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Feb 23, 2021 at 12:39 comment added YCor Yes, sorry, I mean "non-injective on $S$". (Alternatively instead of $S$, I might require the existence of $s_0\in S-\{1_P\}$ such that every homomorphism from $P$ to any finite group maps $s_0$ to $1$, and existence of a homomorphism $P\to G$ not mapping $s_0$ to $1_G$).
Feb 23, 2021 at 12:18 comment added frafour @YCor in your previous comment, do you mean "every homomorphism from $P$ to any finite group is non-injective on $S$"?
Feb 23, 2021 at 10:33 comment added YCor I don't know a reference (it's quite a restatement of the definition) but I can post a proof in an answer.
Feb 23, 2021 at 10:18 answer added markvs timeline score: 1
Feb 23, 2021 at 9:22 comment added frafour Thank you for the comment @YCor ! Could you give a reference for the last statement?
Feb 23, 2021 at 9:15 comment added YCor Anyway, these non-residually finite f.p. subgroups are artifacts. What's needed is the following: a group $G$ is non-LEF iff there is a finitely presented group $P$, a finite subset $S\subset P$ such that every homomorphism from $P$ to any finite group is non-injective, and a homomorphism $P\to G$ that is injective on $S$.
Feb 23, 2021 at 9:11 comment added YCor Each group with a non-LEF subgroup is non-LEF, which makes immediate to produce infinitely presented ones (just take direct product of lamplighter with a finitely presented non-RF group). Examples with no LEF quotient (you probably mean no nontrivial LEF quotient): there's a result saying that every f.g. group embeds in a simple f.g. group, and the proof always produces infinitely presented groups. The last question is too vague to have a definite answer.
Feb 23, 2021 at 8:56 history asked frafour CC BY-SA 4.0