# 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[edit | edit source]

**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 . ∎

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[edit | edit source]

**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[edit | edit source]

**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[edit | edit source]

### Further reading[edit | edit source]

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[edit | edit source]

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[edit | edit source]

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[edit | edit source]

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[edit | edit source]

**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 .

Coming soon. ∎