# Definition of presentation of a group.

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).

• The definition given here is intuitive. Formally we construct a group $G$ with set of generators $X$ and relations $R_i$ among the $X$ as a quotient of the free group on $X$. Specifically, if $F$ is a free group on $X$, then $G$ is defined to be $F/K$, where $K$ is the normal subgroup generated by the relations $R_i$. In the first example, $X = \{x_1, x_2\}$ and $R_1 = x_1^2$, $R_2 = y_1^2$, $R_3 = (x_1y_1)^2$ are the relations. – spin Apr 26 '13 at 13:22
• When giving a "relation" in a group presentation, one is asserting the given word is equal to the identity, e.g. $(x_1 y_1)^2 = 1$ in one of the examples you cite. Think of the relations as generating a normal subgroup of the free group on whatever generators are used. – hardmath Apr 26 '13 at 13:23
• @spin yes, and thanks. But I ask if it has some utility present such an advanced topic at the very beginning... the reader (not possessing like me that notion), in my humble opinion, should be astonished at least... – Bento Apr 26 '13 at 16:16

## 5 Answers

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].

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:

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

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

• thank you, I suppose this is the formal definition, but please consider my comment to Loki. Again, I will reread your answer afterwards. I will procede with order following the order of the book, without putting the cart before the horse. – Bento Apr 26 '13 at 17:30

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 $$\frac{1}{l}+\frac{1}{m}+\frac{1}{n}>1$$

• this is interesting, and I think it's not present in the text (for the first one there's an exercise in which it is used the fact that the group presented is the dihedral, the second is just like an "ipse dixit"). – Bento Apr 26 '13 at 16:23
• You know your groups $\checkmark^{+1}$ – Namaste Apr 27 '13 at 1:18

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.

• some results you propose are not in my algebrical background, and I am at the present more interested in the question if the exposition is formally sensible. Also the connection of the two examples with the preceding assertions of the author, for which connection It seems to me that your answer gives indications. I think I'd better read your reply later (perhaps after finishing the group theory at least) for having plain consciousness of the matter. Eventually, why do you affirm the group to be abelian? I noticed (by an exercise) in the book that this is the dihedral, which is not abelian. – Bento Apr 26 '13 at 17:03
• $D_2=\langle x,y : x^2=y^2=(xy)^2=1 \rangle$ is abelian, since it is generated by $x,y$ and $xyxy=1 \Rightarrow yx=x^{-1} y^{-1}=xy$. For $n>2$, the group $D_n = \langle x,y : x^n=y^2=(xy)^2=1 \rangle$ is not abelian. – Martin Brandenburg Apr 26 '13 at 17:22
• In order to understand my answer, no special group theory is necessary, only basic notions from category theory, namely the Yoneda Lemma and universal properties. For the deep understanding of algebra (and all related areas of mathematics) one needs to learn category theory. Even for basic notions in algebra, such as here the presentation of group, category theory helps a lot to organize the ideas. Feel free to ask questions! – Martin Brandenburg Apr 26 '13 at 17:28
• Thank you very much! Casually I found this site, and I'm going to post questions anytime I need help:-) This is a good site, I have found a lot of competence, and helpful people. Sometime there's a little temptation of not thinking enough, since then someone will certainly give the solution, but in my little experience I try to get it by myself... at least until I must hoist the white flag. I choose the Dummit-Foote also because at the end of the book I noticed an Appendix about Category. (continue) – Bento Apr 26 '13 at 18:55
• (continue) It was also vying Artin's book, in fact, reading comments on the site, it was even preferred by some, but I opted for DF because it contains more material. I have also learned that there is a classic of Birkhoff and Mac Lane that treats the matter from a point of view categorial. I've gave a look just now at the appendix and it seems to me, however, that there is not much (I do not see the lemma). – Bento Apr 26 '13 at 19:03

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.

• I suspected something in such a sense. Thank you for replying, this make me more "relaxed" about the framework of the text... I think that I have just to wait for next chapters (well, I hope it will). – Bento Apr 26 '13 at 16:03

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.

• what I said above I repeat it here. I am at the beginning of the text, I have the basics of abstract algebra, but not all there in the book (which is why I'm reading it). The free groups are among them, and I noticed that there are about at the end of group theory as well as the theory of representation (who I miss too). Thanks for the answer, I need it anyway to give me an idea, albeit approximate. – Bento Apr 26 '13 at 17:18
• The free groups could just as easily be at the beginning of a group theory textbook. See if you can figure out what algebra has words made only from the symbols 0, 1 +, and parentheses, and where no words identify the same element if associativity and 0+x=x don't give you a rule for identifying them. – Loki Clock Apr 26 '13 at 17:31
• I looked at the wiki-link hardmath gave, and I think I understand that, in summary, we proceed in this way: we consider the elements of a group and build a group formed by the expressions of the group (= sequences of elements) in relation to this group I suppose it happens the definition of presentation of the starting group. – Bento Apr 26 '13 at 20:37
• But there are a lot of notions that are involved, so I remain with my opinion that it is premature to expose the concept just after the axioms of a group. (correction of the previous comment: ... AND in relation to this group I suppose it happens the definition of presentation of the starting group. ) – Bento Apr 26 '13 at 20:43
• It's a lot less complicated than what hardmath is talking about. I think this is what you were saying about building a group formed by the expressions of the group: If you want $\Bbb{Z}_3$, you could present it as $<\{a,b\} | a+a=b, a+b=0>$ or as $<\{a\}|a+a+a=0>$. You could find out those expressions, $a+a=b$ and so on, from inspecting $\Bbb{Z}_3$. But say you don't apply any expressions, you just know there's this element $a$ and you assume it's in a group. That assumes it's closed, so $a+a$ is also in the group. You know $(a+a)+a=a+(a+a),$ because you assumed it's associative. What else? – Loki Clock Apr 26 '13 at 21:00