Kourovka Notebook Problem 21.25

Definitions and mathematical statement

A finite group $G$ is simple if it is nontrivial and its only normal subgroups are $1$ and $G$. For a prime $p$ dividing $|G|$, write $|G|=p^a m$ with $p\nmid m$. A Sylow $p$-subgroup is a subgroup of order $p^a$. For subgroups $U,V\le G$, the notation $\langle U,V\rangle$ denotes the smallest subgroup containing both, consisting of finite products of elements from $U\cup V$ and their inverses.

The question is whether for every finite simple group $G$ and every pair of prime divisors $p_1,p_2$ of $|G|$, one can choose Sylow $p_i$-subgroups $H_i$ with \[ \langle H_1,H_2\rangle=G. \] The primes are allowed to coincide. In that case the two Sylow subgroups may be different conjugates; the assertion does not restrict them to be the same subgroup. When the primes are different, no compatibility between the initial Sylow choices is prescribed.

The desired conclusion is generation by the union of the two subgroups, not the stronger set-product equality $G=H_1H_2$. The choices may depend on both primes and on $G$, and the assertion does not claim that every pair of Sylow subgroups works. Cyclic simple groups of prime order are included and give immediate cases; the substantive question also covers all nonabelian finite simple groups. A negative answer needs one group and prime pair for which every eligible choice generates a proper subgroup.

Short English statement

Can every finite simple group be generated by suitably chosen Sylow subgroups for any two prescribed prime divisors of its order, including equal primes?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2534 (KOU-21.25), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 The Kourovka Notebook, 21st edition, new problems, Problem 21.25. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.