Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm just starting to learn a little group theory, so please forgive any ignorance I demonstrate in the following.

I'm trying to understand the concept of a group being defined based on its presentation $G = \langle S \mid R \rangle$ as the quotient group of the free group $F_S$ generated by $S$, and the normal subgroup $R^{F_S}$ generated by $R$ in $F_S$.

I understand the basic idea that the normal subgroup $R^{F_S}$, is gathering up all the elements of $F_S$ that are to be defined as equivalent to $1_G$ (the identity in group $G$). I also understand that dividing $F_S$ by $R^{F_S}$ (is "dividing" the correct term?) partitions $F_S$ and that each of these partitions will essentially end up as a different element of $G$, and further, since $R^{F_S}$ is a normal subgroup, the partitions can be turned into a group by inheriting the group operation of $F_S$.

But I'm having trouble wrapping my head around exactly how the relators set $R$ partitions $F_S$. Not mechanically, that part I understand in terms of finding the cosets $gR^{F_S}$, but in a more intuitive sense, how do we tell how these relators define the structure of the group $G$?

Taking the dihedral group $D_4$, for instance, presenting it as $\langle r, v | r^4 = v^2 = rfrf = 1 \rangle$, how do we understand from this that $fR^{F_S}$, and $r^3R^{F_S}$, and $rfrR^{F_S}$, will all produce their "own" cosets, instead of being equal to $R^{F_S}$, or equal to one another?

My gut feel is that it has something to do with being "less" than the relators, or in some sense "comprime" to the relators, but I don't think I have the vocabulary to describe it correctly, and I may, for that matter, be completely off base.

share|improve this question
    
This is a hard question, see the en.wikipedia.org/wiki/Word_problem_for_groups -- in your particular case you have what is (nearly) called a "polycyclic presentation" and so every element has a unique expression as $r^i v^j R^{F_S}$ with $0 \leq i < 4$ and $0 \leq j < 2$. (so your third element is not in canonical form). –  Jack Schmidt Feb 28 at 20:45
    
I recommend Holt's Handbook of Computational Group Theory and Sims's Computations with Finitely Presented Groups for a thorough treatment. –  Jack Schmidt Feb 28 at 20:46
    
There are two ways of interpreting this question. Going from group to presentation is easy compared to going from presentation to group. That is, given a known group, the problem of finding a presentation for it is relatively easy, including understanding why the presentation is indeed a presentation for the original group, whereas given a known presentation, finding what kind of group it describes is very hard - indeed it involves the word problem (which is impossible to solve uniformly), as Jack mentions above, of determining when a given word is trivial modulo a given set of relations. –  blue Mar 1 at 3:40

2 Answers 2

up vote 7 down vote accepted

Here is an explanation of shotrevlex Knuth-Bendix rewriting for your group. It is long, but if you can understand how I make $R_4$, then the rest of it is just more of the same.


Let me change your group just a tiny bit, $e=rf$ and $r=ef$ so the group is now generated by $T=\{e,f\}$ $$\langle e,f \mid (ef)^4=f^2=e^2=1\rangle$$

For each element of $F_T/R^{F_T}$ I want to find the shortest way of writing it down, and amongst shortest ways, I'll choose the one that comes last alphabetically. If I use a relation to replace a long way with a shorter way (for the same element), I'll call that simplifying.

To keep things easy, I'll start out by only using relations the obvious way: \begin{array}{ll} R_1: & ee & \mapsto 1 \\ R_2: & ff & \mapsto 1 \\ R_3: & efefefef & \mapsto 1 \\ \end{array}

Now I want to make sure I don't miss any simplifications. Notice that $eefefefef$ can be viewed as both $(ee)fefefef \stackrel{R_1}{\mapsto} fefefef$ and as $e(efefefef) \stackrel{R_3}{\mapsto} e$. Clearly $e$ is simpler, but $fefefef$ doesn't match any of my rules. I'll go ahead and add the rule that says $fefefef$ and $e$ are the same but $e$ is simpler:

\begin{array}{ll} R_1: & ee & \mapsto 1 \\ R_2: & ff & \mapsto 1 \\ R_3: & efefefef & \mapsto 1 \\ R_4: & fefefef & \mapsto e \\ \end{array}

Notice the third rule is now redundant: the left hand sides probably should be as simple as possible considering the other rules, but the left hand side of $R_3$ simplifies to $e(fefefef) \stackrel{R_4}{\mapsto} ee$ and so simplified $R_3$ is the same as $R_1$.

I look again for double-rule opportunities: $ffefefef$ is both $(ff)efefef \stackrel{R_2}{\mapsto} efefef$ and $f(fefefef) \stackrel{R_4}{\mapsto} fe$, but again none of the rules do that directly, so I add a rule that says $efefef$ and $fe$ are the same, but $fe$ is simpler:

\begin{array}{ll} R_1: & ee & \mapsto 1 \\ R_2: & ff & \mapsto 1 \\ R_4: & fefefef & \mapsto e \\ R_5: & efefef & \mapsto fe \\ \end{array}

This time $R_4$ is redundant, the left hand side is $f(efefef) \stackrel{R_5}{\mapsto} f(fe) = (ff)e \stackrel{R_2}{\mapsto} e$, so we can skip it too.

Again $eefefef$ is both $(ee)fefef \stackrel{R_1}{\mapsto} fefef$ and $e(efefef) \stackrel{R_5}{\mapsto} e(fe)$ so we add rule 6, and notice rule 5 is redundant:

\begin{array}{ll} R_1: & ee & \mapsto 1 \\ R_2: & ff & \mapsto 1 \\ R_6: & fefef & \mapsto efe \\ \end{array}

This continues one more time until we get:

\begin{array}{ll} R_1: & ee & \mapsto 1 \\ R_2: & ff & \mapsto 1 \\ R_7: & efef & \mapsto fefe \\ \end{array}

Now at this point we have two double rule opportunities $eefef$ and $efeff$. Both work about the same, so I'll show the first: $(ee)fef \stackrel{R_1}{\mapsto} fef$ is the same as $e(efef) \stackrel{R_7}{\mapsto} e(fefe) = (efef)e \stackrel{R_7}{\mapsto} (fefe)e = fef(ee) \stackrel{R_1}{\mapsto} fef$. Well, that's clear, $fef=fef$. The other double rule opportunity is similar.

It is a theorem that because of this lack of exciting double rule opportunities, there are NO exciting opportunities for using rules in different ways. If you apply the rules mechanically until they cannot be applied anymore, then the resulting answer is always the same, no matter which order you choose to use the rules in, or if they match in multiple places, which place you decide to use the rule on first.

This also tells us all of the different elements: we just starts with the identity, and multiply by $e$ and $f$ until we can apply a rule. That would give us a shorter answer so we don't need to consider it.

Doing this we get $\hat G = \{1,e,f,ef,fe,efe,fef,fefe\} \subset F_T$ and that is it. We get that if $u,v \in \hat G$ and $u \neq v$ then $uR^{F_T} \neq vR^{F_T}$ (because no rule applies) and $F_T/R^{F_T} = \{ u R^{F_t} : u \in \hat G \}$ (because $\hat G$ is “closed” under multiplication, after applying rules, and contains the generators $e$ and $f$).

share|improve this answer
    
Wow, nice answer, thanks for putting in the time! I'm going to have to put in some time studying it some more to understand it, but I think I was able to follow each step, just haven't quite pulled it all together into a conclusion yet. Also, I think there's a typo in the last paragraph: you have $fef$ listed twice in $\hat{G}$. –  sh1ftst0rm Mar 1 at 0:21
    
Thanks, fixed! The books I mentioned should explain the ideas behind it. The particular order I chose is not always the best order (it is usually bad for solvable groups, but dihedral groups are an exception). Also infinite groups don't necessarily have finitely many rules. In some groups, you can just keep doing what I did, making infinitely many rules (and whether there are infinitely many or finitely many rules can even change based on the ordering!). I tend to study finite groups, where the choice of ordering is more of an efficiency problem. –  Jack Schmidt Mar 1 at 3:13

Jack Schmidt's answer is about how to compute effectively with group presentations, so let me try and give a more intuitive or conceptual answer.

The first point to note is that, although the definition is "correct", it is not particularly helpful - some definitions are like that. People who work a lot with group presentations do not typically think in terms of normal closures of sets of defining relators. When teaching introductory courses on group theory, I introduce presentations very early, because they are so useful for giving precise but compact descriptions of small examples, but I do this even before the students have encountered quotient groups (let alone free groups), so there is no question of giving the formal definition.

I define the group $\langle X \mid R \rangle$, where $X$ is a set of generators, and $R$ is usually given as a set of equations among words in the generators (like $\langle x,y \mid x^2=y^3=(xy)^2=1 \rangle$) as the "largest" group that is generated by $X$ in which the equations in $R$ hold. Of course I point out that I have not said what "largest" means for infinite groups, and I have not attempted to justify the implied claim that there is always a unique largest such group, but I give some examples (such as the one above, which defines the dihedral group of order 6), in which it can easily be proved that there is a unique largest such group.

Although the above definition is not formal, I think it expresses intuitively what we are trying to define. From the formal definition of $G = \langle X \mid R \rangle$, as a quotient $F(X)/\langle R \rangle^{F(X)}$ of a free group by the normal closure of the defining relators, you can go on to prove some basic properties such as

  • The group $G$ is generated by the images in the quotient group of the generators $X$ of the free group. It is customary to abuse notation and denote these generators also by $X$, although you need to be aware that two elements of the original $X$ could have the same image in $G$.
  • There is a fundamental result expressing the universal property of $G$: if $H$ is any group, and $\phi:X \to H$ is any map with the property that the images of the elements of $R$ under $\phi$ (you need to say exactly what that means) are all equal to the identity in $H$, then $\phi$ extends uniquely to a group homomorphism $G \to H$. This is not hard to prove from the definition of $G$.
  • You should learn the basic Tietze transformations for manipulating group presentations. Again, the proof that these work can be proved easily from the definition.

    I have found that once you are familiar with the above three properties, there is rarely any need to go back the definition of $G$ as a quotient of a free group.

  • share|improve this answer

    Your Answer

     
    discard

    By posting your answer, you agree to the privacy policy and terms of service.

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