Computable two-variable laws for finite groups with abelian Sylow subgroups

Problem

For a prime $p>2$ , give an explicit sequence of two-variable words that eventually characterizes finite groups with abelian Sylow $p$ -subgroups.

\operatorname{Syl}_p(G)\text{ abelian} \quad\Longleftrightarrow\quad \exists N\ \forall n\ge N,\ G\models w_{p,n}(x,y)=1.

Here explicit means a total computable word construction. The exhaustive search below does not claim a practical running-time bound or a short power-commutator formula.

The sequence and its scope

Fix a prime $p$ . The existence of a sequence of two-variable laws eventually characterizing finite groups with abelian Sylow $p$ -subgroups is already known. Brandl (2) gives an explicit power-commutator sequence for $p=2$ , and for arbitrary $p$ within the class of finite $p$ -soluble groups. Problem 11.15 of the Kourovka Notebook (4) asks for an explicit sequence when $p>2$ . Related commutator characterizations are studied in (3).

We give a total algorithm, with a uniform threshold equal to the order of the target group. The algorithm enumerates finite words and tests them in a finite symmetric group and a finite direct product. This is a search construction. Explicit recursive formulas, such as those considered for finite solubility in (1), impose an additional form on the sequence which is not part of our assertion.

We use $[x,y]=x^{-1}y^{-1}xy$ . A word is represented by a finite string in $x,x^{-1},y,y^{-1}$ . Its value at $(g,h)\in G^2$ is denoted by $w(g,h)$ . Thus $G\models w=1$ means that $w(g,h)=1$ for every $(g,h)\in G^2$ .

Theorem 1. There is a total computable function $(p,n)\mapsto a_{p,n}$ from $\mathbb N^2$ to the set of two-variable words with the following properties. If $p$ is prime and $G$ is finite, then

\tag{1} n\geq |G| \quad\Longrightarrow\quad \bigl(G\models a_{p,n}=1 \ \Longleftrightarrow\ \text{every Sylow }p\text{-subgroup of }G\text{ is abelian}\bigr).

For every $p$ -subgroup $H\leq G$ and every $g,h\in H$ ,

\tag{2} a_{p,n}(g,h)=[g,h]\qquad(n\geq |G|).

In particular, if some Sylow $p$ -subgroup is nonabelian, there are fixed $g,h\in G$ such that $a_{p,n}(g,h)\ne1$ for every $n\geq |G|$ .

It follows that the Sylow $p$ -subgroups of $G$ are abelian if and only if, for every pair $(g,h)\in G^2$ , one has $a_{p,n}(g,h)=1$ for all sufficiently large $n$ . In this formulation the threshold may initially depend on the pair; Theorem 1 supplies the uniform threshold $|G|$ whenever the condition holds.

Finite subgroup tests

Let $S$ be a finite group. For a subset $T\subseteq S$ , write

\mathcal C(T)\quad\Longleftrightarrow\quad 1\in T,\quad TT\subseteq T,\quad T^{-1}\subseteq T.

These are precisely the subgroup conditions. Define

\tag{4} \mathcal P_p(a,b) \quad\Longleftrightarrow\quad \exists T\subseteq S\ \bigl(\mathcal C(T),\ a,b\in T,\ |T|=p^j\text{ for some }0\leq j\leq |T|\bigr).

Lemma 2. The predicate $\mathcal P_p$ is decidable from the finite multiplication table of $S$ . If $p$ is prime, then $\mathcal P_p(a,b)$ holds if and only if $a,b$ lie in a common $p$ -subgroup of $S$ .

Proof. Enumerate the subsets of $S$ , test the three subgroup conditions, and test the finitely many integers $0\leq j\leq |T|$ . A successful subset is a subgroup of prime-power order. Conversely, a finite $p$ -subgroup $T$ has order $p^j$ for some $j\geq0$ . Since $p\geq2$ , induction gives $j\leq p^j=|T|$ , so this subgroup occurs in the finite test. ◻

Universal evaluations and Sylow lifts

For a finite group $S$ and $F=F(x,y)$ , put

D_S=\prod_{(a,b)\in S^2}S,\qquad \eta:F\longrightarrow D_S,\qquad \eta(w)_{(a,b)}=w(a,b).

Write $Q=\eta(F)$ . The group $Q$ is finite and generated by $\eta(x)$ and $\eta(y)$ .

A pair of words $(u,v)$ is called suitable for $(p,S)$ when

\begin{align*} &\mathcal P_p\bigl(\eta(u),\eta(v)\bigr) &&\text{in }D_S,\tag{6a}\\ &\mathcal P_p(a,b)\Longrightarrow u(a,b)=a,\quad v(a,b)=b &&\text{for every }(a,b)\in S^2.\tag{6b} \end{align*}

Here the first occurrence of $\mathcal P_p$ uses the finite group $D_S$ . By Lemma 2, suitability is a decidable predicate on finite word pairs.

Lemma 3. If $p$ is prime, then every finite group $S$ admits a suitable pair of words.

Proof. Let $I$ be the set of pairs $(a,b)\in S^2$ satisfying $\mathcal P_p(a,b)$ . For each $i=(a,b)\in I$ , choose a $p$ -subgroup $H_i\leq S$ containing $a,b$ . Every word evaluated at $(a,b)$ lies in $H_i$ , so restriction to these coordinates defines a homomorphism

\psi:Q\longrightarrow \prod_{i\in I}H_i.

The product is a $p$ -group. Indeed, for an element $(h_i)$ , choose integers $e_i$ such that $h_i^{p^{e_i}}=1$ ; its $p^{\sum_i e_i}$ -th power is then $1$ . Thus $L=\psi(Q)$ is a $p$ -group.

Choose a Sylow $p$ -subgroup $P$ of $Q$ . Surjective homomorphisms of finite groups map Sylow subgroups to Sylow subgroups. Applying this to $\psi:Q\to L$ gives $\psi(P)=L$ , since a $p$ -group is its own Sylow $p$ -subgroup. There are therefore $s,t\in P$ with

\psi(s)=\psi(\eta(x)),\qquad \psi(t)=\psi(\eta(y)).

Choose word representatives $u,v\in F$ such that $\eta(u)=s$ and $\eta(v)=t$ . The subgroup $P$ , viewed inside $D_S$ , verifies (6a). For $i=(a,b)\in I$ , equality of the $i$ -th coordinates gives $u(a,b)=a$ and $v(a,b)=b$ , which is (6b). ◻

The word-producing algorithm

We specify a coding so that the search determines a single sequence. Define the pairing function

\pi(r,s)= \begin{cases} s^2+r,&r<s,\\ r^2+r+s,&r\geq s. \end{cases}

This is a bijection $\mathbb N^2\to\mathbb N$ . Give the letters $x^{-1},x,y^{-1},y$ the codes $\epsilon(\ell)=0,1,2,3$ , respectively. For a finite word $\ell w$ , with first letter $\ell$ , define

\tag{10} E(\varnothing)=0,\qquad E(\ell w)=1+\pi(\epsilon(\ell),E(w)),\qquad E(u,v)=\pi(E(u),E(v)).

Pairing inversion and recursive list decoding give a computable partial inverse. Let $\delta:\mathbb N\to\mathcal W^2$ be its total extension, assigning $(\varnothing,\varnothing)$ to invalid codes, where $\mathcal W$ is the set of finite words. Then

\tag{11} \delta(E(u,v))=(u,v).

For prime $p$ and $n\geq0$ , put $S_n=\operatorname{Sym}(\{0,\ldots,n-1\})$ . Starting at $m=0$ , test whether $\delta(m)$ is suitable for $(p,S_n)$ , and increase $m$ by one until the test succeeds. Equivalently, set

\tag{12} m(p,n)=\min\{m\in\mathbb N:\delta(m)\text{ is suitable for }(p,S_n)\}, \qquad (u_{p,n},v_{p,n})=\delta(m(p,n)).

Define

\tag{13} a_{p,n}=u_{p,n}^{-1}v_{p,n}^{-1}u_{p,n}v_{p,n}.

For nonprime $p$ , set $a_{p,n}=\varnothing$ .

Each suitability test terminates by finite enumeration. Lemma 3 and (11) show that a successful code exists for every prime $p$ and every $n$ . Thus (12) is a terminating sequential search, and (13) defines a total computable function. The subgroups selected in Lemma 3 are used to prove termination; the search itself uses only the finite predicates (6a) and (6b).

The uniform tail

Proof of Theorem 1. Let $G$ be finite and $n\geq |G|$ . The regular action of $G$ , extended by fixed points, gives an injective homomorphism $\iota:G\to S_n$ . Write $u=u_{p,n}$ and $v=v_{p,n}$ .

Suppose first that every Sylow $p$ -subgroup of $G$ is abelian. For $g,h\in G$ , evaluate the coordinates of $D_{S_n}$ at $(\iota(g),\iota(h))$ . By (6a), there is a $p$ -subgroup $T\leq D_{S_n}$ containing $\eta(u)$ and $\eta(v)$ . Its image under this coordinate projection is a $p$ -subgroup $M\leq S_n$ containing $u(\iota(g),\iota(h))$ and $v(\iota(g),\iota(h))$ . The subgroup $K=\iota^{-1}(M)$ embeds in $M$ , and is therefore a $p$ -group. Word evaluation commutes with $\iota$ , so $u(g,h),v(g,h)\in K$ . Every $p$ -subgroup of $G$ is contained in a Sylow $p$ -subgroup; hence $K$ is abelian. Consequently

a_{p,n}(g,h)=[u(g,h),v(g,h)]=1.

Next let $H\leq G$ be a $p$ -subgroup and let $g,h\in H$ . The pair $(\iota(g),\iota(h))$ lies in the $p$ -subgroup $\iota(H)$ of $S_n$ . Equation (6b) gives

\iota(u(g,h))=\iota(g),\qquad \iota(v(g,h))=\iota(h).

Injectivity of $\iota$ implies $u(g,h)=g$ and $v(g,h)=h$ , proving (2).

If a Sylow $p$ -subgroup is nonabelian, choose a noncommuting pair $g,h$ in it. This pair is independent of $n$ . Equation (2) gives

a_{p,n}(g,h)=[g,h]\ne1 \qquad(n\geq |G|),

which proves the converse in (1) and the persistent failure assertion. ◻

The pointwise eventual formulation follows as well. In the nonabelian case, the fixed pair just obtained fails at every stage beyond $|G|$ . In the abelian case, the first part of the proof gives a law at every such stage.

References

Preprint · Lean (GitHub)

T. Bandman, G.-M. Greuel, F. Grunewald, B. Kunyavskiı̆, G. Pfister and E. Plotkin, Two-variable identities for finite solvable groups, C. R. Math. Acad. Sci. Paris 337 (2003), no. 9, 581–586, doi:10.1016/j.crma.2003.09.003.

R. Brandl, Finite varieties and groups with Sylow $p$ -subgroups of low class, J. Austral. Math. Soc. Ser. A 31 (1981), no. 4, 464–469, doi:10.1017/S1446788700024265.

R. Brandl, Groups with abelian Sylow subgroups, J. Aust. Math. Soc. 88 (2010), no. 1, 43–47, doi:10.1017/S1446788709000184.

E. I. Khukhro and V. D. Mazurov (eds.), The Kourovka Notebook: Unsolved problems in group theory, 21st ed., September 2026 update, Problem 11.15, arXiv:1401.0300v46.