Equation satisfiability in solvable groups
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Equation satisfiability in solvable groups
Pawe l Idziak · Piotr Kawa lek · Jacek Krzaczkowski · Armin Weiß
Received: date / Accepted: date
Abstract
The study of the complexity of the equation satisfiability problemin finite groups had been initiated by Goldmann and Russell in [10] wherethey showed that this problem is in P for nilpotent groups while it is NP -complete for non-solvable groups. Since then, several results have appearedshowing that the problem can be solved in polynomial time in certain solvablegroups G having a nilpotent normal subgroup H with nilpotent factor G/H .This paper shows that such normal subgroup must exist in each finite groupwith equation satisfiability solvable in polynomial time, unless the ExponentialTime Hypothesis fails.
Keywords equations in groups · solvable groups · exponential timehypothesis · Fitting length
The research of the first three authors was partially supported by Polish NCN Grant no2014/14/A/ST6/00138. The fourth author has been funded by DFG project DI 435/7-1.Pawe l IdziakJagiellonian University, Faculty of Mathematics and Computer Science,Theoretical Computer Science DepartmentE-mail: [email protected] Kawa lekJagiellonian University, Faculty of Mathematics and Computer Science,Theoretical Computer Science DepartmentE-mail: [email protected] KrzaczkowskiMaria Curie-Sklodowska University, Faculty of Mathematics, Physics and Computer Science,Institute of Computer ScienceE-mail: [email protected] WeißUniversit¨at Stuttgart, Institut f¨ur Formale Methoden der Informatik (FMI)Universit¨atsstr. 38, 70569 Stuttgart, GermanyE-mail: [email protected] Pawe l Idziak et al.
The study of equations over algebraic structures has a long history in mathe-matics. Some of the first explicit decidability results in group theory are dueto Makanin [25], who showed that equations over free groups are decidable.Subsequently several other decidability and undecidability results as well ascomplexity results on equations over infinite groups emerged (see [5,9,23,29]for a random selection). Also the famous 10th Hilbert problem on Diophantineequations, that asks whether an equation of two polynomials over the ring ofintegers has a solution, was shown to be undecidable [26].One can treat polynomials over a ring R to be terms over R with somevariables already evaluated by elements of R . The same can be done withgroups to define polynomials over a group G . Now the problem PolSat ( G )takes as input an equation of the form t ( x , . . . , x n ) = s ( x , . . . , x n ) (or equiv-alently t ( x , . . . , x n ) = 1, by replacing t = s by ts − ), where s ( x ) and t ( x )are polynomials over G , and asks whether this equation has a solution in G .Obviously working with terms t , s rather than polynomials this problem triv-ializes by setting all the x i ’s to 1. Likewise PolEqv ( G ) is the problem ofdeciding whether two polynomials t ( x ) , s ( x ) are equal for all evaluations ofthe variables x in G .While for infinite groups G the problems PolSat ( G ) and PolEqv ( G )may be undecidable, they are solvable in exponential time in finite realms.In fact, PolSat ( G ) is in NP , whereas PolEqv ( G ) is in coNP . Actually thehardest possible groups that lead to NP -complete PolSat and coNP -complete
PolEqv are all groups that are not solvable [10,13]. On the other hand it iseasy to see that both these problems can be solved in a linear time for all finiteabelian groups.Also in nilpotent groups both
PolSat and
PolEqv can be solved in poly-nomial time. While the running time of the first such algorithm for
PolSat ,due to Goldmann and Russell [10], is bounded by a polynomial of very highdegree (as this bound was obtained by a Ramsey-type argument), the firstalgorithm for
PolEqv (due to [3]) is much faster. For polynomials of length ℓ the running time for PolEqv ( G ) is bounded by O (cid:0) ℓ k +1 (cid:1) , where k log | G | isthe nilpotency class of the group G . Very recently two much faster algorithmsfor PolSat ( G ) have been described. One by [7] runs in O (cid:16) ℓ | G | log | G | (cid:17) steps.The other one, provided in [21], runs even faster for all but finitely many nilpo-tent groups, i.e. in O (cid:16) ℓ | G | +1 (cid:17) steps. The very same paper [21] concludes thisrace by providing randomized algorithms for PolSat and
PolEqv workingin linear time for all nilpotent groups.However, the situation for solvable but non-nilpotent groups has been al-most completely open. Due to [15] we know that
PolSat and
PolEqv forthe symmetric group S (and some others) can be done in polynomial time.More examples of such solvable but non-nilpotent groups are provided in [12,8]. Actually already in 2004 Burris and Lawrence [3] conjectured that PolEqv quation satisfiability in solvable groups 3 for all solvable groups is in P . In 2011 Horv´ath renewed this conjecture andextended it to PolSat [11]. Actually these conjectures have been stronglysupported also by recent results in [8], where many other examples of solvablenon-nilpotent groups are shown to be tractable.Up to recently, the smallest solvable non-nilpotent group with unknowncomplexity was the symmetric group S . One reason that prevented exist-ing techniques for polynomial time algorithms to work for S is that S doesnot have a nilpotent normal subgroup with a nilpotent quotient. Somewhatsurprisingly, in [17] the first three authors succeeded to show that neither PolSat ( S ) nor PolEqv ( S ) is in P as long as the Existential Time Hypoth-esis holds. Simultaneously, in [30] the fourth author proved super-polynomiallower bounds on PolSat and
PolEqv for a broad class of finite solvablegroups – again unless ETH fails. Both the lower bounds in [17] and [30] de-pended on the so-called Fitting length, which is defined as the length d of theshortest chain 1 = G G . . . G d = G of normal subgroups G i of G with all the quotients G i +1 /G i being nilpotent.Indeed, the lower bounds in [30] apply to all finite solvable groups of Fittinglength at least four and to certain groups of Fitting length three. However,this class of groups does not include S – although its Fitting length is three.The present paper extends these results by showing super-polynomial lowerbounds for the complexity of PolSat ( G ) and PolEqv ( G ) – again dependingon the Fitting length. It strongly indicates that the mentioned conjectures byBurris and Lawrence and by Horv´ath fail by showing the following result. Theorem 1 If G is a finite solvable group of Fitting length d > , then both PolSat ( G ) and PolEqv ( G ) require at least Ω (log d − ℓ ) steps unless ETHfails. The paper [2] contains all necessary pieces to provide for
PolSat ( G ) anupper bound of the form 2 O (log r ℓ ) with r > G whenever G is afinite solvable group. This upper bound relies on the AND -weakness conjecturesaying that each CC circuit for the n -input AND function has at least 2 n δ gates. Thus, the AND -weakness conjecture implies that the lower bounds inTheorem 1 cannot be improved in an essential way.Finally, we note that allowing to use definable polynomials as additionalbasic operations to built the input terms t , s we may substantially shorten thesize of the input. For example with the commutator [ x, y ] = x − y − xy theexpression [ . . . [[ x, y ] , y ] , . . . , y n ] has linear size, while when presented in thepure group language it has exponential size. In this new setting PolSat (and
PolEqv ) have been shown [14,22] to be NP -complete (or coNP -complete,respectively) for all non-nilpotent groups. Actually our proof of Theorem 1shows this as well.Moreover, the paper [19] shows (in a very broad context of an arbitraryalgebra) that allowing such definable polynomials can be simulated by circuitsover this algebra. Pawe l Idziak et al.
Complexity and the Exponential Time Hypothesis.
We use standard notationfrom complexity theory as can be found in any textbook on complexity, e.g.[27].The Exponential Time Hypothesis (ETH) is the conjecture that there issome δ > needs time Ω (2 δn ) in theworst case where n is the number of variables of the given instance. Bythe Sparsification Lemma [20, Thm. 1] this is equivalent to the existence ofsome ǫ > needs time Ω (2 ǫ ( m + n ) ) in theworst case where m is the number of clauses of the given instance (seealso [4, Thm. 14.4]). In particular, under ETH there is no algorithm for running in time 2 o ( n + m ) .Another classical NP -complete problem is C -Coloring for C >
3. Givenan undirected graph Γ = ( V, E ) the question is whether there is a valid C -coloring of Γ , i.e. a map χ : V → [1 .. C ] satisfying χ ( u ) = χ ( v ) whenever { u, v } ∈ E . Moreover, by [4, Thm. 14.6], 3 -Coloring cannot be solved intime 2 o ( | V | + | E | ) unless ETH fails. Since 3 -Coloring can be reduced to C -Coloring for fixed C > C > C -Coloring cannot be solved in time 2 o ( | V | + | E | ) unless ETH fails. Groups and Commutators
Throughout, we only consider finite groups G . Wefollow the notation of [28]. For a groups G and H we write H G if H is asubgroup of G , or H < G if H is a proper subgroup of G . Similarly we write H P G (or H ⊳ G ) if H is a normal subgroup of G (or a proper normalsubgroup). For a subset X ⊆ G we write h X i for the subgroup generated by X , and hh X ii = h x g | x ∈ X, g ∈ G i for the normal subgroup generated by X .We write [ x, y ] = x − y − xy for the commutator and x y = y − xy for theconjugation. Moreover, we write [ x , . . . , x n ] = [[ x , . . . , x n − ] , x n ] for n > X, Y, X , . . . , X k ⊆ G defined by [ X, Y ] = h [ x, y ] | x ∈ X, y ∈ Y i and[ X , . . . , X k ] = [[ X , . . . , X k − ] , X k ]. Note here that the commutator [ H, K ] isa normal subgroup of G whenever H and K are. Finally, we put [ x, k y ] =[ x, y, . . . , y | {z } k times ] and [ X, k Y ] = [ X, Y, . . . , Y | {z } k times ].We will also need the concept of a centralizer of a subset X in G , whichis defined as C G ( X ) = { g ∈ G | [ g, h ] = 1 for all h ∈ X } . If N is a normalsubgroup, then C G ( N ) is a normal subgroup as well.Below we collect some basic facts about commutators of elements andsubgroups.(2.1) For g, x, y, z, x , . . . x n , y , . . . y n ∈ G and normal subgroups K , K , M, N of a group G and we have(i) [ xy, z ] = [ x, z ] y [ y, z ] and [ x, yz ] = [ x, z ][ x, y ] z .(ii) [ K , K ] = [ K , K ] K ∩ K and [ K K , N ] = [ K , N ][ K , N ]. quation satisfiability in solvable groups 5 (iii) If x ≡ y mod N and g ∈ M , then for all k ∈ N we have[ x, k g ] ≡ [ y, k g ] mod [ N, k M ] . (iv) If g ∈ M and x i ≡ y i mod N , then[ g, x , . . . , x n ] ≡ [ g, y , . . . , y n ] mod [ M, N ] . (v) For all f ∈ C G ( N ), g ∈ G , h ∈ N and k ∈ N we have[ hf , k g ] = [ h, k g ] [ f, k g ] . Proof
For (i), see [28, 5.1.5]. The first part of (ii) is clear from the definition,while the second one follows immediately from (i). To see (iii) and (iv), let g ∈ M , x, y ∈ G and h ∈ N with hx = y to see that[ hx, g ] = [ h, g ] x [ x, g ] ∈ [ N, M ][ x, g ] and[ g, hx ] = [ g, x ][ g, h ] x ∈ [ g, x ][ M, N ] . Then our statements follow by induction.Finally, for (v), let f ∈ C G ( N ) = { g ∈ G | [ f, h ] = 1 for all h ∈ N } and g ∈ G , h ∈ N . Then we have[ hf, g ] = [ h, g ] f [ f, g ] = [ h, g ][ f, g ] . Since C G ( N ) is a normal subgroup, also [ f, g ] ∈ C G ( N ) so that we can theninduct on k . (cid:3) Since G is finite, for all x, y ∈ G , there are i < j such that [ x, i y ] = [ x, j y ].Writing k = j − i , we get [ x, i y ] = [ x, i + k y ] for all sufficiently large i ’s. Foreach choice of x and y we might get a different value for k ; yet, by taking acommon multiple of all the k ’s, we obtain some ω ∈ N such that for all x, y ∈ G and all i > ω we have [ x, i y ] = [ x, i + ω y ].Since for normal subgroups M, N of G we have M > [ M, N ] > [ M, N ] > . . . > [ M, i N ] > [ M, i +1 N ] > . . . , the finiteness of G ensures us that there is some k ∈ N such that [ M, k N ] =[ M, k N ] for all k > k and all normal subgroups M, N of G . We can assumethat ω > k . It is clear that ω = | G | ! is large enough, but typically muchsmaller values suffice. Thus, we have:(2.2) For x, y ∈ G , M, N P G and i, j > ω we have – [ x, i y ] = [ x, i + ω y ], – [ M, i N ] = [ M, j N ].We fix ω = ω ( G ) throughout. Be aware that it depends on the specific group G . Pawe l Idziak et al.
Nilpotency and Fitting series.
The k -th term of the lower central series is γ k ( G ) = [ G, k G ]. The nilpotent residual of G is defined as T k > γ k ( G ) = γ ω ( G ) where ω is as above (i.e., γ ω ( G ) = γ i ( G ) for every i > ω ). Recall thata finite group G is nilpotent if and only if γ ω ( G ) = 1.The Fitting subgroup Fit( G ) is the union of all nilpotent normal sub-groups. Let G be a finite solvable group. It is well-known that Fit( G ) itself is anilpotent normal subgroup (see e.g. [16, Satz 4.2]). We will need the followingcharacterization of the Fitting subgroup due to Baer (see [1, Satz L’] for theproof).(2.3) Fit( G ) = { g ∈ G | [ h, ω g ] = 1 for all h ∈ G } . Now we define the upper Fitting series U ( G ) ⊳ U ( G ) ⊳ · · · ⊳ U k ( G ) = G by U i +1 ( G ) / U i ( G ) = Fit( G/ U i ( G )). If the group is clear, we simply write U i for U i ( G ). The number of factors k is called the Fitting length of G (denoted byFitLen( G )). The following fact can be derived by a straightforward inductionfrom the characterization of Fit( G ) as largest nilpotent normal subgroup.(2.4) For H P G and g ∈ G we have – U i ( H ) = U i ∩ H , for all i , – FitLen( H ) i if and only if H U i , – FitLen hh g ii = i if and only if g ∈ U i U i − . Equations in Groups. A term (in the language of groups) is a word over analphabet X ∪ X − where X is a set of variables. A polynomial over a group G is a term where some of the variables are replaced by constants – i.e., aword over the alphabet G ∪ X ∪ X − . Since we are dealing with finite groupsonly, a symbol X − ∈ X − for X ∈ X can be considered as an abbreviationfor X | G |− . We write s ( x , . . . , x n ) or short s ( x ) for a polynomial (resp. term) s with variables from { x , . . . , x n } . There is a natural composition of termsand polynomials: if r ( x , . . . , x n ) , s , . . . , s n are polynomials (resp. terms), wewrite r ( s , . . . , s n ) for the polynomials (resp. terms) obtained by substitutingeach occurrence of a variable x i by the polynomial (resp. term) s i .A tuple ( g , . . . , g n ) ∈ G n is a satisfying assignment for s if s ( g , . . . , g n ) =1 in G . The problems PolSat ( G ) and PolEqv ( G ) are as follows: for bothof them the input is a polynomial s ( x , . . . , x n ). For PolSat ( G ) the questionis whether there exists a satisfying assignment, for PolEqv ( G ) the questionis whether all assignments are satisfying. Note here that these problems havemany other names. For example in in [30,10], PolSat is denoted by EQN-SATand
PolEqv by EQN-ID.
Inducible subgroups.
According to [10], we call a subset S ⊆ G inducible if S = { s ( g , . . . , g n ) | g , . . . , g n ∈ G } for some polynomial s ( x , . . . , x n ) of G .The importance of inducible subgroups lies in the observation that one canrestrict variables in equations to inducible subgroups (simply by replacing each quation satisfiability in solvable groups 7 variable by the polynomial defining the inducible subgroup). This immediatelygives the following lemma. Lemma 1 ([10, Lemma 8], [14, Lemma 9, 10]) If H is an induciblesubgroup of G , then – PolSat ( H ) is polynomial time many-one reducible to PolSat ( G ) , – PolEqv ( H ) is polynomial time many-one reducible to PolEqv ( G ) . We will use this lemma to restrict our consideration for an appropriatesubgroup of the form γ k ( G ). We will see that such subgroups are inducible. The proof of the theorem is based on coding (by group polynomials) functionsthat imitate the behaviour of conjunctions. Unfortunately, the lengths of such n -ary conjunction-like group polynomials are not bounded by any polynomialin n and, therefore, they cannot be used to show NP -completeness of PolSat . However, the group polynomials we are going to produce have length boundedby 2 O ( n d − ) where d = FitLen( G ). Given such relatively short conjunction-likegroup polynomials we reduce graph coloring or , depending on whether | G/H | > H of G . In any case suchreduction, together with the ETH, would give the lower bound 2 Ω (log d − ℓ ) for PolSat ( G ).To see how to produce such relatively short conjunction-like polynomials,we start with the upper Fitting series of G U ⊳ U ⊳ · · · ⊳ U d = G to go downwards along this series and consecutively carefully choose h α ∈U α U α − on each level α = d, d − , . . . , U α − and h α · U α − which are supposed to simulate false andtrue values, respectively.The conjunction-like polynomials are based on the terms ˜q ( k ) ( z, x , . . . , x k )and q ( k ) ( z, x , . . . , x k , w ) for k > ˜q (0) ( z ) = z, ˜q ( k ) ( z, x , . . . , x k ) = h ˜q ( k − ( z, x , . . . , x k − ) , ω x k i , for k > , and q ( k ) ( z, x , . . . , x k , w ) = ˜q ( k +1) ( z, x , . . . , x k , w ) , for k > . Note that our definition of the q ( k ) ’s immediately yields q ( k +1) ( z, x , . . . , x k , w, w ) = q ( k ) ( z, x , . . . , x k , w ) (1) In fact, the mentioned
AND -weakness conjecture prevents the existence of such short –polynomial size – “conjunction-like” expressions. On the other hand, our construction alsoshows that the strongest version of the
AND -weakness conjecture – a 2 Ω ( n ) lower bound –does not hold. Pawe l Idziak et al. The conjunction-like behaviour of the q ( k ) ’s on the U α -cosets is precisely de-scribed in the following lemma. Lemma 2
For any level α d − and h α +1 ∈ U α +1 U α there is some h α ∈ U α U α − such that for each k ∈ N we have q ( k ) ( h α , x , . . . , x k , h α +1 ) ∈ ( h α · U α − , if x i ∈ h α +1 · U α for all i , U α − , if x i ∈ U α for some i. Proof
In this proof we may, without loss of generality, factor out our group G by U α − , or equivalently assume that α = 1. This means that U α = Fit( G )and so, by Baer’s theorem (2.3), there is some a ∈ G with [ a, ω h α +1 ] = 1.Let β ∈ N be maximal such that [ a, ω h α +1 ] ∈ γ β ( U α ) { } for some a ∈ G .Now, we simply put h α = [ a, ω h α +1 ], to observe that h α = [ h α , ω h α +1 ] and hh h α ii γ β ( U α ). The last inclusion gives that for all x , . . . , x k +1 ∈ G we have q ( k ) ( h α , x , . . . , x k +1 ) ∈ γ β ( U α ).Suppose now that one of the x i ’s is in U α . Then ˜q ( i ) ( h α , x , . . . , x i ) = (cid:2) ˜q ( i − ( h α , x , . . . , x i − ) , ω x i (cid:3) ∈ [ U α , ω U α ] = γ ω ( U α ) = { } . Hence, also q ( k ) ( h α , x , . . . , x k , h α +1 ) = 1.On the other hand, if all the x i ’s are in the coset h α +1 U α , then, by (2.1.iv),we have q ( k ) ( h α , x , . . . , x k , h α +1 ) ≡ q ( k ) ( h α , h α +1 , . . . , h α +1 , h α +1 ) = h α mod-ulo [ hh h α ii , U α ] [ γ β ( U α ) , U α ] γ β +1 ( U α ). Hence, q ( k ) ( h α , x, h α +1 ) = h α f forsome f ∈ γ β +1 ( U α ). Thus, all we have to show is that f ∈ U α − , or – in oursetting – that f = 1. To do this we induct on j > β +1 to show that f ∈ γ j ( U α )for all j ’s.Starting with f ∈ γ j ( U α ) γ β +1 ( U α ), we also have [ f, ω h α +1 ] ∈ γ β +1 ( U α ).But now, maximality of β ensures us that[ f, ω h α +1 ] = 1 . (2)Obviously [ f, g ] ∈ γ j +1 ( U α ) whenever f ∈ γ j ( U α ) and g ∈ U α . This simplymeans that f ∈ C G/γ j +1 ( U α ) ( U α /γ j +1 ( U α )), and by (2.1.v) we obtain[ h α f , ω h α +1 ] ≡ [ h α , ω h α +1 ] [ f, ω h α +1 ] mod γ j +1 ( U α ) . (3)Summing up we get h α f = [ h α f , ω h α +1 ] (by (1)) ≡ [ h α , ω h α +1 ] [ f, ω h α +1 ] mod γ j +1 ( U α ) (by (3))= h α · , (by (2))so that f ∈ γ j +1 ( U α ).Going along the j ’s we arrive to a conclusion that f ∈ γ ω ( U α ) = { } , aspromised. (cid:3) quation satisfiability in solvable groups 9 Now, picking h d ∈ G U d − , the consecutive use of Lemma 2 supplies uswith elements h d − , . . . , h that allow us to define conjunction-like polynomials q ( k ) α ( x , . . . , x k ) = q ( k ) ( h α − , x , . . . , x k , h α ). Note here that, since the terms q ( k ) use iterated commutators ( ω · ( k + 2) times), their sizes are exponentialin k . However, to get a conjunction on n = k d − elements we first split theseelements into k d − groups, each having k elements. If there were only two cosetsof G of U d − , then applying to each such k element group the polynomial q ( k ) d everything would be send into U d − ∪ h d − · U d − . Now, we group the obtained k d − values into k d − groups, each of size k and apply q ( k ) d − to each suchgroup. Repeating this procedure we finally arrive into U ensuring that theappropriate composition of the q ( k ) α ’s returns either the value 1 or h . One caneasily notice that the size of such composed polynomial is 2 O ( k ) = 2 O ( n d − ) .Unfortunately, the behaviour of the q ( k ) d ’s and the entire long compositionscan be controlled only on two cosets of U d − . This requires | G/ U d − | = 2 –which very seldom is the case. Thus, the very top level requires a very carefultreatment. First, we replace the group G with a smaller subgroup G of thesame Fitting length but such that G is abelian over its U d − . Then we finda normal subgroup U d − H ⊳ G so that we will be able to control thebehaviour of the q ( k ) ’s on all cosets of H in G . The first step towards realizingthis idea is described in the next observation. Lemma 3
In each finite solvable group G there is a subgroup G satisfying: – G is inducible, – FitLen( G ) = FitLen( G ) = d , and – G / U d − ( G ) is abelian.Proof We simply set G = γ m ( G ) where m is maximal with γ m ( G ) U d − ( G ).This secures FitLen( G ) = d . To see that all groups γ j ( G ) in the lower centralseries G = γ ( G ) > γ ( G ) > . . . > γ ω ( G )are inducible, we induct on j and argue like in [10, Lemma 5]. Let γ j ( G ) bethe image of the polynomial p ( x ). Every element in γ j +1 ( G ) = [ γ j ( G ) , G ] is aproduct of at most | G | elements of the form [ z, y ], where z ranges over γ j ( G )and y over entire G . Thus, introducing new sequences of pairwise differentvariables x , . . . , x | G | we can produce γ j +1 ( G ) as the image of the polynomial Q | G | i =1 [ p ( x i ) , y i ].Finally, G / U d − ( G ) is abelian as we have [ G , G ] = [ γ m ( G ) , γ m ( G )] [ γ m ( G ) , G ] = γ m +1 ( G ) U d − , where the last inclusion is the consequence ofthe maximality of m . (cid:3) From now on we simply change notation and replace our starting group G by G , or in other words we assume that G/ U d − ( G ) is abelian. Now, toconstruct (and control) the promised normal subgroup H first we pick K P G among the minimal (with respect to inclusion) normal subgroups satisfying: – [ K, G ] = K and – FitLen( K ) = d − γ ω ( G ) satisfies both above conditions, such K indeed exists.(3.1) K is indecomposable, i.e. if K = K K for some K , K P G then K = K or K = K . Proof
Suppose that ( K , K ) is a minimal pair (coordinatewise) with K = K K . Since K = [ K, G ] = [ K K , G ] = [ K , G ][ K , G ] and [ K i , G ] K i , weimmediately get [ K i , G ] = K i for both i = 1 ,
2. Now if K i < K , then minimal-ity of K gives FitLen( K i ) d −
2. If this happens for both K and K , then d − K ) = FitLen( K K ) = max { FitLen( K ) , FitLen( K ) } d −
2, a contradiction. (cid:3)
By (3.1) we know that there exists the unique K P G with K < K andsuch that there is no normal subgroup of G that lies strictly between K and K . Note that, if a ∈ K K , we cannot have hh a ii K . This gives(3.2) For all a ∈ K K we have hh a ii = K .The other consequence of the fact that the solvable group G has no normalsubgroups strictly between K and K is the following.(3.3) K/K is abelian.We will also need:(3.4) [ K , ω G ] U d − ( K ). Proof
By our choice of ω , we have [[ K , ω G ] , G ] = [ K , ω G ]. Since [ K , ω G ] K is strictly contained in K and K was chosen to be minimal with [ K, G ] = K and FitLen( K ) = d −
1, we must have FitLen([ K , ω G ]) d − (cid:3) Now we are ready to define the normal subgroup H of G . We simply put H to be the centralizer in G of K modulo K , i.e the largest normal subgroupwith [ H, K ] K . Then obviously H = { g ∈ G | [ K, g ] K } .(3.5) U d − H < G . In particular,
G/H is abelian.
Proof
To see that
H < G suppose otherwise, i.e. [
K, G ] K . This, however,contradicts our choice of K to satisfy [ K, G ] = K .The first inclusion is simply equivalent to [ K, U d − ] K . Indeed, sinceFitLen( K ) = d −
1, we have [ K, ω U d − ] γ ω ( U d − ) U d − and, thus,[ K, U d − ] < K . Since we assumed G/ U d − to be abelian, the second partof the statement follows. (cid:3) Directly from our definitions, we know that [ x, y ] ∈ K whenever x ∈ K and y ∈ H . But the reason for our careful choice of K and then H was to havea precise control over the behaviour of [ x, y ] for y in other cosets of H (and x still in K .) quation satisfiability in solvable groups 11 Thus, for any g ∈ G we define a map ϕ g : K → K/K by ϕ g ( x ) =[ x, g ] · K . Since by (3.3) is K/K is abelian, using (2.1.i), one can easily checkthat ϕ g is a group homomorphism for all g ∈ G . Also we have ϕ g ( K ) K , i.e. the kernel of this homomorphism contains K so that ϕ g actuallyinduces a homomorphism K/K → K/K . We also write ϕ g for this inducedhomomorphism.(3.6) If g ∈ G H , then ϕ g : K/K → K/K is an isomorphism. Proof
We start with showing that for g ∈ Gϕ g ( x b ) = ϕ g ( x ) b (4)whenever x ∈ K and b ∈ G . Indeed, by (3.5), we can write bg = hgb for some h ∈ H . Then we have ϕ g ( x b ) = [ x b , g ] · K = ( x b ) − g − b − xbg · K = ( x b ) − b − g − h − xhgb · K = ( x b ) − b − g − xgb · K (since h ∈ H )= ( x − g − xg ) b · K = ϕ g ( x ) b . To see that the kernel of the original ϕ g is K , pick a ∈ K K , so that,by (3.2), every element x ∈ K can be represented as x = a g · · · a g n for some g , . . . , g n ∈ G . Now, if ϕ g ( a ) = K , then (4) gives ϕ g ( x ) = K for all x ∈ K .This would however put g into the centralizer H , contrary to our assumption. (cid:3) Note that (4) means that ϕ g is not only a group homomorphism but ac-tually a homomorphism of G -modules. Here K/K is a G -module under theaction of G on K/K via conjugation. In terms of modules the proof of (3.6) isstated even easier: The kernel of ϕ g has to be a submodule of K/K . However,by (3.2) K/K is generated, as a G -module, by any of its non-trivial elements. Remark 1
Notice, that for (3.6), we need
G/H to be abelian. Indeed, in gen-eral, if N is a minimal (and, thus, indecomposable) normal subgroup with[ N, G ] = N , the map N → N defined by x [ x, g ] is not necessarily bijectivefor all g C G ( N ). For instance take the semidirect product ( C × C ) ⋊ D where D = (cid:10) a, b (cid:12)(cid:12) a = b = ( ab ) = 1 (cid:11) is the dihedral group of order 8 and a acts by exchanging the two components of C × C and b by inverting thesecond one. Then, N = C × C is an indecomposable normal subgroup and[ N, G ] = N but a C G ( N ) and [(1 , , a ] = [(2 , , a ] = 1, so x [ x, g ] is notbijective on N (here we use an additive notation for C = { , , } ).We summarize our observations in the following claim. (3.7) For all x ∈ K we have ˜q (1) ( x, y ) ∈ ( xK , if y H,K , if y ∈ H. Proof
Note first that ω was chosen to satisfy [ x, ω y ] = [ x, ω y ]. Moreover, fora fixed g ∈ G the unary polynomial ˜q (1) ( x, g ) acts on K as the composition ϕ ωg of ϕ g with itself ω times. Now, if g H , then (3.6) yields that ϕ ωg is theidentity on the quotient K/K . Moreover, ϕ ωg is constant K for g ∈ H . (cid:3) With claim (3.7) we are ready to construct polynomials that will allow tocode coloring or at the very top level.
Lemma 4
There is h ∈ K U d − and families of polynomials r ( k ) ( y , . . . , y k ) and s ( k ) ( y , , y , , y , . . . , y k, , y k, , y k, ) of length O ( k ) such that r ( k ) ( y ) ∈ ( h · U d − , if y i H for all i, U d − , if y i ∈ H for some i, (5) and s ( k ) ( y ) ∈ ( h · U d − , if for all i there is some j with y i,j ∈ H, U d − , if y i, , y i, , y i, H for some i. (6) Proof
First, we use (3.7) and induct on k in order to see that for all a ∈ K K we have ˜q ( k ) ( a, y , . . . , y k ) ∈ ( aK , if y i H for all i,K , if y i ∈ H for some i. Now we fix some arbitrary a ∈ K K and g ∈ G H . Then obviously also h = [ a, ω g ] ∈ K K . Actually h
6∈ U d − , as otherwise h ∈ U d − ∩ K K .Now, by (3.4) we know that M := [ K , ω G ] U d − ( K ). By (2.1.iii) itfollows that q ( k ) ( a, y , . . . , y k , g ) ∈ ( hM, if y i H for all i,M, if y i ∈ H for some i. Thus, r ( k ) ( y , . . . , y k ) = q ( k ) ( a, y , . . . , y k , g ) satisfies (5). Clearly, its length isin 2 O ( k ) .To construct the polynomials s ( k ) , we first define p ( x, y , y , y ) = x · ˜q (3) ( x, y , y , y ) − . quation satisfiability in solvable groups 13 Then for all x ∈ K , by (3.7), we have p ( x, y , y , y ) ∈ ( K , if y j H for all j,xK , if y j ∈ H for some j. Now, with a, g, h and M as above, we proceed as with the r ( k ) ’s to define˜ s ( k ) ( y ) = p ( · · · p ( a, y , , y , , y , ) , . . . , y k, , y k, , y k, )and s ( k ) ( y ) = h ˜ s ( k ) ( y ) , ω g i . As previously, (6) follows from (2.1.iii). (cid:3)
Our next claim summarizes Lemma 2 and Lemma 4.
Lemma 5
For α d − there are elements h α = 1 and families ofpolynomials r ( m ) α ( y , . . . , y m ) and s ( m ) α ( y , , y , , y , . . . , y m, , y m, , y m, ) of length O ( m d − α ) such that r ( m ) α ( y ) ∈ ( h α · U α − , if y i H for all i, U α − , if y i ∈ H for some i, and s ( m ) α ( y ) ∈ ( h α · U α − , if for all i there is some j with y i,j ∈ H, U α − , if y i, , y i, , y i, H for some i. Proof
We induct downwards on α = d − , . . . , ,
1. To start with we refer toLemma 4 to set h d − = h while r ( m ) d − ( y ) = r ( m ) ( y ) and s ( m ) d − ( y ) = s ( m ) ( y ).Now let α < d − k = ⌈ d − α √ m ⌉ and ℓ = (cid:6) mk (cid:7) . By possiblyduplicating some of the variables we may assume that m = kℓ .To define r ( m ) α ( y ) = r ( m ) α ( y , . . . , y m ) we first refer to Lemma 2 to get h α from h α +1 and then we set r ( m ) α ( y ) = q ( k ) (cid:16) h α , r ( ℓ ) α +1 ( y , . . . , y ℓ ) , . . . , r ( ℓ ) α +1 ( y m − ℓ +1 , . . . , y m ) , h α +1 (cid:17) , where the polynomial r ( ℓ ) α +1 is supplied by the induction hypothesis. FromLemma 2 it should be clear that r ( m ) α satisfies the condition claimed for it.Also its length can be bounded inductively. Substituting to the polynomial q ( k ) ( h α , x , . . . , x k , h α +1 ) of length 2 O ( k ) (by Lemma 2) the k = m d − α copies of the polynomial r ( ℓ ) α +1 of length 2 O (cid:0) ℓ d − α − (cid:1) and using ℓ = m d − α − d − α we arriveat the following bound for the length of r ( m ) α O ( k ) · O ( ℓ d − α − ) = 2 O m d − α + (cid:18) m d − α − d − α (cid:19) d − α − ! = 2 O (cid:18) m d − α + m d − α − d − α · d − α − (cid:19) = 2 O (cid:18) m d − α (cid:19) . In a very similar way we produce s ( m ) α ( y ) from the s ( ℓ ) α +1 ’s by simply putting s ( m ) α ( y ) = q ( k ) (cid:0) h α , s ( ℓ ) α +1 ( y , , y , , y , , . . . , y ℓ, , y ℓ, , y ℓ, ) , . . .. . . , s ( ℓ ) α +1 ( y m − ℓ +1 , , y m − ℓ +1 , , y m − ℓ +1 , , . . . , y m, , y m, , y m, ) , h α +1 (cid:1) . (cid:3) Now we are ready do conclude our proof of Theorem 1. Recall that due toLemma 3 we are working in the group G in which G/ U d − G is abelian. Weare going to reduce or C -Coloring to PolSat ( G ) and PolEqv ( G )depending on whether C = | G/H | > C -Coloring to PolSat ( G ) and PolEqv ( G ) works; however, the case C = 2 has to be treated in a different way since 2 -Coloring is decidable inpolynomial time.In our reduction the formula Φ from (or a graph Γ from C -Coloring )is transformed to a polynomial s Φ (or r Γ ) and a group element h so that thefollowing will hold:(A) the length of s Φ (resp. r Γ ) is in 2 O ( d − √ m ) where m is the number of clauses(resp. the number of edges),(B) s Φ (resp. r Γ ) can be computed in time 2 O ( d − √ m ) (i.e., polynomial in thelength of s Φ (resp. r Γ )),(C) if Φ is satisfiable (resp. Γ has a valid C -coloring), then s Φ = h (resp. r Γ = h ) is satisfiable, and,(D) if Φ is not satisfiable (resp. Γ does not have a valid C -coloring), then s Φ = 1 (resp. r Γ = 1) holds under all evaluations.The latter two points imply that s Φ = h (resp. r Γ = h ) is satisfiable if andonly if Φ is satisfiable (resp. Γ has a valid C -coloring) and s Φ = 1 (resp. r Γ = 1)holds identically in G if and only if Φ is not satisfiable (resp. Γ does not havea valid C -coloring).Now, if ℓ denotes the input length for PolSat or PolEqv (i.e. the size of s Φ or r Γ ), then an algorithm for PolSat or PolEqv working in 2 o (log d − ℓ ) -time would solve (resp. C -Coloring ) in time2 O ( d − √ m ) + 2 o (log d − (2 d − √ m )) = 2 o ( m ) , contradicting ETH. quation satisfiability in solvable groups 15 We start with describing the reduction from C -Coloring to PolSat ( G )and PolEqv ( G ) where C = | G/H | . The quotient | G/H | serves as the set ofcolors. For a graph Γ = ( V, E ) with E ⊆ (cid:0) V (cid:1) , | V | = n and | E | = m , we usevariables x v for v ∈ V . For an edge { u, v } ∈ E the value of x u x − v (modulo H ) decides whether the vertices u, v have the same color. To control whetherthe coloring of Γ is proper we define the polynomial r Γ by putting r Γ (( x v ) v ∈ V ) = r ( m )1 (cid:0) ( x u x − v ) { u,v }∈ E (cid:1) where r ( m )1 and h are supplied by Lemma 5 – and, thus, meet the lengthbound (A). Point (B) is clear from the definition of the polynomial. Noticethat the edges can be fed into r ( m )1 in any order without affecting the finalvalue of such polynomials. Every evaluation of the variables x v by elements of G defines a coloring χ : V → G/H in a natural way. If this coloring is valid(i.e. χ ( u ) χ ( v ) mod H for every edge { u, v } ∈ E ), then all the expressions χ ( u ) χ ( v ) − are not in H and Lemma 5 ensures us that r Γ (( x v ) v ∈ V ) = h . Thisshows (C).Conversely, by Lemma 5, for every evaluation of the x v ’s by elements of G that does not satisfy the equation r Γ (( x v ) v ∈ V ) = 1, we have x u x − v H forall edges { u, v } . This obviously yields a valid coloring of Γ – hence, it proves(D).As 2 -Coloring is solvable in polynomial time in the case | G/H | = 2, weinterpret and use the two cosets of H in G as the true/false booleanvalues. We start with the formula Φ = ( A , ∨ A , ∨ A , ) ∧ · · · ∧ ( A m, ∨ A m, ∨ A m, ) , where each literal A i,j is either one of the boolean variables X , . . . , X n or itsnegation. First, we transform the literals A i,j into the expressions x i,j that aresupposed to range over G by picking g ∈ G H and then setting x i,j = ( gx k , if A i,j = X k ,x k , if A i,j = ¬ X k . Finally, we set s Φ ( x , . . . , x n ) = s ( m )1 ( x , , x , , x , , . . . , x m, , x m, , x m, )where again s ( m )1 is supplied by Lemma 5.Now, given an assignment to the boolean variables X , . . . , X n , we obtainan assignment for x , . . . , x n by setting x i = g if X i is true and x i = 1 if X i is false . It can be easily checked using Lemma 5 that the original assignmentwas satisfying for Φ if and only if s Φ ( x ) = h is satisfied (notice that g ∈ H ).This shows (C). On the other hand, if s Φ ( x ) = 1, then, by Lemma 5, for all i there is some j with x i,j ∈ H . Hence, if we assign true to X k if and only if x k H , we obtain a satisfying assignment for Φ – proving (D). (cid:3) With Theorem 1 in mind, one could suspect that finite solvable groups ofFitting length 2 have polynomial time algorithms for
PolSat . As we havealready mentioned, the very recent paper [8] shows that
PolSat is in P formany such groups, in particular, for all semidirect products G p ⋊ A , where G p is a p -group and A is abelian. This, however, does not cover e.g. the dihedralgroup D . In fact, in [18] PolSat ( D ) is shown not to be in P , unless ETHfails. On the other hand, PolEqv ( D ) ∈ P . Actually from [8] we know that PolEqv ( G ) ∈ P for each semidirect product G = N ⋊ A where N is nilpotentand A is abelian. In fact, D is the first known example of a group withpolynomial time PolEqv and non-polynomial (under ETH)
PolSat . Theconverse situation cannot happen as, for a group G , PolSat ( G ) ∈ P implies PolEqv ( G ) ∈ P . Indeed, to confirm that t ( x ) = 1 holds for all possible valuesof the x ’s, we check that for no g ∈ G { } the equation t ( x ) = g has asolution.We conclude our paper with two obvious questions. Problem 1
Characterize finite solvable groups (of Fitting length 2) with
PolSat decidable in polynomial time.
Problem 2
Characterize finite solvable groups (of Fitting length 2) with
PolEqv decidable in polynomial time.Finally, we want to point out the consequences of our main result to anotherproblem: For a finitely generated (but possibly infinite) group with a finite setof generators Σ the power word problem is as follows: The input is a tuple( p , x , p , x , . . . , p n , x n ) where the p i are words over Σ and the x i are integersencoded in binary. The question is whether p x · · · p x n n evaluates to the identityof the group. The complexity of the power word problem in a wreath product G ≀ Z where G is a finite group has a similar behaviour as PolEqv : if G isnilpotent, the power word problem of G ≀ Z is in polynomial time [6] (actuallyeven in TC ) and, if G is non-solvable, it is coNP -complete [24]. Indeed, in[6] a surprising connection to PolEqv has been pointed out: if G is a finitegroup, then PolEqv ( G ) can be reduced in polynomial time to the power wordproblem of the wreath product G ≀ Z . In particular, Theorem 1 implies thatthe power word problem of G ≀ Z where G is a finite solvable group of Fittinglength at least three is not in P assuming ETH. References
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