Describing $\langle x,y : x^{2} , y^{3} , [x,y] , x^{6}y^{6}\rangle$. I am trying to identify the following presentation 

$\langle x,y : x^{2} , y^{3} , [x,y] , x^{6}y^{6}\rangle$

I substituted  the first relation in the final one and got $x^{6}=1$ so the group is cyclic of order 6
The problem is why we can't say that the schreier transversal is $\{x^{0}.....x^{5}\}$ so we omit the other generator $y$.
The correct answer in fact is claiming that the transversal is $\{x^{i}y^{j}\}$. But in another case for The presentation of the cyclic group of order 7 we omit the other generator.
 A: The relators $x^2$, $y^3$ and $[x, y]$ combine to imply the third relator, $x^6y^6$. So your group is:
$$
\langle x, y; x^2, y^3, [x, y]\rangle
$$
Note that the cross product $\mathbb{Z}_2\times\mathbb{Z}_3$ satisfies this presentation (why?). It then follows that your group is precisely this presentation (why?).
This means that your group is cyclic of order 6 (why? - this is where the transversal being $\{x^iy^j\}$ comes from).
Note that although the group is cyclic of order 6, it is not for the reasons which you give in your question. In particular, in your question you imply that $x$ has order 6 as $x^6=1$. However, $x$ has order 2.
A: If you examine $$(xy)^0, (xy)^1, (xy)^2, (xy)^3, (xy)^4, (xy)^5, (xy)^6 \dots$$ you reduce to $$1, xy, y^2, x, y, xy^2, 1 \dots$$ and the sequence repeats. It is just that the presentation you are given does not exhibit a generator of the whole group. There are non-trivial elements of the group such as $x, y, y^2$ which are not generators.
As for the cyclic group of order $7$, since $7$ is prime every non-identity element generates the group.
Is this what is causing your confusion?
