1
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.