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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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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.] –  user641 Oct 8 '13 at 6:23
    
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. –  Derek Holt Oct 8 '13 at 7:59
    
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. –  user1729 Oct 8 '13 at 9:41
    
(Sorry, an arbitrary balanced presentation of the trivial group. It is an east exercise to prove that if $\mathcal{P}=\langle x_1,\ldots,x_n;R_1\ldots,R_n\rangle$ 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 then the presentation $\mathcal{P}$ must define the trivial group.) –  user1729 Oct 8 '13 at 9:48
    
@SteveD I was not aware of that result, where could I find a reference/proof of that? –  Paul Plummer Oct 8 '13 at 12:30

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