\{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}$ under composition. It has order $n!$. We write permutations in disjoint cycle notation: $(1\ 2\ 3)(4\ 5)$ denotes the permutation sending $1 \mapsto 2 \mapsto 3 \mapsto 1$ and $4 \mapsto 5 \mapsto 4$, with $6, 7, \ldots$ fixed. Since permutations are functions, composition is right-to-left: to compute $(1\ 2\ 3) \circ (1\ 2)$, first apply $(1\ 2)$, then $(1\ 2\ 3)$. Tracing each element: $1 \overset{(1\ 2)}{\mapsto} 2 \overset{(1\ 2\ 3)}{\mapsto} 3$, then $2 \overset{(1\ 2)}{\mapsto} 1 \overset{(1\ 2\ 3)}{\mapsto} 2$, then $3 \overset{(1\ 2)}{\mapsto} 3 \overset{(1\ 2\ 3)}{\mapsto} 1$. The composition sends $1 \mapsto 3$, $2 \mapsto 2$, $3 \mapsto 1$, which is the transposition $(1\ 3)$. This right-to-left co

align*}

[/definition]

We usually suppress the operation and write $ab$ for $a \cdot b$. Observe that neither the identity nor inverses are a

vention is essential to keep in mind.

*General linear group.* The set $\mathrm{GL}_n(\mathbb{R})$ of invertible $n \times n$ real matrices forms a group under matrix multiplic

tion. It is non-abelian for $n \geq 2$.

*Cyclic group.* For $n \geq 1$, the cyclic group $C_n = \mathbb{Z}/n\mathbb{Z}$ is the group of integers modulo $n$ u

ssumed to be 

unique in the 

nder additi

efinition — but both turn out to be. If $b$ is also an identity, then $e = eb = b$; if $b$ is also an inverse of $a$, then $b = be = b(aa^{-1}) = (ba)a^{-1} = ea^{-1} = a^{-1}$. So the structure is more rig

n. It is abeli

d than it first appears.

n of order $n$.
[/example]

### Subgroups

Not every subset of a group is itself a group under the inherited operation. We want those subsets that are closed under the operation and under taking inverses — that i

, subsets which are gr

 = ba$ for all

oups in their own right.

[definition:Subgroup]
Let $(G, \cdot, e)$ be a group. A subset $H \subs

$a, b \in G$.
[/definition]

The distinction between abelian and non-abelian groups is fundamental: in an abelian group, the ord

ant, while in a non-abelian group it is not. The symmetry grou

subgroup**, written $H \l

eq G$, if:
\begin{align*}
&\text{(i) } e \in 

p of the equil

teral triangle is non-abelian: rotating by $120°$ and then reflecting across a fixed axis gives a different result than refl

H, \\
&\text{(ii) } a, b \in H \implies ab \i

cting first and then rotating.

[example

n H, \\
&\te

:Foundational Examples of Groups]
The following are the core examples that

xt{(iii) } a \

n H \implies a^{-1} \in H.
\end{align*}
[/definition]

A convenient criterion collapses these three conditions into one. If $H$ is non-empty and closed under the operation $h_1 h_2^{-1}$, then $H$ is a subgroup: taking $h_1 = h_2$ gives $e \in H$; then $h_1 = e$ gives $h_2^{-1} \in H$ for any $h_2 \in H$; and combining, $H$ is closed

will recur throughout the course.

*Additive number groups.* The sets $(\mathbb{Z}, +, 0)$, $(\mathbb{Q}, +, 0)$, $(\

athbb{R}, +, 0)$, $(\mathbb{C}, +, 0)$ are

under products. This "subgroup lemma" is u

 all abelian groups. The key point is that subtraction

sed constantly.

[example:A Subgroup Computation in $S_3$]
Consider $S_3$, the symmetric gr

oup on three el

is always possible: $

 + (-a) = 0$.

*Symmetric group.* The **symmetric group** $S_n$ is the group of all bijections $\{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}$ under composition. It has order $n!$. We write permutations in disjoint c

ements. Its six elements in cycle notation are
\begin

{align*}
e,\

 (1\ 2),\ (1\ 3),\ (2\ 3),\ (1\ 2\ 3),\ (1\ 3\ 2).
\end{align*}
Let $H = \{e, (1\ 2\ 3), (1\ 3\ 2)\}$. We verify $H \leq S_3$. The set is non-empty. Computing products: $(1\ 2\ 3)(1\ 2\ 3) = (1\ 3\ 2)$ and $(1\ 2\ 3)(1\ 3\ 2) = e$, so $H$ is closed under products. Inverses: $(1\ 2\ 3)^{-1} = (1\ 3\ 2)$ and vice versa

vention is essential to keep in mind.

*General linear group.* The set $\mathrm{GL}_n(\mathbb{R})$ of invertible $n \times n$ real matrices forms a group under matrix multiplic

tion. It is non-abelian for $n \geq 2$.

*Cyclic group.* Fo

 So $H \leq S_3$. This is the alternating group $A_3 \cong C_3$.

Now let $K = \{e, (1\ 2)\}$. Then $K \leq S_3$ as well, since $(1\ 2)^{-1} = (1\ 2)$ (a transposition is its own inverse) and $(1\ 2)(1\ 2) = e \in K$. But $\{(1\ 2), (1\ 3)\}$ is not a su

 $n \geq 1$, the cyclic group $C_n = \mathbb{Z}/n\mathbb{Z}$ is the group of integers modulo $n$ u

bgroup, sin

nder additi

1\ 2), (1\ 3)\}$.
[/example]

## Cosets and the Counting Principle

Lagrange's theorem is the first genuinely surprising result of group theory. It says that the order of any subgroup must divide the order of the whole group — a powerful constraint that immediately rules out many potential subgroup orders. The proof is elegant and rests o

n. It is abeli

n of order $n$.
[/example]

### Subgroups

Not eve

 a single observation about how a subgroup partitions its parent group.

Before stat

y subset of a group is itself a group under the inherited operation. We want those subsets that are closed under the operation and under taking inverses — that 

ng it, we make precise the not

i

ion of "size" for both groups and elements.

[definition:Order of a Group]
The **ord

er** of a grou

, subsets 

 $G$, written $|G|$, is the cardi

nality of the underlying set.
[/definition]

[definition:Order of an Element]
The **order** of an element $g \in G$, written $\mathrm{ord}(g)$, is the smallest positive in

oup

teger $n$ such

 in their own right.

[definition:Subgroup]
Let $(G, \cdot, e)$ be a group. A subset $H \subs

that $g^n = e$. If no such $n$ exists, $g$ has **infinite order**.
[/definition]

The cosets of a subgroup 

subgroup**, written $H \l

re the key to Lagrange's

eq G$, if:
\begin{align*}
&\text{(i) } e \in 

H, \\
&\text{(ii) } a, b \in H \implies ab \i

 proof. They partition the group into pieces of equal size.

[definition:Left Cos

et]
Let $H \le

n H

q G$ and $g \in G$. The **left cose

 \\
&\te

xt{(iii) } a 

t** of $H$ wi

n H \implies a^{-1} \in H.
\end{align*}
[/definition]

A convenient criterion collapses these t

written $G/H =

\{gH : g \in G\}$, and the number of left cosets is the **index** $|G : H|$.
[/definition]

Two cosets are either identical or completely disjoint: if $g_1H \cap g_2H \neq \varnothing$, pick a common element $g_1h_1 = g_2h_2$, so $g_2^{-1}g_1 = h_2h_1^{-1} \in H$, and one verifies $g_1H = g_2H$. Since every $g \in G$ belongs to the coset $gH$ (as $e \in H$), the cosets partition $G$. Moreover, the map $h \mapsto gh$ is a bijection $H \to gH$, so 

hree conditions

ll cosets have the 

 into one. If $H$ is non-

empty and closed under the operation $h_1 h_2^{

ame size $|H|$. Together, these two observations give Lagrange's theorem.

[quotetheorem:841]

The power of [Lagrange's Theorem](/theorems/841) is its scope: it applies to every finite group and every subgroup, with no additional hypotheses. Its first consequence is that the order of any element divides the group order, since the cyclic subgroup $\langle g \rangle = \{e, g, g^2, \ldots, g^{n-1}\}$ has order $\mathrm{ord}(g)$, which must divide $|G|$. Less obviously, it implies that every group of prime order $p$ is cyclic: 

ny non-identity element has order dividing $p$, hence order $p$, so it generates the whole group.

What Lagrange's theorem does *not* say is the converse: if $k \mid |G|$, there need not be a subgroup of order $k$. The group $A_4$ (alternating group on four letters, order $12$) has no subgroup of order $6$, even though $6 \mid 12$. The Sylow theo

-1}$, then $H$ is a subgroup: taking $h_1 = h_

2$ gives $e \

ems, proved later

 give the sharpest partial converse: subgrou

in H$; then $h

ps of prime-power order always exist.

[quoteproof:841]

[example:Coset Partition of $S_3$ by $A_3$]
We compute the lef

1 = e$ gives $h_2^{-1} \in H$ for any $h_2 \in H$; and combining, $H$ is closed

 $A_3$-cosets in $S_

under products. This "subgroup lemma" is u

3$ explicitly. 

Here $A_3 = \{e, (1\ 2\ 3), (1\ 3\ 2)\}$ and $|S_3 : 

oup on three el

ements. Its six elements in cycle notation are
\begin

A_3| = 6/3 = 2$.

The two cosets are:
\begin{align*}
e \cdot A

{align*}
e,\

_3 &= \{e, (1

\ 2\ 3), (1\ 3\ 2)\} = A_3, \\
(1\ 2) \cdot A_3 &= \{(1\ 2), (1\ 2)(1\ 2\ 3), (1\ 2)(1\ 3\ 2)\}.
\end{align*}
We compute each product using right-to-left composition. For $(1\ 2)(1\ 2\ 3)$: apply $(1\ 2\ 3)$ first, then $(1\ 2)$. Tracing elements: $1 \overset{(1\ 2\ 3)}{\mapsto} 2 \overset{(1\ 2)}{\mapsto} 1$, then $2 \overset{(1\ 2\ 3)}{\mapsto} 3 \overset{(1\ 2)}{\mapsto} 3$, then $3 \overset{(1\ 2\ 3)}{\mapsto} 1 \overset{(1\ 2)}{\ma

 (1\ 2),\ (1\ 3),\ (2\ 3),\ (1\ 2\ 3),\ (1

 3\ 2).
\end{align*}
Let $H = \{e, (1\ 2

\ 3), (1\ 3\ 2)\}$. We verify $H \leq S_3$. The set is non-empty. Computing products: $(1\ 2\ 

(1\ 2)

}{\mapsto} 2$, 

3)(1\ 2\ 3) = (

then $3 \overset{(1\ 3\ 2)}{\mapsto} 2 \overset

1\ 3\ 2)$ and $(1\ 2\ 3)(1\ 3\ 2) = e$, so $H$ is clos

{(1\ 2)}{\map

sto} 1$. The result is $(1\ 3)$.

Thus:
\begin{align*}
(1\ 2) \cdot A_3 = \{(1\ 2), (2\ 3), (1\ 3)\}.
\end{

ucts. Inverses: $(1\ 2\ 3)^{-1} = (1\ 3\ 2)$ and vice versa

align*}
Th

 So $H \leq S_3$. This is the alternating group $A_3 \cong C_3$.

Now let $K = \{e, (1\ 2)\}$. Then $K \leq S_3$ as well, since $(1\ 2)^{-1} = (1\ 2)$ (a transposition is its own inverse) and $(1\ 2)(1\ 2) = e \in K$. But $\{(1\ 2), (1\ 3)\}$ is not a su

 two cosets are exactly the even and odd

bgrou

, sin

permutations, partitioning $

1\ 2), (1\ 3)\}$.
[/example]

## Cosets and the Counting Principle

Lagrange's theorem is the first genuinely surprising result of group theory. It says that the order of any subgroup must divide the order of the whole group — a powerful constrai

_3$ into two equal halves of size $3$.
[/example]

## Normal Subgroups and Quotient Groups

### The Problem with Cosets

The coset partition $G/H = \{gH :

 g \in G\}$ is 

nt that imm

a beautiful object. A natural quest

diately rules out many potential subg

oup orders. The proof is elegant and rests o

ion is: does 

 a single observation about how a subgroup partitions its parent group.

Before stat

it form a group in its own right, with multiplication defined by
\begin{align*}
(g_1 H) \cdot (g_2 H) = g_1 g_2 H?
\end{align*}
The answer is: not always. The formula must be *well-defined* — it must give the same answer regardless of which representative we choose for each coset. Changing $g_2$ to $g_2' = g_2 h$ (another representat

 gives no trouble: $g_1 g_2 h H 

ion of "size" for both groups and elements.

[definition:Order of a Group]
The **ord

er** of a grou

= g_1 g_2 H$ 

 $G$, written $|G|$, is the cardi

since $h \in H$. But changing $g_1$ to $g_1' = g_1 h$ causes a problem:
\begin{align*}
(g_1 h)(g_2) H = g_1 (h g_2) H.
\end{align*}
For this to equal $g_1 g_2 H$, we n

ed $g_2^{-1} h g_2 \in H$. Th

nality of the underlying set.
[/definition]

[de

inition:Order of an Element]
The **order** of an element $g \in G$, written $\mathrm{

is must hold for *every* $h \in H$ and every $g_2 \in G$. Subgroups with this property are called normal.

[de

finition:Normal

rd}(g)$, is the smallest posit

ive in

 Subgroup]
A subg

teger $n$ such

roup $H \leq 

that $g^n = e$. If no such $n$ exists, $g$ has **infinite order**.

G$ is **normal**, written $H \trianglelefteq G$, if for every $h \in H$ and ever

y $g \in G$,

[/definition]

begin{align*}
g^{-1} h g \in H.
\end{align*}
Equivalently, $gH = Hg$ for all $g \in G$, i.e. left and right cosets coincide.
[/definition]

Every subgroup of an abelian group is normal, since $g^{-1}hg = g^{-1}gh = h \in H$ trivially. In non-abelian groups, normality is a genuine constraint. In $S_3$, the subgroup $A_3 = \{e, (1\ 2\ 3), (1\ 3\ 2)\}$ is normal (as we will see from the coset computa

re th

ion above — it equals its ow

e key to Lagrange's

n right coset), but $\{e, (1\ 2)\}$ is not: one checks that $(1\ 3)^{-1}(1\ 2)(1\ 3) = (2\ 3) \notin \{e,

 (1\ 2)\}$.

 proof. They partition the group into pieces of equal size.

[definition:Left Cos

et]
Let $H \le

\triangleleft

q G$ and $g \in G$. The **left cose

eq G$, the **quotient group** $G/H$ is the set of left $H$-cos

t** of $H$ wi

ets with multi

written $G

lication
\begin{align*}
(g_1 H) \cdot (g_2 H) = g_1 g_2 H,
\end{align*}
identity element $eH = H$, and inverse $(gH)^{-1} = g^{-1}H$.
[/definition]

The normality condition is exactly what makes this

/H =

well-defined. The group axioms are 

\{gH : g \

nherited from $G$: associativity follows from associativity in $G$, and the identity and inverse checks are immediate.

### Homomorphisms and Isomorphisms

Subgroups and quotient groups describe

the internal structure of a sing

ical or completely disjo

le group. We now want to understand how different groups relate to each other, and for that we need structure-preser

int: if $g_1H \cap g_2H \neq \varnothing$, pick a common element $g_1h_1 = g_2h_2$,

ving maps.

[

 so $g_2^{-1}g_

1 = h_2h_1^{-1} \in H$, and one ver

definition:Group Homomorphism]
Let $(G, \cdot, e_G)$ and $(H, \ast, e_H)$ be groups. A functi

on $\varphi :

ifies $g_1H =

 g_2H$. Since every $g \in G$ belongs to the coset $gH$ (as $e \in H$), the cosets partition $G$. Moreover, the map $h \mapsto gh$ is

 a bijection $

**group homomorphism** if
\begin{align*}
\varphi(g_1 \cdot g_2) = \varphi(g_1) \ast \varphi(g_2) \quad \text{for all } g_1, g_2 \in G.
\end{align*}
[/definition]

From the homomorphism property alone one can deduce $\varphi(e_G) = e_H$ (apply the condition with $g_1 = g_2 = e_G$, then cancel $\varphi(e_G)$)

and $\varphi(g^{-1}) = \varphi(g)^{-1}$ (apply with $g_2 = g^{-1}$ and use $\varphi(e_G) = e_H$). So a homomorphism automatically preserves the entire group structure.

Every homomorphism carries two pieces of information: what it hit

 \to gH$, so 

ll cosets have the 

 (the image) and what it collapses (th

e kernel). Understanding this decomposition — what is pr

m{ord}(g)$, which m

eserved versus 

what is identified — is the key to the isomorphi

st divide $|G|$. Less obviously, it implies that every group of prime order $p$ is cyclic: 

sm theorems.

ny non-identity element has order dividing $p$, hence order $p$, so it generates the whole group.

What Lagrange's theorem does *not* say is the converse: if $k \mid |G|$, there need not be a subgroup of order $k$. The group $A_4$ (alternating group on four letters, order $12$) has no subgroup of order $6$, even though $6 \mid 12$. The Sylow theo

[definition

ems, proved later

Kernel of a Homomorphism]
The **kern

 give the sharpest partial converse: subgrou

ps of prime-power order always 

el** of a homomorphism $\varphi : G \to H$ is
\begin{a

lign*}
\ker(\v

xist.

[quoteproof:841]

[example:Coset Partition of $S_3$ by $A_3$]
We compute the lef

arphi) = \{g \in G : \varphi(g) = e_H\}.
\end{align*}
[/definition]

[definition:

 $A_3$-cosets in $S_

Image of a Ho

3$ explicitly. 

momorphism]
T

e **image** of a homomorphism $\varphi : G \to H$ is
\begin{align*}
\operatorname{im}(\varphi) = \{h \in H : h = \varphi(g) \text{ for some } g \in G\}.
\end{align*}
[/definition]

The kernel is always a normal subgroup of $G$, and the image is always a subgr

A_3| = 6/3 = 2$.

The two cosets are:
\begin{align*}
e \cdot A

_3 &= \{e, (1

up of $H$ — both follow from st

\ 2\ 3), (1\ 3\ 2)\} = A_3, \\
(1\ 2) \cdot A_3 &= \{(1\ 2), (1\ 2)(1\ 2\ 3), (1\ 2)(1\ 3\

2)\}.
\end{align

n:Group Isomor

}
We compute each product using right-to-le

hism]
A **group isomorphism** is a bijective group homomorphism. Two groups $G$ and $H$ are **isomorphic**, written $G \cong H$, if there exists an isomorphism between them.
[/definition]

Isom

rphic groups are, for all al

ft composition. For $(1\ 2)(1\ 2\ 3)$: apply $(1\ 2\ 3)$ first, then $(1\ 2)$. Tracing elements: $1 \overset{(1\ 2\ 3)}{\map

to} 2 \overset{(1\ 2

ebraic purposes, identical: they have the same order, the same subgroup lattice, the same element orders. We regard them as the same group, presented differently.

## The Isomorphism Theorems

The three isomorphism theorems are the primary tools for understanding quotient groups. Their comm

n theme: given a homomorphism, the quotient of the domain by the kernel is isomorphic to the image. This single fact, once understood deeply, makes most computations with quotient groups routine.

The problem that the first isomorphism theorem solves is this: we have a homomorphism $\varphi : G \to H$, and we want to understand $\operatorname{im}(\varphi)$ — but $\operatorname{im}(\varphi)$ lives inside $H

)}{\mapsto} 1$,

, which may be comp

 then $2 \overset{(1\ 2\ 3)}{\mapsto} 3 \overset{(1\ 

icated. The theorem says we can instead study $G/\ker(\varphi)$, which lives inside $G$ and is often simpler. More importantly, it gives an explicit isomorphism between the two.

[quotetheorem:842]

The [First Isomorphism Theorem for Groups](/theorems/842) is used in virtually every computation involving quotient groups. Its key feature is that it transforms questions about subgroups of $H$ (the target) into questions about quotients of $G$ (the source), where we have more control. Notice that the theorem does *not* require $\varphi$ to be surjective — if it is, then $\operatorname{im}(\varphi) = H$ and we get $G/\ker(\varphi) \cong H$ directly. The injectivity 

1\ 2)}{\ma

f the induced map

$\bar{\varphi} : G/\ker(\varphi) \to \

(1\

 2)

operatorname{im}(

\varphi)$ is au

}{\mapsto} 2$, 

tomatic: two cosets are identified precisely when they have the same im

then $3 \overset{(1\ 3\ 2)}{\mapsto} 2 \overset

{(1\ 2)}{\map

age, which is ex

actly the cos

sto} 1$. The result is $(1\ 3)$.

Thus:
\begin{align*}
(1\ 2) \cdot A_3 = \{(1\ 2), (2\ 3), (1\ 3)\}.
\end{

align*}
Th

et condition.

[quoteproof:842]

[example:The Exponential Isomorphism]
Consider the map
\begin{align*}
\varphi : (\mathbb{C}, +) &\to (\mathbb{C} \setminus \{0\}, \times) \\
z &\mapsto e^z.
\end{align*}
The identity $e^{z + w} = e^z e^w$ says exactly that $\varphi$ is a group homomorphism. What is the kernel? We need $e^z = 1$, which holds precisely when $z = 2\pi i k$ for some $k \in \mathbb{Z}$, so $\ker(

 two cosets are exactly the even and odd

 i \mathbb{Z}$. The image is all of $\mathbb{C} \setminus \{0\}$, since the com

_3$ into two equal halves of size $3$.
[/example]

## Normal Subgroups and Quotient Groups

### The Problem with Cosets

The coset partition $G/H = \{gH :

plex logarith

 g \in G\}$ i

m exists (the exponential is surjective onto non-zero complex numbers). By the first isomorphism theorem:
\begin{align*}
(\mathbb{C}/2\pi i\mathbb{Z},\ +) \cong (\mathbb{C} \setminus \{0\},\ \times).
\end{align*}
This is a remarkable identification: the additive quotient of $\mathbb{C}$ by an arithmetic progression equals the multiplicative group of non-zero 

complex num

a beautiful object. A natural quest

ers. Geometrically, $\mathbb{C}/2\pi i \mathbb{Z}$ wraps the complex plane into a cylinder by identifying $z$ with $z + 2\pi i$, and $e^z$ wraps that cylinder further to fill $\mathbb{C} \setminus \{0\}$.
[/example]

The second isom

rphism theorem addr

ion is: does 

it form a group in its own right, with multiplication defined by
\begin{align*}
(g_1 H) \cdot (g_2 H) = g_1 g_2 H?
\end{align*}
The answer is: not always. The formula must be *well-defined* — it must give the same answer regardless of which repr

sses a subtler situation: we have two subgroups $H$ and $K$ of $G$, with $K$ normal, and we want to understand how their interaction $H \cap K$ relates to the coset spaces $HK/K$ and $H/(H \cap K)$.

[quotetheorem:843]

The [Second Isomorphism Theorem for Groups](/theorems/843) is best understood as saying: the "$K$-shadow" of $H$ inside $G/K$ is $HK/K$, and the kernel of the restriction of the quotient map $G \to G/K$ to $H$ is exactly $H \cap K$. The proof works by writing down the obvious map $h \mapsto hK$ and applying 

sentat

he first isomorph

ive we choose f

sm theorem — the content is entirely in identifying the kernel and image correctly. A key consequence: $|HK| = |H||K|/|H \cap K|$ (when $G$ is finite), which counts elements in a

or each coset. Changing $g_2$ to $g_2' = g_2 h$

product of two subg

oups.

[quoteproof:843]

The third isomorphism theorem is sometimes called the "cancellation rule" for quotients, and it says that quotienting in stages gives the same result as quotienting all at once.

[quotetheorem:844]

The [Third Isomorphism Theorem for Groups](/theorems/844) is the group-theoretic analogue of the arithmetic identity $(n/m)/1 = n/m$: if we have already taken the quotient by $K$, and we further quotient by $L/K$, we get the same thing as quotienting $G$ by $L$ directly. The pr

resentat

of is a one-line 

 gives no trouble: $g_1 g_2 h H 

pplication of the first isomorphism theorem to the obvious surjective map $G/K \to G/L$. Its main use is in induction arguments, where one wants to

= g_1 g_2 H$ 

since $h \in H$. But changing $g_1$ to $g_1' = g_1 h$ cause

roup and then apply an inductive hypothesis.

[quoteproof:844]

Finally, the correspondence theorem records how the subgroup lattice of $G/K$ mirrors the part of the subgroup lattice of $G$ that lies above $K$.

[quotetheorem:854]

The [Correspondence Theorem for Groups](/theorems/854) is indispensable whenever we need to enumerate or classify subgroups of a quotient group. Since $K \trianglelefteq G$, every normal subgroup of $G$ containing 

K$ descends to a 

s a problem

ormal subgroup of

\begin{align*}
(g_1 h)(g_2) H = g_1 (

 g_2) H.
\end{align*}
For 

$G/K$, and conversely. This will be used repeatedly in Sylow theory and in the classification of simple groups: to show $G/K$ is simple, we show $G$ has no normal subgroups strictly between $K$ and $G$.

[quoteproof:854

## Group Actions

Ev

his to equal $g_1 g_2 H$, we n

ery abstract group arises in practice as a group of symmetries of

ed $g_2^{-1} h g_2 \in H$. Th

 something. A g

al.

[de

roup action formalizes this:

finit

 it is a way of letting $

G$ "act on" a

ion:Normal

 Subgroup]
A subg

 set $X$ by 

roup $H

permuting its e

 \leq 

lements, compatibly with the group structure.

[definition:Group Action]
An **action** of a group $(G, \cdot

G$ is *

)$ on a set $X$ is a function
\begin{align*}
\ast : G \t

imes X &\to X

*normal**, written $H \trianglelefteq G$, if for every $h \in H$ and ever

y $g \in G$,

begin{align*}
g^{-1} h g \in H.
\end{align*}
Equivalently, $gH = Hg$ for all $g \in G$, i.e. left and right cosets coincide.
[/definition]

Every subgroup of an abelian group is normal, since $g^{-1}hg = g^{-1}gh = h \in H$ trivially. In non-abel

apsto g \ast x
\end{align*}
satisfying:
\begin{align*}
&\text{(i) } g_1 \ast (g_2 \ast x) = (g_1 \cdot g_2) \ast x \quad \text{for all } g_1, g_2 \in G,\ x \in X, \\
&\text{(ii) } e \ast x = x \quad \text{for all } x \in X.
\end{align*}
[/definition]

There is a cleaner way to think about this: an action of $G$ on $X$ is the same as a group homomorphism $\varphi : G \to \mathrm{Sym}(X)$. Given an action, define $\varphi(g) = (x \mapsto g \ast x)$; this is

a bijection (with inverse $\varphi(g^{-1

mality is a genuine constraint. 

})$), and condition (i) says $\varphi(g_1) \circ \varphi(g_2) = \varphi(g_1 g_2)$. Conversely, given $\varphi : G \to \mathrm{Sym}(X)$, define $g \ast x = \varphi(g)(x)$. The two constructions are inverse to each ot

her.

[defin

In $S_3$, the

tion:Permutation Re

 subgroup $A_3 = \{e, (1\ 2\ 3), (1\ 3\ 2)\}$ is normal (as we will see from the coset computa

presentation]
A **permutation representation** of $G$ is a group homomorphism $\varph

ion above — it equals its ow

n right coset), but $\{e, (1\ 2)\}$ is not: one c

i : G \to \math

ecks that $(1\ 3)^{-1}(1\ 2)(

rm{Sym}(X)$ for some set $X$. The **kernel** of 

1\ 3) = (2\ 3) \notin \{e,

the action is

 $\ker(\varphi

 (1\ 2)\}$.

$ and the **image** is $

\triangleleft

eq G$, the **quotient group** $G/H$ is the set of left $H$

\operatorname{im}(\varphi)$.
[/definition]

[definition:Orbit]
For a group $G$ 

acting on a set

-cos

ets with mu

 $X$ and an element $x \in X$, the **orbi

lti

\begin{align

lication
\begi

}
G \cdot x = \{g \ast x : g \in G\} \subseteq X.
\end{align*}
[/definition]

[definition:Stabilizer]
The **stabilizer** (or **isotropy group**) of $x \in X$ under the action of $G$ is
\begin{align*}
G_x = \{g \in G : g \ast x = x\} \leq G.
\end{align*}
[/definition]

The

_1 H) \cdot (g_2 H) = g_1 g_2 H,
\end{align*}
identity element $eH = H$, and i

orbits partition $X

 (they are the equivalence classes of the relation $x \sim y \iff y \in G \cdot x$). The stabilizer is always a subgroup of $G$. These two facts combine into the orbit-stabilizer theorem, which is the counting engine behind most applications of group actions.

[quotetheorem:845]

The [Orbit-Stabilizer Theorem](/theorems/845) is the group-action version of Lagrange's theorem. The bijection $G/G_x \leftrightarrow G \cdot x$ says: elements of $G$ that send $x$ to the same image are exactly the elements of the same coset of $G_x$. The finite version $|G|

nverse $(gH)^{

1} = g^{-1}H$.
[/definition]

The normality condition is exactly what makes this

= |G_x| \cdot |G 

well-defined. The group axioms are 

cdot x|$ is the key formula: to count the orbit, divide $|G|$ by the stabilizer size (which is computable if we understand which elements fix $x$). 

his will be used to

nherited from $G$: associativity follows from associativity in $G$, and the identity and inverse checks are immediate.

### Homomorphisms and Isomorphisms

Subgroups and quotient groups describe

count conjugacy classes, to count Sylow subgroups, and to determine the structure of $p$-groups.

[quoteproof:845]

Every group can be viewed as a group of symmetries — not just an abstract algebraic system — by letting it act on itself. This is Cayley's theorem.

[quotetheorem:846]

[Cayley's Theorem](/theorems/846) has a philosophical message as much as a practical one: groups are not just abstract algebraic systems; they are all, in principle, groups of permutations. In practice, the embedding $G \hookrightarrow \mathrm{Sym}(G)$ is rarely the most effi

the internal structure of a sing

le group. We now want to understand how different groups r

ient way to study

late to each other, and for 

$G$ (since $\mathrm{Sym}(G)$ has order $|G|!$, which grows mu

that we need structure-preser

ch faster than $|G|$). But having a concrete realization of every group as a permutation group proves existence results and connects abstract group theory to the older tradition of studying permutations direct

ving maps.

[

y.

[quoteproof:846]

[example:Counting Symmetries of a Cube via Orbit-Stabilizer]
Let $G$ be the rotation group of th

definition:Group Homomorphism]
Let $(G, \cdot, e_G)$ and $(H

 cube, and let $X$ be the set of four main diagonals of the cube (the lines connecting opposite vertices: there are four pairs of opposite vertices, giving four diagonals).

Every rotation of the cube

, \ast, e_H)$ b

e groups. A functi

angular cross-section perpendicular to $d$ are cyc

lically permute

on $\varphi :

d). In addition, there are three rotations by $18

0°$ around ax

oup homomorph

s that pass through midpoints of opposite edges and are perpendicular to $d$ — each of these swaps the two endpoints of $d$ while fixing $d$ as a set. This gives $|G_d| = 6$.

By the [Orbit-Stabilizer Theorem](/theorems/845):
\begin{align*}
|G| = |G_d| \cdot |G \cdot d| = 6 \times 4 = 24.
\end{align*}

As a consistency check, we can 

ism** if
\begin{align*}
\varphi(g_1 \cdot g_2) = \varphi(g_1

nstead let $G$ act on the set of six faces. The orbit of any face $f$ has size $6$ (rotations can send any face to any other), and the stabilizer of $f$ consists of rotations by $0°, 90°, 180°

) \ast \varphi

g_2) \quad \text{for all } g_1, g_2 \in G.
\end{align*}
[/definition]

From the homomorphism property alone one can deduce $\varphi(e_G) = e_H$ (apply the condition with $g_1 = g_2 = e_G$, then cancel $

, 270°$ aro

nd the axis through $f$ and the oppo

varphi(e_G)$)

and $\varphi(g^{-1}) =

ite face, giving $|G_f| = 4$. This confirms $|G| = 4 \times 6 = 24$.

Now $\varphi : G \to S_4$ is injective (a non-identity rotation must move at least one diagonal, so the kernel of the action on diagonals is trivial) and $|G| = 24 = |S_

\varphi(g)^{-1}$ (apply with $g_2 = g^{-1}$ and use $\varphi(e_G) = e_H$). So a homomorphism automatically preserves the entire group structure.

Every homomorphism carries two pieces of informati

|$, so $G \cong S_4$.
[/exam

ple]

## Conjugacy and the Class Equat

n: what it hit

 (the image) and w

ion

Among al

hat it collapses (th

l group actions, the action of a group on it

self by conju

e kernel). Understanding this decomposition — what is pr

eserved versus 

s. The orbits 

what is identified — is t

re the conjugacy classes,

somorphi

 which encode fundamental informatio

sm theorems.

n about the gro

up's structure.

[definition:Conjugacy Class]
The **conjug

[definition

Kernel of a Homomorphism]
The **kern

acy class** o

el** of a homomorphism

f $g \in G$ is
\begin{align*}
\mathrm{ccl}_G(g) = \{hgh^{-1} : h \in G\},
\end{

align*}
i.e. 

 $\varphi : G

he orbit of $g$ unde

 \to H$ is
\b

gin{a

r the conjugation action 

of $G$ on itsel

lign*}
\ker(\v

arphi) = \{g \in G : \varphi(g) = e_H\}.
\end{align*}
[/definition]

[definition:

f.
[/definition]

[definition:Centralizer]
The **centralizer** of $g \in G$ is

Image of a Ho

\begin{alig

n*}
C_G(g) = 

momorphism]
T

e **image** of a homomorphism $\varphi : G \to H$ is
\begin{align*}
\operatorname{im}(\varphi) = \{h \in H : h = \varphi(g) \text{ for some } g \in G\}.
\end{align*}
[/definition

 G} C_G(g).
\e

nd{align*}
[/definition]

Elements of $Z(G)$ commute with everything; they form conjugacy 

up of $H$ — both follow from st

n:Group Isomor

classes of si

hism]
A **group isomorphism** is a bijective group homomorphism. Two groups $G$ and $H$ are **isomorphic**, 

ze $1$ (since $hgh^{-1} = g$ for all $h$ iff $g \in Z(G)$). By the [Orbit-Stabilizer Theorem](/theorems/845), $|\mathrm{ccl}_G(g)| = |G|/|C_G(g)|$. Since the conjugacy class

ritten $G \cong H$, if there exists an

s partition $G$, we get the **class equation**:
\begin{align*}
|G| = |Z(G)| + \sum_{\substack{g \notin Z(G) \\ \text{one per class}}} \frac{|G|}{|C_G(g)|}.
\end{align*}
Every term in the sum on the right is greater than $1$ (since $g \notin Z(G)$ means the centralizer is a proper subgroup). This equation is the key to analyzing $p$-g

oups.

In symmetr

 isomorphism between them.
[/definition]

Isom

rphic g

c groups, conjugacy classes are determined entirely by cycle type: two permutations in $S_n$ are conjugate if and only if they have the same cycle structure. This is because $\sigma (a_1\ a_2\ \cdots\ a_k) \sigma^{-1} = (\sigma(a_1)\ \sigma(a_2)\ \cdots\ \sigma(a_k))$ — conjugation simply relabels the elements being permuted.

[quotetheorem:848]

The [p-Group Has Nontrivial Center theorem](/theorems/848) is one of the most consequential in all of group theory. Its immediate corollary is that a $p$-group of order $p^n$ with $n \geq 2$ is never simple: $Z(G)$ is a non-trivial normal subgroup. More subtly, it enables inductive arguments: since $Z(G)$ is a normal subgroup of $G$, we can form the quotient $G/Z(G)$, which is a strictly smaller $p$-group, 

nd apply the theo

roups are, for 

em again. This "center-killing" induction is the basis for the cla

all al

ebraic purposes, identical: they have the 

ssification of $p$-groups of small order. Notice the role of $p$-divisibility in the proof: the class equation forces $p \mid |Z(G)|$ b

same order, t

cause $p$ divides $|G|$ and $p$ divide

 every non-singleton conjugacy class size; this is where the prime-power hypothesis is used.

[quoteproof:848]

[example:Conjugacy Classes in a Non-Abelian Group of Order $p^3$]
Let $p$ be an odd prime and let $G$ be a non-abelian group of order $p^3$. We determine the complete conjugacy class structure of $G$.

**Step 1: The centre has order $p$.**

By [p-Group Has Nontrivial Center](

he same subgro

p lattice, the same element orders. W

theorems/848), $|Z(G)| \geq p$. If $|Z(G)| = p^3$,

then $G$ is abelian, contradicting our assumption. If $|Z(G)| = p^2$, then $|G/Z(G)| = p$, which is cyclic. But if $G/Z(G)$ is cyclic, then $G$ is abelian — a standard lemma (if every element of $G$ has the form $g^r z$ with $z \in Z(G)$, then any two elements commute) — c

e regard them as the same group, presented differently.

## The Isom

orphism Theorems

The three isomorphism theorems are the primary tools for understa

G)| = p$.

**Step 2: Centralizers of non-central elements.**

For $g \notin Z(G)$, we have $Z(G) \subseteq C_G(g)$ (the centre commutes with everything) and $C_G(g) \subsetneq G$ (since $g \notin Z(G)$)

 By Lagrange, $|C_G(g)

t groups. Thei

$ divides $p^3$ and satisf

ies $p \leq |Z(

 comm

G)| \leq |C_G(g)| < p^3$. The only possibility is $|C_G(g)| = p^2$.

n theme: given a homomorphism, the quotient of the domain by the kernel is isomorphic to the image. This single fact, once understood deeply, makes most computations with quotient groups routine.

The problem that the first isomorphism theorem solves is this

**Step 3:

 we have a homomorphism $\varph

 Class sizes.**

By the orbit-stabilizer theorem, $|\mathrm{ccl}_G(g)| = |G|/|C_G(g)|$. For $g \in Z(G)$: $|

i : G \to H$, and we want to understand $\operatorname{im}(\varphi)$ — but $\operatorname{im}(\varphi)$ lives inside $H

, which may be comp

_G(g)| = |G| = p^3$, so the class has size $1$. For $g \notin Z(G)$: $|C_G(g)| = p^2$, so the class has size $p$.

**Step 4: Counting.**

The class equation gives:
\begin{align*}
p^3 = \underbrace{p}_{\text{elements of } Z(G)} + \lambda \cdot p,
\end{align*}
where $\lambda$ is the number of non-singleton conjugacy classes. Solving: $\lambda = (p^3 - p)/p =

icated. The theorem says we ca

 p^2 - 1$.

So $G$ has exa

n instead stud

 $G/\ker(\varphi)$, which lives inside $G$ and is often simpler. More importantly, it gives an explicit isomorphism between the two.

[quotetheorem:842]

The [First Isomorphism Theorem for Grou

tly $p$ conjugacy classes of size $1$ (the centre) and $p^2 - 1$ conjugacy classes of size $p$. The total number of conjugacy classes is $p + (p^2 - 1) = p^2 + p - 1$. For $p = 2$, this gives $2^2 + 2 - 1 = 5$ conjugacy classes in a non-abelian group of order $8$ (both $D_8$ and $Q_8$ have exactly $5$ conjugacy classes, consistent with this count).
[/example]

## Sylow Theory

agrange's theorem tells us that 

s](/theorems/842) is used in

virtually every computation involving quotient groups. Its key feature is that it transforms questions about subgroups of $H$ (the target) into questions about quotients of $G$ (the source), where we have more control. Notice that the theorem does *not* require $\varphi$ to be surjective — if i

the order of any subgroup divides $|G|$ — but it says nothing about which divisors actually arise. In general, the converse fails: $A_4$ has order $12$ but no subgroup of order $6$. The Sylow theorems provide the strongest result in the other 

 is, then $\operatorname{im}(\varphi) = H$ and we get $G/\ker(\varphi) \cong H$ directly. The injectivity 

f the induced map

every prime power $p^a$ dividing $|G|$ *to its full extent*, a subgroup of that order exists.

[definition:Sylow $p$-Subgroup]
Let $G$ be a finite group with $|G| = p^a m$ where $p$ is prime and $p \nmid m$. A **Sylow $p$-subgroup** of $G$ is a subgroup of order $p^a$. The set of all Sylow $p$-subgroups is denoted $\mathrm{Syl}_p(G)$, and $n_p = |\mathrm{Syl}_p(G)|$.
[/defi

ition]

Why shoul

$\bar{\varphi} : G/\ker(\varphi) \to \

 such subgroups exist? The naive expectation from Lagrange would be that we need to build them up from smaller subgroups, but there is no obvious reason they should piece together correctly. The key insight in Sylow's proof is to act on the set of all $p^a$-element subsets of $G$ — a combinatorial object whose size is coprime to $p$ — and extract an invariant subset.

[quotetheorem:847]

The three parts of [Sylow's Theorems](/theorems/847) answer three different questions about the Sylow $p$-subgroups. The first guarantees existence. The second — that all Sylow $p$-subgroups are conjugate — is the deeper result: it says the subgroup is essentially unique, up to the internal symmetry of $G$. The third gives arithmetic constraints on $n_p$: the cong

uence $n_p \equiv

\varphi)$ is au

tomatic: two cosets are identified precisely when they have the same im

1 \pmod{p}$ and the divisibility $n_p \m

age, which is ex

id m$ together severely restrict how many Sylow subgroups the

e can be. In practice, one combines $n_p \equiv 1 \pmod{p}$ with $n_p \mid m$ to narrow $n_p$ to a small list of candidates, and then either forces $n_p = 1$ (giving a *unique*, hence normal, Sylow subgroup) or derives a contradiction to prove simplicity or non-simplicity.

[quoteproof:847]

[exampl

actly the cos

et condition.

[quoteproof:842]

[example:The Exponential Isomorphism]
Consider the map
\begin{align*}
\varphi 

:No Simple Group of Order 1000]
Let $|G| = 1000 = 2^3 \cdot 5^3$. We show $G$ is not simple.

Apply [Sylow's Theorems](/theorems/847) with $p = 5$. We need $n_5 \equiv 1 \pmod{5}$ and $n_5 \mid 2^3 = 8$. The divisors of $8$ that are congruent to $1 \pmod{5}$ are: $1$ (since $8 \equiv 3 \pmod{5}$, $4 \equiv 4 \pmod{5}$, $2 \equiv 2 \pmod{5}$, $1 

\equiv 1 \p

 (\mathbb{C}, +) &\

od{5}$). The only option is $n_5 = 1$.

o (\mathbb{C} \setminus \{0\}, \times) \\
z &\mapsto e^z.
\end{align*}
The identity $e^{z + w} = e^z e^w$ says exactly that $\varphi$ is a group homomorphism. What is the kernel? We need $e^z = 1$, which holds precisely when $z = 2\pi i k$ for some $k \in \mathbb{Z}$, so $\ker(

Since there is exactly one Sylow $5$-subgroup $P$, it must be normal in $G$: any conjugate $gPg^{-1}$ is again a Sylow $5$-subgro

 i \mathbb{Z}$. The image is all of $\mathbb{C} \setminus \{0\}$, since the com

plex logarith

p (it has the same order $5^3 = 125$), and since there is only one, $gPg^{-1} = P$ for all $g \in G$. So $P \trianglelefteq G$ with $P \neq \{e\}$ and $P \neq G$ (since $|P| = 125 < 1000 = |G|$). Thus $G$ is not simple.
[/example]

[example:No Simple Group of Order 13

m exists (the exponential is surjective onto non-zero complex numbers). By the first isomorphism theorem:
\begin{align*}
(\mathbb{C}/2\pi i\mathbb{Z},\ +) \cong (\mathbb{C} \setminus \{0\},\ \times).
\end{align*}
This is a remarkable identification: the additive quotient of $\mathbb{C}$ by an ari

]
Let $|G| = 132 = 2^2 \cdot 3 \cdot 11$. We show $G$ is not simple by deriving an element-counting contradiction if we assume it is.

**Step 1: Sylow 11-subgroups.** We have $n_{11} \equiv 1 \pmod{11}$ and $n_{11} \mid 12$ (since $132 = 11 \cdot 12$). The divisors of $12$ are $1, 2, 3, 4, 6, 12$. Those congruent 

hmetic progressio

o $1 \pmod{11}$ are: $1$ and $12$. If $G$ were simple, $n_{11} \neq 1$, so $n_{11} = 12$.

**Step 2: Sylow 3-subgroups.** We have $n_3 \equiv 1 \pmod 3$ and $n_3 \mid 44$ (since $132 = 3 \cdot 44$). Divisors of $44$: $1, 2, 4, 11, 22, 44$. Those $\equiv 1 \pmod 3$: $

 equals the multiplicative group of no

n-zero 

_3 = 4$, then $

G$ acts on $\mathrm{Syl}_3(G)$ by conjugation, giving a 

m

homomorphism 

ers. Geometric

$\varphi : G \to S_4$. Since $G$ is simple, $\ker \varphi = \{e\}$, so $G \cong \operatorn

ally, $\mathbb{C}/2\pi i \mathbb{Z}$ wraps the complex plane into a cyl

ame{im}(\varphi

) \leq S_4$. But $|G| = 132 > 24 = |S_4|$, a contradi

inder by identif

ying $z$ with

ction. So $n_

 $z + 2\pi i$, and $e^z$ wraps that cylinder further to fill $\mathbb{C} \setminus \{0\}$.
[/example]

The second isom

3 = 22$.

**Step 3: Element count.** Each Sylow $11$-subgroup has order $11$ (prime)

rphism theorem addr

 so any two distinct Sylow $11$-subgroups intersect trivially. This gives
\begin{align

*}
12 \tim

sses a subtler situation: we have two subgroups $H$ and $K$ of $G$, with $K$ normal, and we want to understand how their interaction $H \cap K$ relates to the coset spaces $HK/K$ and $H/(H \cap K)$.

[quotetheorem:843]

The [Second Isomorphism Theorem for Groups](/theorems/84

3) is best unde

s (11 - 1) = 120 \text{ elements of order } 11.
\end{align*}
Ea

h Sylow $3$-subgroup has order

rstood as saying: the "$K$-shadow" of $H$ inside $G/K$ is $HK/K$, and the kerne

$3$ (prime), so distinct ones also intersect trivially:
\begin{align*}
22 \times (3 - 1) = 44 \text{ elements of order } 3.
\end{align*}
Total elements accounted for: $120 + 44 = 164$. But $|G| = 132 < 164$. Contradiction.

Since assuming $G$ is simple leads to a contradiction, $G$ cannot be si

l of the rest

ple. $\square$
[/example]

riction of the quotient map $G \to G/K$ to $H$ is exactly $H \cap K$. The proof works by writing down the obvious map $h \mapsto hK$ and applying 

he first isomorph

sm theorem — the content is entirely in identifying the kernel and image correctly. A key consequence: $|HK| = |H||K|/|H \cap K|$ (when $G$ is finite), which counts elements in a

Theory

The 

ylow theorems let us decompose many groups by finding normal Sylow subgroups. The groups that resist any such decomposition — groups with no proper non-trivial normal subgroups — are called simple. They are the "atoms" of group theory, in the same way that primes are the atoms of number theory.

[definition:Simple Group]
A non-trivial group $G$ is **simple** if its only normal subgroups are $\{e\}$ and $G$ itself.
[/definition]

Among abelian groups, the simple ones are easy to characterize: $C_p$ for prime $p$ is simple (since its

oups.

only subgroups have order $1$ or $p$, and all subgroups are normal), and conversely, any abelian simple group must be cyclic of prime order (a non-cyclic abelian group has proper non-trivial subgroups, and an infinite cyclic group $\mathbb{Z}$ is not simple since $2\mathbb{Z} \trianglelefteq \mathbb{Z}$). Non-abelian simple groups are far more subtle. The smallest is $A_5$, the alternating group on five letters, which has order $60$.

Why do we care about simple groups? Finite group theory has a decomposition theorem: every finite group $G$ has a *composition series* — a chain $\{e\} = G_0 \trianglelefteq G_

[quotep

 \trianglelefteq \cdots \trianglelefteq G_k = G$ in which each quotient $G_{i+1}/G_i$ is simple. The simple groups are therefo

e the building bloc

oof:843]

The third isomorphism theorem is sometimes called the "cancellation rule" for quotients, and it says that quotienting in stages gives the same result as quotienting all at once.

[quotetheorem:844]

The [Third Isomorphi

s, and the Jordan-Hölder theorem guarantees the list of quotients is an isomorphism invariant of $G$. Classifying all finite simple groups — the Classification of Finite Simple Groups (CFSG), completed in the early 2000s — is one of the greatest achievements of twentieth-century mathematics.

The infinite family of non-abelian simple groups most relevant to this course is the alternating groups $A_n$ for $n \geq 5$.

[quotetheorem:849]

The proof of the [Alternating Groups Are Simple theorem](/theorems/849) uses a beautiful three-step strategy. First, one shows $A_n$ is generated by 3-cycles. Second, one shows any normal subgroup containing *any* 3-cycle must contain *all* 3-cycles, hence all of $A_n$ — this uses conjugation to transport a 3-cycle to any other 3-cycle, with the $n \geq 5$ hypothesis needed to make the con

m Theorem for Group

](/theorems/844) is the group-theoretic analogue of the arithmetic identity $(n/m)/1 = n/m$: if we have already taken the quotient by $K$, and we further quotient by $L/K$, we get the same thing as quotienting $G$ by $L$ directly. The pr

ugating permutati

of is a one-line 

n even. Third, one shows that any non-

rivial normal subgroup must contain a 3-cycle, by examining all possible cycle structures and deriving a 3-cycle in each case. The condition $n \geq 5$ appears twice in the proof, both times to ensure there are enough indices to construct the needed permutations. The result fai

pplication of the first isomorphism theorem to the obvious surjective map $G/K \to G/L$. Its main use is in induction arguments, where one wants to

s for $n = 4$: the 

roup and then apply an inductive hypothesis.

[quoteproof:844]

Finally, the correspondence theorem records how the subgroup lat

ice of $G/K$ mirr

roup $V = \{e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$ is a normal subgroup of $A_4$.

[quoteproof:849]

### Classifying Finite Abelian Groups

While non-abelian simple groups are enormously complex, finite abelian groups have a complete and explicit classification. The key insight is that every finite abelian group can be written uniquely as a product of cyclic groups, with the order of each factor dividing the next.

[quotetheorem:850]

The [Classification of Finite Abelian Groups](/theorems/850) reduces the study of finite abe

rs the part of the subgroup lattice of $G$ that lies above $K$.

[quotetheorem:854]

The [Correspondence Theorem for Groups](/theorems/854) is indispensable whenever we need 

ian groups to purely combinatorial data: the invariant factors $d_1, \ldots, d_r$ with $d_1 \mid d_2 \mid \cdots \mid d_r$. Its proof (which uses the [Structure Theorem for Finitely Generated Modules over Euclidean Domains](/theorems/857) from Chapter 3, applied to $\mathbb{Z}$-modules) is one of the finest examples of how algebraic machinery can produce a complete classification. For now, the theorem should be taken as given; Chapter 3 wi

l prove it.

Wh

o enumerate or clas

ify subgroups of a quotient group. Since $K \trianglelefteq G$, every normal subgroup of $G$ containing 

t the theorem does *not* say is that the decomposition into

K$ descends to a 

ormal subgroup of

re $d_1 \mid d_2 \mid \cd

ts \mid d_r$. There is 

## Group Actions

Ev

lso a *primary decompos

tion*, where each $C_{d_i}$ is further broken into prime-power cyclic pieces via the Chinese remainder theorem; that form is not unique in terms of the ordering, but the multiset of prime powers that appears is unique.

[quoteproo

ery abstract group arises in practice as a group of symmetries of

 something. A g

:850]

[example:Classifying Finite Abelian Groups of Small Order]
We list all finite abelian groups of order $\leq 16$ using the classification theorem.

Order $1$: only $\

roup action formalizes this:

e\}$.

Order $2$: only $C_2$.

Order $3$: only $C_3$.

Order $4$: $d_1 \mid d_2$ with $d_1 d_2 = 4$. Options: $(d_1, d_2) = (2, 2)$ or a single factor $(d_1) = (4)$. This gives $C_2 \times C_2$ and $C_4$. These are distinct: $C_4$ has an element of order $4$; $C_2 \times C_2$ does 

 it is a way of l

ot.

Order $8$: single factor $C_8$; two factors with $d_1 \mid d_2$, $d_1 d_2 = 8$: $(2, 4)$ giving $C_2 \times C_4$; three factors: $(2, 2, 2)$ giving $C_2 \times C_2 \times C_2$.

Order $12$: $C_{12}$; and $C_2 \times C_6$ (note $2 \mid 6$). These are the only two since $(2, 6)$ is the only factorization with $d_1 \mid d_2$ and $d_1 d_2 = 12$ and $d_1 > 1$, $d_2 > 1$. (The optio

tting $

n $(3, 4)$ 

G$ "act on

ails: $3

 a

\nmid 4$.) So there are exactly two abelian groups of order $12$.

Order $16$: $C_{16}$; $C_2 \times C_8$ (since $2 \mid 8$); $C_4 \times C_4$ (since $4 \mid 4$); $C_2 \times C_2 \times C_4$ (since $2 \mid 2 \mid 4$); $C_2 \times C_2 \times C_2 \times C_2$. That is five abelian groups of order $16$, corresponding to the five partitions of $4$: $(4), (1,3), (2,2), (1,1,2), (1,1,1,1)$ translated as exponents in the prime-power decomposition at $p = 2$.
[/example]

# Rings

Groups are powerf

 set $X$ by 

l, but they model only one operation. The integers $\mathbb{Z}$ support two: addition and multiplication, linked by distributivity. A ring is the abstraction of this two-operation structure. The central question driving ring theory is the same one that makes number theory rich: when, and in what form, does factorization work? The integers have unique prime factorization — but most rings do not, and understanding exactly which rings do, and why, will occupy this entire chapter.

The path goes: rings → ideals 

permuting its e

lements, compatibly with the group structure.

[definition:Group Action]
An **action** of a group $(G, \cdot

the ring-theoretic analogue of normal 

)$ on a

ubgroups) → quotient rings and isomorphism theorems → integral domains and fields of fractions → the hierarchy of increasingly well-behaved rings (Euclidean domains, PIDs, UFDs) → polynomial rings and their f

set $X$ is a functi

n
\begin{align*}
\ast : G \t

 → Noetherian rings and the Hilbert basis theorem. Each step generalizes the integers in a precise direction, revealing which properties of $\mathbb{Z}$ are ro

bust and which 

are special.

## Rings and Their Arithmetic

### The 

apsto g \ast x
\end{align*}
satisfying:
\begin{align*}
&\text{(i) } g_1 \ast (g_2 \ast x) = (g_1 \cdot g_2) \ast x \quad \text{for all } g_1, g_2 \in G,\ x \in X, \\
&\text{(ii) } e \ast x = x \quad \text{for all } x \in X.
\end{align*}
[/definition]

There is a cleaner way to think about this: an action of $G$ on $X$ is the same as a group homomorphism $\varphi : G \to \mathrm{Sym}(X)$. Given an ac

ion, define $\var

Definition and First Examples

A ring keeps both operations of $\mathbb{Z}$ but drops the requirement tha

hi(g) = (x \mapst

t multiplication have inverses. This is deliberate: it is the absence of multiplicative inverses that makes

 divisibility

 g \ast x)$; this is

a bijection (with inverse $\varphi(g^{-1

 interesting.

[definition:Ring]
A **ring** is a quintuple $(R, +, \cdot, 0_R, 1_R)$ where $R$ is a set, $+, \cdot : R \times R \to R$ are binary operations, and $0_R, 1_R \in R$, satisfying:
\begin{align*}
&\text{(i) } (R, +, 0_R) \text{ is an abelian group}, \\
&\text{(ii) multiplication is associative: } a(bc) = (ab)c, \text{ and } 1_R \cdot r = r \cdot 1_R = r, \\
&\text{(iii) multiplication distrib

})$), and condition (i) says $\varphi(g_1) \circ \varphi(g_2) = \varphi(g_1 g_2)$. Conversely, given $\varphi : G \to \mathrm{Sym}(X)$, define $g \ast x = \varph

(g)(x)$. The two construct

tes over addition: } r(s + t) = rs + rt \text{ and } (r+s)t = rt + st.
\end{align*}
[/definition]

Note that $0_R \neq 1_R$ unless $R = \{0\}$ (the **zero ring**). Indeed, if $1_R = 0_R$ and $r \in R$, then $r = r \cdot 1_R = r \cdo

ions are inverse to each ot

 0_R = 0_R$, so every

her.

[defin

 ring in which $1_R = 0_R$ is the trivial one-element ring. One basic consequence of the axioms: $r \cdot 0_R = 0_R$ for all $r$, since $r \cdot 0_R =

 r \cdot (0_R 

tion:Permutation Re

 0_R) = r \cdot 0_R + r \cdot

prese

 0_R$, and cancelling gives $r \cdot 0_R = 0_R$.

From now 

ntation]
A **p

on, all rings a

ermutation representation** 

re **commutative**: $ab = ba$ for all $a, b \in R$. This is 

of $G$ is a group homomor

i : G \to \m

lmost all rings arising in number theory and geometry, and the commutativity hypothesis is essential for the ideal theory developed below.

[definition:Subring]
A subset $S \subseteq R$ is a **subring**, written $S \leq R$, if $0_R, 1_R \in S$ and $S$ is closed under addition, subtraction, and multiplication.
[/definit

ion]

[ex

ath

mple:The Ring Hierarchy]
The familiar number systems form a nested chain of subrings:
\begin{align*}
\mathbb{Z} \leq \mathbb{Q} \leq \mathbb{R} \leq \mathbb{C},
\end{alig

rm{Sym}(X)$ 

*}
under the usua

for some set $X$. The **kernel** of 

l $0, 1, +, \cdot$. The Gaussian integers $\mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z}\} \leq \mathbb{C}$ are another subring. Notice that $\ma

the action is

 $\ker(\varphi

thbb{Z}[i]$ co

tains elements (lik

$ and the **image** is $

e $1 + i$) that have no multiplicative inverse within $\mathbb{Z}[i]$, j

\operatorname{im}(\varph

i)$.
[/definition]

[definition:Orbit]
For a group $G$

ust as $2 \in 

mathbb{Z}$ has no inverse in $\mathbb{Z}$.
[/example]

The elements that do have multiplicative inverses are called units. They are the "small" elements from the perspective of divisibility, analogous to $\pm 1$ in $\mathbb{Z}$.

[definition:Unit]
An element $u \in R$ is a **unit** if there exists $v \in R$ with $uv = 1_R

acting on a 

. The set of all units forms a group $R^\times$ under multiplication.
[/definition]

set

definition:Field]
A non-zero

 ring $R$ is a **field** if every non-zero element is a unit.
[/definition]

The fields $\mathbb{Q}, \mathbb{R}, \mathbb{C}$ are exactly those number systems where division is always possible (by non-zero elements). The integers $\mathbb{Z}$ are not a field — only $\pm 1$ are units. This is what makes $\mathbb{Z}$ arithmetically rich: not everything 

 $X$ and an

element $x \in X$, the **orbi

ing else, so divisibility has content.

Rings support a natural polynomial construction, essential for everything that follows.

[definition:Polynomial Ring]
Let $R$ be a ring. The **polynomial ring** $R[X]$ is the set of all expressions $f = a_0 + a_1 X + \cdots + a_n X^n$ with $a_i \in R$, where $X$ is a formal symbol. Addition and 

}
G \cdot x = \{g \ast x : g \in G\} \subseteq X.
\end{align*}
[/definition]

[definition:Stabilizer]
The **stabilizer** (or **isotropy group**) of $x \in X$ under the action of $G$ is
\begin{align*}
G_x = \{g \in G : g \ast x = x\} \leq G.
\end{align*}
[/definition]

The

ultiplication are defined 

by the usual rules for polynomials. The **degree** $\deg f$ is the largest $n$ with $a_n \neq 0$; $f$ is **monic** if its leading coefficient is $1$.
[/definition]

A crucial subtlety: a polynomial is a sequence of coefficients, not a function. In the ring $\mathbb{Z}/2\mathbb{Z}$, the polynomial $X^2 + X$ 

orbits partition $X

 (they are the equivalence classes of the relation $x \sim y \iff y \in G \cdot x$). The stabilizer is always a subgroup o

is non-zero (i

 $G$. These two facts combine into the o

s coefficients are not all ze

o), but $f(0) = f(1) = 0$, 

rbit-stabilizer theorem, which is the counting engine behind most applications of group actions.

[quotetheorem:845]

The [Orbit-Stabilizer Theorem](/theorems/845) is the group-action version of Lagrange's theor

o it defines the zero function. Identifying polynomials with functions would collapse this distinction and lose information.

[definition:Power Series]
The ring $R[[X]]$ of **formal power series** over $R$ consists of infinite expressions $f = \sum_{n=0}^\infty a_n X^n$ with $a_n \in R$, with the obvious addition and Cauchy-product multiplication. The polynomial $1 - X$ is not a unit in $R[X]$, but it is a unit in $R[[X]]$: $(1-X)(1 + X + X^2 + \cdots) = 1$.
[/definition]

em. The biject

# Ideals and Quotie

on $G/G_x \leftrigh

tarrow G \cdot x$ says: elements of $G$ that send $x$ to the same image are exactly th

nt Rings

### Why Not Just Subrings?

In group theory, to form a quotient

 $G/H$ we neede

e elements of t

he same coset of $G_x$. The finite version $|G|

d $H$ to be normal — the condition that makes coset multiplication 

=

well-defined. In ring theory, the analogous condition is being an **ideal**. The difference from a su

 |G_x| \cdot 

bring is tell

|G 

ing: a subring is closed under multiplica

tion by its ow

cdot x|$ is

e)}
\end{align*}
$I$ is a *

t the orbit, divide $|G|$ by the stabilizer size (which is computable if we underst

and which eleme

*proper ideal** if $I \neq R$.
[/definition]

Condition (ii) is strictly stronger than being a subring: a subring requires closure under multip

nts fix $x$). 

lication of two

 elements both from $S$, whereas an ideal requires closure even w

his will be used to

count co

hen only one 

njugacy class

factor is from

$I$. An immediate consequence: if

es, to count S

low subgroups, and to determine the structure of $p$-groups.

[quoteproof:845]

Every group can be viewed as a group of symmetries — not just an abstract algebraic system — by letting it act on itself. This is Cayley's theorem.

[quotetheorem:846]

[Cayley's Theorem](/theorem

 a proper ideal $I$ contained a unit $u$, then $1_R = u^{-1}u \in I$ (by strong closure), so $r = r \cdot 1_R \in I$ for all $r$, giving $I = R$ — a contradiction. So **proper ideals never contain units**, and in particular $1_R \notin I$.

[definition:Generated Ideal]
For $a \in R$, the **principal ideal** $(a) = aR = \{ar : r \in R\}

 is the ideal generated by $a$. More generally, for $a_1, \ldots, a_k \in R$:
\begin{align*}
(a_1, \ldots, a_k) = \{a_1 r_1 + \cdots + a_k r_k : r_i \in R\}.
\end{align*}
[/definition]

[example:Ideals in $\mathbb{Z}$]
Every ideal of $\mathbb{Z}$ is principal. Given $I \trianglelefteq \mathbb{Z}$, if $I = \{0\}$ then $I = (0)$. Otherwise, let $n$ be the smallest positive element of $I$. For any $m \in I$, write $m = qn + r$ with $0 \leq r < n$; since $

/846) has a philoso

r = m - qn 

hical message as much as a practical one: groups are not just abstract algebraic systems; they are all, in principle, groups of permutations. In practice, the embedding $G \hookrightarrow \mathrm{Sym}(G)$ is rarely the most effi

ient way to study

in I$ and $r < n$, minimali

$G$ (since $\mathrm{Sym}(G)$ has order $|G|!$, which grows mu

ty forces $r = 0$, so $n \mid m$. Thus $I = n\mathbb{Z} = (n)$.

The ring $\mathbb{Z}[X]$ is not like this: the ideal $(2

ch faster than $|G|$). But having a concrete realization of every group as a permutation group proves existence results and connects abstract group theory to the older tradition of studying permutations direct

, X) = \{2f + X

mmetries of a Cub

l. Suppose $(2, X) = (h)$. Since $2

 via Orbit-Stabilizer]
Let $G$ be the rotation group of th

 \in (h)$, we

 have $h \mid 2$, so $h$ is a constant $\pm 1$ or $\

 cube, and let $X$ be the set of four main diagonals of the cube (the lines connecting op

osite vertices: the

pm 2$. If $h =

e are four pairs of opposite vertices, giving four diagonals).

Every rotation of the cube

\pm 1$, then $(h) = \mathbb{Z}[X]$, but $1 \notin (2, X)$ (any element of $(2, X)$ evaluated at $0$ is even). If $h = \pm 2$, then $X \in (2, X) = (\pm 2)$ would require $2 \mid X$ in $\mathbb{Z}[X]$, which is false. Contradiction.
[/example]

[definition:Quotient Ring]
Let $I \trianglelefteq R$. The **quotient ring** $R/I$ is the set of additive cosets $\{r + I : r \in R\}$ with operations

angular cross-section perpendicular to $d$ are cyc

\begin{align*}
(r_1 + I) + (r_2 + I) &= (r_1 + r_2) + I, \\
(r_1 + I)(r_2 + I) &= r_1 r_2 + I.
\end{align*}
The zero is $0_R + I = I$ and the one is $1_R + I$.
[/definit

lically permute

on]

Multiplication is well-defined p

ecisely because of the strong closure property of $I$: if $r_1' = r_1 + a_1$ and $r_2' = r_2 + a_2$ with $a_1, a_2 \in I$, then $r_1' r_2' = r_1 r_2 + r_1 a_2 + a_1 r_2 + a_1 a_2$, and the last three terms

d). In addition, there are three rotations by $18

0°$ around ax

all lie in $I$ by s

s that pass through midpoints of opposite edges and are perpendicular to $d$ — each of these swaps the two endpoints of $d$ while fixing $d$ as a set. This gives $|G_d| = 6$.

By the [Orbit-Stabilizer Theorem](/theorems/845):
\begin{align*}
|G| = |G_d| \cdot |G \cdot d| = 6 \times 4 = 24.
\end{align*}

As a consistency check, we can 

rong closure. Just as in group theory, the condition we imposed on $I$ is exactly the condition needed to make the quotient well-defined.

The quotient map $\pi : R \to R/I$ sending $r \mapsto r + I$ is a surjective ring homomorphism with kernel $I$. This is the ring-theoretic analogue of the quotient group map.

### The Isomorphism Theorems for Rings

The isomorphism theorems carry over from groups to rings almost verbatim, since rings are abelian groups under addition, and the additional multiplicative structure is preserved by the same constructions.

[quotetheorem:851]

The [First Isomorp

ism Theorem for R

nstead l

ngs](/theorems/851) has the sam

t $G$ act on the 

et of six faces. The orbit of any face $f$ has size $6$ (rota

e shape as its group-theoretic counterpart: a ring homomorphism $\varphi : R \to S$ factors as a surjection onto its image $\operatorname{im}(\varphi)$, followed by an isomorphism from $R/\ker(\varphi)$. The proof is the same as for groups, with one additional check: the quotient map respects multiplication, which follows immediately from the homomorphism property of $\varphi$. The theorem is used constantly to identify quotient rings: to show $R/I

tions can send any face to any other), and the stabilizer of $f$ consists of rotations by $0°, 90°, 180°

 \cong T$, it s

hism $R \to T

nd the axis through $f$ and the oppo

ite face, giving $|G_f| = 4$. This confirms $|G| = 4 \time

$ with kernel $I$.

[quoteproof:851]

[example:Polynomial Quotients]
The evaluation homomorphism $\varphi : \mathbb{R}[X] \to \mathbb{C}$ defined by $\varphi(f) = f(i)$ (where $i = \sqrt{-1}$) is a surjective ring homomorphism, since every complex number $a + bi = \varphi(a + bX)$. Its kernel consists of all $f \in \mathbb{R}[X]$ with $f(i) = 0$, i.e. all polynomials divisible by the minimal polynomial of $i$

 over $\mat

 6 = 24$.

Now $\varphi : G \to S_4$ is injective (a non-identity rotation must move at least one diagonal, so the kernel

of the action on diagonals is trivial) and $|G| = 24 = |S_

bb{R}$, which is $X^2 + 1$. So $\ker(\varphi) = (X^2 + 1)$. The first isomorphism theorem gives:
\begin{align*}
\mathbb{R}[X]/(X^2 + 1) \cong \mathbb

|$, so $G \cong S_4$.
[/exam

C}.
\end{align*}
More explicitly, every element of $\mathbb{R}[X]/(X^2 + 1)$ has a unique representative $a + bX$ (reduce modulo $X^2 + 1$ using the Euclidean algorithm in $\mathbb{R}[X]$), and multiplication in this rin

ple]

## Conjugacy and the Class Equat

 satisfies $X^2 \equiv -1$, 

ion

Among al

recovering the standard multiplication rule for complex numbers. The abstract machinery has constructed $\mathbb{C}$ from $\mathbb{R}$ purely algebraically, with no appeal to geometry.
[/example]

The correspondence between ideals of $R/I$ and ideals of $R$ 

containing $I$

l group actions, the action of a group on it

holds in rings exactly as it held for normal subgroups of quotient groups.

There is also a useful parallel to the **characteristic** of a ring. For any ring $R$, the unique ring homomorphism $\iota : \mathbb{Z} \to R$ (sending $1 \mapsto 1_R$) has kernel $n\mathbb{Z}$ for some unique $n \geq 0$.

[definition:Characteristic]
The **characteristic** $\operatorname{char}(R)$ 

self by conju

s. 

f a ring $R$ is the unique non-negative inte

he orbits 

er $n$ such that $\ker(\iota : \mathbb{Z} \to R) = n\mathbb{Z}$. Equivalently, it is the smallest positive $n$ with $n \cdot 1_R = 0_R$, or $0$ if no such $n$ exists.
[/definition]

The rings $\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$ all have characteristic $0$. The ring $\mathbb{Z}/n\mathbb{Z}$ has characteristic 

re the conjugacy classes,

n$. When $R$ is an integra

l domain, its characteristic is either $0$ or a prime (since $\mathbb{Z}/\operatorname{char}(R)\mathbb{Z}$ emb

 which encode fundamental informatio

eds into $R$ a

n about the gro

up's structure.

[definition:Conjugacy Class]
The **conjug

 forcing $\operatorname{char}(R)\mathbb{Z}$ to be prime).

## Integral Domains and Fields of Fractions

Most rings that arise naturally

 — $\mathbb{Z}

f $g \in G$ is
\begin{align*}
\mathrm{ccl}_G(g) = \{hgh^{-1} : h \in G\},
\end{

align*}
i.e. 

, $\mathbb{Z}[i]$, polynomial rings over fields — share a key property of $\mathbb{Z}$: the product of two non-zero elements is non-zero. Rings with this property are called integral domains, and they are the setting in which a meaningful theory of divisibility can be developed.

[definition:Zero Divisor]
An element $x \in R$ (with $x \neq 0$) is a **zero divisor** if there exists $y \neq 0$ in $R$ with $xy = 0$.
[/definition]

[def

he orbit of $g$ unde

nition:Integral Domain]
A non-zero commutative ring $R$ is an **integral domain** if it has no zero divisors: whenever $ab = 0$ in $R$, either $a = 0$ or $b = 0$.
[/definition]

The rings $\mathbb{Z}/6\mathbb{Z}$ fail this: $2 \cdot 3 = 6 = 0$ in $\mathbb{Z}/6\m

thbb{Z}$, so $2$ and $3$ are zero divisors. But $\mathbb{Z}/p\mathbb{Z}$ for prime $p$ is an integral domain (in fact a field). The polynomial ring $R[X]$ over an integral domain $R$ is again an integral domain, since the leading coefficient of $fg$ is the product of the leading coefficients of $f$ and $g$, which is non-zero if both factors are non-zero.

An integral domain satisfies the **cancellation law**: if 

r the conjugation action 

of $G$ on itsel

ba = bc$ and $b \neq 0$, then $a = c$ (since $b(a-c) = 0$ and $b \neq 0$ forces $a - c = 0$). This is the algebraic form of "dividing both sides by $b$" — valid in 

f.
[/definition]

[definition:Centralizer]
The **centralizer** of $g \in G$ is

ntegral domains but

not in general rings.

Every field is an integral domain: if $ab = 0$ and $b \neq 0$, then $a = a \cdot (bb^{-1}) = (ab)b^{-1} = 0$. The converse fails in general ($\mathbb{Z}$ is an integral domain but not a field), but holds for *finite* integral domains: any finite integral domain is a field, since the map $x \mapsto ax$ (for $a \neq 0$) is injective (by cancellation) and hence bijective on a finite set, giving $ab = 1$ for some $b$.

The construction of $\mathbb{Q}$ from $\mathbb{Z}$ — taking formal fractions $a/b$ and identifying $a/b = c/d$ when $ad = bc$ — generalizes to any integral domain.

[quotetheorem:866]

The [field of fractions construction](/theorems/866) is one of the most powerful tools in ring theory: it lets us e

\begin{alig

bed any integral 

n*}
C_G(g) = 

omain into a field, unlocking 

 G} C_G(g).
\e

nd{align*}
[/definition]

Elements of $Z(G)$ commute with everything; they form conjugacy