An exercise to understand free group I am new to the concept of free groups, reading Artin's Algebra but completely lost. So I hope I can learn from concrete examples instead of theorems corollaries. So I jumped to the exercise, and here is the fist one. I suppose this is an easy one but I am sure I can have a better understand if I can solve it. So the exercise is

Let $F$ be the free group on $\{x,y\}$. Prove that the three elements $u=x^2$, $v=y^2$, and $z=xy$ generate a subgroup isomorphic to the free group on $u$, $v$, and $z$.

Anyone can give some help? Thanks!
 A: Suppose you have two elements $a,b$ of a group. You can write words in these letters and their inverses, and each such word can be multiplied out using the group law.
Consider for example the word
$$w = a b b^{-1} a b a^{-1} a b^{-1} a^{-1} a^{-1}
$$
This word $w$ represents the identity no matter what the group is, as can be seen by repeatedly applying the inverse law. Any word with this property, namely that it represents the identity by repeated application of the inverse law, is said to be a "freely reducible" word in the elements $a$ and $b$. For instance, the word $aba^{-1} b^{-1}$ is not freely reducible (even though it would represent the identity if the group were known to be abelian). 
And although I have only discussed words in two letters, the concept of freely reducible obviously generalizes to words in any number of letters.
Consider now a specific group $G$ and a generating set, perhaps a finite generating set such as $a_1,a_2,...,a_n$. Obviously every freely reducible word $w$ in this generating set represents the identity element of $G$. The group $G$ is a free group (on the generating set $a_1,..,a_n$) if the only words representing the identity are the freely reducible words.
To put it another way, free groups are those groups which are "as free as is possible", meaning that they have the fewest words expressing the identity as is possible.
A: Think geometrically! Consider the following two graphs:

Now convince yourself of the following facts:


*

*The free subgroup of $F(x,y)$ generated by $\{x^2,y^2,xy\}$ can be described exactly as the set of closed paths at the origin of the graph on the left (of course assuming any sequence of a letter followed by its inverse as irrelevant).

*The free group $F(u,v,z)$ can be described exactly as the set of closed paths at the origin of the graph on the right (of course assuming any sequence of a letter followed by its inverse as irrelevant).

*Those subsets are isomorphic (you can just rename the "loops" accordingly).

A: First, define the first morphism that comes to mind that looks like it might work.

 $$\varphi:\left\{\begin{array}{lll}Free(u, v, w) & \to & Free(x, y)\\u&\mapsto &x^2\\v&\mapsto&y^2\\z&\mapsto&xy\end{array}\right.$$

And then try to prove that it's an isomorphism when restricted properly:

 $Im(\varphi)=\langle x^2,y^2,xy\rangle$.

and

 $Ker(\varphi)=\{1\}$. Let $x\in Free(u, v, w)$. $x=y_1\dots y_n$ where $y_i\in\{u,v,w\}$ and $\varphi(x)=z_1\dots z_n$ where $z_i\in \{x^2,y^2,xy,x^{-2},y^{-2},y^{-1}x^{-1}\}$ and $\varphi(y_i)=z_i$. If $\varphi(x)=1$, then $z_1\dots z_n=1$. Since we're in a free group, it means that somewhere, we have $xx^{-1}$ or $x^{-1}x$ or $yy^{-1}$ or $y^{-1}y$. it can't be in a $z_i$ so you have $z_iz_{i+1}=l_1xx^{-1}l_2$ where $l_i\in\{x,x^{-1},y,y^{-1}\}$ (or some other case). Now, by checking all the cases, you can see that (I didn't check but it should be true), there is a $z'=l_1l_2\in \{x^2,y^2,xy,x^{-2},y^{-2},y^{-1}x^{-1}\}$. Take $y' \in \{u,v,w\}$ so that $\varphi(y')=z'$. Then $\varphi(x)=\varphi(y_1\dots y_{i-1}y_iy_{i+1}y_{i+2}\dots y_n)=\varphi(y_1\dots y_{i-1})\varphi(y_iy_{i+1})\varphi(y_{i+2}\dots y_n)=\varphi(y_1\dots y_{i-1})z_iz_{i+1}\varphi(y_{i+2}\dots y_n)=\varphi(y_1\dots y_{i-1})l_1l_2\varphi(y_{i+2}\dots y_n)=\varphi(y_1\dots y_{i-1})z'\varphi(y_{i+2}\dots y_n)=\varphi(y_1\dots y_{i-1})\varphi(y')\varphi(y_{i+2}\dots y_n)=\varphi(y_1\dots y_{i-1}y'y_{i+2}\dots y_n)$ and so you've reduced the number of generators used by one. You can then juste prove that $\varphi(x)=1\implies x=1$ by recurrence on the number of generators used to write $x$.

