Eventual centrality and Hopficity of direct products of finitely generated soluble groups

Problem

Is the direct product of two finitely generated soluble Hopfian groups Hopfian?

A,B\text{ finitely generated, soluble, Hopfian} \quad\stackrel{?}{\Longrightarrow}\quad A\times B\text{ Hopfian}.

A group $G$ is Hopfian if every surjective endomorphism $G\to G$ is injective.

The product theorem

A group $G$ is Hopfian if every surjective endomorphism of $G$ is injective. We write $G'=[G,G]$ , $Z(G)$ for the center, and $G_{\mathrm{ab}}=G/G'$ . Composition is read from right to left.

Theorem 1. If $A$ and $B$ are finitely generated soluble Hopfian groups, then $A\times B$ is Hopfian.

Theorem 1 answers Problem 5.38 of the Kourovka Notebook (1). The diagonal components of an epimorphism of a direct product need not be surjective. Earlier work on Hopficity of direct products includes Hirshon (2). A recent factorisation theorem of Deré and Vandermeersch (3) concerns monomorphisms of virtually soluble minimax groups under an indecomposability condition on their algebraic hulls. Here the factors may have arbitrary soluble length and need not be minimax or directly indecomposable.

For $f\colon A\times B\to A\times B$ , write

\tag{1} f(a,b)=\bigl(f_{11}(a)f_{12}(b),\ f_{21}(a)f_{22}(b)\bigr).

The two images in each coordinate commute elementwise.

Theorem 2. Let $A,B$ be finitely generated soluble groups and let $f$ be a surjective endomorphism of $A\times B$ . There is an integer $m>0$ such that $F=f^m$ satisfies

\tag{2} F_{12}(B)\le Z(A),\qquad F_{21}(A)\le Z(B).

Proposition 9 proves injectivity under (2) for finitely generated Hopfian factors. Its proof uses homomorphisms $H_{\mathrm{ab}}\to Z(H)$ to correct the diagonal endomorphisms.

A bound on nonabelian component paths

For a finitely generated abelian group $M$ , put

b(M)=\operatorname{rank}_{\mathbb Z}M+ |\operatorname{Tor}(M)|-1.

Then $b(M)>0$ for $M\ne0$ , and

\tag{4} b(M\oplus N)\ge b(M)+b(N).

Indeed, ranks add, torsion orders multiply, and $(s-1)(t-1)\ge0$ for positive integers $s,t$ . For a finitely generated group $H$ , define

w(H)=b\bigl((H/Z(H))_{\mathrm{ab}}\bigr).

If $H$ is soluble and nonabelian, then $w(H)>0$ : a nontrivial soluble group cannot have trivial abelianization, since a perfect soluble group is trivial.

Lemma 3. Suppose $H=PQ$ , where $P,Q$ are commuting subgroups. Then

H/Z(H)\cong P/Z(P)\times Q/Z(Q).

If $P,Q$ are finitely generated, then $w(P)+w(Q)\le w(H)$ .

Proof. Since $P$ commutes with $Q$ and they generate $H$ , $Z(P)=P\cap Z(H)$ , and similarly for $Q$ . Multiplication induces a surjection from the displayed direct product to $H/Z(H)$ . If $pq$ is central, commuting it with $P$ shows $p\in Z(P)$ , and commuting it with $Q$ shows $q\in Z(Q)$ . The map is therefore injective. Abelianization commutes with direct products, so (4) gives the inequality. ◻

Fix a surjective $f\colon G\to G$ , where $G=A\times B$ . Let $e_0(a,b)=(a,1)$ and $e_1(a,b)=(1,b)$ . For a finite binary word $v=i_0\cdots i_{n-1}$ , set

\tag{7} p_v=e_{i_0}f\cdots e_{i_{n-1}}f,\qquad P_v=p_v(G),\qquad p_{\varnothing}=1_G.

Here $1_G$ denotes the identity map. At every fixed length $n$ , the subgroups $P_v$ commute pairwise, and

\tag{8} f^n(x)=\prod_{|v|=n}p_v(x),\qquad G=\prod_{|v|=n}P_v.

To see commutativity, take two distinct words and remove their common prefix. Their next projections have commuting images, and a homomorphism preserves commutation. The first identity follows by repeatedly using $x=e_0(x)e_1(x)$ ; surjectivity of $f^n$ gives the second. Also

\tag{9} P_v=P_{v0}P_{v1},\qquad [P_{v0},P_{v1}]=1.

Surjectivity of the final $f$ in each child map is used here.

Lemma 4. Assume $A,B$ are finitely generated and soluble. There is $m>0$ such that every word $v$ of length greater than $2m$ with $P_v$ nonabelian satisfies $v_m=v_{2m}$ , with indices starting at zero.

Proof. Every $P_v$ is a finitely generated soluble group. Lemma 3 and (9) give

w(P_{v0})+w(P_{v1})\le w(P_v),\qquad \sum_{|v|=n}w(P_v)\le w(G).

Thus at most $w(G)$ words at each length have nonabelian image. Every prefix and every suffix of such a word also has nonabelian image: a prefix has an image containing $P_v$ , and $P_v$ is a homomorphic image of the image of a suffix. Moreover, a word with nonabelian image has at least one child with nonabelian image, by (9).

Let $X$ be the set of infinite binary sequences all of whose finite prefixes have nonabelian image. Each finite word with nonabelian image extends to an element of $X$ , by successive choices of such a child. The set $X$ has at most $w(G)$ elements. Otherwise, $w(G)+1$ distinct sequences would have distinct prefixes at some common length, contradicting the bound above.

Deleting the first letter defines a map $\sigma\colon X\to X$ , by the suffix property. Choose $m>0$ at least as large as every preperiod length of $\sigma$ and divisible by every cycle length. Then $\sigma^{2m}=\sigma^m$ . Hence $x_m=x_{2m}$ for every $x\in X$ . Extend the given finite word to $X$ to obtain the result. ◻

Lemma 5. Under the hypotheses of Lemma 4, some $m>0$ satisfies

\tag{11} f^m e_j f^m e_k(G)\le Z(G)\qquad(j,k\in\{0,1\},\ j\ne k).

Proof. If $P_v$ is abelian, it is central in $G$ : it commutes with itself and with the other factors in (8). Choose $m$ as in Lemma 4. Expanding both copies of $f^m$ in (11) gives factors of the form

p_u e_j p_v e_k(x),\qquad |u|=|v|=m.

If the first letter of $v$ differs from $j$ , this factor is $1$ . Otherwise $e_jp_v=p_v$ , and the image of this term equals the image of $p_{uvk}$ : appending the last $f$ does not change the image. The word $uvk$ has letters $j,k$ at positions $m,2m$ respectively, so its image is abelian by Lemma 4. Every factor in the expansion is therefore central. ◻

From mixed components to central components

Lemma 6. Let $H$ be finitely generated, $q\colon H\to H$ surjective, and $i\colon T\to H$ , $r\colon H\to T$ homomorphisms with $ri=1_T$ . If $T$ is soluble and $q\circ i$ is the trivial homomorphism, then $T=1$ .

Proof. The surjection induced by $q$ on the finitely generated abelian group $H_{\mathrm{ab}}$ is injective. Thus $\ker q\le H'$ . Consequently $i(T)\le H'$ and

T=ri(T)\le r(H')\le T'.

The soluble group $T$ is perfect and hence trivial. ◻

Proof of Theorem 2. Choose $m$ from Lemma 5 and set $F=f^m$ . Put $P=F_{11}(A)$ and $Q=F_{12}(B)$ . The subgroups $P,Q$ commute and $A=PQ$ . Lemma 3 gives

\tag{14} G/Z(G)\cong \bigl(P/Z(P)\times Q/Z(Q)\bigr)\times B/Z(B).

In particular $T=Q/Z(Q)$ is a soluble retract of $G/Z(G)$ , embedded by $qZ(Q)\mapsto(q,1)Z(G)$ .

Every surjective endomorphism preserves the center, so $F$ induces a surjection $\overline F$ on $G/Z(G)$ . Equation (11) says that $\overline F$ kills this copy of $T$ . Lemma 6 implies $T=1$ . Hence $Q$ is abelian; since it commutes with $P$ , $Q\le Z(A)$ . Interchanging $A,B$ proves $F_{21}(A)\le Z(B)$ . ◻

Integral corrections through the center

Abelian groups in this section are written additively.

Lemma 7. Let $M$ be a finitely generated abelian group, $Z$ an abelian group, and $c\colon Z\to M$ a homomorphism. There is an integer $d\ne0$ such that, for every endomorphism $u$ of $M$ satisfying

\tag{15} u(cZ)\subseteq cZ,\qquad M=uM+cZ,\qquad (u-1)M\subseteq dM,

there is $h\colon M\to Z$ for which $u+ch$ is surjective.

Proof. Let $T=\operatorname{Tor}(M)$ , let $E=M/T$ , and write $q\colon M\to E$ . The group $E$ is free of finite rank. Put

D=qc(Z),\qquad S=\{x\in E: nx\in D\text{ for some }n\in\mathbb Z\setminus\{0\}\}.

The quotient $E/S$ is torsion-free and finitely generated, hence free; therefore $S$ is a direct summand of $E$ . Choose a retraction $p\colon E\to S$ . The finitely generated torsion group $S/D$ is finite. Choose $d\ne0$ annihilating both $S/D$ and $T$ . In particular $dS\subseteq D$ .

Let $\bar u$ be the map induced by $u$ on $E$ . It preserves $D$ and $S$ . For $x\in S$ , congruence in (15) and saturation give

x-\bar u(x)\in dE\cap S=dS\subseteq D.

Thus $\delta=(1-\bar u)|_S$ is a homomorphism $S\to D$ . Since $S$ is free, $\delta$ lifts through the surjection $qc\colon Z\to D$ : there is $k\colon S\to Z$ with $qck=\delta$ . Define

\bar h=kp\colon E\to Z,\qquad \bar v=\bar u+qc\bar h.

Then $\bar v|_S=1_S$ and $(\bar v-\bar u)E\subseteq S$ . Also $E=\bar u(E)+D$ , so $\bar v$ is onto $E/S$ . These two facts make $\bar v$ surjective: for $y\in E$ , choose $x$ with $y-\bar v(x)\in S$ and observe $\bar v(x+y-\bar v(x))=y$ .

Set $h=\bar hq$ and $v=u+ch$ . Then $qv=\bar vq$ and $h(T)=0$ . For $t\in T$ , write $u(t)-t=dy$ . The left side is torsion, so $y$ is torsion, since $d\ne0$ and $M/T$ is torsion-free. As $dT=0$ , we obtain $u(t)=t$ , and hence $v(t)=t$ . Surjectivity on $M/T$ and the identity on $T$ now give surjectivity of $v$ on $M$ . ◻

Proposition 8. For every finitely generated group $H$ there is $d\ne0$ with the following property. Suppose $u\colon H\to H$ is an endomorphism, $u(Z(H))\le Z(H)$ , and

\tag{19} H=u(H)Z(H),\qquad u(H')=H',\qquad (u_{\mathrm{ab}}-1)H_{\mathrm{ab}}\subseteq dH_{\mathrm{ab}}.

There is a homomorphism $c_u\colon H\to Z(H)$ such that $x\mapsto u(x)c_u(x)$ is surjective and agrees with $u$ on $H'$ . If $H$ is Hopfian, $u|_{H'}$ is injective.

Proof. Apply Lemma 7 to $M=H_{\mathrm{ab}}$ and the natural map $c\colon Z(H)\to M$ . The resulting $h\colon M\to Z(H)$ defines $c_u=h\pi$ , where $\pi\colon H\to M$ is abelianization. Centrality makes $v(x)=u(x)c_u(x)$ a homomorphism. It induces the surjective map $u_{\mathrm{ab}}+ch$ on $M$ , and $v|_{H'}=u|_{H'}$ is surjective onto $H'$ . To lift $y\in H$ , choose $x$ with $yv(x)^{-1}\in H'$ and then $z\in H'$ with $v(z)=yv(x)^{-1}$ ; thus $v(zx)=y$ . If $H$ is Hopfian, $v$ is injective, so its restriction to $H'$ is injective. ◻

Hopficity with central off-diagonal components

Proposition 9. Let $A,B$ be finitely generated Hopfian groups. Every surjective endomorphism $F$ of $A\times B$ satisfying (2) is injective.

Proof. Choose the moduli $d_A,d_B$ from Proposition 8. The endomorphism induced by $F$ on the finite abelian group

(A\times B)_{\mathrm{ab}}/ d_A d_B(A\times B)_{\mathrm{ab}}

is surjective, hence a permutation. A positive power is the identity. For the same $n>0$ , put $U=F^n$ . Its diagonal maps satisfy

\tag{21} (U_{11,\mathrm{ab}}-1)A_{\mathrm{ab}}\subseteq d_AA_{\mathrm{ab}}, \qquad (U_{22,\mathrm{ab}}-1)B_{\mathrm{ab}}\subseteq d_BB_{\mathrm{ab}}.

Indeed, project the congruence for $U$ after inserting either factor into the product.

Each diagonal component of $F$ preserves the center of its factor. Indeed, $F_{11}(Z(A))$ commutes with $F_{11}(A)$ and $F_{12}(B)$ , whose product is $A$ ; the argument for $F_{22}$ is identical. If an endomorphism $V$ has central off-diagonal components, then

(FV)_{12}(b)=F_{11}(V_{12}(b))F_{12}(V_{22}(b))\in Z(A) \qquad(b\in B).

The corresponding formula gives $(FV)_{21}(A)\le Z(B)$ . Induction therefore proves (2) for $U=F^n$ .

Surjectivity and centrality give

A=U_{11}(A)Z(A),\qquad B=U_{22}(B)Z(B).

Taking commutators yields $U_{11}(A')=A'$ and $U_{22}(B')=B'$ . Thus Proposition 8, with (21), makes $U_{11}|_{A'}$ and $U_{22}|_{B'}$ injective.

The map induced by $U$ on the finitely generated abelianization is surjective and hence injective. Therefore

\ker U\le (A\times B)'=A'\times B'.

The central maps $U_{12},U_{21}$ vanish on $B',A'$ respectively. If $(a,b)\in\ker U$ , then $a\in A'$ , $b\in B'$ and $U_{11}(a)=U_{22}(b)=1$ . Diagonal injectivity gives $a=b=1$ . Hence $U=F^n$ , and therefore $F$ , is injective. ◻

Proof of Theorem 1. Let $f$ be a surjective endomorphism of $A\times B$ . Theorem 2 gives a positive iterate $f^m$ satisfying (2). Proposition 9 makes $f^m$ injective, which implies that $f$ is injective. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st ed., 2026, Problem 5.38. arXiv:1401.0300.

R. Hirshon, On Hopfian groups, Pacific J. Math. 32 (1970), 753–766. doi:10.2140/pjm.1970.32.753.

J. Deré and K. Vandermeersch, Automorphisms and monomorphisms of direct products of virtually solvable minimax groups, Transform. Groups (2026). doi:10.1007/s00031-026-09994-8.