Computing the order and the index of a subgroup in a polycyclic group
aa r X i v : . [ m a t h . G R ] F e b Computing the order and the index ofa subgroup in a polycyclic group
Bettina EickFebruary 9, 2021
Abstract
This contains a new version of the so-called ’non-commutative Gauss’ algorithm forpolycyclic groups. Its results allow to read off the order and the index of a subgroupin an (possibly infinite) polycyclic group.
Practical algorithms to compute with finite polycyclic groups have been described by Laue,Neub¨user & Schoenwaelder in [3]. A main basis for most of these algorithms is the so-called ’non-commutative Gauss’ algorithm. Given a finite polycyclic group G and a finiteset of generators for a subgroup U , this computes a so-called induced polycyclic generatingsequence for U . In turn, this allows to read off | U | and [ G : U ] and forms a basis for manyother algorithms.Eick [2] introduced various practical algorithms to compute with possibly infinite poly-cyclic groups. These also included an extended version of the ’non-commutative Gauss’algorithm. The proof of this extended version has gaps. It is the main aim here to cover thisopen problem and to introduce a practical and reliable version of the ’non-commutativeGauss’ algorithm for possibly infinite polycyclic groups.A GAP implementation of the code described here is available online, see [1]. We first introduce the setting of the algorithm. We assume that the group G is given bya consistent polycyclic presentation; that is, it has generators g , . . . , g n and non-negativeintegers r , . . . , r n and relations of the form g i g j = g j g a i,j,j +1 j +1 · · · g a i,j,n n for 1 ≤ j < i ≤ ng i g − j = g − j g b i,j,j +1 j +1 · · · g b i,j,n n for 1 ≤ j < i ≤ n with r i = 0 g r i i = g c i,i +1 i +1 · · · g c i,n n for 1 ≤ i ≤ n with r i > , where a i,j,k , b i,j,k , c i,k are integers that are contained in { , . . . , r k − } if r k >
0. Addi-tionally, it is required that for each element g of G there exists a unique ( e , . . . , e n ) ∈ Z n ≤ e k < r k if r k > g = g e · · · g e n n . We call g e · · · g e n n the normal form of the element g . It can be computed readily for anyarbitrary word in the generators by iteratedly applying the relations of the group.Let G i = h g i , . . . , g n i for 1 ≤ i ≤ n . The relations of G imply that G i +1 ✂ G i for1 ≤ i ≤ n − G i /G i +1 is cyclic. The consistency of the presentation implies that[ G i : G i +1 ] = r i if r i > G i : G i +1 ] = ∞ if r i = 0.We introduce some further notation for elements in a group G given by a consistentpolycyclic presentation. Let g = g e · · · g e n n be a normal form and assume that g = 1.Then • the depth of g is d if e = . . . = e d − = 0 and e d = 0. Write d ( g ). • the leading exponent of g is e d . Write l ( g ). • the relative order of g is | gG d +1 | . Write r ( g ).If g = 1, then we say that g has depth n + 1 and leading exponent or relative order of g do not exist. If g = 1, then r ( g ) = ∞ if r d = 0 and r ( g ) | r d otherwise. Let G be given by a consistent polycyclic presentation. A generating set u , . . . , u m of asubgroup U is an igs if the series U i = h u i , . . . , u m i with 1 ≤ i ≤ m coincides with theseries G i ∩ U for 1 ≤ i ≤ n where duplicates have been removed. The following has beenproved in [2]. Let u , . . . , u m be a generating set for U . Then u , . . . , u m is an igs for U ifand only if • u u j i ∈ U j +1 for ≤ j < i ≤ m , • u r ( u i ) i ∈ U i +1 for ≤ i ≤ m with r ( u i ) > , • d ( u ) < d ( u ) < . . . < d ( u m ) .Proof : We include a proof for completeness.(1) First assume that u , . . . , u m is an igs. Choose j maximal with U i = G j ∩ U . Then U i +1 = G j +1 ∩ U and thus U i +1 ✂ U i with U i /U i +1 of order r ( u i ). Thus all three itemsfollow.(2) Now assume that the three items are satisfied. By item (a) it follows that U i +1 ✂ U i and by construction and item (b) the quotient U i /U i +1 is cyclic of order r ( u i ) if r ( u i ) > ∞ if r ( u i ) = 0. Induction now yields the desired result. • We assume that generators h , . . . , h l for a subgroup U of G are given. Our aim is todetermine an igs for U . 2 .1 Normalisations of elements Let g ∈ G , g = 1, with depth d , leading exponent a and relative order r d . If r d = 0,then let e = sign ( a ) and call g e the normalisation of g . If r d >
0, then write a = xy with x = gcd ( a, m ) and y = a/x . Note that z = y − mod m exists and call g z the normalization of g . Remark:
Let g ∈ G , g = 1, with depth d and normalisation h .(a) d ( h ) = d ( g )(b) l ( h ) | l ( g ).(c) h g, G d +1 i = h h, G d +1 i .Remark 2 indicates why the normalisation of an element is of interest: in the cyclic group G d /G d +1 it yields the unique generator of h gG d +1 i with smallest leading exponent. A partial igs is a list of length n (the number of generators of the parent group G ) whose i th entry is either empty or a normalised element g in G of depth i .The following function takes a partial igs I and an arbitrary element g ∈ G and determinesa new partial igs J so that h J i = h I, g i . AddGenToPIgs ( I , g ):(1) Initialise L as the list with a single entry g .(2) While L is not empty do:(a) Take an element h from L and eliminate it in L .(b) Let d = d ( h ). If d > n then go back to (2).(c) If I [ d ] is empty then:(i) Insert the normalisation of h at position d in I .(ii) If r d > h · I [ d ] − l ( h ) /l ( I [ d ]) to L .(d) If I [ d ] is not empty then:(i) Let k = I [ d ] and b = l ( k ) and a = l ( h ).(ii) Let e = gcd ( a, b ) = ua + vb and w = h u k v .(iii) Insert the normalisation of w at position d in I .(iv) Add h · I [ d ] − l ( h ) /l ( I [ d ]) to L .(v) Add k · I [ d ] − l ( k ) /l ( I [ d ]) to L .First, note that in Steps (2d)(iv) and (2d)(v), the quotients l ( h ) /l ( I [ d ]) and l ( k ) /l ( I [ d ])are integers, since l ( I [ d ]) | l ( w ) = e = gcd ( a, b ) and a = l ( h ) and b = l ( k ). Hence theseSteps yield elements of G that are added to L .Second, in the Steps (2c)(ii), (2d)(iv) and (2d)(v) there are elements of G added to L . Allof these elements have depth greater than d . This implies that the algorithm terminateseventually. 3 Lemma:
Let J = AddGenT oP Igs ( I, g ) . Then J is a partial igs satisfying h J i = h I, g i .Proof : J is a partial igs, since we only add normalised elements at the places associatedwith their depth. It remains to prove h J i = h I, g i . ⊆ : Each element that is inserted into I during the algorithm is a product of elements of I ∪ { g } . Hence J ⊆ h I, g i and this part follows. ⊇ : We show that h L, I i does not change in the course of the algorithm. Since I ∪{ g } = I ∪ L to begin with and J = I ∪ L at the end, this yields the desired result. We consider thechanges made to I and L in the course of the algorithm. In Step (2a) we take an element h from L . There are several cases:(Case 1): I [ d ] is empty and r d = 0. Then we add h or h − to I and the result follows.(Case 2): I [ d ] is empty and r d >
0. Then we add the normalization h z to I and h · ( h z ) − q to L for some q . Hence h = h · ( h z ) − q · ( h z ) q can be obtained from L and I and the resultfollows.(Case 3): I [ d ] = k . Then we add the normalisation of ( h u k v ) to I and suitable quotientsof h and k to L . As in Case 2, the quotients yield that h and k can be recovered from L and I [ d ]. Hence the result follows in this case also. • We note two obvious improvements of the algorithm.(1) If there exists l ∈ { , . . . , n } so that l ( I [ d ]) = 1 for l ≤ d ≤ n , then we can improve thebreak in Step (2b) to: ’If d ≥ l then go back to (2)’.We can also replace the elementsin I so that I [ d ] = g d for d ≥ l .(2) In Step (2d) we insert the normalisation of w only if its leading exponent is not equalto b . Further, if the leading exponent of the normalisation of w equals either a or b ,then only one left quotient needs to be added to L . The following algorithm takes a list L of elements of G and determines an igs for thesubgroup they generate. The algorithm is based on Lemma 1. IgsByGenerators ( L ):(1) Initialise I as a list of length n with empty entries.(2) While L is not empty do:(a) Take an element g from L and eliminate it in L .(b) Run AddGenToPIgs ( I , g ).(c) Let N denote the list of changes to I in (2b).(d) For g in N do:(i) If r ( g ) is finite, then add g r ( g ) to L .(ii) For h in I with h = g add [ g, h ] to L .(3) Return I . 4he algorithm terminates, since the depths of the elements in L increases in each step.The algorithm determines an igs for h L i , since it returns a list that generates h L i andsatisfies the conditions of Lemma 1. Suppose that a subgroup U of G is given by a set of generators. Then U and [ G : U ] canboth be read off from an igs of U . Let u , . . . , u m be an igs for U , let D = { d ( u i ) | ≤ i ≤ m } and let D = { , . . . , n } \ D . (a) | U | = r ( u ) · · · r ( u m ) . (b) [ G : U ] = l ( u ) · · · l ( u m ) · Q d ∈ D r d . Suppose that two subgroups U and V of G are given. We would like to have an effectivetest for U = V . We say that an igs u , . . . , u m is canonical if the normal forms u i = g e i · · · g e in n satisfy that if d ( u k ) = d then e id ∈ { , . . . , l ( u k ) − } for 1 ≤ i ≤ m . It is not difficult todetermine a canonical igs from an arbitrary one by replacing u i by u i · u − qk for all i and k where e id = l ( u k ) q + r with 0 ≤ r < l ( u k ) is determined by division with remainder. Two subgroups U and V are equal if and only if their canonical igs coincide. References