Abstract Algebra/Group Theory/Products and Free Groups

During the preliminary sections we introduced two important constructions on sets: the direct product and the disjoint union. In this section we will construct the analogous constructions for groups.

Product Groups

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Definition 1: Let   and   be groups. Then we can define a group structure on the direct product   of the sets   and   as follows. Let  . Then we define the multiplication componentwise:  . This structure is called the direct product of   and  .

Remark 2: The product group is a group, with identity   and inverses  . The order of   is  .

Theorem 3: Let   and   be groups. Then we have homomorphisms   and   such that   and   for all  . These are called the projections on the first and second factor, respectively.

Proof: The projections are obviously homomorphisms since they are the identity on one factor and the trivial homomorphism on the other.

Corollary 4: Let   and   be groups. Then   and  .

Proof: This follows immediately from plying the first isomorphism theorem to Theorem 3 and using that   and  .

Theorem 5: Let   and   be groups. Then   and   are normal subgroups of  .

Proof: We prove the theorem for  . The case for   is similar. Let   and  . Then  .

 
Commutative diagram showing the universal property satisfied by the direct product.

We stated that this is an analogous construction to the direct product of sets. By that we mean that it satiesfies the same universal property as the direct product. Indeed, to be called a "product", a construction should have to satisfy this universal property.

Theorem 6: Let   and   be groups. Then if   is a group with homomorphisms   and  , then there exists a unique homomorphism   such that   and  .

Proof: By the construction of the direct product,   is a homomorphism if and only if   and   are homomorphisms. Thus   defined by   is one homomorphism satisfying the theorem, proving existence. By the commutativity condition this is the only such homomorphism, proving uniqueness.

Products of Cyclic Groups

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Theorem 7: The order of an element   is  .

Proof: The lowest positive number   such that   is the smallest number such that   and   for integers  . It follows that   divides both   and   and is the smallest such number. This is the definition of the least common divider.

Theorem 8:   is isomorphic to   if and only if   and   are relatively prime.

Proof: We begin with the left implication. Assume  . Then   is cyclic, and so there must exist an element with order  . By Theorem 7 we there must then exist a generator   in   such that  . Since each factor of the generator must generate its group, this implies  , and so  , meaning that   and   are relatively prime. Now assume that   and   are relatively prime and that we have generators   of   and   of  . Then since  , we have   and so  . this implies that   generates  , which must then be isomorphic to a cyclic group of order  , im particular  .

Theorem 9 (Characterization of finite abelian groups): Let   be an abelian group. Then there exists prime numbers   and positive integers  , unique up to order, such that

 

Proof: A proof of this theorem is currenly beyond our reach. However, we will address it during the chapter on modules.

Subdirect Products and Fibered Products

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Definition 10: A subdirect product of two groups   and   is a proper subgroup   of   such that the projection homomorphisms are surjective. That is,   and  .

Example 11: Let   be a group. Then the diagonal   is a subdirect product of   with itself.

Definition 12: Let  ,   and   be groups, and let the homomorphisms   and   be epimorphisms. The fiber product of   and   over  , denoted  , is the subgroup of   given by  .

In this subsection, we will prove the equivalence between subdirect products and fiber products. Specifically, every subdirect product is a fiber product and vice versa. For this we need Goursat's lemma.

Theorem 13 (Goursat's lemma): Let   and   be groups, and   a subdirect product of   and  . Now let   and  . Then   can be identified with a normal subgroup of  , and   with a normal subgroup of  , and the image of   when projecting on   is the graph of an isomorphism  .

Proof:

Semidirect Products

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Further reading

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More on the automorphism groups of finite abelian groups. Some results require theory of group actions and ring theory, which is developed in a later section.

http://arxiv.org/pdf/math/0605185v1.pdf

Free Groups

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In order to properly define the free group, and thereafter the free product, we need some preliminary definitions.

Definition 10: Let   be a set. Then a word of elements in   is a finite sequence   of elements of  , where the positive integer   is the word length.

Definition 11: Let   and   be two words of elements in  . Define the concatenation of the two words as the word  .

Now, we want to make a group consisting of the words of a given set  , and we want this group to be the most general group of this kind. However, if we are to use the concatenation operation, which is the only obvious operation on two words, we are immediately faced with a problem. Namely, deciding when two words are equal. According to the above, the length of a product is the sum of the lengths of the factors. In other words, the length cannot decrease. Thus, a word of length   multiplied with its inverse has length at least  , while the identity word, which is the empty word, has length  . The solution is an algorithm to reduce words into irreducible ones. These terms are defined below.

Definition 12: Let   be any set. Define the set   as the set of words of powers of elements of  . That is, if   and  , then  .

Definition 13: Let  . Then we define a reduction of   as follows. Scan the word from the left until the first pair of indices   such that   is encountered, if such a pair exists. Then replace   with  . Thus, the resulting word is  . If no such pair exists, then   and the word is called irreducible.

It should be obvious if   with length  , then   will be irreducible. The details of the proof is left to the reader.

Definition 14: Define the free group   on a set   as follows. For each word   of length  , let the reduced word  . Thus   is the subset of irreducible words. As for the binary operation on  , if   have lengths   and   respectively, define   as the completely reduced concatenation  .

Theorem 15:   is a group.

Proof:

Example 16: We will concider free groups on 1 and 2 letters. Let   and  . Then

  with  .
  such that   for any   and   for any  . Example product:  .

Group Presentations

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In this subsection we will briefly introduce another method used for defining groups. This is by prescribing a group presentation.

Definition 17: Let   be a group and   a subgroup. Then define the normal closure of   in   as the intersection of all normal subgroups in   containing H. That is, if   is the normal closure of  , then

 .

Definition 18: Let   be a set and  . Let   be the normal closure of   in   and define the group  . The elements of   are called generators and the elements of   are called relators. If   is a group such that  , then   is said to be a presentation of  .

The Free Product

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Using the previously defined notion of a group presentation, we can now define another type of group product.

Defintion : Let   and   be groups with presentations   and  . Define the free product of   and  , denoted  , as the group with the presentation  .

Remark : Depending on the context, spesifically if we only deal with abelian groups, we may require the free product of abelian groups to be abelian. In that case, the free product equals the direct product. This is another example of abelian groups being better behaved than nonabelian groups.

Lemma : The free product includes the component groups as subgroups.

Remark : The free product is not a product in the sense discussed previously. It does not satifsy the universal property other products do. Instead, it satisfies the "opposide", or dual property, obtained by reversing the direction of all the arrows in the commutative diagram. We usually call a construction satisfying this universal property a coproduct.

Problems

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Problem 1: Let   and   be groups of relatively prime orders. Show that any subgroup of   is the product of a subgroup of   with a subgroup of  .

Answer

Coming soon.