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A group is LEF (locally embeddable in the class of finite groups) if it embeds into an ultraproduct of finite groups. Residually finite groups are LEF and finitely presented LEF groups are residually finite. So if a finitely presented group is not residually finite, then it is not LEF either.

Most of the literature seems to be concerned with positive results. All non-examples I could find use the criterion above, in particular they are all finitely presented. What are some examples of finitely generated, infinitely presented groups that are not LEF? Among these, are there examples that have no LEF quotient? And if not, can the largest LEF quotient be determined explicitly?

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    $\begingroup$ 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. $\endgroup$ – YCor Feb 23 at 9:11
  • $\begingroup$ 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 $S$. $\endgroup$ – YCor Feb 23 at 9:15
  • $\begingroup$ Thank you for the comment @YCor ! Could you give a reference for the last statement? $\endgroup$ – frafour Feb 23 at 9:22
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    $\begingroup$ I don't know a reference (it's quite a restatement of the definition) but I can post a proof in an answer. $\endgroup$ – YCor Feb 23 at 10:33
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    $\begingroup$ 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 $1_G$). $\endgroup$ – YCor Feb 23 at 12:39
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Any finitely presented infinite simple groups is not LEF. For example, Thompson groups $T, V$. Take a free product of such a group with an infinitely presented group. This will be an infinitely presented non-LEF group.

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