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I am looking at page 26 of Dummit and Foote, and I see the following statement discussing the presentation of $D_{2n} = \langle r, s\mid r^2=1, s^n=1, rs=sr^{-1} \rangle$.

...$D_{2n}$ has the relations $r^2=1, s^n=1, rs=sr^{-1}$. Moreover, these relations have the property that any other relation among the elements of $S = \{r, s \}$ can be deduced from these three.

My question is, is this true for all group presentations? That is, can any relation between elements of the generators be determined from the relations in the presentation?

My inclination is no; because a few lines below, Dummit and Foote say

...in an arbitrary presentation it may be extremely difficult (or even impossible) to tell when two elements of the group (expressed in terms of the given generators) are equal.

Bernard
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Blue
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2 Answers2

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Can any relation between elements of the generators be determined from the relations in the presentation?

Yes! Sometimes, a group is defined by one of its presentations, so there's no other choice when that happens.

My inclination is no; because a few lines below, Dummit and Foote say

...in an arbitrary presentation it may be extremely difficult (or even impossible) to tell when two elements of the group (expressed in terms of the given generators) are equal.

This is known as a decidability result. It's the word problem. Generally speaking, given two elements of a group given by a presentation, it is literally impossible to decide whether or not one is equal to another using a Turing machine in finite time.

But this does not mean that any and all relations between elements of a group cannot be determined by the relations of a presentation of that group.

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Shaun
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  • Thanks for your answer! But suppose that we have two expressions $A,B$ (where $A, B$ are in terms of the given generators) which are equal, but it is impossible to derive their equality from the given relations. We now have a relation between the generators (namely $A=B$) which cannot be derived from the given relations. How doesn't this contradict the fact that we can derive any relation from the given relations? – Blue Nov 22 '19 at 19:05
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    Oh I guess I sort of see the answer to my comment; so in general, you cannot give a finite algorithm to determine if two expressions are equal. BUT, if I give you a concrete presentation, and two concrete elements (expressed in terms of the generators), then you will $100%$ be able to tell me if they are equal or not. Is that correct? – Blue Nov 22 '19 at 19:08
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    You're welcome! No, undecidable in general does not mean impossible for every element of a group. There are classes of groups in which the word problem is decidable (I think). And yes, you got it in your second question, although (I think) there are classes of groups for which this is not the case, even for two "concrete" elements of a "concrete" group. The details are covered in Magnus et al.'s "Combinatorial Group Theory [. . .]". – Shaun Nov 22 '19 at 19:12
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    @Blue No, that is not correct. There are specific finite group presentations that can be written down explicitly (there are examples in the literature) for which the problem of deciding whether two words in the group generators represent the same element is undecidable. But even so, the answer to your original question is yes. If two words represent the same element, then it is possible to prove that they do (although it may be very difficult). – Derek Holt Nov 22 '19 at 21:27
  • @DerekHolt So suppose I have a group $\langle x, y|r_1, r_2 \rangle$, and it is true that $x^2yx=yx^2y^3$, but this cannot be proved from $r_1, r_2$. Haven't I just found a relation, namely $x^2yx=yx^2y^3$, which cannot be proved from $r_1, r_2$? – Blue Nov 23 '19 at 16:51
  • @Blue, if you cannot derive it from $r_1$ and $r_2$ in principle, then it cannot be true in the group. – Shaun Nov 23 '19 at 16:54
  • @Shaun Ahhh I think I understand. So if I give you that group again and I ask you weather $x^2yx=yx^2y^3$, you can tell me that $if$ it is true, then it should be derivable from $r_1, r_2$. But you don't have a good way to determine weather $x^2yx=yx^2y^3$ is derivable from $r_1, r_2$. All you can do is try random manipulations (of course I don't literally mean random); but if you never arrive at $x^2yx=yx^2y^3$, you can never know weather that equation is true or not. – Blue Nov 23 '19 at 17:24
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    @Blue That is correct. If that relation holds in the group, then there exists a proof of it assuming only the defining relations. But it's not a good idea to try random manipulations. You can look systematically for a proof - probably the best way in practice is to use the Knuth-Bendix procedure for which there are good computer implementations. And if the proof exists then you would find if you searched for long enough - the big problem is that (except in special cases) you cannot predict in advance how long you might need to search. – Derek Holt Nov 23 '19 at 17:52
  • @DerekHolt Thank you so much! – Blue Nov 23 '19 at 17:53
  • @DerekHolt If I could ask you one last thing; suppose that I have a group $G$, generators $a_i$, and some relations $r_i$. Suppose that I know that any relation in $G$ can be obtained using the $r_i$. Does this mean that $\langle a_i|r_i \rangle$ is a presentation of $G$? Is this a correct (slightly informal) definition of a presentation? – Blue Nov 23 '19 at 19:29
  • Yes it does (provided that the $r_i$ really are relations of $G$). – Derek Holt Nov 23 '19 at 19:31
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It's all about the difference between "being mathematically determined" and "being computable". All relations between elements are logically obtainable from the relations in the presentation, but it is not a computable problem in general (of course for some groups it is, like for finite groups, since at worst you can just write the whole multiplication table).

Captain Lama
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