A nonsplit extension of a product of free groups by the additive group of a countable division ring

Problem

Let $D$ be a division ring and $G\le D^\times$ . Must every extension by $D^+$ split when the conjugation action is left multiplication in $D$ ?

1\longrightarrow D^+\longrightarrow E\xrightarrow{\pi}G\longrightarrow1, \qquad\exists s:G\to E:\ \pi s=1_G\;?

The prescribed module structure is $g\cdot d=gd$ ; replacing it by the trivial action changes the problem.

The extension problem

Let $D$ be a division ring and $G\leq D^\times$ . Regard the additive group $D^+$ as a left $G$ -module by multiplication in $D$ . Kourovka Problem 18.76 concerns splitting of extensions with this action (1). The action is fixed throughout: replacing it by the trivial action changes the question.

Theorem 1. There exist a countable division ring $D$ of characteristic zero, an embedding $F_2\times F_2\hookrightarrow D^\times$ , and a countable torsion-free group $E$ fitting into a nonsplit exact sequence

\tag{1} 1\longrightarrow D^+\xrightarrow{\iota}E\xrightarrow{\pi}F_2\times F_2\longrightarrow1.

Conjugation by any lift of $g\in F_2\times F_2$ acts on $D^+$ as $u\mapsto gu$ .

We write $P=F(a,b)$ , $Q=F(c,d)$ and $G=P\times Q$ . Elements of the two factors commute in $G$ and therefore in every division ring containing its group algebra. Commutativity within a factor is neither assumed nor used.

A countable division ring containing the group algebra

Lemma 2. The group algebra $\mathbb Q[G]$ embeds in a countable division ring of characteristic zero.

Proof. Free groups are bi-orderable, and a lexicographic product of bi-orders orders $G$ . One construction for a free group uses the Magnus expansion in noncommuting variables: send a free generator $x_i$ to $1+X_i$ and use

(1+X_i)^{-1}=1-X_i+X_i^2-X_i^3+\cdots.

The expansion is injective. Ordering monomials first by degree and then lexicographically orders a nonidentity series by the sign of its first nonzero coefficient after the constant term. Multiplication and conjugation preserve that first term, so the induced order is invariant on both sides. This is the standard bi-order on a free group used in the ordered-series construction (2).

For a bi-ordered group, the Mal’cev–Neumann series ring $\mathbb Q((G))$ consists of formal sums

\sum_{g\in G}r_g g\qquad(r_g\in\mathbb Q)

with well-ordered support. Convolution is defined because only finitely many products from two well-ordered supports contribute to a given coefficient. This ring is a division ring (2; 3): after extracting the least nonzero term, a nonzero series has the form $r g(1-u)$ with $\operatorname{supp}(u)>1$ , and its inverse is given by the summable geometric series $\sum_{j\geq0}u^j$ . Finite sums identify $\mathbb Q[G]$ with a subring.

Let $D_0=\mathbb Q[G]$ , and let $D_{r+1}$ be the subring generated by $D_r$ and the inverses of all its nonzero elements inside $\mathbb Q((G))$ . Each $D_r$ is countable. The union $D=\bigcup_{r\geq0}D_r$ is countable and closed under inversion of nonzero elements. It is the required division ring. ◻

Fix such an embedding and suppress it from the notation. In particular,

\tag{4} a-1\neq0,\quad b-1\neq0,\quad c-1\neq0,\quad d-1\neq0

in $D$ . Every element of $\mathbb Q[P]$ commutes with every element of $\mathbb Q[Q]$ ; inverses of nonzero elements retain this commutation property.

Derivations and the factor set

Lemma 3. There are maps $\alpha:P\to\mathbb Q[P]$ and $\beta:Q\to\mathbb Q[Q]$ satisfying

\begin{align*} \alpha(xx')&=\alpha(x)+x\alpha(x'),& \alpha(a)&=0,&\alpha(b)&=1,\tag{5a}\\ \beta(yy')&=\beta(y)+y\beta(y'),& \beta(c)&=0,&\beta(d)&=1.\tag{5b} \end{align*}

Both maps vanish at the identity.

Proof. For a free group, prescribe values on the free generators and put $\alpha(x^{-1})=-x^{-1}\alpha(x)$ for their inverses. If $x_1\cdots x_r$ is a word, set

\alpha(x_1\cdots x_r)=\sum_{j=1}^r x_1\cdots x_{j-1}\alpha(x_j).

Inserting or deleting adjacent inverse letters changes this sum by $w\bigl(\alpha(x)+x\alpha(x^{-1})\bigr)=0$ . The value is therefore well-defined on the free group. Concatenating words proves (5a). Apply the same construction to $Q$ . ◻

For $g=(x,y)$ and $h=(x',y')$ , define

\tag{7} C(g,h)=\alpha(x)y\beta(y')\in D.

Lemma 4. The map $C$ is normalized and satisfies

\tag{8} C(g,h)+C(gh,k)=gC(h,k)+C(g,hk) \qquad(g,h,k\in G).

Proof. Normalization follows from $\alpha(1)=\beta(1)=0$ . Write $k=(x'',y'')$ . The left side of (8) expands as

\alpha(x)y\beta(y')+\alpha(x)yy'\beta(y'') +x\alpha(x')yy'\beta(y'').

The right side has the same three terms, since $y$ commutes with $\alpha(x')$ and $\beta(y'y'')=\beta(y')+y'\beta(y'')$ . ◻

Define a multiplication on $E=D\times G$ by

\tag{10} (u,g)(v,h)=\bigl(u+gv+C(g,h),gh\bigr).

Equation (8) proves associativity, and $(0,1)$ is the identity. The inverse is

\tag{11} (u,g)^{-1}=\bigl(-g^{-1}(u+C(g,g^{-1})),g^{-1}\bigr).

It is a right inverse by direct substitution. The cocycle equation with $(g,g^{-1},g)$ gives $C(g,g^{-1})=gC(g^{-1},g)$ , which also verifies the left inverse.

Set $\iota(v)=(v,1)$ and $\pi(u,g)=g$ . These are homomorphisms with $\iota$ injective, $\pi$ surjective and $\ker\pi=\iota(D^+)$ . Moreover, normalization gives

(u,g)(v,1)=(gv,1)(u,g),

which proves the prescribed conjugation action.

The obstruction to a section

Lemma 5. The projection $\pi:E\to G$ has no group-homomorphic section.

Proof. Suppose a section exists. Write its values on the four free generators as

(u_a,a),\quad(u_b,b),\quad(v_c,c),\quad(v_d,d).

Here a generator denotes its canonical image in $P\times Q$ . Each of $a,b$ commutes with each of $c,d$ . The corresponding lifts must therefore commute. From (7),

C(a,c)=C(a,d)=C(b,c)=0,\quad C(b,d)=1,

whereas $C(c,a)=C(d,a)=C(c,b)=C(d,b)=0$ . Substitution in (10) gives

\begin{align*} (c-1)u_a&=(a-1)v_c,\tag{15a}\\ (d-1)u_a&=(a-1)v_d,\tag{15b}\\ (c-1)u_b&=(b-1)v_c,\tag{15c}\\ (d-1)u_b-(b-1)v_d&=1.\tag{15d} \end{align*}

By (4), equations (15a)–(15c) imply

\begin{align*} v_c&=(a-1)^{-1}(c-1)u_a,\\ v_d&=(a-1)^{-1}(d-1)u_a,\\ u_b&=(b-1)(a-1)^{-1}u_a. \end{align*}

For the last equality, commute $c-1$ across $b-1$ and $(a-1)^{-1}$ and cancel it on the left. Similarly $d-1$ commutes with both of these factors, so the left side of (15d) is

(d-1)(b-1)(a-1)^{-1}u_a -(b-1)(a-1)^{-1}(d-1)u_a=0.

This contradicts (15d). ◻

Proof of Theorem 1. The construction above gives the exact sequence and its action, and Lemma 5 proves that it does not split. Both $D$ and $G$ are countable, hence so is $E$ . Free groups are torsion-free: a nonempty cyclically reduced word has nonempty powers, and every nonidentity word is conjugate to one. Therefore $P\times Q$ is torsion-free. If an element of $E$ has finite order, its image in $G$ is trivial. It belongs to $D^+$ , which is torsion-free since $\operatorname{char}D=0$ . Thus $E$ is torsion-free. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, Problem 18.76. arXiv:1401.0300.

B. H. Neumann, On ordered division rings, Trans. Amer. Math. Soc. 66 (1949), 202–252. doi:10.1090/S0002-9947-1949-0032593-5.

B. Poonen, Units in Hahn–Mal’cev–Neumann rings. https://math.mit.edu/~poonen/papers/malcev.pdf.