Consider the set A={x,y}. We are considering the group F_{2}=(F[A], *), where * refers to the operation of concatenation followed by reduction. An example of * is as follows: y^{-4}*y^{7}=y^{-4}y^{7}=y^{3}. From this point forward, we will abuse notation and suppress the * when referring to this group. We will define F[A] soon.

What does it mean for a group to be free?

For our purposes, our group is free in that it has no constraints. For instance, in class we considered the group generated by two elements, *a* and *b*, where a^{2}=e, b^{2}=e, and (ab)^{4}=e. The free group with two generators has no such constraints, and is consequently very large.

Terminology:

Now, we will introduce several bits of terminology.

A **syllable** is anything of the form x^{n}, y^{n}, where n is an element of $\Bbb{Z}$.

A **word** is a string of syllables

A **reduced word** is a string of syllables where all possible cancellations have been made, i.e. x^{-4}x^{12}y^{4}y^{-5}=x^{8}y^{-1}.

What we are really studying is the set of all reduced words generated by x and y added to the empty word 1. We shall denote this set by F[A].

We will now prove that F_{2} is a group.

Associativity:

Let w_{1}=a_{1}^{n1}…a_{m}^{nm}, and let w_{2}=b_{1}^{p1}…b_{k}^{pk}, and let w_{3}=c_{1}^{l1}…c_{q}^{lq}

It is clear that (w_{1}w_{2})w_{3}=w_{1}(w_{2}w_{3}), because the "timing" of when certain words are concatenated is not important, so long as they appear in the same order. After words are concatenated (regardless of "when" they are concatenated) they can then be reduced as much as possible. Consider that (w_{1}w_{2})w_{3}=(a_{1}^{n1}…a_{m}^{nm}b_{1}^{p1}…b_{k}^{pk})c_{1}^{l1}…c_{q}^{lq}=a_{1}^{n1}…a_{m}^{nm}b_{1}^{p1}…b_{k}^{pk}c_{1}^{l1}…c_{q}^{lq}=a_{1}^{n1}…a_{m}^{nm}(b_{1}^{p1}…b_{k}^{pk}c_{1}^{l1}…c_{q}^{lq})=w_{1}(w_{2}w_{3}).

Note that these words are as reduced as possible unless a_{m} = b_{1} or b_{k} = c_{1}.

Identity:

By our definition of F[A], the identity element is the empty word, 1. To be sure that it is the identity, we can check that for w $\in$ F[A], 1(w)=(w)1=w, and as such, 1 is the identity. (Note that concatenating 1 with any word does not change the word in any way).

Inverses:

For any w $\in$ F[A], w^{-1} is formed simply by reversing the order of the syllables of w, and then inverting every syllable. Concatenating w with w^{-1} allows for the pairwise cancellation of every syllable of n_{i} with its inverse n_{i}^{-1}.

Since F[A] is associative under concatenation, since it has an identity, and since all of its elements have inverses, F[A] is a group.

**Subgroup**

Note that a subgroup of the free group with two generators is the free group with one generator.

Consider the set P={x^{n} |n $\in$ $\mathbb{Z}$}. Note that every element of this set is also in the set of the free group generated by two elements, because every element can be written as x^{n}y^{0}, where 0 is fixed.

We shall now prove that this set is indeed a subgroup.

**Closure**: Let *a* and *b* be elements of P. Then, a=x^{r} and b=x^{s}. Then, ab=x^{r+s}, which is an element of P, since r+s is an integer.

**Identity**: x^{0} is an element of P because 0 is an integer.

**Inverses**: Let c be an element of P. Then c=x^{f} where f is an integer. Note that x^{-f} =g is also in P, since -1 is an integer. Then, notice that fg=e. As such, inverses exist for all elements of P.

Since P is closed, contains the identity, and contains inverses, P is a subgroup.

**Coset**

Since P is infinitely large, and since there are infinitely elements that are in F_{2} and not in P, there are infinitely many cosets of P. Let us consider left cosets, in particular. All left cosets of P will be of the form k(P), where k is an element of F_{2} but not in P. Then, the cosets are all elements of P multiplied by some k. For example, consider some k_{i}. k_{i}(P)=(…k_{i}x^{-2}, k_{i}x^{-1}, k_{i}k^{0}, k_{i}x^{1}, k_{i}x^{2},…).