# A presentation of the infinite dihedral group.

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

Some pages below he gives an example: 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 $$Y$$?

• Good question! I'm not sure. Are the isomorphism theorems covered in the book so far? – Shaun Jan 9 at 20:49
• Yes, they are in the preceding chapter, chapter one, which is a summary of material normally presented in a whole book on group theory. – stf91 Jan 9 at 21:26
• $Y = \{x,y\}$ 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 $G$. – Derek Holt Jan 9 at 21:27
• 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 $<Y>$! – stf91 Jan 9 at 21:38
• 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 $g_i \in G$. – Derek Holt Jan 9 at 22:22

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 $$G$$.
• 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). – stf91 Jan 9 at 21:55
• @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 $y$. – Arturo Magidin Jan 9 at 22:30
• 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$? – stf91 Jan 10 at 16:32 • According to Rylee Lyman (see answer)$\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\$? – stf91 Jan 10 at 16:39