S_3 is a quotient of the free group F({x,y})

Question

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:

1. 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 $R$?
2. 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.
3. 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?

Answer 1

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.