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# Groups, subgroups and constructions

**Definition (group)**:

A **group** is a set together with three operations, namely

- a nullary operation called the identity,
- a unary operation called inversion
- and a binary operation called the group operation (also often referred to simply as the "operation" of the group)

such that the following axioms are satisfied for all elements of :

- (identity axiom)
- (inverse axiom)
- (associativity axiom)

**Definition (abelian group)**:

An **abelian group** is a group , whose operation (which is usually denoted by ) is commutative.

**Proposition (elementary rules for computing in groups)**:

Let be a group with group operation and let . Then we have the following rules of computation:

- and ("cancellation laws")

**Proof:** First note that when , then we may multiply by on the left using in order to get by using the identity axiom and to get left cancellation, and right cancellation is proved similarly. Then, we note that by multiplying the inverse axiom by on the left and using the identity axiom, we get and hence, since by the identity axiom, we may apply cancellation to infer , proving 2. Finally, .

Henceforth, we shall sometimes refer to the group operation of a group simply as the "operation".

**Definition (opposite group)**:

Let be a group, where denotes the operation of . Then the **opposite group** of is defined as the group that has the same underlying set as , but whose group operation is given by for (inversion and identity are carried over from ).

**Proposition (opposite groups are groups)**:

Whenever is a group, is also a group.

**Proof:** We have to check that the axioms for the given operations are satisfied. First note that and that by the rules that were derived above. Finally, .

**Definition (subgroup)**:

Let be a group. A subset is called a **subgroup** of iff together with the group law of it is itself a group (in particular, for all we must have .

Note: Often, the explicit notation for the group operation is omitted and the product of two elements is denoted solely by juxtaposition.

Subgroups with the inclusion map represent subobjects of a group.

**Proposition (subgroup criterion)**:

Let be a group and a subset. Then is a subgroup of if and only if .

**Proof:** Suppose first that is a subgroup. Then the condition follows immediately from being closed under inversion and the group law. On the other hand, if the condition is satisfied, setting shows that is closed under inversion, and setting shows that it is closed under the group law. Finally, setting shows that contains the identity. Associativity and the law that the identity must satisfy are automatically inherited from the group operation of .

## Exercises

edit- Make explicit the proof of right-cancellation ("right-cancellation" means ).
- Let be a group, and let be subgroups such that neither nor . Prove that is not a subgroup of .
- Let together with the operation , .
- Prove in detail that , together with the operation , is a group.
- Prove that in , there exists a subgroup which is not equal to with subgroups .

# Representations

**Definition (representation)**:

Let be a group, a category, an object of . Then a **representation** (also called **action**) of on by automorphisms of is a group homomorphism

- .

**Example (symmetric group acting on a product set)**:

Let be a set and let be the product of copies of . The symmetric group acts on via

- .

Note that in this notation, we identified the element of with the automorphism of to which the element is sent via the homomorphism of the representation. This convention is followed throughout group theory and will be understood by every mathematician.

**Definition (equivalence of representations)**:

Let be a group, a category, objects of and , two representations of on resp. . Then an **equivalence of representations** is an isomorphism such that

- .

**Proposition (inverse of equivalence of representations is equivalence of representations)**:

Let be a group, a category, objects of and , two representations of on resp. . Let be an equivalence of representations. Then is also an equivalence of representations.

**Proof:** We have

- ,

since is an equivalence of representations.

# Cosets and Lagrange's theorem

**Definition (left coset)**:

Let be a group, and a subgroup, and . Then the **left coset** of represented by is the set

- .

Right cosets are defined in an analogous fashion:

**Definition (right coset)**:

Let be a group, and a subgroup, and . Then the **right coset** of represented by is the set

- .

For both of these, we have the following proposition:

**Proposition (being in the same left coset is an equivalence relation)**:

Let be a group, and define a relation on by

- .

Then is an equivalence relation, and we also have the formula

- .

**Proof:** We first prove the alternative formula for being in the same coset. If and , we find from the latter equation an so that , so that and hence , the latter identity because is a group and in particular closed under inversion. On the other hand, if , then for some , and hence (because the identity is in ) and then .

Hence, the two formulae for the relations coincide, and it remains to check that we're dealing with an equivalence relation. Indeed, suppose and . Then there exist so that and so that . Then , so that , ie. , proving transitivity. Reflexivity follows since the identity is in , and symmetry follows because implies and hence .

Analogously, we have the following proposition:

**Proposition (being in the same right coset is an equivalence relation)**:

Let be a group, and define a relation on by

- .

Then is an equivalence relation, and we also have the formula

- .

**Proof:** Consider the opposite group of . There is a bijection , given by , under which two elements belonging to the same right coset of correspond to elements belonging to the same *left* coset of . But the relation defined by the latter was seen to be an equivalence relation.

**Definition (index)**:

Let be a group, and let be a subgroup. Then the **index** of is defined to be the number

- .

That is, the index is precisely the number of left cosets.

**Proposition (Lagrange's theorem)**:

Let be a group, and let be a finite subgroup. Then

- .

In particular, the order of divides the order of .

**Proof:** We have seen that being in the same left coset is an equivalence relation, so that the equivalence classes partition . Moreover, every equivalence class (ie. coset) has the same cardinality as via the bijection .

**Proposition (number of right cosets equals number of left cosets)**:

Let be a group, and a subgroup. Then the number of right cosets of equals the number of left cosets of .

**Proof:** By Lagrange's theorem, the number of left cosets equals . But we may consider the opposite group of . Its left cosets are almost exactly the right cosets of ; only the orders of the products are interchanged. But in particular, the number of left cosets of , which, by Lagrange's theorem equals , is equal to , which is what we wanted to prove.

Hence, we may also use the notation for the number of *right* cosets.

**Proposition (degree formula)**:

Let be a group, and let . Then

- .

**Proof:** We may partition into a family of left cosets , where for all we have . Moreover, may be partitioned into a family of left cosets of . Then is a family of left cosets of that partitions (since each element is in one left coset of , and then is in a unique coset , and then is the unique coset in which is), and the cardinality of this family, which is , is the number of left cosets of in .

## Exercises

edit- Prove that , thus establishing another formula for the equivalence relation of being in the same coset.
- Formulate Lagrange's theorem for right cosets, without using index notation.

# Normal subgroups and the Noether isomorphism theorems

**Definition (normal subgroup)**:

Let be a group. A subgroup is called a **normal subgroup** if and only if for each we have .

**Proposition (elements of disjoint normal subgroups commute)**:

Let be a group, and let so that . Then .

**Proof:** Let and . Since is a normal subgroup, , so that there exists such that . Similarly, since , there exists so that . Then , hence and since we get , so that and since and were arbitrary, .

**Proposition (characterisation of direct products within groups)**:

Let be a group, and let be subgroups of . Consider the group generated by these groups. The following are equivalent:

- The function is an isomorphism
- For we have and
- Each element in can be uniquely written as a product , where for we have

**Proof:** Certainly 1. 2., since for all , the subgroup of corresponds via to the subgroup , which is certainly normal. For 2. 3., observe that any element of *may* be written as a product , where and each is an element of some . But by 2., the are pairwise disjoint, so that any elements in distinct commute. Hence, we may sort the product so that the first few entries are in , the following entries are in and so on. The products of the entries that are contained within then form the element as required by the decomposition in 3. Finally, for 3. 1., observe that 3. implies that the given function is bijective. But it is also a homomorphism, because 3. immediately implies that the are disjoint, and we may use that elements of disjoint normal subgroups commute to obtain that indeed commutes with the respective group laws.

**Definition (subgroup product)**:

Let be a group, and let be subgroups. Then the **subgroup product** of and is defined to be

- .

In general, the subgroup product is not a subgroup. However, if one of the subgroups involved in the product is a normal subgroup, then it is:

**Proposition (subgroup product of subgroup and normal subgroup is subgroup)**:

Let be a group, and let be subgroups such that one of is normal. Then and are subgroups of .

**Proof:** Without loss of generality assume that is normal. Let , where and . Then

for some because is normal (here we applied the subgroup criterion). Similarly, is a subgroup.

**Proposition (product of normal subgroups is normal)**:

Let be a group, and let be normal subgroups of . Then , ie. is a normal subgroup of , and the same holds for .

**Proof:** By symmetry, it suffices to prove that is a normal subgroup. Indeed, let and , where and . Then

for some , , since are normal.

## Exercises

edit- Prove that the intersection of normal subgroups is again normal.
- Let be a group, and let such that are pairwise coprime. Prove that .

# Cardinality identities for finite representations

**Definition (permutation representation)**:

Let be a group and let be a set. A **permutation representation** of on is a representation , where the automorphisms of are taken in the category of sets (that is, they are just bijections from to itself).

**Definition (pointwise stabilizer)**:

Let be a group, let be an algebraic variety and let be an instance of . Suppose that is a representation in the category defined by . Let . Then the **pointwise stabilizer** of is given by

- .

**Proposition (transitive permutation representation is equivalent to right multiplication on quotient by stabilizer)**:

Let be a group, let be a set and suppose that we have a permutation representation which is transitive. Let be arbitrary and let be the pointwise stabilizer of . Consider the action by left multiplication, where is the set of left cosets of (which is in fact never a normal subgroup in this situation, unless the action is trivial, because ). Then there exists a -isomorphism from to .

**Proof:** We define as follows: shall be mapped to . First, we show that this map is well-defined. Indeed, suppose that we take . Then is mapped to . Then we note that the map is surjective by transitivity. Finally, it is also injective, because whenever , we have by applying to both sides and using a property of a group action, and thus , that is to say . That follows immediately from the definition, so that we do have an isomorphism of representations.

We are now in a position to derive some standard formulae for permutation representations.

**Theorem (orbit-stabilizer theorem)**:

Let be a group, and let be a permutation representation on a set . Then

- .

**Proof:** acts transitively on . The above -isomorphism between and is bijective as an isomorphism in the category of sets. But the notation stood for .

**Theorem (class equation)**:

Let be a finite group and let be a permutation representation on the finite set . Then

- ,

where are the orbits of , and for . (We also say that are a system of representatives for the orbits of .)

**Proof:** acts transitively on each orbits, and the orbits partition . Hence, by the orbit-stabilizer theorem,

- .

**Definition (fixed point set)**:

Let be a group that acts on a set , and let be a subset of . Then the **fixed point set** of is defined to be

- .

# Free products and amalgamated sums

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**Definition (reduced word)**:

Let be any set, and define the set to be the set of *formal* inverses to the elements of ; that is, , so that ; for example, we could define . Let denote the empty tuple. Then a **reduced word** over is either

- the empty tuple , or
- a finite tuple of elements of such that whenever are two adjacent elements, then neither and nor and .

**Definition (empty word)**:

The empty tuple is also called the **empty word**

**Proposition (reduction of tuples to reduced words)**:

Let be any set, and let be the set of formal inverses. Suppose that is any tuple (not necessarily a reduced word). Then in finitely many steps, one may obtain a reduced word from by removing adjacent elements such that either and or and .

**Proof:** This follows immediately since the length of the tuple is an integer, which is reduced by 2 whenever adjacent elements that contradict the definition of a reduced word are eliminated. Doing this elimination repeatedly until it is no longer possible will hence lead to a reduced word in a finite number of steps.

Note that when is odd, then the reduced word obtained in this way will not be the empty tuple. Otherwise, the empty tuple may result.

**Definition (free group)**:

Let be any set. Then the **free group** over is defined to be the group whose elements are the reduced words over and whose group operation is given by first concatenation and then reduction to a reduced word.

**Proposition (the free group is a group)**:

Let be a set. Then is a group.

**Proof:** The empty tuple serves as an identity. Associativity holds because if are three reduced words, then

Finally, whenever is a reduced word, we claim by induction on that it has an inverse. Certainly the empty word has Indeed, suppose that ; then , which has an inverse by the induction hypothesis, so that by associativity is an inverse of .

## Exercises

edit- Prove that when is a set such that , then is not an abelian group.

# Abelian groups and the Grothendieck group of a monoid

**Definition (abelian)**:

Let be a group. We call an **abelian** group if and only if for all , we have (where we denote the group operation by juxtaposition).

**Definition (cyclic group)**:

A **cyclic group** is a group that is generated by a single of its elements, ie. for a certain .

**Proposition (cyclic group is abelian)**:

Let be a cyclic group. Then is abelian.

**Proof:** Indeed, write any two elements as , , where is such that . Then , using associativity.

# The action by conjugation and p-groups

**Definition (global stabilizer)**:

Let be a group that acts on , where belongs to some algebraic variety . Let be a subset. Then the **global stabilizer** of is the set

- ,

where the notation stands for the set .

**Definition (p-group)**:

Let be a prime number. Then a ** -group** is a group of order for some .

**Proposition (cardinality of fixed point set of a p-group equals cardinality of set mod p)**:

Let be a -group that acts on a set . Then

- .

**Proof:** By the class equation,

- ,

where for each orbit of the action of on we pick one representative of that orbit. Since is a -group, whenever is not , it is divisible by by Lagrange's theorem. Hence, by taking the above equation , we get

- ,

where is the number of those for which . But means precisely that the orbit of is trivial, that is, that is fixed by all of .

**Proposition (p-groups have nontrivial center)**:

Let be a -group. Then , where denotes the identity.

**Proof:** acts on itself via conjugation. Furthermore,

- ,

so that is precisely the fixed point set of under that action. But since the cardinality of the fixed point set of a p-group equals the cardinality of the whole set mod p, we get that

- ,

which would be impossible if .

# Simple groups and Sylow's theorem

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**Definition (Sylow p-subgroup)**:

Let be a group and let be a prime number such that . Then a **Sylow -subgroup** of is a subgroup such that , where is maximal such that .

**Theorem (Cauchy's theorem)**:

Let be a group whose order is divisible by a prime number . Then contains an element of order .

**Proof:** acts on itself via conjugation. Let be a system of representatives of cojugacy classes. The class equation yields

- .

Either, there exists such that is both not and not divisible by , in which case we may conclude by induction on the group order, noting that divides and , or for all the number is either or divisible by ; but in this case, by taking the class equation , we obtain that is nontrivial and moreover that its order is divisible by . Hence, it suffices to consider the case where is an abelian group. Take then any element . If has order divisible by , raising to a sufficiently high power will produce an element of order . Otherwise, the order of is divisible by , and by induction we find an element whose order is divisible by . Then the order of will also be divisible by , because otherwise, passing to the quotient, for some not divisible by .

**Theorem (Sylow's theorem)**:

Let be a finite group, such that with . Then the following hold:

- has a Sylow subgroup
- The action of by conjugation on the Sylow subgroups is transitive
- If is the number of Sylow -subgroups, then and
- Every -group of is contained within some Sylow -group of

**Definition (simple group)**:

A group is a **simple group** if and are the only normal subgroups of (where denotes the identity).

# Subnormal subgroups and series

{{definition|subnormal subgroup|Let be a group. A subgroup is called **subnormal subgroup** if and only if there exists

**Definition (subnormal series)**:

Let be a group. Then a **subnormal series** is a finite family of subgroups such that

- ,

where is the identity.

**Definition (composition series)**:

Let be a group. A **composition series** of is a subnormal series

of such that for all the quotient group is simple.

**Theorem (Schreier refinement theorem)**:

Let be a group, and let

be a subnormal series of .

# Commutators, solvable and nilpotent groups

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**Definition (commutator)**:

Let be a group and let . Then the **commutator** of and is defined to be the element

- .

{{definition|commutator

**Proposition (commutators form a subgroup)**:

Let be a group. Then the set forms a subgroup of .

**Proof:** By the subgroup criterion, it is sufficient to show that for , the element is of the form for suitable . Indeed,

**Failed to parse (syntax error): {\displaystyle \begin{equation*} t \end{equation*}}**

**Definition (commutator subgroup)**:

Let be a group. Then the **commutator subgroup** of is defined to be .

**Definition (perfect)**:

A group is called **perfect** if and only if .

**Proposition (subdirect normal product of perfect groups is direct)**:

Let be perfect groups, and let

be a subdirect product which is simultaneously a normal subgroup of their outer direct product. Then in fact .

**Proof:** It suffices to show that whenever and , then

- ,

since is a perfect group. Thus, let be arbitrary, and pick , where for all , such that . Since is a subgroup and normal, the element

is in .

**Definition (solvable)**:

**Proposition (group is solvable iff maximal normal subgroup is solvable)**:

Let be a group, and let be a maximal normal subgroup. Then is solvable if and only if is solvable.

# Characteristic subgroups

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**Definition (characteristic subgroup)**:

Let be a group. A **characteristic subgroup** of is a subgroup such that for all .

**Proposition (characteristic subgroups are normal)**:

Any characteristic subgroup of a group is a normal subgroup of .

**Proof:** This follows since the map is a group automorphism of .

**Definition (characteristically simple)**:

A group is called **characteristically simple** if and only if its only characteristic subgroups are and , where denotes the identity of .

**Proposition (characteristically simple groups)**:

Let be a characteristically simple finite group, and let be any of its minimal normal subgroups. Then is isomorphic to a product of copies of , that is, , where is an index set (of finite cardinality).

**Proof:** Let be a subgroup of maximal subject to the following two conditions:

- is the direct sum of images of under a
- is normal

Suppose that . Note that the group is characteristic, so that it equals all of . Hence, we find such that is not a subgroup of . Since is an automorphism, , so that . Since the product of normal subgroups is normal, we conclude that the product subgroup is a normal subgroup that is a direct product of homomorphic images of in , in contradiction to the maximality of with these properties. Hence, and we are done.

**Proposition (minimal normal subgroups of a characteristically simple groups are simple)**:

Let be a characteristically simple group, and let be a minimal normal subgroup of . Then is simple.

**Proof:**

**Proposition (powers of characteristically simple groups are characteristically simple)**:

We conclude:

**Theorem (structure theorem of finite, characteristically simple groups)**:

The finite, characteristically simple groups are precisely the powers of simple groups.

**Proof:** We have seen that each characteristically simple finite group is the direct product of copies of isomorphic images of any of its minimal normal subgroups, and that the latter are always simple in characteristically simple groups. We conclude that each finite, characteristically simple group is a power of simple groups. Conversely, let be a simple group, , and set

- .

## Exercises

edit- Prove that all subgroups of are characteristic.
- Let be two finite simple groups such that is divisible by a prime number that does not divide . Use the structure theorem for characteristically simple groups to prove that is not characteristically simple.
- Prove that a subgroup of a characteristically simple group need not be characteristically simple.
- Prove that the product of characteristically simple subgroups whose minimal normal subgroups are not isomorphic is not characteristically simple.

# The symmetric group

**Definition (symmetric group)**:

Let be a set. Then the **symmetric group** of is defined to be

- ;

that is, it is the set of all bijective functions from to itself with composition as operation.

**Definition (permutation)**:

A **permutation** is, by definition, an element of .

**Proposition (symmetric group essentially depends only on the cardinality of the underlying set)**:

Let be sets of the same cardinality. Then there exists a group isomorphism

- .

**Proof:** Suppose that is a bijective function. Then the group isomorphism is given by

- ;

indeed, an inverse is given by

- .

**Definition (finite symmetric group)**:

Let . Then the **symmetric group of order **, denoted , is defined to be

- .

**Theorem (Cayley's theorem)**:

Let be a finite group, and set . Then there exists a subgroup of which is isomorphic to .

**Proof:** acts transitively on itself by left multiplication in the category of sets. This means that we have a group homomorphism . Moreover, this morphism is injective; indeed, only the identity element of induces the identity element in . Hence, the claim follows from the first Noether isomorphism theorem.

**Definition (matrix representation of a permutation)**:

The representation of in the category of vector spaces over a field given by

is called the **matrix representation** of the permutations contained within .

**Definition (sign)**:

Let be a permutation. Then the **sign** of , written , is defined to be , where there exist transpositions such that .

The following proposition shows that this notion is well-defined:

**Proposition (equivalent characterisations of the sign of a permutation)**:

**Definition (alternating group)**:

Let . Then the **alternating group**, a subgroup of , is defined to be

- .

**Proposition (alternating group is maximal and normal in the symmetric group)**:

Let . Then , and further is a maximal subgroup of .

In particular, is a maximal normal subgroup in (ie. maximal among the normal subgroups).

**Proof:** Note first that is normal as the kernel of a group homomorphism. We then have that is a group homomorphism from to , and by the first Noether isomorphism theorem, . In particular, there are only two cosets of . Suppose that there existed a subgroup . Then by the degree formula, we would have , so that either or . In both cases, one of the inclusions is not strict, a contradiction.

**Proposition (conjugation in the symmetric group is re-labeling)**:

Let be a cycle, and let be any element. Then

- .

**Proposition (in degrees 5 or larger all three-cycles are conjugate in the alternating group)**:

Let , and let and be any two three-cycles in . Then there exists such that .

**Proof:** Since being in the same conjugacy class is an equivalence relation, assume .

**Theorem (in degrees 5 or larger the alternating group is simple)**:

Let . Then is a simple group.