Divisible subgroups as exact equalizers in torsion-free nilpotent groups

Problem

If $H$ is a divisible subgroup of a torsion-free nilpotent group $G$ of class at most $c$ , is its dominion in that class equal to $H$ ?

\operatorname{dom}^{\mathcal N_c^0}_G(H)\stackrel{?}{=}H.

The dominion contains the elements on which every pair of homomorphisms into a group of $\mathcal N_c^0$ agree whenever they agree on $H$ . Neither finite generation nor normality of $H$ is assumed.

Dominions and exact equalizers

Let $\mathcal N_c^0$ be the class of torsion-free nilpotent groups of class at most $c$ , where $c\geq1$ . We use $\gamma_1(G)=G$ and $\gamma_{r+1}(G)=[G,\gamma_r(G)]$ , so membership in $\mathcal N_c^0$ means torsion-freeness and $\gamma_{c+1}(G)=1$ . For $H\leq G\in\mathcal N_c^0$ , its dominion is

\tag{1} \begin{aligned} \mathop{\mathrm{dom}}_G^{\mathcal N_c^0}(H)=\{g\in G:\ &f_0(g)=f_1(g)\text{ whenever }K\in\mathcal N_c^0,\\ &f_0,f_1:G\to K\text{ are homomorphisms and }f_0|_H=f_1|_H\}. \end{aligned}

Kourovka Problem 17.34 asks whether this set equals $H$ when $H$ is divisible (1). We prove the following stronger separation statement.

Theorem 1. Let $G\in\mathcal N_c^0$ and let $H\leq G$ be divisible. There exist $K\in\mathcal N_c^0$ and injective homomorphisms $f_0,f_1:G\to K$ such that

\tag{2} f_0(g)=f_1(g)\quad\Longleftrightarrow\quad g\in H.

In particular, $\mathop{\mathrm{dom}}_G^{\mathcal N_c^0}(H)=H$ .

The construction permits $G$ and $H$ to have arbitrary cardinality. A divisible subgroup need not be normal, so a quotient of $G$ by $H$ cannot in general serve as the separating target.

A filtered derivation

Let $L$ be a Lie algebra over $\mathbb Q$ with $\gamma_{c+1}(L)=0$ , and let $S\leq L$ be a Lie subalgebra. Put

U=U(L),\qquad I=\ker(U\xrightarrow{\varepsilon}\mathbb Q),\qquad J=US.

Here $J$ is the left ideal generated by $S$ . The quotient

\tag{4} M=U/(J+I^{c+1})

is a left $U$ -module, and hence an $L$ -module. Write $V$ for the image of $I$ in $M$ , and $V_r$ for the image of $I^r$ , for $r\geq1$ . Thus $V=V_1$ and $V_{c+1}=0$ . Let

\tag{5} \delta:L\longrightarrow V,\qquad \delta(x)=x+(J+I^{c+1}).

Lemma 2. One has

L\cap(J+I^{c+1})=S.

Consequently $\ker\delta=S$ .

Proof. Choose a basis of $S$ adapted to the finite filtration $S\cap\gamma_r(L)$ , and extend it to a basis $\mathcal B$ of $L$ adapted to $\gamma_r(L)$ . For $b\in\mathcal B$ , let $w(b)$ be the largest $r$ such that $b\in\gamma_r(L)$ . Every weight lies in $\{1,\ldots,c\}$ . Order the complementary basis elements before all the basis elements of $S$ , and otherwise choose any total order.

By the Poincaré–Birkhoff–Witt theorem, the ordered monomials in $\mathcal B$ , together with $1$ , form a basis of $U$ . Give a monomial $b_1\cdots b_t$ weight $w(b_1)+\cdots+w(b_t)$ . We claim that

\tag{7} I^r=\mathop{\mathrm{span}}_{\mathbb Q}\{\text{ordered PBW monomials of weight at least }r\}.

Indeed, $\gamma_j(L)\subseteq I^j$ , by induction using $[x,y]=xy-yx$ . Thus a monomial of weight $w$ belongs to $I^w$ . Conversely, reordering products uses

yx=xy+[y,x],\qquad [\gamma_r(L),\gamma_s(L)]\subseteq\gamma_{r+s}(L).

Neither term has lower weight than the unreordered product. A product of $r$ elements of $I$ therefore has PBW expansion supported in weights at least $r$ . This proves (7).

The left ideal $J$ is spanned by the ordered PBW monomials containing a basis element of $S$ . Each such monomial ends in an element of $S$ , so belongs to $US$ . Conversely, expand a product $us$ with $s\in S$ into PBW monomials. Since the basis of $S$ comes last, only the terminal $S$ -factors require reordering with $s$ , and their brackets still lie in $S$ . Every resulting monomial retains an $S$ -factor.

It follows that $J+I^{c+1}$ is spanned by the union of two subsets of the PBW basis: monomials containing an $S$ -factor, and monomials of weight at least $c+1$ . A complementary basis element of $L$ is a one-letter monomial of weight at most $c$ in neither subset. Thus the intersection with $L$ is precisely $S$ . ◻

Lemma 3. For all $x,y\in L$ ,

\tag{9} \delta([x,y])=x\delta(y)-y\delta(x).

Moreover, $\gamma_r(L)V_s\subseteq V_{r+s}$ for $r,s\geq1$ .

Proof. Equation (9) is the image of $[x,y]=xy-yx$ in $M$ . For the filtration assertion, use $\gamma_r(L)\subseteq I^r$ and $I^rI^s=I^{r+s}$ . ◻

Separation without increasing nilpotency class

Form the semidirect Lie algebra $N=L\ltimes V$ , with $V$ abelian and bracket

\tag{10} [(x,u),(y,v)]=([x,y],xv-yu).

Define

\tag{11} F_0(x)=(x,0),\qquad F_1(x)=(x,\delta(x)).

Proposition 4. The Lie algebra $N$ is nilpotent of class at most $c$ . Both $F_0,F_1$ are injective Lie homomorphisms, and

F_0(x)=F_1(x)\quad\Longleftrightarrow\quad x\in S.

Proof. The first-coordinate projection makes both maps injective. Equation (9) proves that $F_1$ preserves brackets; this is immediate for $F_0$ . The equalizer statement is Lemma 2.

For the nilpotency bound, put $N_r=\gamma_r(L)\oplus V_r$ . Then $N_1=N$ and $N_{c+1}=0$ . Lemma 3 and (10) give

[N_r,N_s]\subseteq N_{r+s}.

Inductively $\gamma_r(N)\subseteq N_r$ , whence $\gamma_{c+1}(N)=0$ . ◻

The choice of $V$ , the image of the augmentation ideal, is essential for this class bound. Using all of $M$ would introduce a degree-zero component on which a product of $c$ elements of $L$ could act nontrivially. The filtration above starts in degree one.

The derivation method is related to the enveloping-algebra separation argument for Lie-algebra epimorphisms (2). The quotient by $I^{c+1}$ and its positive-degree submodule supply a separating Lie algebra within the prescribed nilpotent class.

Passage to groups

We recall the rational Mal’cev correspondence in the form used here (3; 4). Every torsion-free nilpotent group $G$ of class at most $c$ embeds in a uniquely divisible nilpotent group $G^{\mathbb Q}$ of the same class bound. Every element of $G^{\mathbb Q}$ has a positive integral power in $G$ . The finite Baker–Campbell–Hausdorff series identifies uniquely divisible nilpotent groups of class at most $c$ with nilpotent Lie algebras over $\mathbb Q$ of class at most $c$ .

These assertions do not require finite generation. The rational completions of finitely generated subgroups form a directed system under their canonical embeddings; their directed union gives $G^{\mathbb Q}$ . All operations in the correspondence are finite expressions at a fixed class bound, so the finite-dimensional correspondence on these subgroups is compatible with that union. Homomorphisms correspond in both directions. In particular, a subgroup closed under all rational powers corresponds to a Lie subalgebra.

Lemma 5. If $H\leq G$ is divisible, its image in $G^{\mathbb Q}$ is a uniquely divisible subgroup. Hence $S=\log H$ is a Lie subalgebra of $L=\log G^{\mathbb Q}$ .

Proof. Given $h\in H$ and $m\geq1$ , divisibility provides $u\in H$ with $u^m=h$ . Roots in the uniquely divisible ambient group are unique, so the root of $h$ in $G^{\mathbb Q}$ belongs to $H$ . Thus $H$ is closed under rational powers. The subgroup part of the rational Mal’cev correspondence gives the conclusion. ◻

Proof of Theorem 1. Apply Proposition 4 to $L=\log G^{\mathbb Q}$ and $S=\log H$ . Let $K=\exp N$ , with the finite Baker–Campbell–Hausdorff product. It is uniquely divisible and nilpotent of class at most $c$ . In particular, it is torsion-free: $(\exp z)^m=\exp(mz)$ and $mz=0$ implies $z=0$ over $\mathbb Q$ .

The Lie embeddings $F_0,F_1$ induce group embeddings $\exp L\to\exp N$ . Restricting these to $G$ gives $f_0,f_1$ . For $g\in G$ ,

\begin{align*} f_0(g)=f_1(g) &\ \Longleftrightarrow\ F_0(\log g)=F_1(\log g)\\ &\ \Longleftrightarrow\ \log g\in\log H \ \Longleftrightarrow\ g\in H. \end{align*}

Every element of $H$ lies in its dominion by definition. Conversely, the single pair $f_0,f_1$ excludes every element of $G\setminus H$ from the set (1). Hence the dominion equals $H$ . ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, Problem 17.34. arXiv:1401.0300.

G. M. Bergman, Epimorphisms of Lie algebras, Theorem 2.1. https://math.berkeley.edu/~gbergman/papers/unpub/LieEpi.pdf.

T. H. Fay, Remarks on the Mal’cev completion of torsion-free locally nilpotent groups, Cahiers Topologie Géom. Différentielle Catég. 35 (1994), 75–84. Numdam.

I. Stewart, An algebraic treatment of Mal’cev’s theorems concerning nilpotent Lie groups and their Lie algebras, Compositio Math. 22 (1970), 289–312. Numdam.