1
$\begingroup$

This is from Robinson, A Course in the Theory of Groups, 2nd edition:

Part 1 Part 2

Some pages below he gives an example:

Part 3

In saying that $G=\langle x,y\mid x^2=1,y^2=1\rangle$, he is using the definition of presentation given in (2) in the picture. But a presentation, according to the author, is an epimorphism $\pi$ from some free group $F$ to $G$. In this example I want to find $\pi$. So, first I have to find the $Y$ and $S$ in definition (1), where $X=Y^{\pi}$ and then $F$ will be the free group on $Y$. Humm... $F$ I can already find it. It will simply be the free group on a set with two elements. But how do I find $Y$?

$\endgroup$
  • $\begingroup$ Good question! I'm not sure. Are the isomorphism theorems covered in the book so far? $\endgroup$ – Shaun Jan 9 at 20:49
  • $\begingroup$ Yes, they are in the preceding chapter, chapter one, which is a summary of material normally presented in a whole book on group theory. $\endgroup$ – stf91 Jan 9 at 21:26
  • 2
    $\begingroup$ $Y = \{x,y\}$ and $S = \{x^2,y^2\}$. Maybe you are confused because it is common practice to regard $x$ and $y$ as elements of $G$, but that is actually abuse of notation. $x$ and $y$ are elements of $F$, and it is their images $\pi(x)$ and $\pi(y)$ that are elements of $G$. $\endgroup$ – Derek Holt Jan 9 at 21:27
  • $\begingroup$ Well then $\pi$ could be the natural homomorphism form $F$ to $Y^F$ (normal closure of $Y$ in $F$). The problem is now to find $Y^F$. But I can't even find $<Y>$! $\endgroup$ – stf91 Jan 9 at 21:38
  • 2
    $\begingroup$ You seem to be confused about the notation. $Y = \langle x,y \rangle = F$ , $G = F/\langle S^F \rangle$, and $\pi:F \to G$ is the natural homomorphism. $\langle S^F \rangle$ consists (by definition) of all product $t_1^{g_1}t_2^{g_2} \cdots t_r^{g_r}$, where $r$ can be any non-negative integer, each $t_i$ is $x^2$ or $y^2$, and each $g_i \in G$. $\endgroup$ – Derek Holt Jan 9 at 22:22
3
$\begingroup$

The free group $F=F(Y)$ on a set $Y$ has the defining property that given a function of sets $f\colon Y \to G$, where $G$ is (the underlying set of) a group, there is a homomorphism $\hat f\colon F \to G$ that extends $f$ in the sense that if $\iota\colon Y \to F$ is the natural inclusion (of sets), then $\hat f\iota = f$. Therefore to define the map $\pi$, it suffices to choose the image of the elements of $Y$ in $G$.

So let $Y = \{a,b\}$ be a two-element set and $G$ the infinite dihedral group as defined by the presentation $\langle x,y \mid x^2 = 1, y^2 = 1\rangle$. I will choose the function $Y \to G$ sending $a$ to $x$ and $b$ to $y$. The resulting homomorphism $\pi \colon F\to G$ is surjective because $\{x,y\}$ is a set of generators for $G$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. So I know the restriction of $\pi$ to $Y$ and that's all. I don't really know what $\pi$ is in this example (infinite dihedral group). $\endgroup$ – stf91 Jan 9 at 21:55
  • 3
    $\begingroup$ @stf91: Yes you do: every element of $F$ is a product of $a$s, $b$s, and their inverses. And a given such word is mapped to the corresponding word in $x$ and $y$. $\endgroup$ – Arturo Magidin Jan 9 at 22:30
  • $\begingroup$ Ah! [EMPTY SPACE] $\endgroup$ – stf91 Jan 9 at 22:43
  • $\begingroup$ According to Rylee Lyman (see answer) $\pi$ is the presentation $<x,y| x^2=1, y^2=1>$. But then Ker $\pi$ is the normal closure of $\{a^2, b^2\}. I can't see how I can prove this. Let $K= Ker \pi, S=\{a^2, b^2\}$ and $S^F$ the normal closure of $S$. To begin with, what is $S^F$? $\endgroup$ – stf91 Jan 10 at 16:32
  • $\begingroup$ According to Rylee Lyman (see answer) $\pi$ is the presentation $<x,y| x^2=1, y^2=1>$. But then Ker $\pi$ is the normal closure of $\{a^2, b^2\}$. I can't see how I can prove this. Let $K=$ Ker $\pi, S=\{a^2, b^2\}$ and $S^F$ the normal closure of $S$. To begin with, what is $S^F$? $\endgroup$ – stf91 Jan 10 at 16:39

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.