1
$\begingroup$

Consider the following general problem. There is a finite group $G$ and $H_1,H_2 < G$. Suppose we know that $\langle H_1, H_2 \rangle = G$, i.e. $G$ is generated by $H_1$ and $H_2$. Denote by $n_0$ the minimal $n \in \mathbb{N}$ such that the following holds: $$ G = \underbrace{H_1 H_2 H_1 H_2 \ldots H_1}_{\text{$2n-1$ subgroups}} $$ Problem: How to estimate $n_0$?

This is, definitely, too general.

Question: Do you know about any results of such kind regarding any specific group $G$ and corresponding specific $H_1$ and $H_2$? Maybe matrix groups?

$\endgroup$
  • 2
    $\begingroup$ We have $|H_1H_2| = |H_1||H_2|/|H_1 \cap H_2|$, so you can determine whether $G = H_1H_2$ just by looking at orders of subgroups. Also, knowing how many steps it takes to cover the commutator group $[H_1,H_2]$ will give a good estimate, since $G = [H_1,H_2]H_1H_2$. $\endgroup$ – Colin Reid Jul 12 '14 at 15:04
3
$\begingroup$

Finite groups for which $n_0=1$ have been studied in various guises by various people:

  • Geometric ABA-groups by Higman and McLauglin. This paper connects the study of groups $G$ that factorize as $G=ABA$, with the study of automorphism groups of designs. This is a famous paper that has been cited many times. Note, though, that to make the connection to designs, one needs to place restrictions on the intersection $A\cap B$. The nature of this restriction varies.

  • Bruhat decompositions. This is due first to Bruhat, and later to Chevalley, and is a result for reductive linear algebraic groups $G$ (and finite groups of Lie type). The precise statement varies depending on what assumptions you are making about your group but, roughly speaking, it says that one can write $G=BNB$ where $B$ is a Borel subgroup, and $N$ is the normalizer of a maximal split torus.

  • Finally, the following paper...

    Alavi, S. Hassan; Praeger, Cheryl E. On triple factorizations of finite groups. J. Group Theory 14 (2011), no. 3, 341–360.

... studies the general question of which groups $G$ can be written as $G=ABA$. The results connect the factorization to the associated permutation representations of $G$ (on the set of cosets of $A$ and $B$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.