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These questions are sort of preliminary questions and reference requests for a project I am doing.

Lets say, for concreteness, that $R$ is a set of words in the free group of rank two and that $\langle a,b \mid R \rangle$ is the trivial group.

How "bad" can $R$ be?

I guess my ideal "bad" is that $R$ is infinite and if $\varnothing \neq T \subseteq R $, then $\langle a,b \mid R \setminus T \rangle$ is not the trivial group. Also how would one go about finding/constructing these "bad" $R$, maybe for finitely generated group in general.

I guess a more general question: Let $R_{\text {fam}}$ be a countable family of disjoint sets of words in the free group generated by the set $S$ such that $\langle S \mid \cup R_{\text{fam}} \rangle $ is the trivial group;

How "bad" can $R_{\text{fam}}$ be?

The "bad" here is essentially the same except looking at $\cup (R_{\text{fam}}\setminus T)$ where $T$ is some non empty subset of $R_\text{fam}$.

I am mostly looking for an answer to the specific example and references for these sorts of questions. There are plenty of variations, maybe looking at finite $\langle S|R \rangle$ and looking at how bad that $R$ can get, also looking at "preloaded" $R$, that is $R$ has to have certain relations.

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    $\begingroup$ Since the trivial group is finitely presented, any presentation with finitely many generators, and infinitely many relators, can be "pruned" down to a finite presentation (that is, by removing a cofinite subset of the relators). So no such R exists. [I am addressing the second paragraph.] $\endgroup$ – user641 Oct 8 '13 at 6:23
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    $\begingroup$ I think you need to make it clearer what you understand by $R$ being "bad". Since the problem of deciding triviality of $\langle X \mid R \rangle$ for finite $X$ and $R$ is known to be undecidable, it follows that the presentation can be bad in the sense that the difficult of proving that the group is trivial (i.e. the length of such a proof) is not a recursive function of the total length of the presentation. $\endgroup$ – Derek Holt Oct 8 '13 at 7:59
  • $\begingroup$ You might be interested in the Andrews-Curtis conjecture. This conjectures that an arbitrary balanced presentation $\langle x_1,\ldots,x_n;R_1\ldots,R_n\rangle$ can be transformed into the presentation $\langle x_1,\ldots,x_n;x_1,\ldots,x_n\rangle$ by a finite sequence of elementary Nielsen transformations (that is, the list $(R_1,\ldots,R_n)$ is in the automorphic orbit of $(x_1,\ldots,x_n)$ in $F_n$). This is a major open problem, and I understand it is widely believed to be false. My point is: if it were true then every balanced presentation of the trivial group would be really rather nice. $\endgroup$ – user1729 Oct 8 '13 at 9:41
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    $\begingroup$ Here's the idea: if you have a presentation $\langle a,b\mid R\rangle$, and it presents the trivial group, then there must be a way to deduce, for example, $a=1$ from the relators in $R$. This just means there is a finite set of relators in $R$ (possibly after conjugation) that multiply to $a$. The same statement holds for $b$. Then the union of these two finite sets is a finite subset of $R$ which is a complete set of relators for the trivial group. $\endgroup$ – user641 Oct 8 '13 at 18:18
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    $\begingroup$ I feel someone should mention "Markov properties" here: A Markov property is a property $P$ of groups which is preserved under group isomorphism, is possessed by some finitely presented group, and there exists some group $G$ which cannot be embedded as a subgroup of some group with property $P$. Clearly, "bring trivial" is a Markov property. Adian and Rabin (independently) proved that if $P$ is a Markov property then you cannot determine if a group presentation defines a group with property $P$. Hence, it is impossible to determine if a group presentation defines the trivial group. $\endgroup$ – user1729 Dec 3 '18 at 13:02
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Expanding on the comments, there is a very strong sense in which every "infinitary" description of the trivial group can be boiled down to a "finitary" one - namely, the compactness theorem for first-order logic. This asserts roughly that any set of axioms which proves some single sentence has a finite subset which proves that sentence.

Finitely generated group presentation fall into this setting: the sentence we're interested in is $$(a_1=e)\,\,\wedge\,\, ...\,\,\wedge\,\, (a_n=e)$$ (where "$\wedge$" is "and" and the $a_i$s are the generators of our group), and our set of axioms consists of the axioms of group theory together with axioms corresponding to each relation. By the compactness theorem, any set of relations which makes the finitely many generators each trivial has a finite subset which does the same thing.

Note that this argument breaks down for infinitely-generated groups - and indeed the claim itself is false, the simplest counterexample having infinitely many generators, each of which is trivialized directly - the point being that to say that each of infinitely many generators is trivial is no longer first-order, since we need an infinite conjunction.

The compactness theorem similarly applies to your more general $R_{fam}$ situation: some finitely many sets in $R_{fam}$ must enforce triviality, as long as $S$ is infinite.

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  • $\begingroup$ Thanks for providing an answer, and giving a nice general way to view it. It is sort of funny that a logician answered it because this question came out of trying to find a little project for a set theory course I took years ago (selection principles in topology). $\endgroup$ – Paul Plummer Dec 3 '18 at 2:23

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