Nonattainment of the p-Jordan exponent in locally finite triangular groups

Problem

If the infimum of the admissible $p$ -Jordan exponents of a group is finite, is that infimum itself admissible?

\begin{gathered} j_p(H)=\min\{[H:N]:N\trianglelefteq H,\ N\text{ abelian},\ p\nmid|N|\},\\ \mathcal E_p(\Gamma)=\{s:\exists J>0\ \forall H\le\Gamma\text{ finite},\ j_p(H)\le J|H|_p^s\},\\ \inf\mathcal E_p(\Gamma)\stackrel{?}{\in}\mathcal E_p(\Gamma). \end{gathered}

The example below has $\mathcal E_p(\Gamma)=(1,\infty)$ for every prescribed prime $p$ .

The exponent and the construction

Fix a prime $p$ . For a finite group $H$ , write $|H|_p$ for the largest power of $p$ dividing $|H|$ , and define

\tag{1} j_p(H)=\min\{[H:N]:N\lhd H,\ N\text{ abelian},\ p\nmid|N|\}.

For a group $\Gamma$ , set

\tag{2} \mathcal E_p(\Gamma)=\left\{s\in\mathbb R: \begin{array}{l} \text{there exists }J>0\text{ such that }j_p(H)\leq J|H|_p^s\\ \text{for every finite subgroup }H\leq\Gamma \end{array}\right\}.

Its infimum, when finite, is the $p$ -Jordan exponent. Kourovka Problem 21.121(a) concerns attainment of this infimum (1).

Theorem 1. For every prime $p$ , there is a countable locally finite subgroup $\Gamma\leq\mathop{\mathrm{GL}}_2(\overline{\mathbb F}_p)$ such that all its matrices are upper triangular, its derived length is exactly two, and

\mathcal E_p(\Gamma)=(1,\infty).

In particular, its $p$ -Jordan exponent is one, but no positive constant works at exponent one.

Put $F=\overline{\mathbb F}_p$ . For $k\geq5$ , define

\tag{4} d_k=k!,\qquad m_k=\Phi_{k!}(p),\qquad M_k=\prod_{i=5}^k m_i,

where $\Phi_n(X)$ is the $n$ th cyclotomic polynomial. Set $M_k=1$ for $k<5$ , and write $\mu_m(F)=\{a\in F^\times:a^m=1\}$ . The group in the theorem is

\tag{5} D=\bigcup_{k\geq5}\mu_{M_k}(F),\qquad \Gamma=\left\{\begin{pmatrix}a&b\\0&1\end{pmatrix}:a\in D,\ b\in F\right\}.

Every $M_k$ is prime to $p$ , as proved below, and $M_k\mid M_{k+1}$ . Thus $D$ is an increasing union of finite cyclic groups.

Cyclotomic values at factorial indices

Lemma 2. For $k\geq5$ , the following hold:

\begin{align*} \gcd(m_k,k!)&=1,\qquad m_k>k!,\tag{6a}\\ \mathop{\mathrm{ord}}_r(p)&=k!\qquad(1<r\mid m_k).\tag{6b} \end{align*}

Here $\mathop{\mathrm{ord}}_r(p)$ is the multiplicative order of $p$ modulo $r$ . The integers $m_k$ are pairwise coprime.

Proof. We first recall the elementary cyclotomic reduction rule. If a prime $q$ divides $\Phi_n(p)$ and $n=q^a f$ with $q\nmid f$ , then the image of $p$ in $\mathbb F_q$ has order $f$ . Indeed, when $a=0$ , the roots of $\Phi_f$ in characteristic $q$ are precisely the primitive $f$ th roots. For $a>0$ , reduction of the cyclotomic factorization, together with Frobenius, gives

\Phi_{q^a f}(X)=\Phi_f(X)^{q^{a-1}(q-1)}\quad\text{in }\mathbb F_q[X].

Thus its roots are the same primitive $f$ th roots. In particular $f\mid q-1$ .

Suppose now that $q\mid m_k$ and $q\mid k!$ . Then $q\leq k$ , and the $q$ -free part $f$ of $k!$ divides $q-1$ . If $q<5$ , the factor $5$ of $k!$ divides $f$ , which is impossible since $f\leq q-1<5$ . If $q\geq5$ , the integer $(q-1)!$ divides $f$ , since it is prime to $q$ and divides $k!$ . But $(q-1)!>q-1$ , again impossible. This proves (6a).

Since $\Phi_{k!}(p)\mid p^{k!}-1$ , every divisor $r$ of $m_k$ is prime to $p$ and $\mathop{\mathrm{ord}}_r(p)\mid k!$ . If $r>1$ , choose a prime $q\mid r$ . By (6a), $q\nmid k!$ , so the reduction rule gives $\mathop{\mathrm{ord}}_q(p)=k!$ . As $\mathop{\mathrm{ord}}_q(p)\mid\mathop{\mathrm{ord}}_r(p)$ , equality follows, proving (6b). A prime dividing two distinct $m_i,m_j$ would give $i!=j!$ , which is impossible for $i,j\geq5$ .

Finally, $m_k>1$ . The complex product for $\Phi_{k!}(p)$ has all factors $p-\zeta$ with $\zeta\neq1$ on the unit circle, so $|p-\zeta|>p-1\geq1$ . The value is positive, hence greater than one. Choose a prime $q\mid m_k$ . Equation (6b) gives $k!\mid q-1$ , and therefore $m_k\geq q>k!$ . ◻

Lemma 3. If $m>1$ divides some $M_j$ , there is a $k\geq5$ such that

\tag{8} \mathop{\mathrm{ord}}_m(p)=k!,\qquad m\mid M_k.

Also $M_k\mid p^{k!}-1$ for every $k\geq5$ .

Proof. For $i\leq k$ , $m_i\mid p^{i!}-1\mid p^{k!}-1$ . Pairwise coprimality gives the assertion for $M_k$ . Choose the least $k$ with $m\mid M_k$ . Then $k\geq5$ and $r=\gcd(m,m_k)>1$ : otherwise $m\mid M_{k-1}m_k$ would imply $m\mid M_{k-1}$ . Thus

k!=\mathop{\mathrm{ord}}_r(p)\mid\mathop{\mathrm{ord}}_m(p)\mid k!,

using Lemma 2 and $m\mid M_k\mid p^{k!}-1$ . ◻

Uniform growth of the scalar orders

Lemma 4. One has

\tag{10} \frac{\log M_k}{k!}\longrightarrow0.

Consequently, for every $\varepsilon>0$ , there is a constant $C_\varepsilon\geq1$ such that

\tag{11} M_k\leq C_\varepsilon p^{\varepsilon k!}\qquad(k\geq5).

Proof. Euler’s formula and divergence of the sum of reciprocal primes give

\frac{\varphi(k!)}{k!}=\prod_{\substack{q\leq k\\q\text{ prime}}}\left(1-\frac1q\right) \leq\exp\left(-\sum_{\substack{q\leq k\\q\text{ prime}}}\frac1q\right)\longrightarrow0.

The complex root product also gives $m_i\leq(p+1)^{\varphi(i!)}$ . Hence

\tag{13} 0\leq\frac{\log M_k}{k!} \leq\log(p+1)\frac{\sum_{i=0}^k\varphi(i!)}{k!}.

Induction gives $\sum_{i=0}^{k-1}i!\leq3(k-1)!$ for $k\geq1$ . Since $\varphi(i!)\leq i!$ ,

\frac{\sum_{i=0}^k\varphi(i!)}{k!} \leq\frac{\varphi(k!)}{k!}+\frac3k\longrightarrow0.

This proves (10). For sufficiently large $k$ , it implies $\log M_k\leq\varepsilon k!\log p$ . Enlarge the multiplicative constant to include the finitely many earlier indices, proving (11). ◻

Finite affine subgroups

Write an affine element as $(a,b)$ , with multiplication $(a,b)(a',b')=(aa',b+ab')$ , and let $V=\{(1,b):b\in F\}$ . For finite $H\leq\Gamma$ , put

P=H\cap V,\qquad B=\{a:(a,b)\in H\text{ for some }b\},\qquad m=|B|.

The group $B$ is a finite subgroup of $F^\times$ , so it is cyclic and has order prime to $p$ . The kernel $P$ is elementary abelian. Consequently

\tag{16} |H|=|P|m,\qquad |H|_p=|P|.

Lemma 5. If $P=1$ , then $j_p(H)=1$ . If $P\neq1$ , then

\tag{17} j_p(H)=|H|,\qquad m\mid |P|-1.

In particular, if $m>1$ and $t=\mathop{\mathrm{ord}}_m(p)$ , then $|P|=p^{tr}$ for some integer $r\geq1$ .

Proof. If $P=1$ , projection embeds $H$ into $B$ , making $H$ itself an admissible normal abelian $p'$ -subgroup in (1).

Suppose $P\neq1$ , and let $N\lhd H$ have order prime to $p$ . Since $P$ is normal, $[N,P]\subseteq N\cap P=1$ . Thus $N$ centralizes every translation in $P$ . For $b\neq0$ ,

(a,u)(1,b)=(1,b)(a,u)\quad\Longleftrightarrow\quad (a-1)b=0 \quad\Longleftrightarrow\quad a=1.

It follows that $N\subseteq P$ , and hence $N=1$ . This proves $j_p(H)=|H|$ .

Identify $P$ with a finite additive subgroup $W\leq F$ and choose a generator $a$ of $B$ . Conjugation by any lift of $a$ permutes $W\setminus\{0\}$ by multiplication by $a$ . The nonzero product of all elements of this set therefore satisfies

\prod_{w\in W\setminus\{0\}}w =\prod_{w\in W\setminus\{0\}}aw =a^{|W|-1}\prod_{w\in W\setminus\{0\}}w.

Cancellation gives $a^{|P|-1}=1$ , so $m\mid|P|-1$ . Write $|P|=p^u$ with $u>0$ . Then $\mathop{\mathrm{ord}}_m(p)\mid u$ , giving the last assertion. ◻

The set of admissible exponents

Proof of Theorem 1. Take $s=1+\varepsilon>1$ and $C_\varepsilon$ from Lemma 4. For finite $H\leq\Gamma$ , the desired bound follows immediately from Lemma 5 if $P=1$ or $m=1$ , since $C_\varepsilon\geq1$ .

In the remaining case, $m>1$ . A generator of the cyclic group $B$ belongs to some $\mu_{M_j}(F)$ , so $m\mid M_j$ . Lemma 3 gives $k\geq5$ with $\mathop{\mathrm{ord}}_m(p)=k!$ and $m\leq M_k$ . Lemma 5 then implies $|P|\geq p^{k!}$ . Therefore

j_p(H)=|P|m\leq |P|M_k\leq C_\varepsilon|P|p^{\varepsilon k!} \leq C_\varepsilon|P|^{1+\varepsilon}=C_\varepsilon|H|_p^s.

The constant is independent of $H$ , so every $s>1$ is admissible.

The field $F$ is the countable union of its finite subfields, and every finite subset lies in one of them. Every finitely generated subgroup of $\Gamma$ therefore lies in a finite affine group. Thus $\Gamma$ is countable and locally finite.

Choose $a_k\in D$ of order $m_k$ . The finite subgroup $H_k=\langle(a_k,0),(1,1)\rangle$ has nontrivial translation subgroup and scalar image of order $m_k$ . Hence $j_p(H_k)=m_k|H_k|_p>k!|H_k|_p$ , excluding a bound at exponent one. Since $|H|_p\geq1$ , a bound at any $s\leq1$ would imply one at exponent one. Thus $\mathcal E_p(\Gamma)=(1,\infty)$ .

Finally, $\Gamma'\subseteq V$ since $\Gamma/V$ is abelian. For $a\in D\setminus\{1\}$ the identity $[(a,0),(1,b)]=(1,(a-1)b)$ , with $b\in F$ , gives $V\subseteq\Gamma'$ . Hence $\Gamma'=V\neq1$ and $\Gamma''=1$ , proving that the triangular group (5) has derived length exactly two. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, Problem 21.121(a). arXiv:1401.0300.