I'm struggling to understand group presentation.
There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $\pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(\pi)$.
Question 1, I know we can denote $G$ as $<X \mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(\pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(\pi)$.
Question 2, $Z_7=\{0,1,2,3,4,5,6\}$ is a cyclic group generated by $\{1\}$, which can also be written as $<x \mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X=\{x\}$, and $R=\{x^7\}$. But then $G=F/ker(\pi)$, what exactly is $F$ and $\pi$ in this example?
In the cyclic group $C_7$ with presentation $\langle x \mid x^7 \rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = \{ x^7 \}$, and $\ker \pi = \langle R \rangle =\langle R^F \rangle$ (the normal closure of $R$ in $F$). So in this example, $\langle R \rangle$ is equal to its normal closure. But that is not usually true.
For another example, consider $G = \langle x,y \mid r \rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/\langle R^F \rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = \{ r \}$.
So $\langle R \rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $\langle R^F \rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j \in {\mathbb Z}$.