Separating soluble subgroups

M. Anagnostopoulou-Merkouri and T. C. Burness ask whether, for all sufficiently large $n$, any two soluble subgroups $H,K$ of $S_n$ or $A_n$ admit a conjugator in the same ambient group with trivial intersection. The answer to this first question is affirmative. The additional question asking whether $n\ge21$ suffices is outside this result. Kourovka Notebook, Problem 21.3.

Theorem. There is one integer $N$ such that, whenever $n\ge N$ and $G\in\{S_n,A_n\}$, every pair of soluble subgroups $H,K\le G$ admits $x\in G$ with

In fact, define

Then $p_n\to e^{-9/2}$. The minimum ranges over all pairs in each degree, so positivity of this limit gives a cutoff independent of the subgroups. A separate parity argument gives the alternating conclusion.

Why the constant is $9/2$

Let $T(H)$ be the subgroup generated by the transpositions in $H$. The graph joining two points when their transposition lies in $H$ has the following familiar property: its edge transpositions generate the full symmetric group on each connected component. Thus $T(H)$ is a product of symmetric groups on disjoint blocks. Every block has size at most four, since a larger block would put a nonsoluble $S_5$ inside $H$.

Take the corresponding block partitions for $H$ and $K$, and relabel the second by a uniformly random permutation. Their symmetric products intersect trivially exactly when every intersection of two blocks has size at most one. Equivalently, no unordered pair of points belongs to a block of both partitions. Call such a pair a collision, and let $Z$ count collisions.

If the two partitions contain $a$ and $b$ within-block pairs, respectively, then

because every point has at most three partners and hence $a,b\le3n/2$.

Counting the mean alone is insufficient. For each fixed $j$, counting families of $j$ collision witnesses gives

uniformly over both partitions. Families on disjoint points give the leading term; overlapping families have a vanishing contribution because block sizes are bounded. Finite Bonferroni inequalities, followed by a uniform bound on the exponential-series remainder, yield

The constant is sharp. Take $H=K$ to be a product of copies of $S_4$ on four-point blocks, with one smaller block if necessary. These groups are soluble and equal their transposition cores. Here $a/n,b/n\to3/2$, so $\mu\to9/2$ and the same moment calculation gives success probability tending to $e^{-9/2}$.

The elements outside the transposition core

The remaining issue is substantial: disjoint cores do not by themselves imply disjoint groups. The structural estimate used to bridge this gap is as follows. Every soluble $H\le S_n$ is contained in a soluble group $M$ for which, with an absolute constant $C$, the numbers $a_s(M)$ of elements moving exactly $s$ points and $b_s(M)$ of such elements outside $T(M)$ satisfy

The estimate is obtained by recursively decomposing the action into orbits and blocks, embedding it in direct and wreath products of soluble primitive actions. The primitive input is elementary: a nontrivial abelian normal subgroup of a faithful primitive soluble group acts regularly. The recursive action is recorded by a forest. Encoding a permutation by its moved locations gives the first support bound; the extra restriction on elements outside the transposition core gives the second. The constant is uniform throughout this construction.

Apply this construction to both groups, obtaining $M,N$. If their cores intersect trivially after conjugation but their groups do not, a common nonidentity element lies outside at least one core. A union bound over its support and conjugacy class, using the two estimates, bounds the probability of this event by

The series with upper limit infinity converges. Combining this error with the collision estimate gives the uniform lower bound for $M,N$, and therefore for $H,K$, since enlarging groups can only reduce the proportion of successful conjugators. The four-point-block example gives the matching upper bound and proves $p_n\to e^{-9/2}$.

Making the conjugator even

For $H,K\le A_n$, enlarge them to maximal soluble subgroups $M,N$ of $S_n$. If $M$ contains an odd permutation, right multiplication by it changes a successful conjugator's parity while preserving success. If $N$ contains one, use left multiplication instead. Thus a successful conjugator can be made even in either case.

If both $M,N$ lie in $A_n$, they contain no transpositions, so their cores are trivial. Maximality makes each equal to a soluble overgroup supplied by the preceding construction. The entire bad-conjugator probability is then $O(n^{-1})$. Eventually it is less than $1/2$, whereas exactly half the permutations are even. Hence some even conjugator is successful. Combining these uniform eventual statements proves the theorem.

The solution and formalization are by Nilradical. The argument uses standard primitive-group theory and the Brun–Bonferroni method; its substantive work is the uniform support estimate and its combination with collision counting and parity. The complete formalization uses Lean and mathlib. Determining an effective cutoff, and in particular settling the proposed threshold $21$, remains beyond this theorem.