# Group Theory/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 | edit source]- Prove that when is a set such that , then is not an abelian group.