Group Theory Note
Date: 2024/11/26Last Updated: 2025-08-11T07:47:24.000Z
Categories: Mathematics
Tags: Group Theory, Algebra, Mathematics, Math, Note
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0.1 Contents
1 Group Theory Note
1.1 Introduction
This is my personal note on group theory.
1.2 Group Actions
1.2.1 Orbit and Stabilizer
1.2.2 Burnside's Lemma
1.2.2.1 Group With
Conjugacy Classes
Group with
conjugacy classes, are particularly interesting.
As only finite groups have
conjugacy classes, well be
.
And any group with
conjugacy classes is simple.
Thus, it may serve as a good example to infinite simple groups.
Theorem 1. If is a group with
conjugacy classes, then
is simple.
Proof: As itself is a conjugacy class, the other conjugacy class must be all the other elements.
Let be a non-trivial normal subgroup of
.
Take
such that
.
Then for any
,
,
there exists
such that
.
Thus,
.■
Theorem 2. If is a finite group with
conjugacy classes, then
is isomorphic to
.
Proof: As itself is a conjugacy class, the other conjugacy class must be all the other elements.
By Burnside's lemma, we have the following equation:
Where , which is the elements that is fixed by
.
As
must fix
and itself, thus, for any
,
.
Therefore, we have the following inequality:
Which then gives us the equality:
■
1.3 Abelian Group
In this section, we will only focus on Abelian groups. And we will use the additive notation for the group operation.
1.3.1 Free Abelian Group
Definition 1. An Abelian group is free Abelian if it is a direct sum of infinite cyclic groups.
More precisely, there is a subset
of elements of infinite order,
called a basis of
, with
; i.e.,
.
Theorem 3. There is an infinite p-primary group
each of whose proper subgroups is finite and cyclic.
Proof: We define a group have generators:
with relations:
We let be the free Abelian group generated by
.
And let
be the subgroup generated by the relations.
Let
, as
form a basis of
,
form a basis of
.
And as
,
,
,
,
for every
, we have
. So
is a p-primary group.
To prove is infinite, we prove that
are distinct.
Assume
for some
.
Then by previous reasoning, we have
,
which implies
.
And previous reasoning also shows that
.
Thus, the order of
must divides
.
Which implies
, which is a contradiction.
Take to be a proper subgroup of
.
By previous argument, for all
,
is in the cyclic group generated by
.
Thus, we have
If is finite, then there must be
such that
.
Then
is a subgroup of finite and cyclic group, thus
is finite and cyclic.
If is infinite, then there must be a sequence of elements in
,
say
, such that
,
and
.
Assume
,
then
must be
,
as otherwise
will be in
.
Then
also generate
,
which implies
.
And thus,
, which implies
.■
1.3.2 Divisible and Reduced Groups
Definition 2. A group is called divisible if for every
and every positive integer
, there exists an element
such that
.
Example 1. The group is divisible.
We define a group have generators:
with relations:
We let be the free Abelian group generated by
.
And let
be the subgroup generated by the relations.
Let
, as
form a basis of
,
form a basis of
.
Also,
,
,
,
,
therefore we have,
for all
,
is in the cyclic group generated by
.
Thus, we have
We take such that
.
Then for any element
,
we have
such that
.
Assume
,
we also take
.
Then, by the relation of generators, we have
.
As
is co-prime with
,
by Bézout's identity, there exists
such that
.
We take
.
Then
Which finish the proof that is divisible.
Theorem 4 (Injective Property, Baer, 1940). Let be a divisible group and let
be a subgroup of a group
.
If
is a homomorphism,
then
can be extended to a homomorphism
;
that is, the following diagram commutes:
Proof: This theorem can be proved by Zorn's lemma, or an equivalent proof of well-ordering principle.
Consider to be the set of all pairs
,
where
is a subgroup of
containing
and
is a homomorphism extending
.
Note that
is non-empty as
.
We define a partial order on by
if
and
.
Then for any chain
in
,
we can define
and
by
for
.
It is easy to check
is well-defined and
is a homomorphism.
Thus,
is an upper bound of the chain.
By Zorn's lemma, there exists a maximal element in
.
If
, then there exists
.
And we discuss whether
.
If , then we can extend
to
by defining
.
Which contradicts the maximality of
.
If , then we can take
to be the smallest positive integer such that
.
Then as
is divisible, we can find
such that
.
Then we can extend
to
by defining
.
And it is easy to check that
is a homomorphism.
Which contradicts the maximality of
.■
Follows from the injective property, we have the following corollary:
Corollary 1. If a divisible group is a subgroup of a group
,
then
is a direct summand of
.
Proof: We consider the following diagram:
As is divisible, we can extend
to
.
By theorem in group extension, we have
.■
Definition 3. If is a group, then
is the subgroup of
generated by all divisible subgroups of
.
Note that is invariant under all automorphism of
.
As image of any divisible subgroup is also divisible.
Lemma 1. For any group ,
is the unique maximal divisible subgroup of
.
Proof: We first prove that is divisible.
For any
and
,
there exists
which is a divisible subgroup of
such that
.
As
is divisible, there exists
such that
.
And as
is a subgroup of
,
.
The maximal is obvious,
as if there exists a divisible subgroup such that
,
then
, by the definition of
.
It is unique as if there exists another maximal divisible subgroup ,
then
is also a divisible subgroup,
then
.■
Definition 4. A group is called reduced if
.
Theorem 5. For every group , there exists a decomposition
,
where
is divisible and
is reduced.
Proof: Since is divisible, by corollary of injective property,
is a direct summand of
. Thus, there exists
such that
.
Next, we prove that is reduced.
If
is not reduced, then there exists a non-zero element
such that
for every
,
is divisible by
.
Then take an element
, then
is also divisible by
for every
.
Which contradicts the maximally of
.■
Every abelian group is an extension of the torsion group
by a torsion free subgroup, but
need not to be a direct summand of
. However,
is a direct summand of
.
Definition 5. Given a group define
.
Lemma 2. If and
are divisible p-primary group, then
if and only if
.
Proof: The "only if" part is obvious.
For the "if" part, we define ,
to be the isomorphism between
and
.
Then
is also a homomorphism from
to
.
By previous lemma, we can extend
to a homomorphism
.
And we prove that
is injective first.
For injectivity, we prove this by induction on ,
that if
, and
, then
.
The base case
is obvious, as now
,
which means
, and as
is isomorphic,
.
Now we assume the statement is true for
,
and we prove it for
.
If
, then
,
also
must implies
,
thus by the induction hypothesis,
,
which implies
by the base case.
As is injective,
is an isomorphism.
And we define
as the inverse of
.
Also, note that the inverse of
extends the inverse of
,
as it is easy to check
.
As
is a homomorphism from a subgroup of
to
,
by previous lemma, we can extend
to a homomorphism
.
As
is a surjection by definition,
is also a surjection.
And also as
extends
,
apply a similar reasoning as above, we have
is injective.
Thus,
is an isomorphism.■
Theorem 6. Every divisible group is a direct sum of copies of
and of copies of
for various primes
.
Proof: It is easy to check that , the torsion group of
is divisible,
so
is a direct summand of
.
Thus, there exists
such that
.
As now
is torsion free, and divisible, it is a vector space over
,
and thus,
is a direct sum of copies of
.
Now we prove that is a direct sum of copies of
,
for various primes
.
For prime
, let
.
Then
is a divisible p-primary group.
It is easy to check that
.
So, we only focus on
.
It is easy to check that the group is now a vector space over
,
let
be the dimension of
.
By previous lemma,
is isomorphic to direct sum of
copies of
.■
There is an analogy between theorems about free Abelian groups and divisible groups,
that may be formalized as follows.
Given a commutative diagram containing exact sequences,
then its dual diagram is the diagram obtained by reversing all the arrows.
For example, the dual diagram of is
,
and this leads on the say that "subgroup" and "quotient" are dual concepts.
Theorem 7. Every group can be embedded in a divisible group
.
Proof: As every Abelian group can be written as ,
where
is a free Abelian group and
is a subgroup of
.
Now
, so that
.
Hence,
.
As
is a quotient of a divisible group, it is divisible.■
Corollary 2. A group is divisible if and only if it is a direct summand of any group containing it.
Proof: By previous theorem, we can embed in a divisible group
.
Then
is a direct summand of
.
And any direct summand of divisible group is divisible.■
1.3.2.1 Exercises
Exercise 1. If
is an exact sequence and if
and
are reduced,
then
is reduced.
Proof: We name the maps in the exact sequence as follows:
As is injective, we now assume
.
Assume
is not reduced, then
.
Then,
is divisible, as it is a quotient of divisible group.
Because
is reduced,
,
which implies,
.
As
is reduced,
.■
Exercise 2.
- Prove that
.
- Prove that
.
Proof: As is a quotient of divisible group, it is divisible.
Also given
,
we have
.
Thus,
,
and
contains no summand of
.
It is easy to check that:
Then it is a cyclic group of order ,
which is isomorphic to
.
Then by previous theorem and lemma, we have,
.
For the second part, we factorize .
Then we have:
■
Exercise 3. Prove that a group is divisible if and only if
has
the injective property.
Proof: The "only if" part is given by previous theorem.
For the "if" part, we prove that if a group contains
,
is a direct summand of
.
And then by previous corollary,
is divisible.
By injective property, we have the following diagram,
where is an extension of
.
Then by previous theorem about group extension,
is a direct summand of
.■
Exercise 4. A group is divisible if and only if
for all prime
.
Proof: If is divisible,
then for any element
, as
is divisible,
there exists
such that
.
Thus, we have
.
And as
obviously, we have
.
Then, if for all prime
,
then for any
,
we do induction on
.
The base case
is obvious,
any element
is divisible by
.
Now we assume that any element
is divisible by all natural numbers less than
,
and we prove that
is divisible by
.
If
is a prime, then as
, there exists
such that
.
If
is not a prime, then there exists
and
such that
.
As
is now divisible by
by induction hypothesis,
there exists
such that
.
And as
is now divisible by
by induction hypothesis,
there exists
such that
.
Thus,
.■
Exercise 5. A -primary group
is divisible if and only if
.
Proof: If is divisible,
then for any element
, as
is divisible,
there exists
such that
.
Thus, we have
.
And as
obviously, we have
.
Then, if ,
then for any
,
we do induction on
.
The base case
is obvious,
any element
is divisible by
.
Now we assume that any element
is divisible by
all natural numbers less than
,
and we prove that
is divisible by
.
If
,
then as
, there exists
such that
.
If
is a prime that is not
,
then we assume the order of
is
,
and by Bézout identity, there exists
such that
.
Then we have
, thus
is divisible by
.
If
is not a prime,
then there exists
and
such that
.
As
is now divisible by
by induction hypothesis,
there exists
such that
.
And as
is now divisible by
by induction hypothesis,
there exists
such that
.■
Exercise 6. The following are equivalent for a group :
is divisible.
- Every non-zero quotient of
is infinite.
has no maximal proper subgroup.
Proof: From 1 to 2:
If is divisible,
then for any non-zero quotient
of
,
is divisible.
And by previous theorem,
is a direct summand of
,
and
,
thus
is infinite.
From 2 to 3:
We assume has
as a maximal proper subgroup,
and derive a contradiction.
As
is maximal,
is simple.
And by 2,
is infinite.
Take
, if
has finite order,
then
is a proper subgroup of
,
contradicts the simplicity of
.
If
has infinite order,
then
is a proper subgroup of
,
contradicts the simplicity of
.
From 3 to 1:
We assume that has no maximal proper subgroup,
and
is now not divisible, and tries to derive a contradiction.
As
is not divisible, then there exists
such that
.
And we assume
is the canonical map.
Then
, obviously.
Thus,
is a vector space over
,
we select a basis of
, namely
.
Then, we define
by
. Then
is a surjection.
And
is a surjection.
Thus,
is a maximal proper subgroup of
,
which contradicts the assumption.■
Exercise 7. If and
are divisible groups each of which is
isomorphic to a subgroup of the other,
then
.
Is this true if we drop the adjective "divisible"?
Proof: Let be the embedding of
to the
subgroup of
that is isomorphic to
.
And let
be the embedding of
to the
subgroup of
that is isomorphic to
.
Then for any prime
,
we can restrict
and
to
and
,
and we have
and
.
As
and
are now vector spaces over
,
and as we have the embedding as above,
the dimension of
is less than or equal to the dimension of
, and vice versa.
Thus, the dimension of
is equal to the dimension of
.
And thus,
is isomorphic to
.
By previous lemma,
is isomorphic to
.
For the torsion free part,
we define and
.
And
by
, and
by
.
To prove
is well-defined, we assume
,
then there exists
such that
,
thus we have:
As ,
,
thus
.
In the same way, we can prove
is well-defined.
We then check that is injective.
Assume
,
then
.
As
is an isomorphism between
and
,
the order of
is same as the order of
in
,
thus
.
In the same way, we can prove
is injective.
Also, and
are divisible,
thus are vector spaces over
,
and as both
and
are injective,
the dimension of
is equal to the dimension of
.
Thus, is isomorphic to
.
If we drop the adjective "divisible",
then the statement is not true.
For instance, we take .
And
.
Then
is obviously divisible, but
is not.
We define by
.
And
by
.
Then both
and
are injective,
which are both embedding.
But
is not isomorphic to
, as one is divisible and the other is not.■
Exercise 8.
- Prove that the following groups are all isomorphic:
.
- The circle group
.
.
.
.
- Prove that
.
Proof: The isomorphism between and
the circle group is obvious.
By consider a real number as decimal representation,
the cardinality of
is the same as
.
And it is also easy to verify that the torsion group of
and
is isomorphic to
.
Thus,
and
must have the same cardinality,
by looking at the cardinal number division.
And we could say
and
have the say dimension, again by cardinal number arithmetic.
Thus, they are isomorphic.
As ,
.
Thus, we only need to verify that
is isomorphic to
,
which is equal to
.
As
is an infinite dimensional vector space over
,
quotient by
will minus one dimension,
thus, the dimension of
is equal
the dimension of
,
which verify that
is isomorphic to
.
As is isomorphic to
,
which
is the multiplicative group of positive real numbers.
And
is isomorphic to
,
by the natural logarithm map.
Thus,
,
and
.
Also
.
By above discussion,
we already now that
,
thus,
.
It is obvious that
is then a
vector space over
,
and the dimension of it is equal to square of the dimension of
.
By the cardinal number arithmetic,
square of infinite cardinal number is still the same as the original infinite cardinal number.
Thus, we have
.
Then by previous result,
we have
.■