Math-1191 · mathematics · Group theory

Math-1191

Kourovka Notebook Problem 21.4

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Statement

Definitions and mathematical statement

For a group $L$, its derived subgroup $L'$ is generated by all commutators $a^{-1}b^{-1}ab$ with $a,b\in L$. Repeatedly taking derived subgroups gives the derived series, starting with $L$ itself. A finite group is soluble when this series eventually becomes trivial. The soluble radical $R(G)$ of a finite group $G$ is its largest soluble normal subgroup. The hypothesis $R(G)=1$ rules out every nontrivial soluble subgroup that is normal in $G$.

Given a finite group with $R(G)=1$ and five soluble subgroups $H_1,\ldots,H_5\le G$, is there a choice $x_1,\ldots,x_5\in G$ for which \[ \bigcap_{i=1}^{5}x_i^{-1}H_i x_i=\{1\}? \] All five subgroups and all finite ambient groups satisfying the radical condition are quantified over. Repeated subgroups and the trivial subgroup are allowed, and no maximality condition is imposed on any $H_i$.

The conjugators may be chosen independently, but they must give one simultaneous trivial intersection. It is not required that every pair of conjugate subgroups have trivial intersection. Conjugating the entire configuration by the same element does not change whether the total intersection is trivial, so one could equivalently fix $x_1=1$ and choose the other four. Normality is imposed only on the radical condition; the given soluble subgroups need not be normal and generally are not. The proposed number five is an absolute bound, independent of $|G|$.

Short English statement

Can five arbitrary soluble subgroups of a finite group with trivial soluble radical be independently conjugated to have trivial total intersection?

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