Bounded-index nilpotent subgroups of rational linear groups with finitely many automorphism orbits

Problem

Must a rational linear group with finitely many automorphism orbits be virtually soluble?

G\le\operatorname{GL}_n(\mathbb Q),\quad \omega(G)=|\operatorname{Aut}(G)\backslash G|<\infty \quad\stackrel{?}{\Longrightarrow}\quad G\text{ virtually soluble}.

The automorphisms are all abstract group automorphisms. The proof yields a torsion-free normal nilpotent subgroup with an explicit index bound.

Statement

Write $\omega(G)$ for the number of orbits of $\mathop{\mathrm{Aut}}(G)$ on $G$ . Here $\mathop{\mathrm{Aut}}(G)$ consists of all abstract group automorphisms; no compatibility with a chosen linear representation is imposed. Kourovka Problem 21.40 asks whether a rational linear group with $\omega(G)<\infty$ is virtually soluble (1). We prove a stronger assertion with an explicit index bound. We use $\gamma_1(U)=U$ and $\gamma_{r+1}(U)=[U,\gamma_r(U)]$ .

Theorem 1. Let $n\geq1$ and $G\leq\mathop{\mathrm{GL}}_n(\mathbb Q)$ , and put $A=\mathop{\mathrm{span}}_{\mathbb Q}G$ and $d=\dim_{\mathbb Q}A$ . If $\omega(G)<\infty$ , there is a normal subgroup $U\lhd G$ such that

\tag{1} U\text{ is torsion-free},\qquad \gamma_n(U)=1,\qquad [G:U]\leq(2n+1)^d\leq(2n+1)^{n^2}.

There is no finite-generation hypothesis. In particular, when $n=1$ the conclusion is $U=1$ . The algebra $A$ is unital: products of elements of $G$ belong to $G$ , and $1\in G$ .

Theorem 2. If $G\leq\mathop{\mathrm{GL}}_n(\mathbb Q)$ has finite trace set $T$ , there is a torsion-free normal subgroup $U$ with $\gamma_n(U)=1$ and

\tag{2} [G:U]\leq |T|^{\dim_{\mathbb Q}\mathop{\mathrm{span}}_{\mathbb Q}G}.

Bounded-degree roots and automorphism orbits

Lemma 3. Let $0\neq\alpha\in\overline{\mathbb Q}$ and $D\geq1$ . There are only finitely many algebraic numbers $\beta$ such that $[\mathbb Q(\beta):\mathbb Q]\leq D$ and $\beta^m=\alpha$ for some $m\geq1$ .

Proof. Let $f\in\mathbb Z[X]$ be the primitive irreducible polynomial of $\alpha$ , with positive leading coefficient $a$ . Put

C=\max\bigl(1,\{|\alpha'|:\alpha'\text{ is a conjugate of }\alpha\}\bigr).

If $\beta^m=\alpha$ , its primitive irreducible polynomial $h\in\mathbb Z[X]$ divides $f(X^m)$ in $\mathbb Z[X]$ , by Gauss’s lemma. Its positive leading coefficient $b$ therefore divides $a$ . Every conjugate $\beta'$ satisfies $(\beta')^m=\alpha'$ for some conjugate $\alpha'$ ; hence $|\beta'|\leq C$ . If $r=\deg h\leq D$ , the coefficient of $X^{r-j}$ in $h$ has absolute value at most

b\binom rj C^j\leq a(1+C)^D.

There are finitely many integral polynomials of degree at most $D$ with this coefficient bound, and each has finitely many roots. ◻

Lemma 4. If $G\leq\mathop{\mathrm{GL}}_n(\mathbb Q)$ and $\omega(G)<\infty$ , every eigenvalue of every element of $G$ is a root of unity.

Proof. Fix $g\in G$ . Two terms of $g,g^2,g^4,\ldots$ lie in the same automorphism orbit. Thus, for some $i<j$ and $\phi\in\mathop{\mathrm{Aut}}(G)$ ,

b=g^{2^i},\qquad q=2^{j-i}>1,\qquad \phi(b)=b^q.

Set $b_k=\phi^{-k}(b)$ . Applying $\phi^{-k}$ to the displayed identity and iterating gives $b_k^{q^k}=b$ .

Let $\alpha$ be an eigenvalue of $b$ . For every $k$ , some eigenvalue $\beta_k$ of $b_k$ satisfies $\beta_k^{q^k}=\alpha$ ; this follows by triangularizing $b_k$ over $\overline{\mathbb Q}$ . Each $\beta_k$ has degree at most $n$ . Lemma 3 makes the set of possible $\beta_k$ finite. Hence $\beta_r=\beta_s$ for some $r<s$ , and

\beta_r^{q^s-q^r}=1.

Consequently $\alpha$ is a root of unity. An eigenvalue $\lambda$ of $g$ has $\lambda^{2^i}$ among the eigenvalues of $b$ , so $\lambda$ is also a root of unity. ◻

Corollary 5. Under the hypotheses of Lemma 4,

\mathop{\mathrm{tr}}(g)\in\mathbb Z\cap[-n,n]\qquad(g\in G).

In particular, the trace set has at most $2n+1$ elements.

Proof. The trace is a rational algebraic integer, because it is a sum of roots of unity. It is therefore an integer. The triangle inequality gives $|\mathop{\mathrm{tr}}(g)|\leq n$ . ◻

The trace radical of a matrix algebra

Let $A\subseteq M_n(\mathbb Q)$ be a unital subalgebra and define

\tag{8} J=\{a\in A:\mathop{\mathrm{tr}}(ab)=0\text{ for every }b\in A\}.

Cyclicity of trace shows that $J$ is a two-sided ideal: for $a\in J$ and $b,c\in A$ ,

\mathop{\mathrm{tr}}((ca)b)=\mathop{\mathrm{tr}}(a(bc))=0,\qquad \mathop{\mathrm{tr}}((ac)b)=\mathop{\mathrm{tr}}(a(cb))=0.

Lemma 6. Every element of $J$ is nilpotent, and $J^n=0$ .

Proof. Take $a\in J$ . The Fitting decomposition for the endomorphism $a$ gives, for sufficiently large $N$ ,

\mathbb Q^n=\ker a^N\oplus\operatorname{im}a^N.

The projection $e$ onto $\operatorname{im}a^N$ along $\ker a^N$ is a polynomial in $a$ with zero constant term. Indeed, write the minimal polynomial as $X^r q(X)$ with $q(0)\neq0$ . When $r>0$ , choose a polynomial congruent to $0$ modulo $X^r$ and to $1$ modulo $q$ by Bezout’s identity. When $r=0$ , the invertibility of $a$ expresses $1$ as $a$ times a polynomial in $a$ . These formulas give the stated projection in both cases. Thus $e\in J$ and $\mathop{\mathrm{tr}}(e)=0$ . An idempotent over $\mathbb Q$ has trace equal to its rank; hence $e=0$ . The image of $a^N$ is zero, so $a$ is nilpotent.

For every $a\in J$ , the finite geometric series in $a$ inverts $1-a$ in $A$ . Consider $W_r=J^r\mathbb Q^n$ . If $JW_r=W_r\neq0$ , choose a minimal finite generating family $w_1,\ldots,w_s$ for $W_r$ as an $A$ -module. Such a family exists since $W_r$ is finite-dimensional over $\mathbb Q$ . The equality $JW_r=W_r$ gives

w_s=a_1w_1+\cdots+a_sw_s\qquad(a_i\in J).

Since $1-a_s$ is invertible, $w_s$ belongs to the $A$ -span of the preceding generators, a contradiction. Thus $W_{r+1}\subsetneq W_r$ whenever $W_r\neq0$ . Since $\dim_{\mathbb Q}W_0=n$ , we have $J^n\mathbb Q^n=0$ . The given action of $A$ is faithful, so $J^n=0$ . ◻

Lemma 7. The group $1+J$ is torsion-free and satisfies $\gamma_n(1+J)=1$ .

Proof. The inverse of $1+a$ is a finite geometric series, so $1+J$ is a group. If $(1+a)^m=1$ with $m\geq1$ , then

0=a\left(\sum_{j=1}^m\binom mj a^{j-1}\right).

The expression in parentheses is $m$ times $1$ plus a nilpotent polynomial in $a$ , and is invertible. Thus $a=0$ .

For $r,s\geq1$ , multiplication modulo $J^{r+s}$ is commutative between $1+J^r$ and $1+J^s$ , giving

[1+J^r,1+J^s]\subseteq1+J^{r+s}.

Induction yields $\gamma_r(1+J)\subseteq1+J^r$ . Apply Lemma 6 with $r=n$ . ◻

Counting cosets by traces

Proof of Theorem 2. Use $A=\mathop{\mathrm{span}}_{\mathbb Q}G$ and its ideal $J$ from (8), and set

U=G\cap(1+J)=\ker\bigl(G\longrightarrow(A/J)^\times\bigr).

It is normal, torsion-free, and satisfies $\gamma_n(U)=1$ by Lemma 7. Choose a basis $g_1,\ldots,g_d$ of $A$ from $G$ . Define

\Psi:G\longrightarrow T^d,\qquad \Psi(g)=\bigl(\mathop{\mathrm{tr}}(gg_1),\ldots,\mathop{\mathrm{tr}}(gg_d)\bigr).

For $g,h\in G$ , linearity and the basis property give

\begin{align*} \Psi(g)=\Psi(h) &\ \Longleftrightarrow\ \mathop{\mathrm{tr}}((g-h)a)=0\quad(a\in A)\\ &\ \Longleftrightarrow\ g-h\in J \ \Longleftrightarrow\ h^{-1}g-1\in J \ \Longleftrightarrow\ h^{-1}g\in U. \end{align*}

Thus the fibres of $\Psi$ are exactly the cosets of $U$ , and $[G:U]\leq|T|^d$ . ◻

Proof of Theorem 1. Corollary 5 gives $|T|\leq2n+1$ . Apply Theorem 2 and use $d\leq n^2$ . ◻

Corollary 8. Let $K$ be a number field of degree $e$ and let $G\leq\mathop{\mathrm{GL}}_n(K)$ with $\omega(G)<\infty$ . There is a torsion-free normal subgroup $U$ with

\gamma_{ne}(U)=1,\qquad [G:U]\leq(2ne+1)^{(ne)^2}.

Proof. Restriction of scalars embeds $\mathop{\mathrm{GL}}_n(K)$ into $\mathop{\mathrm{GL}}_{ne}(\mathbb Q)$ . Apply Theorem 1 to the resulting faithful representation of the same abstract group. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, Problem 21.40. arXiv:1401.0300.