Primitive normalisers in quasipolynomial time

Abstract

The normaliser problem has as input two subgroups H and K of the symmetric group $$\mathrm {S}_n$$ S n , and asks for a generating set for $$N_K(H)$$ N K ( H ) : it is not known to have a subexponential time solution. It is proved in Roney-Dougal and Siccha (Bull Lond Math Soc 52(2):358–366, 2020) that if H is primitive, then the normaliser problem can be solved in quasipolynomial time. We show that for all subgroups H and K of $$\mathrm {S}_n$$ S n , in quasipolynomial time, we can decide whether $$N_{\mathrm {S}_n}(H)$$ N S n ( H ) is primitive, and if so, compute $$N_K(H)$$ N K ( H ) . Hence we reduce the question of whether one can solve the normaliser problem in quasipolynomial time to the case where the normaliser in $$\mathrm {S}_n$$ S n is known not to be primitive.

Computing normalisers of intransitive groups

“…The fastest known bound for this problem, in general, is Wiebking's simply exponential bound 2 O(n) [20]. Better bounds are known for various cases: for example, quasipolynomial 2 O(log 3 n) if N Sn (H) is primitive [16,5], and polynomial if G has restricted composition factors by work of Luks and Miyazaki [13]. In terms of practical computation, there are many algorithms to solve special cases of the normaliser problem (see [8,11,14] for examples).…”