The Power Word Problem in Graph Products · arXivDesk
2201.06543Jan 17, 2022Version 3 fixes a mistake in the previous versions. There our results on graph products were wrongly stated without the additional assumption that the base groups do not have elements of order two. Version 3 also includes some content from arXiv:1904.08343
The power word problem for a group G asks whether an expression u1x1⋯unxn
Nearby in the stack
, where the
ui
are words over a finite set of generators of
G
and the
xi
binary encoded integers, is equal to the identity of
G
. It is a restriction of the compressed word problem, where the input word is represented by a straight-line program (i.e., an algebraic circuit over
G
). We start by showing some easy results concerning the power word problem. In particular, the power word problem for a group
G
is
NC1
-many-one reducible to the power word problem for a finite-index subgroup of
G
. For our main result, we consider graph products of groups that do not have elements of order two. We show that the power word problem in a fixed such graph product is
AC0
-Turing-reducible to the word problem for the free group
F2
and the power word problems of the base groups. Furthermore, we look into the uniform power word problem in a graph product, where the dependence graph and the base groups are part of the input. Given a class of finitely generated groups
C
without order two elements, the uniform power word problem in a graph product can be solved in
AC0(C=LUPowWP(C))
, where
UPowWP(C)
denotes the uniform power word problem for groups from the class
C
. As a consequence of our results, the uniform knapsack problem in right-angled Artin groups is NP-complete. The present paper is a combination of the two conference papers. In [StoberW22] and previous iterations of this paper our results on graph products were wrongly stated without the additional assumption that the base groups do not have elements of order two. In the present work we correct this mistake. While we strongly conjecture that the result as stated previously is true, our proof relies on this additional assumption.