Is there a finitely presented sofic group which is not residually finite, but all of its finitely generated subgroups are Hopf groups?

It seems like the Baumslag Solitar groups $BS(m,n)$ don't work (i.e. for $|m|=1$ or $|n|=1$ or $|m|=|n|$ they are residually finite, and otherwise they contain a non-Hopf finitely generated subgroup).

Note: Thanks to YCor for pointing out that in the initial formulation I said "subgroups" instead of "finitely generated subgroups". The latter is my intention.

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    $\begingroup$ What's a non-Hopfian subgroup in $BS(2,4)$? recall that for $|m|,|n|>1$, $BS(m,n)$ is Hopfian iff $m,n$ have the same prime divisors, but is residually finite only when $|m|=|n|$. In particular $BS(2,4)$ is Hopfian and non-RF, but I'm not sure whether all its subgroups are Hopfian. $\endgroup$ – YCor Oct 1 '16 at 14:51
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    $\begingroup$ Also you should clarify if you really mean "all its subgroups are Hopf groups". Because whenever a group admits a free subgroup, then it admits a free subgroup of infinite rank, which is clearly not Hopfian. Maybe you mean, "all finitely generated subgroups are Hopfian"? $\endgroup$ – YCor Oct 1 '16 at 14:57
  • $\begingroup$ @YCor: Yes. I meant finitely generated subgroups. Thanks for the correction! As for non-Hopfian subgroup of $BS(2,4)$, the subgroup $\langle a,b^{-1}ab,b^2\rangle$ works by Meskin's "Nonresidually finite one-relator groups" / TAMS 1972, Vol. 164. (see Lemma 2.2 and the reference within). $\endgroup$ – Mike Brezkin Oct 1 '16 at 15:58

Houghton's group $H_3$ (see Section 5.3 here for a definition) is (locally finite)-by-$\mathbf{Z}^2$, which easiy implies that all its finitely generated subgroups are Hopfian. (Not all its subgroups are Hopfian: it admits an isomorphic copy of $F^{(\mathbf{N})}$ as a subgroup for every finite group $F$.) It is finitely presented (K. Brown 1987, reference at the above link). It is not residually finite because it has the finitary symmetric group $S_\infty$ as a subgroup. It is amenable hence sofic.


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