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These are Zixuan’s notes for Part IA – Groups at the University of Cambridge in 2025. The notes are not endorsed by the lecturers or the University, and all errors are my own.

The latest version of this document is available at academic.micfong.space. Please direct any comments to my CRSid email or use the contact details listed on the site.

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Lecture 1 · 2025-10-10

1 Introduction on Groups

One can think about groups in two ways:

on the one hand, they are related to algebra,
on the other hand, they are related to symmetry.

1.1 Motivation

An equilateral triangle has rotational symmetry, reflective symmetry, and the identity symmetry.

Let us list these symmetries:

So an equilateral triangle has exactly 6 symmetries.

Exercise. How many symmetries does a square have? What about a regular pentagon? A regular n-gon?

Things get more interesting when we start to compose (or multiply) symmetries. Note that, after composition of symmetries, our result must also be a symmetry. To find out exactly which one is the end result, we can label the vertices on the triangle.

Here are some important features of symmetries:

symmtries can be composed,
there is an identity,
every symmetry has an inverse,
comoposition of symmetries is associative,

Important. The symmetries may not necessarily commute.

We can extend this method to algebra to handle more complex cases, e.g. for a 17-gon.

1.2 Introduction on Groups

Definition 1.1 (Binary Operation)

A binary operation on a set is a function .

Definition 1.2 (Group)

A group is a triple where

is a set
is a binary operation on ,

that satisfies the following four axioms:

Closure. For all , . [This can be deduced by definition of binary operations.]
Associativity. For all , .
Identity. For all , .
(Right) Inverse. For all , there exists such that .

Example 1.3

We noticed earlier that the symmetries of an equilateral triangle form a group.

We can also think about the definitions as encompassing algebra with one operation, as shown in the following example.

Exercise. Show that forms a group.

The definition has some important consequences.

Proposition 1.4

Let be a group, and .

If , then . [right inverses are left inverses.]
[left identities are right identities.]
If , then . [inverses are unique.]
If , then . [the identity is unique.]

Proof.

Using Definition 1.2, we have

By the inverse axiom, there is a such that . Multiplying both sides on the right by , we get
Using Definition 1.2, and the previous part, there exists such that .

Now
We have
Using the inverse axiom from Definition 1.2, and part (1), there is a such that .

Multiplying by on both sides of gives

Since , we have .

Notation. By Proposition 1.4, inverses are unique. Therefore, if , we may write

Lecture 2 · 2025-10-13

Part (1) of Proposition 1.4 tells us that

which tells us the following.

Corollary 1.5

For in a group , we have

Notation. It makes sense to extend this notation. For any ,

for any
for any

Exercise. Show that, if is a group, then for and ,

Recall that it is not necessarily true that in a group . Hence, it is also not necessarily true that

Proposition 1.6

Let be a group and . Then

Proof. We have

Since inverses are unique, the result follows.

Definition 1.7 (Abelian Group)

If is a group and for all , then the group is called abelian.

Definition 1.8 (Trivial Group)

If and , then is called the trivial group.

1.3 Familar Examples from Arithmetic

Example 1.9

are all abelian groups. The inverse of is in each case.
is not a group due to a lack of inverses.
is an abelian group. Note that is not a group since has no inverse.

Similarly, and are abelian groups.

1.4 Finite Groups

Most of the groups above are infinite.

Definition 1.10 (Order of a Group)

The order of a group is the number of elements of , denoted by .

If , then is finite.

Example 1.11

For a specific , let

then is an abelian group.
Let . For , let . Then is an abelian group.

Lecture 3 · 2025-10-15

1.5 Symmetric Groups

We need to introduce some definitions before we introduce the notion of a symmetric group.

Definition 1.12 (Bijection)

Let be sets. A bijection is a map that has an inverse such that

i.e. for all , and for all .

Definition 1.13 (Permuation)

A bijection is called a permutation.

Definition 1.14 (Symmetric Group)

is defined to be the set of permutations of a set .

We will prove that this is a group in Proposition 1.16.

Recall that .

Lemma 1.15 (Composition Is Associative)

Consider the following maps of sets:

Then .

Proof. For any ,

This makes it easy to see that is a group.

Proposition 1.16

For any set , is a group.

Proof.

Closure is automatic, since the composition of two permutations is still a permutaion.
Associativity follows from Lemma 1.15.
The identity map is the identity element.
Inverses exist by definition of a bijection.

Therefore the result follows.

Definition 1.17

If , then we write

Example 1.18

is the group of ways to rearrange three flower pots on a windowsill.
is the group of ways to shuffle a deck of cards.

Proposition 1.19

The order of is .

Notation. Writing is cumbersome. Henceforth, we will just write .

If we want to emphasize that is the identity element of , we will write for . Likewise, we can also write for the operation on .

1.6 Subgroups

Sometimes, we want to restrict our attention to smaller groups. For instance, inside , the rotations of a triangle instead of all symmetries.

Definition 1.20 (Subgroup)

Let be a group and . If we have

,
for all ,
for all .

Then we say that is a subgroup of . We write .

Remark. If is a group, and , then is also a group.

Example 1.21 (Example of Subgroups)

Every group is a subgroup of itself.
For any group , is the trivial subgroup of .
.
For , let .

Now we have
- Let . Then
- Let . Then
Therefore, . In fact, these are all of the subgroups .

Definition 1.22 (Proper Subgroup)

Let be a group. Then is a proper subgroup of if , and .

Proposition 1.23

If , then for some .

Proof. For the trivial case , we can construct it by .

Otherwise, if , we may choose to be the smallest positive . [Note that, if and , then and . Therefore, unless , contains a positive element.]

By induction, we see that for all . By the closure of inverses, we conclude that for all . Hence .

It remains to prove that . We shall prove this by contradiction. Suppose that , so there exists some such that . Dividing by and taking remainders, we get

for some and .

But now , so . However, we have which contradicts with our construction of .

Lecture 4 · 2025-10-17

Proposition 1.24

If , then

Similarly, for any family of subgroups , we have

Definition 1.25 (Generated Subgroup and Generated Set)

Let be a group, and be a subset of . Then

which is the intersection of all the subgroups of that contain , and is called the subgroup generated by .

If , we say generates , or is a generating set for .

Intuitively, is the smallest subgroup containing . If generates , it means that every can be written as

for some , where all . [Note that and can be repetitive elements from .]

1.7 Geometric Examples of Subgroups

Let be the plane, equipped with the usual notion of distance.

Definition 1.26 (Isometry)

For any an isometry of is a bijection

that preserves distance:

Proposition 1.27 (Isometry Groups In )

Let . The set of isometries of , , is a subgroup of . In particular, is a group.

Proof. Let us check Definition 1.20. Let and .

Clearly, .
Since and are isometries,

So .
Let . Then

because is an isometry. So

and as required.

Definition 1.28 (Dihedral Groups)

Let be the -gon with vertices for .

Define the th dihedral group to be

Example 1.29

is the symmetry group of an equilateral triangle, as seen in Section 1.1.

We can fast-forward to take a look at Theorem 1.32:

For a dihedral group, we have for .

The proof will also give us a good desciption of all the elements. But first, we need some geometric lemmas.

Lemma 1.30 (Kite Lemma)

Let . If

then is perpendicular to .

Proof. By symmetry,

But is a straight line, so .

Lecture 5 · 2025-10-20

Lemma 1.31 (3 Point Lemma)

Let and . If there are non-collinear points such that for , then .

Proof. The proof is by contradiction. Suppose that for some . Then

for .

Now apply Lemma 1.30, with , , then we get

Similarly,

Hence, are collinear, contradicting the hypothesis.

Remark. There is equally an -point lemma valid in .

Now, to prove the theorem we stated above, we define two elements of :

Theorem 1.32

For a dihedral group, we have for , and we have

In particular, generates .

Proof. This will be a long proof.

Outline.

We show that .
We show that by showing
- that , and
- that .
We show that there are no duplicate elements in .

STEP 1:

Let the polygon be . First, we show that .

For :

Indeed, For any , then

so is indeed an isometry. Also,

so sends vertices of to vertices, and hence preserves .
For :

Similarly for any , we have

so is indeed an isometry. Also,

so sends vertices of to vertices, and hence preserves .

STEP 2:

We have shown that . Therefore, by induction,

To see that this is all the elements, let . We aim to prove that . We shall apply Lemma 1.31 to complete this proof.

Consider , and .

For :

Since , is a vertex of . So

for some . We will try to undo using and .

Therefore,
For and :

Now, , so

and hence .

For the same reasons, . Therefore, there are two cases:
1. and ,
2. and .
- Case 1. forms .
  
  Since are not collinear,
  
  by Lemma 1.31. Hence, .
- Case 2. We have
  
  Therefore,
  
  Simiarly to Case 1, by the Lemma 1.31, we get
  
  Hence, as required.

This proves that

STEP 3: DUPLICATE CHECK

Now, we need to check that this list does not contain duplicate elements, so that

First, if such that

then

so .

Now, if then

so and .

But then , which is a contradiction. Therefore, for any .

Now, if , then , contradicting the previous case.

Finally, if there exists such that , then multiplying by inverses to the right by gives

as above.

To understand the group operations on , we need to understand the different ways to multiply and .

Lemma 1.33 (Dihedral Relation)

For where represents the rotation and represents the reflection, we have

Proof. By Lemma 1.31, it suffices to check that the two expressions do the same thing to . Indeed, with , and ,

Lecture 6 · 2025-10-22

2 Homomorphism and Isomorphism

2.1 Introduction on Homomorphisms and Isomorphisms

Some groups are not set-theoretically equal, but nonetheless have the same structure. e.g. and . We wish to encompass this underlying strucutre.

Definition 2.1 (Homomorphism)

A map between groups

is called a homomorphism if

for all .

Example 2.2

For any two groups and , the map

for all is a homomorphism, called the trivial homomorphism.
If , then the map

is the inclusion homomorphism.
Recall that .

Exercise. Show that if , then

is a homomorphism.
Since , the determinant function

is a homomorphism.

Lemma 2.3

If is a homomorphism, then

.
for all .

Proof.

We have

Note that , being multiplied by a group element and yielding itself, must be the identity element . This is by Proposition 1.4 (4).
We have

Thus, by Proposition 1.4 (3), we have .

To discuss the notion of two groups being “structurally the same”, we can do this using isomorphisms.

Definition 2.4 (Isomorphism)

If a homomorphism

is a bijection, then is an isomorphism. In this case, we write

From the perspective of group theory, isomorphic groups are considered the same.

Example 2.5

Recall that with operation , and with operation .

Let

Clearly, is bijective. Further more, for any ,

for some , so

Therefore, is indeed an homomorphism, and hence an isomorphism. That is,

for all .
The exponential map

is a homomorphism, because

Since , is also a bijection. Hence is an isomorphism.

The following lemma justifies the claim that we may think of isomorphic groups as “the same”.

Lemma 2.6

If is a isomorphism, so is .
If are homomorphisms, so is .
is an equivalence relation.

Exercise. Prove this lemma.

We have seen that every subgroup leads to an inclusion homomorphism. This is conversely true, that homomorphism leads to subgroups.

Definition 2.7 (Image and Kernel)

Let be a homomorphism.

The image of is
The kernel of is

Proposition 2.8

If is a homomorphism, then

, and
.

Proof.

- In Lemma 2.3, we showed that .
- For , we have
- For ,
Therefore .
- For ,
  
  So .
- For , we have
  
  So .
Therefore .

Proposition 2.9

Let be a homomorphism.

is surjective if and only if .
is injective if and only if .

Lecture 7 · 2025-10-24

Proof.

This is immediate from the definition of surjectivity and image.
Indeed, if is injective, then

This has at most one element, and since , we have .

Conversely, suppose . If , then

This calculation shows that , so by assumption, and hence . Thus is injective.

Proposition 2.10

By Proposition 2.9, a homomorphism is an isomorphism if and only if and .

2.2 Cyclic Groups

The groups that we have seen are examples of cyclic groups.

Definition 2.11 (Cyclic Group)

A group is cyclic if there exists an element such that

Such an element is called a generator of .

Example 2.12

is cyclic, with generator .
is cyclic, with generator .
is cyclic, since .

Theorem 2.13

If is cyclic, then either

for some , or
.

Proof. Let be a cyclic group with generator . Let

and let

Case 1. If , Define

We need to show that is an isomorphism. Since

is certainly a homomorphism.

By the definition of cyclic groups, is surjective.

To prove that is injective, for the purpose of contradiction, suppose that . Since , we may replace by if necessary, and assume . Then , which is a contradiction because .

Therefore, so is injective. Hence, is an isomorphism and .

Case 2. If , define

Since for some , we have

Thus, is a homomorphism.

To prove that is surjective, since is cyclic, every element can be written as for some . By the division algorithm, we can write for some and . Therefore,

This proves that is surjective.

To prove injectivity, suppose that for some [this is equivalent to saying ]. Then or . Since is minimal in , and that and , it follows that , because . Therefore, and is injective.

Therefore,

Because of this theorem, we will write for any cyclic group of order , and .

Definition 2.14 (Order of an Element of a Group)

For any group , and element , let

Note that is cyclic, so for some . This number is called the order of , denoted by .

2.3 Dihedral Groups, Revisited

Remark. Whenever satisfy the dihedral relation

Then, for any , we have

by induction on .

If , we also have

In summary, we have shown that

for all .

Lemma 2.15

Let be a group, that for some , the following relations hold:

for some

Then and defines a homomorphism that sends the generators of to respectively.

Proof. There are 4 cases to check:

Proposition 2.16

Suppose has generating set

satisfying:

for some

Then

Proof. By Lemma 2.15, there is a homomorphism

sending and . Since and generate , is surjective. Also, since , is bijective. Therefore, is an isomorphism, and

Lecture 8 · 2025-10-27

3 Lagrange’s Theorem

This very important theorem helps us to think about the orders of a groups and subgroups.

Theorem 3.1 (Lagrange's Theorem, Weak Version)

If and , then .

The idea of this theorem is to somehow partition the group into cosets of .

3.1 Cosets

Definition 3.2 (Left Cosets)

Let and . The corresponding left coset is

The set of all left cosets of in is denoted by

Remark. We may similarly define right cosets

The set of right cosets of in is denoted by

Lemma 3.3

If , the left cosets partition . That is,

If , then for any

Proof.

For any , we have since . Thus, . Hence, . The reverse inclusion is obvious. Hence the equality holds.
Suppose , so there is a in the intersection. Then,

for some . Thus,

Since is a group, . Thus

Furthermore, for any ,

Hence . By symmetry, we have the reverse inclusion. Thus, the equality holds.

A schematic picture of the lemma above is shown below.

However, we can say more about the sizes of the cosets.

Lemma 3.4

Let , then there is a bijection

for any . In particular, .

Proof. The map defined by has inverse defined by . Hence it is a bijection.

So the schematic picture above can be redrawn, where each coset has the same size as .

Definition 3.5 (Index of a Subgroup)

Let . The index of in is defined as

Theorem 3.6 (Lagrange's Theorem)

If and , then

Proof. Since left cosets partition ,

3.2 Consequences of Lagrange’s Theorem

Corollary 3.7

If and , then

Proof. Recall that and by Lagrange’s Theorem.

Corollary 3.8

If and , then .

Proof. Corollary 3.7 says that for some , so

Corollary 3.9

If is prime, then is cyclic, and any element generates it.

Proof. Choose any . Then

So, since is prime, either or . Since , we must have . Therefore,

So generates , and hence is cyclic.

3.3 Applications of Lagrange’s Theorem

Lagrange’s theorem 3.6 implies an important result in number theory.

Definition 3.10 (Euler Totient Function)

The Euler totient function

Let denote multiplication modulo on , which is associative with identity .

Lecture 9 · 2025-10-29

Recall that by the division algorithm, has a multiplicative inverse modulo , iff there are such that

Hence,

is a group with operation .

Theorem 3.11 (Fermat-Euler Theorem)

Let . If , then

Proof. By Corollary 3.8, we have

And by definition

4 Group Actions

Groups become groups of symmetries when they act.

4.1 Introduction

Definition 4.1 (Group Action)

An action of a group on a set is a function

with

such that

for all ,
for all and .

Notation. We write to indicate that acts on .

Example 4.2

For any group , there is a trivial action of on any set defined by for all and .
by .
If and then by restriction.

In particular, since .
Similarly, dihedral groups acts on (the regular -gon). It also acts on the set of vertices of the regular -gons.
Every group acts on itself by left multiplication: for all .

This is called the left regular action of .

Theorem 4.3

An action of a group on a set is the same as a homomorphism .

Proof. We will show the two directions separately.

Suppose . Consider defined by for all . Then

So . Similarly, . Thus is invertible, so . Therefore, we can define a homomorphism by .

We shall now prove that is a homomorphism. For any and , we have

Thus, , so . Hence, is a homomorphism.

Conversely, given a homomorphism

we may define an action by

Let us check the two properties of an action.

For any ,

For any and ,

Thus, the two properties hold, so we have an action of on .

Theorem 4.4 (Cayley's Theorem)

Every group is isomorphic to a subgroup of some .

Furthermore, if , we may choose with .

Proof. Let . Consider the left regular action of on itself. By the previous theorem, this corresponds to a homomorphism

Let . We may think of as a surjective homomorphism from to .

Claim. .

Proof. Indeed, if for all . In particular,

Hence, is bijective, and thus an isomorphism. So so required.

Automatically, if , then .

A lot of important results in group theory come from studying group actions.

Definition 4.5 (Orbits and Stabilisers)

Let , and let .

The orbit of is the set

The stabiliser of is the set

Notation. Some sources write for , which we will avoid to prevent confusion with subscript notation.

Definition 4.6 (Action Transitivity and Faithfulness)

If , we say that the action is transitive.

If every element (except ) has such that , then is faithful.

Lecture 10 · 2025-10-31

Remark. An action is faithful if and only if the corresponding homomorphism is injective.

4.2 Orbit-Stabiliser Theorem

Proposition 4.7

Suppose . Then

For any , .
The orbits form a partition of .

Remark. (2) means that iff there is only one orbit. Therefore, transitivity is independent of the choice of .

Proof.

We need to check satisfies the definition of a subgroup.
- If , then
  
  so .
- since .
- If , then
  
  so .
So all subgroup criteria are satisfied.
Similarly to the proof of Lemma 3.3,
- so orbits cover .
- If , then for some . Hence,
  
  Moreover, any can now be written as
  
  Hence . By symmetry, we have the reverse inclusion. Thus, the equality holds.

Example 4.8

Consider , where is the set of the regular -gon.

For any ,

Therefore,

The above calculation also shows that . Hence,

Theorem 4.9 (Orbit-Stabiliser Theorem)

Suppose acts on a set . Then for any , the formula

defines a well-defined bijection

Corollary 4.10

If and , then

Proof. Theorem 4.9 gives

Example 4.11

Consider again as in the previous example. We saw that, if is a vertex, then

This gives us an easy (but circular for now) proof that

Proof. [For Theorem 4.9]

For notational convenience, let , and define

We have to check several things about .

well-defined. That is, for any such that , we have , i.e. .

Now, means that there is such that . Then,

Thus, is well-defined.
is surjective. For any , we have

Thus, is surjective.
is injective. Suppose for some . By definition, this means that

We need to show that . Let . Now,

Thus, , so

Since cosets paritition, it follows that . Hence, is injective.

Example 4.12 (Symmetries of a Cube)

Let be the group of isometries of a cube, and let be the centre of a face. Then,

since it is just the number of faces of the cube.

Now, the face is essentially a square, so since the stabiliser must permute the four edges of the face. Thus,

Therefore, by the orbit-stabiliser theorem,

Lecture 11 · 2025-11-03

The next theorem is a different kind of application of the orbit-stabiliser theorem.

Theorem 4.13 (Cauchy's Theorem)

If and is a prime that divides , then there is such that .

Proof. Consider the set of -tuples (distinct entries not required)

Define an action of on as follows.

If , let , which is just a cyclic rotation of the -tuple. We need check that this is a well-defined action.

It is easy to see that

We also need to check that

Suppose, therefore, that

For convenience, let and . Then, we know that . So, and therefore . Hence,

Thus, the action is well-defined.

Now compute . For any choices of , there is a unique choice of such that

as inverses are unique. Thus, there are choices for , so

The action on partitions into orbits. So let

By orbit-stabiliser, for each ,

so either or . Let be the number of orbits with size . After renumbering, we may assume that

Since the orbits partition , we have

Since, by our assumption, divides , it follows that divides .

Now, note that has an orbit of size if and only if . In particular, has an orbit of size . Thus, . Since divides and , we must have

so there is at least one more orbit with .

The definition of implies that , so , whence as required.

4.3 Conjugation

Definition 4.14 (Conjugation)

Let be a group and . For any , the element is called the conjugate of by .

Intuition. One way to think about conjugation is that has the same shape as .

Another way is to think about it is that corresponds to changing the coordinates of .

Example 4.15

If is an abelian group, then for all . So the only conjugate of any element is itself.

Definition 4.16 (Conjugacy Class)

The conjugacy class of an element is the set

Exercise. acts on itself by conjugation:

defines . This is very different from the left regular action. Then is just the orbit of under this action.

Example 4.17

In Example Sheet 2 Q6, it is proven that has 1 conjugacy class of reflections if is odd, and 2 conjugacy classes if is even.

Note that the red reflections cannot be obtained by conjugating the blue reflections, and vice versa.

Definition 4.18 (Centraliser)

The centraliser of is defined to be

which is just the stabiliser of under the conjugation action.

Remark. Note that

so is the set of elements that commutes with .

Definition 4.19 (Centre)

The centre of is defined to be

This is exactly the set of elements that commutes with every element of .

Lecture 12 · 2025-11-05

5 The Möbius Group

This chapter is about an interesting group. It is almost a group of bijections of , but we need to add a formal point at .

5.1 Riemann Sphere and Möbius Transformations

Definition 5.1 (Riemann Sphere)

The Riemann sphere is the set

This set is called a sphere because we can visualize it as follows.

We can define a map called the stereographic projection. It identifies with all the points on the sphere except the north pole at .

Definition 5.2 (Möbius Transformation)

Let be complex numbers with . If , then the corresponding Möbius transformation is the map defined by, if ,

If , then is defined by

We usually just write and interpreting the cases of and appropriately.

Then

together with composition of functions forms a group, called the Möbius group.

Proposition 5.3

then

Remark. Compare this with matrix multiplication:

5.2 The Möbius Group

Theorem 5.4 (Möbius Group)

is a group.

Proof.

Composition of functions is associative.
The identity map is a Möbius transformation with and .
For , let .

Then is also a Möbius transformation and and . Hence is the inverse of .

One important way to study Möbius transformations is via their fixed points.

Definition 5.5

Suppose is a permutation. Any such that is called a fixed point of .

Lemma 5.6 (Three-Point Lemma For )

If fixes three distinct points of , then .

Proof. Let . A fixed point satisfies

Case 1. If there is a fixed point , WLOG let , then . So and satisfy

This is a linear equation with at least 2 roots, so we cannot have two distinct fixed points and unless and . Then .

Case 2. If none of the fixed points is , then we have three distinct complex numbers satisfying

This is a quadratic equation with at least 3 roots, so we cannot have three distinct fixed points unless , , and . Then .

Exercise. Show that every has at least one fixed point in .

Example 5.7

Consider . Then .
Consider . Then .

Lemma 5.8 (Triple Transitivity)

For any triples of distinct points and , there exists a such that for .

Proof. Instead of constructing directly, we construct two auxiliary Möbius transformations and such that and . Then we can let .

Let be defined by

[Modify appropriately if any of is .]

Then , , and .

Similarly, let be defined by

[Modify appropriately if any of is .]

Then , , and .

Note that and are both in . Then let

Then for .

Remark. The three-point lemma implies that in Lemma 5.8 is unique. We say that is sharply triply transitive.

Definition 5.9 (Cross Ratio)

Let be distinct points. Because sharply triply transitively, there exists a unique such that

The cross-ratio is defined as

Note that we saw that can be computed by

by the proof of Lemma 5.8.

Lecture 13 · 2025-11-07

Just like for dihedral groups, we can use the 3-point lemma to find a generating set for .

Proposition 5.10

is generated by the set of elements of the following 3 forms:

, where
, where

Proof. Let be arbitrary, and let , , .

Step 1. Construct such that .

Either and or

where . Then

Let and .

Step 2. Let , and let , i.e.

Note that and . By construction,

Let

Step 3. Let and

By construction,

By the three-point lemma, , so

5.3 Circles

Definition 5.11 (Circle In )

A circle in is either

a Euclidean circle in , or
where is a Euclidean straight line in .

Euclidean circles are described by the equation

for some and ,

while lines are described by

Theorem 5.12 (Circles and Möbius Transformations)

Möbius transformations map circles to circles. Formally, if is a circle, then then is also a circle.

Proof. Recall that is generated by

So it suffices to show that each of these generators maps circles to circles.

It is clear that and map circles to circles. So we just need to show that maps circles to circles.

If is a Euclidean circle in , then has equation

If , then we have

This is the equation of a line. Otherwise, , and the equation becomes

which is the equation of a circle.

We are left with the case where is a line, which follows a very similar calculation to the above and is left as an exercise.

Corollary 5.13

Four points lie on a circle if and only if their cross-ratio .

Proof. Let such that , , , and so

[] If lie on a circle , then by the previous theorem, is also a circle containing . The only such circle is the real line , so .

[] If , then lie on the circle . So by the previous theorem, their preimages also lie on a circle.

6 Finite Groups

We have already seen some nice theorems about finite groups, including Lagrange’s, orbit-stabilizer, Cayley’s, and Cauchy’s theorems.

Lecture 14 · 2025-11-10

We will now develop some small examples of finite groups. We shall proceed naively, trying to list examples by order.

Now, if , we know that is always an option. But is there another?

Definition 6.1 (Direct Product)

If are groups, the direct product is

with operation

Note that is a identity, and .

Example 6.2 (Klein 4-Group)

The Klein 4-group is . It can also be thought of as . Note that, for every , we have . So every element has order at most 2. In particular, .

Theorem 6.3 (Direct Product Theorem)

If and

,
,
, i.e. for every , there exist such that ,

then .

Proof. Define with . We need to show that is an isomorphism.

IS A HOMOMORPHISM

For any , , by definition,

IS SURJECTIVE

This is immediate from (3).

IS INJECTIVE

Recall that we need to show that . Suppose that . Then , so . But also, so by (1), we must have and . Thus , as required.

Remark. If , then . In particular, if , we can conclude that (1) implies (3).

Lemma 6.4 (Groups of Order 4)

If , then or .

Proof. By Lagrange’s theorem, every non-trivial element of has order 2 or 4. If there is a such that , then .

Otherwise, every non-trivial element has order 2. Let be distinct elements such that . Let and . It is immediate that [this can be seen by writing out the elements explicitly], which gives us (1). Following the remark, (3) holds.

Finally, since , we have , so , giving us (2) [this was mentioned in Example Sheet 1, Q11]. Thus by the direct product theorem, , as required.

Another application of the direct product theorem 6.3 is to find out when a product of two cyclic groups is cyclic.

Theorem 6.5 (Chinese Remainder Theorem)

If , then .

Proof. Let . Set

We will check against DPT 6.3.

Note that

Similarly,

Therefore, iff divides . Since , this happens iff . Thus, as required.
Since is abelian, this is immediate.
This follows from (1) and the remark after DPT.

We shall now move on to groups of order 5 and above.

Lemma 6.6 (Groups of Order 6)

If , then or .

Proof. By Cauchy’s theorem 4.13, there exist such that and . Since , , and . Therefore,

since the other coset must be . So for some .

If , then , so , a contradiction.
If , then . Then and satisfy the conditions of DPT 6.3, so
If , then . In this case, with relations , , which is satisfies the dihedral relation 1.33. Thus, .

Remark. is a non-abelian group of order 6, so by the above lemma, .

Continuing on,

Lecture 15 · 2025-11-12

Consider .

We have and and as abelian groups of order 8.

We also have as a non-abelian group of order 8.

Exercise. None of the groups , , , and are isomorphic to each other.

Definition 6.7 (Quaternion Group)

Let

It is easy to check that these form a group. We usually use Hamilton’s notation:

So the elements of are , with relations

,
, etc.
, , ,
commutes with everything.

We call this group the quaternion group.

Since , is non-abelian and so

By considering the orders of elements, we can also see that . has 5 elements of order 2 (one 180° rotation and the 4 reflections), whereas has only one (the element ).

It can be shown that these are all the groups of order 8.

Lemma 6.8 (Groups of Order 8)

If , then is isomorphic to one of , , , , or .

Proof. By Lagrange’s theorem, the possible orders of elements in are 1, 2, 4, or 8. We have the following cases:

If there is an element of order 8, then .
If every non-trivial element has order 2, then by a similar argument as in Lemma 6.4. We will start by choosing non-trivial elements with and . We can see that

by using Theorem 6.3 twice. [Check Example Sheet 1, Q11 for related details.]
If there is an element of order 4, but no element of order 8, let with . Let . By Lagrange’s theorem,

so

In particular, for some .
- If , then , so .
- If , then , so for all , then is abelian. [Case A.]
- If , then , so But by Example Sheet 2, Q1. However, .
- If , then , which looks similar to the dihedral relation. [Case B.]
Next, note that if , then and so .

Thus . We have several more cases:
- , which we will handle later. [Case I.]
- If , then , so .
- , which we will handle later. [Case II.]
- If , then , so .
There are now four subcases to consider:
- [Case A, I.] If is abelian and , then , so by Theorem 6.3,
- [Case A, II.] If is abelian, , so
  
  Again, by Theorem 6.3.
- [Case B, I.] We have and with relation . These are exactly the relations for (using Proposition 2.16), so .
- [Case B, II.] We have and with .
  
  Let , , , .
  
  Then, the elements of are which are in Hamilton’s notation.
  
  It is easy to check that the relations in match those in . This defines an isomorphism between and .
  
  Therefore , as required.

In summary, the groups of order up to 8 are as follows:

,
,
, , , ,

7 Quotient Groups

7.1 Normal Subgroups

Let be a group homomorphism.

is always a special kind of subgroup.

Definition 7.1 (Normal Subgroup)

is called a normal subgroup if for all , we have . If so, we write .

Example 7.2

, for any group .
If is abelian and , then .
since

by the dihedral relation [we don’t have to check since it is in ]. But is not normal since
Suppose is a homomorphism. If and then

so . Thus .

Lecture 16 · 2025-11-14

Lemma 7.3

Suppose . Then iff

for all .

Proof.

[] Let and . Since , we have . Therefore

Therefore . Similarly, for any , we have so

Thus , and so .

[] Suppose for all . Let . Then

so there exists such that .

Thus .

7.2 Quotient Groups

Theorem 7.4 (Quotient Group Is a Group)

If , the set of (left) cosets is a group with operation

Proof. We need to check that the operation is well-defined and satisfies the group axioms.

WELL-DEFINEDNESS

Suppose and since .

That is, there are such that

Therefore,

by Lemma 7.3, for some . Thus,

since cosets partition.

GROUP AXIOMS

Associativity. Immediate from associativity in .
Identity. The identity is .
Inverses. The inverse of is since
Closure. Immediate from the definition of the operation.

Therefore is a group.

Definition 7.5 (Quotient Group)

If , the group provided by Theorem 7.4 is called the quotient of by .

Example 7.6

, .
Since is abelian, for any . Thus, for any , we have the quotient group

with generator of order .
Let be a group, , and suppose . Then, for any ,

and . So by Lemma 7.3, . Furthermore, since its order is 2.
An example of (3) is

and hence .

Important. It is a common error that one might think we can multiply groups using direct products and get back to the original group.

Note that

But since is cyclic but is not. This shows that quotient groups do not undo direct products. In particular, for groups ,

7.3 The Isomorphism Theorem

Theorem 7.7 (Isomorphism Theorem)

If is a homomorphism, then

Proof. Since , the quotient is a group. Let us define

We first check that is a well-defined isomorphism.

WELL-DEFINEDNESS

Suppose . We need to show that .

We have

for some since . Then,

HOMOMORPHISM

For ,

INJECTIVITY

Recall that a homomorphism is injective iff its kernel is trivial.

If , then

so , and hence .

That is, , so is injective.

SURJECTIVITY

A typical element of is for some . But

so is surjective.

Example 7.8

Because defined by is a homomorphism with image and kernel , by isomorphism theorem 7.7, we have
Similarly, defined by is a homomorphism with

so by isomorphism theorem 7.7, we have

Lecture 17 · 2025-11-17

Definition 7.9 (Simple Group)

A group is simple if the only normal subgroups are and itself. Thus every homomorphism is either trivial or injective. [Since , so either or .]

Example 7.10

is simple whenever is a prime.

An important question in group theory is to find and understand examples of non-abelian simple groups.

8 Permutations

8.1 Permutations and Cycle Notation

Recall from Definition 1.13 that a permutation of a set is a bijection , and from Definition 1.14 that is the set of all permutations of .

Example 8.1

If , then . So examples of permutations include

We can compute by representing permutations as lists.

Example 8.2

This is nonetheless a bit cumbersome. A more compact notation is to write permutations in cycle notation.

Definition 8.3 (Cycle)

Any list of distinct elements

defines a -cycle:

which sends

and leaves all other elements fixed.

Example 8.4

For Example 8.1, we have

The important rule about cycle multiplication is that the rightmost cycle acts first.

Example 8.5

For Example 8.1, we have

Another example is

Remark. .

Definition 8.6 (Disjoint Cycles)

Cycles and are disjoint if the sets and are disjoint. Note that disjoint cycles commute.

Theorem 8.7 (Disjoint Cycles)

Every can be written as a product of disjoint cycles. This expression is unique up to

shifting the elements within each cycle, and
reordering the cycles.

Proof. The action of on partitions into orbits. Let

Let . We see that

which proves existence of the decomposition.

The choices we have made are on the representatives of each orbit, and the order of the orbits, which proves uniqueness up to the stated conditions.

Example 8.8

Consider

Definition 8.9 (Cycle Type)

then is called a -cycle.

The (multi)set of numbers is called the cycle type of . We often omit singletons from the cycle type.

Remark. If is a -cycle, then . More generally, if is a -cycle, then .

8.2 Transpositions and the Sign Homomorphism

Definition 8.10 (Transposition)

A transposition is a -cycle.

Theorem 8.11 (Transpositions Generate)

The set of transpositions generates for any finite .

Proof. We shall prove by induction on .

Base case. Consider . Then is generated by the transposition .

Inductive step. Assume that is generated by transpositions. Consider . Let .

If , then , so by the inductive hypothesis is generated by transpositions.
Otherwise, let . Then

So , so by the inductive hypothesis is generated by transpositions. Since

we see that is also generated by transpositions.

Lecture 18 · 2025-11-19

Definition 8.12 (Adjacent Transpositions)

A transposition of the form is called adjacent.

Lemma 8.13

Any transposition can be written as a product of an odd number of adjacent transpositions.

Proof. Assume . Then the proof is by induction on .

Base case. If , then is already an adjacent transposition.

Inductive step. Assume that we can write as a product of an odd number of adjacent transpositions. Then

and since can be written as a product of an odd number of adjacent transpositions by the inductive hypothesis, so can .

In particular, is generated by adjacent transpositions.

This discussion leads to a notion of parity for permutations.

Lemma 8.14

If are all transpositions and

then is even.

Proof. By Lemma 8.13, we may assume that all are adjacent transpositions.

We say that a pair is called an inversion of a permutation if but .

Claim. For any , the number of inversions of has the same parity as .

Proof. We prove this by induction on .

Base case. If , then has inversions, which is even.

Inductive step. Let

Since are all adjacent transpositions, we have for some . Consider which pairs would be inversions of but not , or vice versa.

The only such pair is such that and . This is because only swaps and , so any other pair would remain an inversion or non-inversion in both and .

For this pair, we have

but

Therefore, if and is an inversion for but not for , while if and is an inversion for but not for .

In either case,

as required.

By the claim, since has inversions, we see that must be even.

This enables us to define the sign homomorphism.

Theorem 8.15 (Sign Homomorphism)

The map

is a well-defined homomorphism.

Proof. To see that this is well-defined, suppose are all transpositions such that

Then

so by the previous lemma, is even. Therefore, , so .

To see that this is a homomorphism, note that

Definition 8.16 (Parity of a Permutation)

If , then is called an even permutation. Otherwise, it is called an odd permutation.

Definition 8.17 (Alternating Group)

The subgroup

is called the alternating group on elements. i.e. it is the set of all even permutations in .

Example 8.18

In , the permutations are

We have and , so the even permutations are

Remark. The cycle type makes it easy to determine the sign of a permutation.

Indeed,

so is even iff is odd.

More generally, a -cycle is even iff is even.

Example 8.19

is even
is odd

8.3 Conjugacy in and

We shall apply what we have obtained so far to study conjugacy in in . Recall that since is a normal subgroup of , the conjugacy class of any element in is contained in .

Theorem 8.20 (Conjugacy In )

Two permutations are conjugate iff they have the same cycle type.

Proof.

[] Suppose

is a product of disjoint cycles. We have

Since the cycle types are the same, can be written as

Now

defines a permutation of . We can compute

Therefore, .

Lecture 19 · 2025-11-21

[] Suppose . The above argument shows that, if

then we can define so that

for all and . Therefore, and have the same cycle type.

This makes it easy to count conjugacy classes in .

Example 8.21

Consider

Then we have

Example 8.22

Consider without knowing its exact elements. Then we have

Recall that Conjugation Classes 4.16 are essentially orbits under the conjugation action of on itself, and the Centraliser 4.18 of an element is its stabiliser under this action.

Then, Orbit-Stabiliser Theorem 4.9 implies that

Therefore, it is also easy to count the sizes of centralisers.

Example 8.23

In , we have

Indeed, we can make a list:

Example 8.24 (Conjugacy At )

We can list all the conjugacy classes in .

We can write out a table:

Typical element

We should verify that the sizes of the conjugacy classes add up to .

Indeed, as expected.

Now, counting conjugacy classes in is slightly more subtle. Recall that is the set of elements of that commute with .

Lemma 8.25 (Conjugacy Classes In )

Let .

If some odd element of commutes with , then
Otherwise, if every element of that commutes with is even, then splits into two:

where is any transposition (or any odd permutation).

Proof. Orbit-Stabiliser Theorem 4.9 gives

Since , this gives

is the even permutations that commute with , and is all permutations that commute with . [So is the even bits of .] Therefore, we can write as the kernel of the sign homomorphism restricted to :

The image of has size or [since it is a subgroup of ], so by the Isomorphism Theorem 7.7, we have

If there is an odd element of that commutes with [, then this element is in but not in ], so

Using Lagrange’s theorem 3.6, [we have , and] Equation * then becomes

Since , as required.
The hypothesis means that

so Equation * becomes

So is half as big as .

Now consider a transposition . Note that .

For the sake of contradiction, suppose . Then there exists such that

But then, by rearranging, we have

which means that commutes with . So . But is odd[, so some odd permutation now commutes with ].

Therefore, , as required.

This makes it possible to determine the conjugacy classes in .

Example 8.26 (Conjugacy In )

Consider . The even elements of are , -cycles and -cycles. Note that commutes with every element, so its conjugacy class in is the same as in .

Since there is an odd number of -cycles in , the conjugacy class of remains intact in [we cannot split it into two equal parts].

Finally, consider a -cycle, say . We have

and since we know that the cyclic group generated by definitely commutes with , we have

and the conjugacy class of splits into two in . In summary,

Typical element

Lecture 20 · 2025-11-24

Finally, let us look at conjugacy in and .

Example 8.27 (Conjugacy In )

We can write out the conjugacy classes in as follows:

Even	Typical element

The sizes of the conjugacy classes add up to as expected.

Example 8.28 (Conjugacy In )

Consider . The even elements of are , -cycles, -cycles and -cycles.

Note that

commutes with every element, so its conjugacy class in is the same as in .
, so the conjugacy class of remains intact in .
Since is odd, the conjugacy class of remains intact in .
, so . Therefore, the conjugacy class of splits into two in .

Typical element

The sizes of the conjugacy classes add up to as expected.

Theorem 8.29

is simple.

Proof. Suppose . By Example Sheet 3 Q5, is a union of conjugacy classes in . At this point, we can list the possible union sizes of conjugacy classes in and see if they divide (by Lagrange’s Theorem 3.6).

The possible union sizes are [note that must be included]:

Therefore, the only possible sizes for are and , so or . Hence, is simple.

9 Matrix Groups

Let be the set of all matrices with real entries.

From IA Vectors and Matrices, we know that matrix multiplication is associative and has an identity element, the identity matrix , though not all matrices have inverses under multiplication.

Lemma 9.1

has an inverse iff .

Definition 9.2 (General Linear Group)

Let . This is a group under matrix multiplication.

Here is another result from IA Vectors and Matrices.

Lemma 9.3

For , .

This implies that is a homomorphism

Definition 9.4 (Special Linear Group)

Let . This is a subgroup of called the special linear group.

By the Isomorphism Theorem 7.7,

For any ,

Hence and so

Remark. We can replace with in the above and get similar results. Therefore we have

9.1 Change of Basis

This is a familiar concept from IA Vectors and Matrices. There is a natural action by conjugation:

Proposition 9.5

Let be an -dimensional vector space over , and a linear map. If that represents in some basis, then the orbit

consists of all matrices that represent in any basis.

Proof. A basis for defines an isomorphism of vector spaces

The claim that represents in this basis means that

and so .

Lecture 21 · 2025-11-26

Likewise, another basis for corresponds to another isomorphism

and a matrix represents in these coordinates if

Therefore,

where [because its inverse exists, namely the matrix representing ] represents the isomorphism in the standard basis. Thus, the set of all matrices representing in any basis is contained in the orbit .

Conversely, if

for some , then setting

we get a basis

for . In this basis, represents .

9.2 Möbius Transformations, Revisited

Recall that multiplication in looked similar to multiplication of matrices.

Proposition 9.6

Identify

then

and

Proof. We can prove both statements by constructing a surjective homomorphism from onto with kernel , by the Isomorphism Theorem 7.7. Consider the map

By our previous computation of multiplication in , we see that is a homomorphism. Also, is surjective since for any Möbius transformation with , the matrix is in .

A matrix iff its image fixes , and by the Three Point Lemma for 5.6. Hence

Thus, which we have identified with .

Therefore, by the Isomorphism Theorem 7.7

9.3 Orthogonal Groups

Let us write for the normal notion of length on , i.e.

Definition 9.7 (Orthogonal Group)

The -dimensional orthogonal group is the subgroup of that preserves distance in :

In fact, the dot product

is often more convenient to work with.

Lemma 9.8 (Polarisation Identity)

For any ,

Proof.

It follows that we can characterise using the dot product.

Lemma 9.9 ( And the Dot Product)

Proof. If for all , then for any ,

Therefore .

Conversely, if , then ,

Hence for all as required.

This quickly leads to a nice characterisations of matrices in .

Lemma 9.10 (Matrices In )

Let . The following are equivalent:

.
The columns of form an orthonormal basis of .
.

Proof. Let .

[(1) (2).] Let be the standard basis for . The th column of is . since

The columns of form an orthonormal basis.

[(2) (3).] As explained above, (2) means that

Since , this means that

But is the th entry of the matrix , so this shows that the th entry of is for all . Therefore .

[(3) (1).]

Suppose . then

Hence as required.

Recall that . Therefore,

So for any .

Lecture 22 · 2025-11-28

Definition 9.11 (Special Orthogonal Group)

The special orthogonal group is the subgroup

Note that

Thus,

Examples of elements of are provided by reflections.

Definition 9.12 (Reflection)

Any defines an orthogonal plane .

The reflection in is defined to be

Remark.

We will sometimes write for the reflection in the plane .
We may replace by and assume that . then

Lemma 9.13

Proof. We may assume that . From the definition, is linear in . So we can think of as a matrix . Now,

So,

So indeed . In particular, is invertible with inverse , So

Finally, for any ,

Hence as required.

Remark. Let , and pick an orthonormal basis for . In the basis for , has matrix

so and hence .

Theorem 9.14 (Reflections Generate )

Every is a product of at most reflections.

Proof. We will prove this by induction on .

Base case. When , . The matrix is the reflection in the origin, so the result holds.

Inductive step. Let be the standard basis for . Let .

Then , [and since is an orthogonal transformation, by Lemma 9.9, dot products are preserved, and hence vectors that are orthogonal to are sent to some vector that is still orthogonal to ,] so preserves .

By induction, there are such that

Since both sides also fix , they also agree on . Therefore,

Orthogonal transformations are especially easy to analyse in low dimensions.

Lemma 9.15 (Elements Of )

Let .

If then is a reflection.
If then is a rotation about .

Proof. Recall that , so . By Theorem 9.14, we may take .

If , then is odd and hence . So is a reflection.
If , then is even, so unless , we can write for some that are not parallel.

We claim that only fixes the origin. [Here, we define a rotation to be an orthogonal transformation that only fixes the origin.] Indeed, for , suppose

But is parallel to and is parallel to , so this implies that is parallel to .

Hence only fixes the origin, and is therefore a rotation about .

Remark. Let . We have seen that the columns form an orthonormal basis of , so we may write

with . Thus, there exists such that and , so

Lemma 9.16 (Elements Of )

If , the is a rotation.

Proof. By Theorem 9.14, is a product of at most reflections. Since , either or for some that are not parallel. Since ,

where is a line through the origin. Since fixes pointwise and also fixes pointwise, their composition also fixes pointwise. [We shall define a rotation in to be an orthogonal transformation that fixes a line pointwise.]

Also , similar to Lemma 9.15, either

, in which case is fixed by , or
is parallel to .

Thus, only fixes the line pointwise, and is therefore a rotation.

Lecture 23 · 2025-12-01

10 Platonic Solids

While there are infinitely many regular 2-dimensional polygons, in three dimensions there are only five regular solids, known as the Platonic solids.

Definition 10.1

A convex polyhedron is a Platonic solid if

every face of is a regular -gon for some ,
acts transitively on the faces.
if is the midpoint of a face, then

Proposition 10.2

There are, up to similarity, exactly five Platonic solids:

the tetrahedron, with 4 triangular faces and 4 vertices,
the cube, with 6 square faces and 8 vertices,
the octahedron, with 8 triangular faces and 6 vertices,
the dodecahedron, with 12 pentagonal faces and 20 vertices,
the icosahedron, with 20 triangular faces and 12 vertices.

Two solids are dual if can be constructed from by putting vertices in the centers of each face, and then joining vertices in adjacent faces by edges.

Example 10.3

The cube and the octahedron are dual.

Example 10.4

The tetrahedron is dual to itself.

Example 10.5

The dodecahedron and the icosahedron are dual.

In particular, if and are dual, then , so we only have three distinct isometry groups of Platonic solids to consider.

Example 10.6 (Tetrahedron)

Let . By definition, acts transitively on the four faces, and the so by the orbit-stabilizer theorem,

where is the center of a face.

Furthermore, the action of on the four vertices defines a homomorphism

We shall prove that is injective. Suppose . Then fixes each vertex of the tetrahedron, which are not coplanar. So by the 4-point lemma. Therefore is injective, and we may identify with a subgroup of .

But and , so in fact .

Let us also identify the group of rotational symmetries [which are the ones we can actually realize by rotating the solid in space]:

where the tetrahedron is centered at the origin.

Lemma 10.7 (Uniqueness Of )

If and , then .

Proof. Because , , and . We therefore have surjective homomorphism with .

Since transpositions generate , there is a transposition with . Because all transpositions are conjugate [they have the same cycle type], so for any other transposition and some . Thus

Therefore , so

Therefore, since , we have .

Example 10.8 (Cube and Octahedron)

Let . By definition, acts transitively on the six faces, and the so by the orbit-stabilizer theorem,

where is the center of a face.

In particular, the index-two rotational subgroup has order 24.

In Example Sheet 2 Q7, we saw that acts on the set of the four long diagonals of the cube, giving a homomorphism

Since both and have order 24, to show that is injective it suffices to show that it is surjective.

Claim. is surjective.

Proof. Since Transpositions Generate 8.11, it suffices to show that all transpositions are contained in . Indeed,

rotation about the axis through the midpoints of an edge maps to a transposition under .

We get different transpositions this way, which is all of them [because it happens that ]. Therefore .

Hence comparing orders, we have .

Now, since commutes with everything in , by Direct Product Theorem 6.3 we have

Lecture 24 · 2025-12-03

Remark. We have and , so

Since cubes have reflectional symmetries, we have in that case. The same applies to the other Platonic solids as well.

Now, for the final two Platonic solids.

Example 10.9 (Dodecahedron and Icosahedron [Non-Examinable])

Let and the rotational subgourp of index two.

By definition, acts transitively on the twelve faces, and the so by the orbit-stabilizer theorem,

where is the center of a face. And so .

By drawing diagonoals on faces, we may inscribe 5 cubes into the dodecahedron.

Since the 5 cubes are built symmetrically from the geometry of the dodecahedron, acts on the set of these 5 cubes, giving a homomorphism

Rotation around the axes through an opposite pair of vertices leads to a 3-cycle. There are 10 diagonals between opposite pairs of vertices, so we get 10 inverse pairs of 3-cycles. So, we get all 10 3-cycles in .

Claim. Let be the set of 3-cycles. Then

Proof. By Example Sheet 4 Q2,

Since is simple, either or . Since , we must have .

In summary, if is the set of 3-cycles, we have seen that

Therefore we have

so is surjective and hence an isomorphism. Therefore .

Finally, since commutes with everything in , by Direct Product Theorem 6.3 we have