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This is likely a pretty basic question, but I couldn't find answers anywhere. It's to do with how group presentations are represented.

It's said that, for example, the cyclic group of order $n$ may be represented as

$$\langle a \mid a^n = e\rangle$$

But, this makes sense only if $n$ is the smallest possible positive integer such that $a^n = e$. Naively, the fact that $a^n = e$ doesn't rule out the possibility that $|a|$ is some divisor of $n$, which would make this an inaccurate presentation of the group.

So, is the convention to assume that $n$ is the smallest possible integer making $a^n = e$ true? Is this 'convention' true for all presentations, where one of the relations given is something like $x^n = e$?

Thank you! Apologies if this is basic.

Shaun
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adam dhalla
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  • Use \langle and \rangle, not < and >. The latter are relational symbols, the former are delimiters (which is what you want). – Arturo Magidin Feb 18 '23 at 23:34
  • Looking at the definition of group presentation as discussed here, the point seems to be that the group $\langle S|R\rangle$ is the universal ("smallest") group which satisfies the relations. That is: If a group satisfies the relations of the presentation, then it either is the group of interest or it contains it as a subgroup. – Semiclassical Feb 18 '23 at 23:48
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    @Semiclassical You've got it absolutely backwards. Universal here means largest group satisfying the relations: Any group having elements satisfying these relations and generated by them is a quotient of the group. – Arturo Magidin Feb 18 '23 at 23:58
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    @Semiclassical. By your reading, the cyclic group of order $5$ would contain the cyclic group of order 10 as a subgroup, since it contains an element $x$ satisfying $x^{10}=1$. It would force every presentation to define the trivial group. – Arturo Magidin Feb 19 '23 at 00:00
  • I'm prepared to accept the point---group theory isn't an area where I can be trusted---but in my defense I took the "smallest" bit from the answers in the linked post. @ArturoMagidin – Semiclassical Feb 19 '23 at 00:06
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    @Semiclassical Yup. Martin Brandenburg goofed there (I've left a comment to that effect). The other instance of "smallest" in that page refer to the smallest normal subgroup of the free group by which you quotient out to get the group being presented; since what is small here is what you mod out by, you get the largest possible result. – Arturo Magidin Feb 19 '23 at 00:14
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    @Semiclassical Actually, no, he's right, but he is defining it differently. Not the smallest group containing elements blah... but the smallest group that maps to every group containing elements satisfying blah. That is, the smallest group having the homomorphism property described in von Dyck's Theorem. "Smallest" there implies the uniqueness clause. But he is degining it in terms of what it maps to, not in terms of what it contains. – Arturo Magidin Feb 19 '23 at 00:31

1 Answers1

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It's not a convention. It's what the meaning of a presentation is.

When we give a presentation, $$\langle x_1,x_2,\ldots\mid R_1,R_2,\ldots\rangle$$ (where $R_i$ are the relations satisfied by the generators), we are describing "the most general group that is generated by elements $x_1,x_2,\ldots$ satisfying the relations $R_1$, $R_2,\ldots$". As such, when we write $$\langle a\mid a^n=e\rangle$$ we are saying "the most general group that is generated by an element $a$ subject to the condition $a^n=e$." The most general such group is the one in which the order is exactly $n$, and not any divisor of $n$.

This idea of "most general group" is captured by von Dyck's Theorem, and by the construction of a group given a presentation.

von Dyck's Theorem. If $G$ is the group given by the presentation $$\langle x_1,x_2,\ldots\mid R_1,R_2,\ldots\rangle$$ then given any group $H$ and any elements $h_1,h_2,\ldots\in H$ such that, replacing $x_i$ with $h_i$ in the relations $R_1,R_2,\ldots$ results in statements that are true in $H$, there exists a unique group homomorphism $\phi\colon G\to H$ such that $\phi(x_i)=h_i$ for $i=1,2,\ldots.$

That is: any group generated by elements satisfying the relations $R_1,R_2,\ldots$ must be a quotient of $G$.

This is true for the cyclic group of order $n$, but not for the cyclic group of order $k$ for $k\neq n$, $k\mid n$: because, for example, the cyclic group of order $10$ is generated by an element satisfying $x^{20}=1$, but so does the cyclic group of order $20$, and the latter is not a quotient of the former.

Construction. To construct the group given by the presentation given above, first we rewrite all relations so that they are in the form $w_i(x_1,x_2,\ldots)=1$; for example, if the first relation is $x_1x_2=x_2x_1$, then we rewrite it as $x_1x_2x_1^{-1}x_2^{-1}=1$. Then we take the free group $F$ on $x_1,x_2,\ldots$, and let $N$ be the smallest normal subgroup of $F$ that contains all the relations $w_1(x_1,x_2,\ldots), w_2(x_1,x_2,\ldots),\ldots$. Then the group we want is $G=F/N$.

Caveat. Note that just because a presentation has a relations of the form $x^n=e$, that by itself does not mean that the order of $x$ is $n$; it is possible that, when combined with other relations, the order of $x$ will be smaller. For example, as discussed here (and you may want to take a look at that answer anyway), the group $$G = \langle a,b\mid a^5 = b^4 = 1, aba^{-1}b=1\rangle$$ will actually yield a group in which the order of $b$ is two, not $4$. That's because the other relations, together with the relation $b^4=1$, imply that $b^2=1$ must also hold. So just because you see $x^n=1$ in the presentation, it does not mean that $x$ will definitely have order $n$.

In the case of the cyclic group, $\langle a\mid a^n=1\rangle$, there are no other relations to interact with $a^n=1$, so that one can show that you get the group of order $n$; but this is not a matter of "convention" or "understanding", but of what the presentation requires and means.

Arturo Magidin
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  • My def of "most general" would be that two words are equivalent iff their equality follows from the relations and the group axioms. If I had to say, I think both your def and mine are equivalent. – Vivaan Daga Sep 08 '23 at 17:00
  • @VivaanDaga First, I did not give a "definition", that's why the term "most general" is in quotation marks and why I invoke and use von Dyck's Theorem. Second, your definition is lacking/imprecise; it's not that the words are "equivalent", it's that they take the same values in the group in question (words are equivalent when they define the same element in the absolutely free group). The definition would be that it is (isomorphic to) the free group on the set of generators, modulo the smallest normal subgroup of that group generated by the relations. – Arturo Magidin Sep 08 '23 at 17:26
  • Right, I meant take the same values, if we use my definition is it equivalent to the def you gave above? – Vivaan Daga Sep 08 '23 at 17:47
  • @VivaanDaga: you would need to then provide an actual formal definition of "their equality follows from the relations and the group axioms"... which you will discover essentially amounts to "lies in the normal closure of the subgroup of the corresponding free group generated by the relators", so once you actually write out an actual definition, you will discover it is not merely equivalent, but in fact identical. – Arturo Magidin Sep 08 '23 at 18:04
  • I am not sure they are identical, although your definition actually proves existence. I would say two words equality follows from the group axioms and the relations if there is a formal proof starting from one word and ending in the other with the steps only using the relations or the group axioms. – Vivaan Daga Sep 08 '23 at 18:16
  • @VivaanDaga I would say you are still being vague. A "formal proof" is a finite sequence of sentences (well-formed formulas without free variables) in the language of the theory, in which every term is either a tautology, an axiom, or follows from previous terms via a valid rule of inference in the theory. What you are envisioning is no doubt more like a sequence of statements that say things like "this expression is equal to this expression by doing this, and that is equal to this expression by doing that" etc., which is not really a "formal proof". – Arturo Magidin Sep 08 '23 at 18:25
  • @VivaanDaga: In short, you are using words to give an appearance of full formalism, but you are not really being fully formal; which seems a bit silly. – Arturo Magidin Sep 08 '23 at 18:26
  • But surely the most direct way to formalise that type of reasoning does not go through quotient groups so we have a different definition? Or are you saying that it is not possible to directly formalise that sort of reasoning without going through quotient groups? – Vivaan Daga Sep 08 '23 at 18:30
  • @VivaanDaga; Again: if you want to formalize, then you are going to have to be formal. You are not being formal, and you are not formalizing anything. You are just saying words that include the adjective "formal" and pretending that this is a "formalization" and makes your definition a "formal" definition. That's not how things work. You can give an informal definition, but you cannot then pretend you gave a formal one. – Arturo Magidin Sep 08 '23 at 18:32
  • So basically my def serves as an intuition but the formalisation comes from the quotient group? – Vivaan Daga Sep 08 '23 at 21:18
  • @VivaanDaga There are many ways to give a formal definition of "group presented by generators and relations". You can define it as a specific object, or as an object with a specific universal property (in the former case you will then want to prove it has the universal property, i.e. von Dyck's theorem; in the latter, you will need to prove it exists, i.e. give an explicit construction). You did not define it. You gave a description of the idea that you hope to capture with an eventual definition/construction. Don't confuse anything you wrote with an actual definition. – Arturo Magidin Sep 08 '23 at 21:57
  • What about this: two words are equivalent if one can be obtained from another by adding or removing relators and words of the form $xx^{-1}$ – Vivaan Daga Sep 09 '23 at 12:33
  • @VivaanDaga How about we stop doing extended discussions in comment threads, which is not what they are for, especially when they have wandered far away from the text of the post? – Arturo Magidin Sep 09 '23 at 16:38
  • Sure, I just feel the def I gave above is a more intuitive one. – Vivaan Daga Sep 09 '23 at 16:58