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?

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    $\begingroup$ I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) $\endgroup$ – Patrick Stevens Apr 18 '17 at 6:37
  • $\begingroup$ Well this is the notation used in my class. $\endgroup$ – W.Scott Apr 18 '17 at 7:02
  • $\begingroup$ Really? They use $<X \mid R>$ rather than $\langle X \mid R \rangle$? That's unusual. $\langle, \rangle$ are angle brackets; $<, >$ are less than/greater than signs. $\endgroup$ – Patrick Stevens Apr 18 '17 at 7:09
  • $\begingroup$ Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one. $\endgroup$ – W.Scott Apr 18 '17 at 7:12

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 $i,j \in {\mathbb Z}$.

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  • $\begingroup$ How exactly do you compute normal closure of $R$? $\endgroup$ – W.Scott Apr 18 '17 at 8:14
  • $\begingroup$ $\langle R^F \rangle$ is the by definition the subgroup $\langle frf^{-1} : r \in R, f \in F \rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"? $\endgroup$ – Derek Holt Apr 18 '17 at 9:48
  • $\begingroup$ So is the normal closure of $R$ always equal to the kernel of $\pi$? Because on one hand, $G$ is isomorphic to $F/ker(\pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$. $\endgroup$ – W.Scott Apr 18 '17 at 11:45
  • $\begingroup$ And for the $Z_7$ example, how can you show that $<R>=<R^F>$? $\endgroup$ – W.Scott Apr 18 '17 at 11:49
  • $\begingroup$ Yes, with your notation $N = {\rm ker} \pi$. In the $Z_7$ example $F$ is cyclic and hence abelian. $\endgroup$ – Derek Holt Apr 18 '17 at 12:13

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