Group Theory/Printable version


Group Theory

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Groups, subgroups and constructions

 

To do:
#Product subgroups

  1. renaming of this section to something more appropriate


Definition (group):

A group is a set   together with three operations, namely

  1. a nullary operation   called the identity,
  2. a unary operation   called inversion
  3. 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  :

  1.   (identity axiom)
  2.   (inverse axiom)
  3.   (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:

  1.   and   ("cancellation laws")
  2.  
  3.  

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

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  1. Make explicit the proof of right-cancellation ("right-cancellation" means  ).
  2. Let   be a group, and let   be subgroups such that neither   nor  . Prove that   is not a subgroup of  .
  3. Let   together with the operation  ,  .
    1. Prove in detail that  , together with the operation  , is a group.
    2. 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

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  1. Prove that  , thus establishing another formula for the equivalence relation of being in the same coset.
  2. 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:

  1. The function   is an isomorphism
  2. For   we have   and  
  3. 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

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  1. Prove that the intersection of normal subgroups is again normal.
  2. 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

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

  1. the empty tuple  , or
  2. 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

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  1. 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

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:

  1.   has a Sylow subgroup
  2. The action of   by conjugation on the Sylow subgroups is transitive
  3. If   is the number of Sylow  -subgroups, then   and  
  4. 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

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,

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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

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:

  1.   is the direct sum of images of   under a  
  2.   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

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  1. Prove that all subgroups of   are characteristic.
  2. 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.
  3. Prove that a subgroup of a characteristically simple group need not be characteristically simple.
  4. 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.