Math-1190 · mathematics · Group theory

Math-1190

Kourovka Notebook Problem 21.3

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Statement

Definitions and mathematical statement

Let $S_n$ be the permutation group of $\{1,\ldots,n\}$ and let $A_n$ be its subgroup of even permutations. For a group $L$, define $L^{(0)}=L$ and $L^{(j+1)}=[L^{(j)},L^{(j)}]$, where the commutator subgroup is generated by $[a,b]=a^{-1}b^{-1}ab$. The group is soluble if $L^{(j)}=1$ for some finite $j$. For a subgroup $K\le G$ and $x\in G$, put $K^x=x^{-1}Kx$.

The first question asks whether there is an integer $N$ such that for every $n\ge N$, for either choice $G=A_n$ or $G=S_n$, and every pair of soluble subgroups $H,K\le G$, there exists $x\in G$ satisfying \[ H\cap K^x=\{1\}. \] The second asks the sharper assertion with the specific threshold $N=21$. Thus it includes every $n\ge21$, not only a sufficiently large subsequence of degrees.

The two soluble subgroups may coincide, and neither is assumed maximal as a subgroup or maximal among soluble subgroups. The conjugating element may depend on both subgroups and on the ambient group. For $G=A_n$ it must be an even permutation; finding a conjugator only in $S_n$ would not satisfy that instance. The intersection is required to be exactly the identity, with no nontrivial element common to the two subgroups. One uniform threshold must handle all allowed pairs simultaneously.

The current primary edition records an affirmative solution of part (a). Both original parts are defined here; part (b), with the explicit threshold twenty-one, is the remaining question.

Short English statement

For large symmetric and alternating groups, can any two soluble subgroups be conjugated to intersect trivially, and does degree twenty-one already suffice?

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