Definition of presentation of a group.

Question

basing myself on suggestions I found in previous discussions, I have opted for the algebra book of Dummit. However, I have now a little problem. On page 26 (3th edition) the authors say that

"presentations" give an easy way of describing many groups, but there are a number of subtleties that need to be considered. One of this is that in an arbitrary presentation it may be difficult (or even impossible) to tell when two elements of the group (expressed in terms of the given generators) are equal. As a result it may not be evident what the order of the presented group is, or even whether the group is finite or infinite.

Then they report two examples

$\qquad$$\qquad$$\qquad$$\qquad$ $<x_1,y_1 \;|\; x_{1}^2=y_{1}^2=(x_{1}y_{1})^2=1>$

saying that one can show this is a presentation of a group of order 4, whereas

$\qquad$$\qquad$$\qquad$$\qquad$ $<x_2,y_2 \;|\; x_{2}^3=y_{2}^3=(x_{2}y_{2})^3=1>$

is a presentation of an infinite group.

The problems I encountered are essentially two:

1) I have found vague the "definition" of "relations"(and consequently that of "presentation"), because it is set metamathematicalwise (at least this way it sounds to me), using the concept of "derivation/deduction". Infact, no method is exhibited to clarify and rigorously formalize the expression used in the "definition", so I find some trouble, the underlying ideas having been left up in the air (what is meant, for example, as they say "it may be difficult (or even impossible) to tell when two elements of the group are equal"? what's the corresponding mental mechanism? and what if the $S$ generators set has an exorbitant cardinality? Is perhaps (also) this a motive of difficulty to which the authors refer?);

2) with reference to the two examples I mentioned, I am not able to understand the link between these and what has been said just before: there may be some problems about assigning the order to the presented group (it has been said), but the two presentations in the examples are well defined in this sense (I say).

Answer 1

Let $\bar x$ be a vector of variables (e.g. $\bar x = a,b,c$).

Let $\bar R$ be a list of words formed from the variables in $\bar x$ (e.g. $aa$, $b^{-1}c^{-1}bc$).

The presentation $\langle \bar x \mid \bar R \rangle$ defines a group that is a quotient of [the free group on the symbols $\bar x$] by [the smallest relation on words of $\bar x$ that makes each word of $\bar R$ the identity].

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A simple example is $\langle a \mid a^3 \rangle$, this defines the group $C_3$ because we start with the free group $\{\ldots, a^{-2}, a^{-1}, a, a^{2}, a^{3}, \ldots\}$ (this is isomorphic to $\mathbb Z$) and then quotient it out by the relation defined by $a^3 = 1$. That relation is the reflexive symmetric transitive closure of:

• $aaa \sim 1$
• $x^{-1} \sim y^{-1} \implies x \sim y$
• $x \sim x'$, $y \sim y' \implies xy \sim x'y'$

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Throughout I used the term 'words', but it's really 'words plus inverses' but I don't have a name for that.

Answer 2

What I know about the two groups is that they are from a family of groups called Von Dyke group $D(l,m,n)$. They can be presented as $$D(l,m,n)=\langle a,b\mid a^l=b^m=(ab)^n=1\rangle$$ Here, you see $m=n=l=2$ and $m=n=3$. You can google to find that this group is finite if and only if $$\frac{1}{l}+\frac{1}{m}+\frac{1}{n}>1$$

Answer 3

Consider the free group on a set S of generators. From a set of simple statements $a \equiv b$ we can generate a congruence relation on the free group - an equivalence relation $\equiv$ such that $a \equiv b \implies ax \equiv bx, a/x \equiv b/x$ for all x in the free group. Congruence relations are homomorphisms with kernel $\{a|a \equiv 0\}$.

The laws the equivalence relation must satisfy depend on the signature of the algebra, and congruence relations can be used to define presentations for arbitrary varieties of algebra.

Answer 4

The difficulty or "impossibility" that Dummit refers to in your quoted material is the undecidability of the word problem for even finitely presented groups. Although for many group presentations the word problem can be solved algorithmically, a general solution was proved impossible by Pyotr Novikov in 1955 and again (using a different approach) by William Boone in 1958.

Answer 5

Here is a more abstract point of view.

The group $G=\langle x,y : x^2 = y^2 = (xy)^2 = 1 \rangle$ can be defined(!) by the universal property $\hom(G,H) \cong \{(a,b) \in H \times H : a^2 = b^2 = (ab)^2 = 1\}$, naturally in $H$, which is an arbitrary group. Thus, $G$ is the universal ("smallest") group containing two elements $x,y$ satisfying the three relations. One can easily check that $\mathbb{Z}/2 \times \mathbb{Z}/2$ solves this universal property, thus $G$ is isomorphic to this group. In particular, $G$ is finite, and abelian.

However, if $p$ is an odd prime, then $G=\langle x,y : x^p = y^p = (xy)^p = 1 \rangle$ is infinite and not abelian. This can be seen using the homomorphism $G \to \mathrm{Sym}(\mathbb{Z})$ corresponding to $a=(-p+1,-p+2,...,-1,0)(1,2,...,p)(p+1,...,2p)... $ and $b=...(-p+2,-p+3,...,0,1)(2,...,p+1)(p+2,...,2p+1)...$ (see MO/22459).

More generally, if $R_1(x_1,\dotsc,x_n),\dotsc,R_m(x_1,\dotsc,x_n)$ are group words in $x_1,\dotsc,x_n$, then

$$G=\langle x_1,\dotsc,x_n : R_1(x_1,\dotsc,x_n),\dotsc,R_m(x_1,\dotsc,x_n)=1 \rangle$$

is defined to be the group satisfying the universal property

$$\hom(G,H) \cong \{a \in H^n : R_1(a_1,\dotsc,a_n)=\dotsc=R_m(a_1,\dotsc,a_n)=1\}.$$

The uniqueness is justified by the Yoneda Lemma, and existence follows by taking $G=F_n/\langle \langle R_1,\dotsc,R_m \rangle \rangle$, where $F_n$ is the free group on $n$ generators and $\langle \langle - \rangle \rangle$ denotes the normal closure. In particular, the elements of $G$ are represented by group words in $x_1,\dotsc,x_n$, and we calculate modulo the relations $R_1,\dotsc,R_n$ and all relations derived inductively from them.

But in my opinion the universal property is far more important and also useful than the element description, which is even intractable in many interesting cases (see also hardmath's answer). For example, try to show that $\langle x,y : x^3 = y^3 = (xy)^3 = 1 \rangle$ is infinite with the element definition. Good luck!

You cannot really understand a group just by looking at its elements. You have to map the whole group it into more concrete groups. This is the whole point of representation theory, and actually is a special case of the philosophy of the Yoneda Lemma.

By the way, presentations are nothing special to groups. They apply to arbitrary algebraic structures. For example in the theory of $\mathbb{R}$-algebras, $\langle x : x^2=0 \rangle$ is the ring of dual numbers.