Let $\Gamma$ be a group, say finitely generated if it helps. Does $\Gamma$ admit a largest Hopfian quotient? That is, does there exist a Hopfian quotient $H$ of $\Gamma$, such that every surjective homomorphism from $\Gamma$ onto a Hopfian group factors through $H$?

I first thought to define $H = \Gamma / K$, where $K$ is the intersection of all kernels of surjective endomorphisms of $\Gamma$, but I cannot see why this should be Hopfian, or why it should have the factorization property.

Then I thought to define $H = \Gamma/K$, where $K$ is the intersection of all kernels of surjective homomorphism from $\Gamma$ onto a Hopfian group, and this clearly has the factorization property but I don't see why it should be Hopfian.

Does such a quotient always exist? Does it exist under some additional conditions? Do you have some explicit example where it exists, and it is not the same as the largest residually finite quotient? In particular I would like to know this for the Baumslag-Solitar group $BS(2, 3)$.

  • 3
    $\begingroup$ Probably not; I'd rather guess that $BS(2,3)$ is residually Hopfian. Rough argument: there's an exact sequence $1\to F\to BS(2,3)\to H\to 1$, where $F$ is free of infinite rank, and $H=\mathbf{Z}[1/6]_{2/3}\mathbf{Z}$ is metabelian. Let $(F^n)$ be the lower central series of $F$. I would find rather likely that for every $n$ there exists a normal subgroup $N$ of $BS(2,3)$, contained in $F^n$, such that $BS(2,3)/N$ is Hopfian. I agree it's speculative. $\endgroup$
    – YCor
    Mar 19, 2021 at 10:58
  • 5
    $\begingroup$ For arbitrary groups there's a trivial counterexample: an infinite $p$-elementary abelian group, for given prime $p$ (indeed its Hopfian quotients are precisely its finite quotients, and the intersection of finite index subgroups is then trivila). $\endgroup$
    – YCor
    Mar 19, 2021 at 11:00

1 Answer 1


The answer is no even for finitely generated groups. Here's a construction of a finitely generated residually Hopfian, non-Hopfian group. It is even solvable (actually center-by-metabelian).

Denote by $M(u,v,x,y,z)$ the matrix $$\begin{pmatrix}u & x & z\\ 0 & v & y \\ 0 & 0 & 1\end{pmatrix}.$$

Fix a prime $p$; let $G$ be the group of matrices $M(0,t^n,x,y,z)$ for $(n,x,y,z)\in\mathbf{Z}\times\mathbf{F}_p[t,t^{-1}]^3$. Note that $G$ is finitely generated (by $\{M(0,1,0,0,0),M(0,0,1,0,0),M(0,0,0,1,0)\}$).

Let $Z$ be the central subgroup of $G$ of those matrices of the form $M(0,0,0,0,z)$ for $z\in\mathbf{F}_p[t]$. Let $\alpha$ be the matrix $M(t,0,0,0,0)$ (i.e., diagonal $(t,1,1)$). Then $\alpha Z \alpha^{-1}$ is contained in $Z$ with index $p$. In particular, $\alpha$ induces a surjective, non-injective endomorphism of $G/Z$, so $G/Z$ is non-Hopfian.

Proposition. $G/Z$ is residually Hopfian.

I will indeed prove that it has two normal subgroup with trivial intersection, each quotient being Hopfian.

Let us "classify" quotients of $G$. Let $Z'$ be its center (the subgroup of those $M(0,0,0,0,z)$ for $z\in\mathbf{F}_p[t,t^{-1}]$.

Claim 1: every normal subgroup $N$ of $G$ is either contained in the center $Z'$, or contains some finite index subgroup of $Z'$.

Proof: suppose that $N$ contains an element not in $Z'$. Write $N'$ for the set of elements in $N$ of the form $M(0,0,x,y,z)$; then $N'$ is also a normal subgroup of $G$. Performing a commutator if necessary, we see that $N'$ is not contained in $G$: it contains an element of the form $M(0,0,x,y;z)$ with $(x,y)\neq (0,0)$; by symmetry we can suppose $x\neq 0$. Write $p_1,p_2,p_3$ for the projections $M(0,0,x,y;z)\mapsto x,\mapsto y,\mapsto z$. Since $N'$ is a normal subgroup, $p_1(N')$ is a nonzero $\mathbf{F}_p[t,t^{-1}]$ submodule of $\mathbf{F}_p[t,t^{-1}]$, that is, an ideal. Taking commutators with $M(0,0,0,1,0)$, we deduce that $p_3(N')$ also contains this finite index ideal. So $N$ contains some finite index subgroup of $Z'$.$\Box$

Claim 2: every finite-by-metabelian f.g. group $G$ is Hopfian.

Proof: every such group satisfies the max-n property (max condition on chains of normal subgroups), an old result of Ph. Hall, and max-n immediately implies Hopfian (for a surjective non-injective endomorphism, the sequence of kernels of iterates is strictly increasing). $\Box$

If $N$ is a normal subgroup of $G$ above, if $N$ contains a finite index subgroup of $Z'$, then $G/N$ is finite-by-metabelian, hence Hopfian by Claim 2.

Otherwise, $N$ is contained in $Z'$, i.e., is central. We now assume so. Since the metabelian group $G/Z'$ has trivial center [this has index 2 in a lamplighter group $C_p\wr\mathbf{Z}$, and is a lamplighter $C_p^2\wr\mathbf{Z}$], the center of $G/N$ is reduced to $Z'/N$. Let $f$ be a surjective endomorphism of $G/N$. By surjectivity, $f$ maps center into the center, and hence induces an endomorphism of the metabelian quotient $G/Z'$, which is Hopfian. Hence the inverse image of the center equals the center, and $f$ has kernel contained in the center $Z'/N$.

Up to replace $f$ with its square, we can suppose that the action of $f$ on $\mathbf{Z}$ (which appears modding out by the set of torsion elements, which is a subgroup) is trivial. Hence the action of $f$ on "$(x,y)$" (formally speaking, its action on $\mathbf{F}_p[t,t^{-1}]^2$ obtained modding out the center and restricting to the torsion subgroup) is by $\mathbf{F}_p[t,t^{-1}]$-module automorphism: it is given by some matrix $q\in\mathrm{GL}_2(\mathbf{F}_p[t,t^{-1}])$. Passing to some proper power if necessary, we can suppose that the determinant $\det(q)$ equals $t^m$ for some $m\in\mathbf{Z}$.

Shaking hands, the action of $q$ on the center $Z'$ (this doesn't make sense!) should be given by multiplication the determinant $t^m=\det(q)\in\mathbf{F}_p[t,t^{-1}]^\times$ and this should restrict the possibilities for $N$.

More precisely, for $s\in\mathbf{F}_p[t,t^{-1}]$, write $X_s=M(0,0,s,0,0)$, $Y=M(0,0,0,1,0)$, so $X_sYX_s^{-1}Y^{-1}=M(0,0,0,0,s)=Z_s$. Let $X'_s$, $Y'$ and $Z'_s$ be their images in $G/N$. So $f(Z'_s)=Z'_{F(s)}$ for some $F(s)$ which is well-defined up to addition by an element of $I=p_3(N)$.

Then computing $f$ on this commutator shows that $\det(q)s-F(s)\in I$ for every $s$. Prescribing $F$ on the canonical basis, we can suppose that $F$ is an additive homomorphism. Let $K$ be the kernel of $f$ and $J=p_3(K)$. Then $I$ is strictly contained in $J$, and $F^{-1}(I)=J$ (essentially by definition).

If $m=0$ by contradiction, we get that $F$ is identity modulo $I$, which contradicts $f(J)\subset I$.

Now suppose that $I$ is has basis $(t^n)_{n\in A}$ for some negatively unbounded subset $A\subset\mathbf{Z}$ containing $\mathbf{N}$ which in the negative is "more and more sparse" (in the sense that $A=\mathbf{N}\cup\{a_k:k\in\mathbf{N}\}$ with $a_k-a_{k-1}\to - \infty$). We combine the conditions $F(I)\subset I$ and $t^ms-F(s)\in I$, to get $t^{m+n}\in I$ for all $n\in A$, which contradicts the sparse condition.

Hence if $N=N_I$ is the group just chosen, then $G/N_I$ is Hopfian. Taking $I,I'$ sparse and with intersection reduced to $\mathbf{N}$, we get $N_I\cap N_{I'}=Z$. So $G/Z$ is residually Hopfian.

  • $\begingroup$ Thank you so much for such a detailed answer! $\endgroup$
    – frafour
    Mar 19, 2021 at 15:59
  • $\begingroup$ Nice! It would still be great to know the answer for $BS(2,3)$ (or any finitely presented group). $\endgroup$
    – HJRW
    Mar 20, 2021 at 9:09
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    $\begingroup$ @HJRW I think I can produce a f.p. solvable group answering this (also with the same idea: non-Hopfian, while the are two central subgroups with trivial intersection, each quotient being Hopfian). But writing details would take at least as much space. $\endgroup$
    – YCor
    Mar 20, 2021 at 9:20
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    $\begingroup$ Every f.g. torsion-free acylindrically hyperbolic group is residually simple, hence residually Hopfian. In particular, $BS(2,3)*BS(2,3)$ or $BS(2,3)*\mathbb{Z}$ are residually Hopfian but non-Hopfian. $\endgroup$ Mar 29, 2021 at 17:06

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