S_3 is a quotient of the free group F({x,y}) I am self-learning Algebra: Chapter 0 by Paolo Aluffi. He defined a presentation of a group $G$ as follows:

So, according to my understanding, $R$ is the kernel (a normal subgroup) of the surjection $\rho$. and $\mathscr{R}$ must be $R$'s generator in a typical group presentation.
Then the author gave a simple example: the symmetric group $S_3$. We know that $S_3 = \{e, x, y, xy, y^2, xy^2 \}$ with relations $x^2=e$, $y^3=e$, and $yx = xy^2$.
The author said $S_3$ is a quotient of the free group $F(x,y)$ (the free group generated by 2 elements) by the smallest normal subgroup containing $x^2$, $y^3$, and $yx = xy^2$.  Then he claimed that $S_3$ can be presented as $(x,y | x^2,y^3,xyxy)$.
I have 3 questions here:

*

*To me it's not obvious that $S_3 \cong \frac{F(x,y)}{R}$. What is the explicit surjection $\rho: F(x,y) \rightarrow S_3$? What is its kernel $R$?

*Why the genrator of $R$( i.e. $\mathscr{R}$) becomes $\{x^2,y^3,xyxy\}$?  Shouldn't it be $\{x^2$, $y^3$, $yx = xy^2\}$ ? For this question, it has been answered here.

*Why didn't the author give any proof except the definition above? Maybe it's really obvious? I've never learned group presentation before. And I found many books including Dummit and Foote only touch a little on this subject. Are there any good notes or books that could help me understand this question better?


Hint
The transposition $\tau = (1 \ 2)$ and the $3$-cycle $\sigma =(1 \ 2 \ 3)$ form a generating set of $\mathfrak S_3$. Therefore if you define $\rho$ by $\rho(x) =\tau$ and $\rho(y) = \sigma$, you get the surjection you're looking for.
As $\tau^2 = \sigma^3 = \tau \sigma \tau \sigma=\text{Id}$ (where $\text{Id}$ is the identity of $\mathfrak S_3$), $x^2, y^3$ and $ xyxy$ are in the kernel $R$.
Now $yx = xy^2$ is equivalent to $xyxy=e$. To convince yourself about that, just multiply the initial equality by $x$ on the left and $y$ on the right and use $x^2=y^3=e$.
A Course in the Theory of Groups from Derek Robinson provides a pretty good coverage on free groups. Also, the questions above are becoming clearer is you have some knowledge on the permutations group.
