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A new notion of rank for finite supersolvable groups and free linear actions on products of spheres For a finite supersolvable group G, we define the saw rank of G to be the minimum number of sections Gk − Gk−1 of a cyclic normal series G∗ such thatGk − Gk−1 owns an element of prime order. The axe rank of G, studied by Ray[10], is the minimum number of spheres in a product of spheres admitting a freelinear action of G. Extending a question of Ray, we conjecture that the two ranksare equal. We prove the conjecture in some special cases, including that wherethe axe rank is 1 or 2. We also discuss some relations between our conjecture andsome questions about Bieberbach groups and free actions on tori.
2000 Mathematics Subject Classification. Primary: 20D15; Secondary: 20J05,57S25.
What is the minimum number a such that a given group has a free linear action on a productof a spheres? Not all finite groups have a free linear action on a product of spheres but anysupersolvable finite group does have such an action. For finite supersolvable groups, we seekan abstract group-theoretic characterization of the number a. We also discuss, in the finalsection, some related cohomological invariants.
Throughout, let G be a finite supersolvable group. An element g ∈ G is said to act freely on a CG-module X provided no non-zero element of X is fixed by g. Given a set X of CG-modules, then G is said to act freely on X provided each non-trivial element of G acts freelyon at least one of the elements of X . Imposing a G-invariant inner product on X, we get a Gaction on the unit sphere S(X) in X. Hence, G acts on the product action of G on a product of spheres can be constructed in this way, we say that the action islinear. Plainly, G acts freely on X if and only if G acts freely on Ray [10] defined a good group to be a finite group that has a free linear action on a product of spheres. She proved that any non-abelian simple factor of a good group is isomorphic toA5 or A6. The solvable group A4 is not good since its involutions do not act freely on anyCA4-module. Ray observed that any finite supersolvable group is good. The proof is easy:Consider a normal cyclic subgroup C of G. Let Y be a faithful CC-module. The non-trivialelements of C act freely on the induced CG-module IndG C (Y ). Meanwhile, by an inductive argument on |G|, each element of G − C acts freely on the inflation of some C(G/C)-module.
As required, we have shown that each non-trivial element of G acts freely on some CG-module.
We define axe(G), called the axe rank of G, to be the minimum size of a set of CG- modules upon which G acts freely. In other words, axe(G) is the minimum number of spheresin a product of spheres admitting a free linear action.
A group element of prime order is called a Cauchy element. Consider a normal series G∗ : 1 = G0 ✁ . ✁ Gt = G whose factors Gk/Gk−1 are cyclic. For 1 ≤ k ≤ t, the factor Gk/Gk−1 is said to be Cauchyprovided the set Gk − Gk−1 owns a Cauchy element. The number of Cauchy factors in G∗ iscalled the rank of G∗, denoted rk(G∗). We define the saw rank of G, denoted saw(G), to bethe minimum rk(G∗) as G∗ runs over the normal series with cyclic factors. Note that, whenG is a p-group, the factor Gk/Gk−1 is Cauchy if and only if the sequence 1 → Gk−1 → Gk →Gk/Gk−1 1 splits.
In the case where G has prime exponent p and order ps, Ray [10, Section 3] asked whether axe(G) is equal to s. In this case, s = saw(G). We extend the question and, in view of theevidence we shall accumulate, we pose it as a conjecture: Conjecture 1.1. For a finite supersolvable group, the axe rank is equal to the saw rank. Thus, we are proposing saw(G) as an abstract group-theoretic characterization of axe(G).
Note that there are examples where the other notions of rank fail to coincide with the axerank. For instance, when G is the non-abelian p-group of order p3 and exponent p, the numberof generators of G and the rank of maximal elementary abelian subgroups in G are both 2,whereas saw(G) = axe(G) = 3. The sectional rank of G (the largest number s such that everysubgroup H ≤ G is generated by s elements) is also 2. Another interesting example is given in[10] where G is the 3-group which is not meta-cyclic, but axe(G) = saw(G) = 2. So, the axerank can be strictly less than the minimum number of cyclic sections.
In the cases where we have resolved the conjecture affirmatively, the equality axe(G) = saw(G) may be of interest in its own right. The resolved cases and the reductions we haveobtained seem to be sufficiently diverse to justify the conjectural status of the equality. Propo-sition 2.5 says that the conjecture holds for all finite abelian groups and, in this case, theaxe rank and saw rank are both equal to the minimum size of a generating set. As a spe-cial case of Corollary 2.6, the conjecture holds for all p-central groups. Theorem 2.8 assertsthat axe(G) = 1 if and only if saw(G) = 1. Theorem 4.1 asserts that, when G is a p-group,axe(G) = 2 if and only if saw(G) = 2. Theorem 5.9 asserts the conjectured equality in the casewhere some maximal subgroup of G is elementary abelian.
It is an immediate consequence of Corollary 2.4 that, if the conjecture holds for all groups of prime power order, then it holds for all finite nilpotent groups. Proposition 5.5 asserts thatif the conjecture holds for all finite groups with exponent p, then it holds for all finite regularp-groups. Half of Conjecture 1.1 is almost trivial: Lemma 2.2 says that axe(G) saw(G).
The axe rank and saw rank are, in different ways, the sizes of minimal covers of the Cauchy elements. When G acts freely on X , each element X ∈ X sweeps out a ragged swath consistingof those Cauchy elements that act freely on X. The axe rank is the minimum number ofswipes needed to reap all the Cauchy elements. On the other hand, a subnormal series G∗ forG cuts the set of non-trivial elements neatly into slices G1 − G0, ., Gt − Gt−1. A Cauchy factorGk/Gk−1 accounts for precisely those Cauchy elements that belong to Gk − Gk−1.
Although it is rather difficult to characterize the groups with a fixed saw rank, some observations can be made in the case of saw rank 2. A brief discussion of classification ofp-groups with saw rank 2 can be found in Section 4. For more general characterizations, it would be desirable to have a way of constructing a cyclic normal series G∗ with the minimalrank rk(G∗) = saw(G). We must leave that as an open problem.
In Section 6, we discuss relations between free linear actions on products of spheres and free actions on tori. The main observation is that, given a free linear action on a product ofk spheres, there is a natural way to construct a free action on a torus, where the group actson the homology of the torus by permuting a basis with k orbits. When G acts freely on atorus X = T n, the fundamental group of orbit space X/G is a Bieberbach group. So thereis an associated Bieberbach group for every free linear action on a product of spheres. Usingthese associations we show that the axe-saw conjecture would follow from affirmative answersto certain questions about Bieberbach groups and free actions on tori.
Finally, we would like to point out that the saw rank conjecture has applications to product actions on products of spheres. Given G-spaces X1, . . . , Xk, the diagonal G-action on X1 ×· · · × Xk is called a product action. Notice that free linear actions are product actions whereG acts on each sphere through a complex representation. Dotzel and Hamrick proved in [6]that when G is a p-group, G-actions on mod p-homology spheres resemble linear actions. Theresemblance is explained through dimension functions. In particular, their result implies thatif G acts on a mod p homology sphere, it acts on a sphere linearly with exactly same groupelements acting freely. So, given a free product action on products of k-spheres, there is a freelinear action on same number of spheres. In this way, for p-groups, the saw-axe conjectureapplies to product actions as well.
Some general properties and easy consequences We begin with two easy but very useful lemmas.
Lemma 2.1. Given H ≤ G, then axe(H) axe(G) and saw(H) saw(G). Proof. For the first inequality, observe that if G acts freely on a set X , then H also acts freely byrestriction. For the second, let G∗ be a normal cyclic-factor series for G with rk(G∗) = saw(G),and let H∗ be the series obtained by intersecting each term with H. Then rk(H∗) rk(G∗) =saw(G).
Lemma 2.2. axe(G) saw(G). Proof. Let G∗ be a normal cyclic-factor series for G with rk(G∗) = saw(G). For each Cauchyfactor Gk/Gk−1, let Yk be a CGk-module such that ker Yk = Gk−1. Let Xk = IndG every Cauchy element of Gk −Gk−1 acts freely on Xk. Hence, G acts freely on the set {Xk}.
We shall consider some special cases. First, we need a lemma: Lemma 2.3. For a direct product G = G × G , we have axe(G) axe(G ) + axe(G ) andsaw(G) saw(G ) × saw(G ). Furthermore, if |G | and |G | are coprime, then axe(G) =max(axe(G ), axe(G )) and saw(G) = max(saw(G ), saw(G )). Proof. The inequality for axe rank holds by considering induction from G and G to G. Theinequality for saw rank is obvious. Now suppose that |G | and |G | are coprime. By Lemma 2.1,we need only show that axe(G) max(axe(G ), axe(G )) and similarly for saw(G). If G andG act freely on {X1, ., Xα} and {X1 , ., Xα}, then G acts freely on {X1 ⊗ X1 , ., Xα ⊗ Xα}.
If G∗ and G∗ are normal cyclic-factor series for G and G , then there exists a normal cyclic- factor series G∗ for G such that the j-th Cauchy factor of G∗ is isomorphic to the direct productof the j-th Cauchy factor of G∗ and the j-th Cauchy factor of G∗.
Corollary 2.4. Suppose that G is nilpotent. Then the axe rank of G is the maximum axe rankof a Sylow subgroup of G. The saw rank of G is the maximum saw rank of a Sylow subgroupof G. Proposition 2.5. If G is abelian, then axe(G) = rk(G) = saw(G). Proof. By the previous two results, axe(G) saw(G) rk(G). Suppose that G acts freelyon a set of irreducibles {X1, ., Xa}, and let χ1, ., χa be the corresponding characters. Thefunction g → (χ1(g), ., χa(g)) is a group monomorphism, so rk(G) axe(G).
Alternatively, for abelian G, we can obtain the inequality axe(G) rk(G) by the following counting argument. Let p be a prime with maximal multiplicity r in |G|, and let E be themaximal elementary abelian p-subgroup of A. Then rk(E) = r = rk(G). By Lemma 2.1,we may assume that E = G. Hence, the kernel of any irreducible CG-module has index pin G. The intersection of a such kernels has index at most pa. Again, we have shown thataxe(G) rk(G).
Corollary 2.6. If all the Cauchy elements of G are central, then axe(G) = axe(Z(G)) =saw(Z(G)) = saw(G). Proof. By considering induction and restriction, it is easy to see that axe(G) = axe(Z(G)). Byextending a normal cyclic factor series for Z(G), we deduce that saw(G) = saw(Z(G)). Themiddle equality holds by Proposition 2.5.
Let us compare the saw rank with some other ranks. Recall that, for a finite group H, the rank of H, written as rk(H), is defined to be the largest r such that (Z/p)r ≤ H for someprime p. The minimal number of generators of H is denoted by d(H). The sectional rank,srk(H), is defined to be the maximal d(K) over all subgroups K ≤ H.
Proposition 2.7. We have rk(G) saw(G). If G is a p-group with p > 2, then d(G) srk(G) saw(G). Proof. The first inequality follows from Lemma 2.1 and Proposition 2.5. When G is a p group,with p > 2, Laffey [9] proves that d(G) max{ logp |K| : K ✂ G, [K, K] ≤ Z(K), exp(K) = p }. Since logp |K| = saw(K) when K has exponent p, and saw(K) saw(G) for all subgroupsK ≤ G, we get d(G) saw(G). Applying this to each subgroup, we get srk(G) saw(G).
In Section 4, we shall see that the following result is an easy consequence of some material in Section 3. Let us also give a direct proof.
Theorem 2.8. The following conditions are equivalent: (a) saw(G) = 1,(b) axe(G) = 1,(c) Every subgroup whose order is a product of two primes is cyclic,(d) The Cauchy elements generate a cyclic subgroup. Proof. By Lemma 2.2, (a) implies (b). A special case of Wolf [13, Theorem 5.3.1] says that(b) implies (c). Supposing that (d) holds, consider a normal series G∗ with cyclic quotientsand such that G1 is the subgroup generated by the Cauchy elements. Then rk(G∗) = 1, andcondition (a) holds.
Before showing that (c) implies (d), let us recall a general property of finite supersolvable groups: Any maximal normal abelian subgroup A of G is maximal as an abelian subgroup ofG. To see this, suppose, for a contradiction, that A is a maximal normal abelian subgroup suchthat A < CG(A). The normal series 1 ✂ A ✂ CG(A) ✂ G refines to a chief series G∗. WritingA = Gk−1, then Gk is a normal abelian subgroup of G. This contradicts the maximality of A.
Now assume (c). Let A be a maximal normal abelian subgroup of G. By the hypothesis, the Sylow subgroups of A are cyclic. To deduce (d), we may assume, for a contradiction, thatan element g of prime order p belongs to G − A. Consider the conjugation action of g on A.
For each prime divisor q of |A|, the conjugation action stabilizes the Sylow q-subgroup Q ofA, and also stabilizes the subgroup Q0 ≤ Q generated by the Cauchy elements. Notice thatsince Q is cyclic, Q0 is cyclic of order q. If q = p, we obtain a contradiction by observing thatg and Q0 generate an elementary abelian p-group of rank 2. Supposing now that q = p, thehypothesis implies that g Q0 is cyclic. If a non-trivial automorphism of a cyclic q-group hasorder coprime to q, then it restricts to a non-trivial automorphism of the subgroup of order q.
Therefore g must centralize Q. We have shown, in fact, that g centralizes A. This contradictsthe condition that A is maximal as an abelian subgroup.
Recall that the quaternion groups and the cyclic groups of prime-power order are the only finite groups with a unique subgroup of prime order. (See, for instance, Ashbacher [2, Exercise8.4]). So, when G is a p-group, the axe rank and saw rank of G are unity if and only if G isquaternion or cyclic.
We have noted that the definition of the saw rank is purely group theoretic. In order to relatethe axe rank and the saw rank, it will help to have a purely group theoretic description of theCauchy elements that act freely on a suitable CG-module. Given subgroups K and H of Gsuch that K ✂ H ✂ G and H/K is cyclic, the subset is called a swath of G. Given a CG-module X, we write C(X) for the set of Cauchy elementsthat act freely on X.
Recall that a CG-module X is said to be monomial provided X is induced from a 1- Lemma 3.1. Let H − K be a swath. Let Y be any 1-dimensional CH-module with kernel K, and let X be the monomial CG-module IndG H (Y ). Then C(X ) is the set of Cauchy elements Proof. As a CH-module by restriction, X is the sum of the 1-dimensional modules of the formY ⊗ g where g ∈ G. The elements of H − Kg are precisely the elements x ∈ G such that x(piecewise) stabilizes Y ⊗ g and x does not (pointwise) fix the elements of Y ⊗ g.
Lemma 3.2. Let X be a monomial CG-module. Then there exists a swath H − K such that |H : K| is square-free and C(X) ⊆ H − K. H Y where Y is a 1-dimensional CH -module. Any Cauchy element not in the core of H must permute the spaces Y ⊗ g non-trivially, and therefore fixes a non-zerovector. So any Cauchy element acting freely on X must belong to the core of H . ReplacingH with its core and X with its restriction to the core, we can assume that X is a monomialCG induced from normal H.
Let K be the kernel of Y . By Lemma 3.1, C(X) is contained in H − K. Assume that H is minimal with this property. Let L be the subgroup of H such that K ≤ L ≤ H and L/K is thesubgroup of H/K generated by the Cauchy elements in H/K. Plainly, |L : K| is square-free.
Any Cauchy element x ∈ H − K also belongs to L − K. The conjugates of x, being Cauchy elements in H − K, also belong to L − K. Therefore By the minimality assumption, H = L. Therefore |H : K| is square-free.
If G acts freely on a set X of CG-modules, then each element X ∈ X can be replaced with an irreducible summand of X; the action will still be free. It is well-known (see Serre[11, Theorem 16]) that every irreducible complex representation of a finite supersolvable groupis monomial. So, we can apply the above two lemmas to the free actions on set of arbitraryCG-modules, and obtain the following alternative characterization for the axe rank.
Proposition 3.3. The axe rank axe(G) is the minimum number a such that all the Cauchyelements belong to a union j of a swaths of G. Furthermore, we may assume that each index |Hj : Kj| is square free.
Proof.
The assertion now follows from Lemmas 3.1 and 3.2.
Lemma 3.4. Suppose that G is a non-trivial 2-group. Given a monomial CG-module X, thenC(X) is contained in a swath H − K such that |H : K| = 2 and K ✂ G. Proof. We may assume that C(X) is non-empty. Let K be the kernel of X, and let H be thesubgroup generated by K and C(X). The elements of C(X) act on X as multiplication by 1,so the product of any two of them belongs to K, so |H : K| = 2.
When G is a p-group with p odd, C(X) need not be contained in a swath H − K with K✂G.
Indeed, let G be the wreath product of C3 C3, let H be the normal subgroup C3 × C3 × C3,and put X = IndG H (Y ) where Y has kernel C3 × C3 × 1. Writing the elements of H as vectors (x, y, z) over the field of order 3, then C(X) consists of the 8 vectors whose coordinates x, y, zare all non-zero. Let K be the subgroup of H consisting of the vectors whose coordinates sumto zero. Then K is the unique index p subgroup of H such that K ✂ G. But H − K does notcontain C(X).
Lemma 3.5. Suppose that G is a p-group with p odd. Let H − K be a swath of G such that K is cyclic. Then the set of Cauchy elements in H − K is contained in a swath H − K of Proof. Let H be the subgroup of H generated by the Cauchy elements, and let K = H ∩ K .
Ashbacher [2, 23.4] says that, for p-groups with class at most 2, the Cauchy elements generatean elementary abelian subgroup. In particular, H is elementary abelian. Since H /K and Kare cyclic, and 1 < K < H, we have |K| = p and |H| = p2.
Supersolvable groups of low axe and saw rank Using swaths, the rank 1 case of Conjecture 1.1 is very easy. Indeed, we can now give a quickerproof of Proposition 2.8. Trivially, (d) implies (c). By Lemma 2.1, (c) implies (a). As notedbefore, (a) implies (b) by Lemma 2.2. Assume (b). By Proposition 3.3, the Cauchy elementsof G all belong to some swath H − K. Since K is trivial, the normal subgroup H of G is cyclic. We have deduced (d), and the argument is complete.
Theorem 4.1. Suppose that G is a p-group. Then axe(G) = 2 if and only if saw(G) = 2. Proof. By Theorem 2.8, it suffices to show that axe(G) 2 if and only if saw(G) 2. Onedirection is immediate from Lemma 2.2. For the other direction, suppose that axe(G) 2.
Let C be the set of Cauchy elements of C. By Proposition 3.3, we can write where |H1 : K1| = p = |H2 : K2|.
First, let us assume that p = 2. If |K1| = 1 = |K2|, then H1 and H2 are normal cyclic groups of order p, and C ∪ {1} = H1 ∪ H2. Hence G is cyclic and saw(G) = 1. Suppose that|K1| = 1 = |K2|. Then |H1| = p. All the Cauchy elements of K2 belong to H1, so H1 is theunique subgroup of K2 with order p. So K2 is cyclic. (See the comment at the end of Section2). By Lemma 3.5, we may assume that H ∼ 2 = Cp × Cp. The normal series 1 ✁ H1 ✁ H2 ✂ G refines to a chief series with rank 2. So saw(G) 2.
Now consider the case where both K1 and K2 are non-trivial. The intersection C ∩ K1 ∩ K2 is empty, so K1 ∩ K2 is trivial. The subgroup H1 ∩ K2 owns all the Cauchy elements of K2 andis isomorphic to a subgroup of the cyclic group H1/K1. Therefore K2 has a unique subgroupof order p. Again K2 is cyclic. Similarly, K1 is cyclic. As before, we may assume that H1and H2 are isomorphic to Cp × Cp. But K1 and K2 are both contained in H1 and are bothcontained in H2, so H1 = K1K2 = H2. The normal series 1 ✁ Z(G) ∩ H1 ✁ H1 ✂ G refines toa chief series with rank 2. The case p = 2 is finished.
Now assume that p = 2. By Lemma 3.4, we may assume that K1 and K2 are normal in G. When both K1 and K2 are trivial, the argument is the same for odd p. Supposing thatonly K1 is trivial, then H1 is the unique subgroup of order 2 in K2. It follows that the normalseries 1 ✁ H1 ✂ K1 ✁ H2 ✂ G refines to a chief series with rank 2.
Finally, supposing that both K1 and K2 are non-trivial, then they own central involutions c1 and c2, respectively. Furthermore, c1 and c2 are distinct because K1 and K2 intersecttrivially. By Lemma 2.1, G does not contain an elementary abelian subgroup of rank 3. SoC ⊆ c1, c2 and, once again, saw(G) = 2.
In the above proof, the separation of the cases p > 2 and p = 2 was necessary because the Cauchy elements do not need to generate a subgroup of exponent 2 when G is a 2-group. Forexample, the group G = D8, the dihedral group of order 8, is such a group and has axe andsaw rank 2.
Let us now discuss the p-groups with saw rank 2. By Lemma 2.7, such groups have rank 2, and when p > 2 all of its subgroups are generated by 2 elements. Using Blackburn’s workon these groups, we prove the following: Proposition 4.2. For odd p, a p-group of saw rank 2 is either meta-cyclic or a 3-group ofmaximal class. Proof. Theorem 4.2 in [4] tells us immediately that, given a p-group G of order greater than orequal to p6 such that each subgroup of order p4 is generated by two elements, then G is eithermeta-cyclic or a 3-group of maximal class. For smaller groups, we use the fact that the groupsof saw rank 2 cannot include a subgroup of order p3 and exponent p. For groups of order p5,this observation disposes of the exceptional cases given in [4, Theorem 4.2]. For p-group G withsaw rank 2 of order less than p5, [4, Theorem 3.2] implies that if G is not meta-cyclic, then Gis a group of order 34 and of maximal class. (This is the group mentioned in the introductionas an example of a non-meta-cyclic group of saw rank 2).
For 2-groups, the situation is more complicated. A simple example, G = Q8, the quaternion group of order 8, shows that the number of generators may be more than the saw rank. (Inthis case saw rank is 1, whereas the number of generators is 2.) By Proposition 2.7 and Theorem 2.8, we know that the groups with saw rank 2 must have rk(G) = 2. On the other hand, there are groups with rank 2 with saw rank strictly bigger than2. We give an example of such group after the following proposition.
Proposition 4.3. Let G be a 2-group with saw(G) = 2. Then the subgroup generated by theCauchy elements has an index 2 subgroup that is cyclic or generalized quaternion. Proof. Let G be a 2-group of saw rank 2, and let Ω(G) denote the subgroup generated by theCauchy elements in G. By Lemma 2.1, Ω(G) has saw rank 1 or 2. If G∗ : 1✁G1✁· · · Gn−1✁Gn =Ω(G) is a cyclic series for Ω(G) with rk(G∗) = saw(Ω(G)), then the top section must includea Cauchy element, so Gn−1 must be cyclic or generalized quaternion.
Example 4.4. Let G be the central product of D8 with Q8. This is the quotient group ofD8 × Q8 with kernel c1c2 where c1 and c2 are central elements of order 2 in D8 and Q8.
Let a and b be involutions generating D
8, and let c and d be generators of Q8. The elementsa, b, abc, abd are Cauchy elements and they generate G. On the other hand, every subgroupH ≤ G of index 2 fits into a central extension of the form 0 Z/2 → H → (Z/2)3 0which shows that H has rk(H) 2. Hence, no index 2 subgroup of G is cyclic or generalizedquaternion. The classification of groups with saw rank two seems to be manageable problem. This might serve as a first step for the classification of groups with rank 2, which is recognized as adifficult problem.
The exponent p case, and related cases In Proposition 5.4, we return to the case of Conjecture 1.1 originally raised (as a question) byRay, namely the case where G has exponent p. But that case is more general than it appearsto be, since Proposition 5.5 says that the case of a regular p-group reduces to the exponentp case. In Theorem 5.9, we show that the conjecture holds for a certain class of non-regularp-groups.
Throughout this section, we let G be a p-group, and we write Our conjectured equality is a = s. Lemma 2.2 already tells us that a ≤ s. We seek to provethe reverse inequality.
Suppose that exp(G) = p. Then |G| = ps. Proposition 5.4, below, implies that if s ≤ p + 2, then a = s. To prove it, we first need a technical definition and some lemmas. Let us saythat G is inductible provided, whenever G acts freely on a set consisting of a irreducibleCG-modules, at least of them is 1-dimensional.
Lemma 5.1. Suppose that exp(G) = p. If G is inductible and axe(H) = saw(H) for everymaximal subgroup H of G, then a = s. Proof. Let G act freely on a set X consisting of a irreducible CG-modules one of which, sayX, is 1-dimensional. The kernel K of X has order ps−1. If s = 1, then K acts freely on theset of restrictions of X − {X}. We have s − 1 = saw(K) = axe(K) ≤ a − 1.
Lemma 5.2. Suppose that exp(G) = p. If every subgroup of G is inductible, then a = s. Proof. This follows from Lemma 5.1 via an inductive argument.
Lemma 5.3. Suppose that exp(G) = p. If a ≤ p + 1, then G is inductible. Proof. It suffices to show that whenever G acts freely on a set X of monomial CG-modules allof dimension greater than unity, we have |X | ≥ p + 2. Consider an element X ∈ X , and letH − K be a swath such that |H : K| = p and C(X) ⊆ H − K. Since H = G, we have |C(X)| ≤ ps−1 − ps−2. C(X) = G − {1}, hence |X |(ps−1 − ps−2) ≥ ps − 1. But G is non-abelian, so s ≥ 3. Therefore |X |(p − 1) ≥ p2, that is to say, |X | ≥ p + 2.
Proposition 5.4. Suppose that exp(G) = p. If a = s, then p + 1 < a < s. Proof. This is immediate from Lemmas 5.2 and 5.3.
Recall that a p-group is said to be regular provided, for all elements x and y, and any n = , we have (xy)n = xnyns where s is a product of n-th powers of elements of the derivedgroup of x, y . Regular p-groups are discussed in Hall [7, Section 12.4]. We mention thatevery p-group with nilpotency class less than p is regular. In particular, every p-group of orderat most pp is regular.
As noted in Hall [7, Theorem 12.4.5], the Cauchy elements of a regular p-group G, together with the identity element, comprise a normal subgroup Gp of G.
Proposition 5.5. Suppose that G is regular. If axe(Gp) = saw(Gp), then a = s. Proof. The normal series 1 ✂ Gp ✂ G refines to a chief series with rank saw(Gp). Hences ≤ saw(Gp). By Lemma 2.1, axe(Gp) ≤ a.
Therefore, if Conjecture 1.1 holds for all groups of exponent p, then it holds for all regular p-groups. Furthermore, Propositions 5.4 and 5.5 imply: Corollary 5.6. If |G| ≤ pp, then a = s. An example of a non-regular p-group is the wreath product Cp Cp. Indeed, Cp Cp is generated by two Cauchy elements but, on the other hand, observing that Cp Cp is isomorphicto the Sylow p-subgroups of the symmetric group Sp2, we see that exp(Cp Cp) = p2. As aspecial case of Theorem 5.9, below, axe(Cp Cp) = p = saw(Cp Cp). Again, we put part of theproof in some preliminary lemmas.
Lemma 5.7. Suppose that G is a semidirect product EC where |C| = p and the normalsubgroup E is elementary abelian. Then any irreducible CG-module of dimension greater thanunity is induced from an irreducible CE-module. Proof. It is well-known that the assertion still holds when E is replaced by any abelian sub-group of index p. We give a short proof for completeness. Let X be a simple CG module ofdimension greater than unity, and let Y be a 1-dimensional summand of ResG nius Reciprocity, X must be summand of IndG E (Y ). But the dimension of X is divisible by p, Lemma 5.8. Suppose that G = EC as in the previous lemma. Regard E as a vector spaceover the field Fp of order p, and hence regard E as an FpC-module. Then G has exponent p ifand only if no direct summand of the FpC-module E is free. Proof. We may assume that E is indecomposable as an FpC-module. It is well-known thatthe free FpC-module of rank unity has a unique composition series 0 < M1 < . < Mp = FpCwhere dim(Md) = d. Furthermore, each Md is indecomposable, and every indecomposableFpC-module is isomorphic to one of the modules Md (see, for instance, Landrock [8, SectionI.8]). Write E ∼ = Md, and identify G with the subgroup MdC of MpC. We are to show that exp(G) = p if and only if d < p.
The elements of MpC can be written in the form (x1, ., xp)with xk ∈ Fp; the group (x1, ., xp)(y1, ., yp)= (x1 + y1+β, ., xp + yp+β)gαgβ. Here, the subscripts are interpreted modulo p. The Frattini subgroup of MpC is the abeliangroup Mp−1, which consists of the elements of the form (x1, ., xp) such that x1 + . + xp = 0.
We have ((x1, ., xp))p = (x, ., x), where x = x1 + . + xp. So MpC − Mp−1C is precisely the set of elements of order p2. Inparticular, the group E = MdC has exponent p if and only if d < p.
Theorem 5.9. Suppose that G has an elementary abelian p-subgroup with index p. Write|G| = pn. If exp(G) = p, then a = n = s, otherwise a = n − 1 = s. Proof. Write n = e + 1. Let E be an elementary abelian subgroup of G with |E| = pe. ByLemmas 2.2, 2.1 and Proposition 2.5, e + 1 ≥ s ≥ a ≥ axe(E) = e = saw(E). Our task is to show that if exp(G) = p then a = e + 1, otherwise s = e. We may assume thatG − E owns a Cauchy element g, since otherwise s = e. Writing C for the subgroup generatedby g, then G = EC as a semidirect product.
Suppose that exp(G) = p. The element g does not act freely on any CG-module induced from E. By Lemma 5.7, G is inductible. By an inductive argument on e, we have axe(H) =saw(H) for every proper subgroup H of G. Lemma 5.1 now yields a = s = e + 1.
Now suppose that exp(G) = p. As in Lemma 5.8, we regard E as an FpC-module. First, consider the case where E is indecomposable. Lemma 5.8 tells us that E ∼ are now dealing with the case G ∼ = Cp Cp.) Let f be any element of G with order p2.
Using the above formula for the group operation, an easy calculation shows that the conjugacyclass of f is the coset Mp−1f. So every element of Mp−1f − Mp−1 has order p2, and rk(M1 ✁ . ✁ Mp−1 ✁ Mp−1f ✁ MpC) = p. We have shown that s ≤ p when E is indecomposable. In fact, we must have equality s = a = p.
For the general case G = EC, Lemma 5.8 allows us to write E = M ⊕ N as a direct sum of FpC-modules, where M is free of rank unity. As we saw above, the quotient group G/N has achief series where one of the factors is non-Cauchy. So the normal series 1 ✂ N ✁ G refines toa chief series with rank e. Again, we have shown that s ≤ e and, again, we must have equalitys = a = e.
Special classes associated to free linear actions Let G be a finite group and M a ZG-lattice (a Z-free ZG-module). Let H∗(G, M ) denote thecohomology of G in twisted coefficients M . In particular, H2(G, M ) denotes the equivalenceclasses of factor sets f : G × G → M . Recall that, for every subgroup H ≤ G, the inclusionmap gives rise to the restriction map ResG H : H ∗(G, M ) → H ∗(H, M ).
Definition 6.1. A cohomology class α ∈ H2(G, M ) is called as a special class if ResG Special classes appear in many contexts. Given a group extension of the form 0 → M → Γ → G → 1, it is known that Γ is torsion free if and only if the associated cohomology class α ∈ H2(G, M )is a special class (see, for instance, [14]). These types of extensions appear as short exactsequences of fundamental groups associated to a free action on a torus.
The most common appearance of special classes is in the study of compact flat manifolds (Riemannian manifolds with zero curvature). It is well known that Γ is isomorphic to thefundamental group of such a manifold if and only if it fits into an extension 0 → M → Γ →G → 1 where G is finite and M is a free abelian and maximal abelian in Γ (see Charlap [5]).
Such a group Γ is called a Bieberbach group. The group G is the holonomy group of thecorresponding manifold. The condition that M is a maximal abelian subgroup is equivalentto M being a faithful ZG-lattice. In fact, given an arbitrary ZG-lattice M with kernel K ≤ Gand a special extension (extension with associated class special) 0 → M → Γ → G → 1, thegroup extension is necessarily abelian (see, for instance, Theorem 5 in [14]), so Γ fits into an extension 0 → A → Γ → G/K → 1 where A is now maximal abelian in Γ. Therefore, for any ZG-lattice M , the extension groupΓ of a special extension is a Bieberbach group.
Proposition 6.2. Let G be a finite group and M a ZG-lattice. Then the following are equiv-alent:(i) There is a special class α ∈ H2(G, M ).
(ii) There is an extension 0 → M → Γ → G → 1 where Γ is torsion free.
(iii) The group G acts freely on a torus X = T n where H1(X, Z) Proof. For (i) (ii) and (iii) (i), see [14]. We only need to show (ii) (iii). By the abovediscussion, Γ is a Bieberbach group, so Γ imbeds into the group of isometries of Rn wheren = dim M (see [3]). Since M acts as translations, Rn/M = T n, and the group G = Γ/M actsfreely on Rn/M .
Unlike the case of free actions on products of spheres, for every finite group G, we can find a free G-torus. In other words, for every group G, there is a suitable M such that H2(G, M )has a special class. In fact, if we take M as the direct sum of all induced modules IndG all cyclic subgroups C ≤ G, we have H1(C, Q/Z) So, by picking nontrivial homomorphisms for each cyclic subgroup, we can form a special classin H2(G, M ). But this is not the most efficient way to get such a class, since M is usuallyvery big. In general, it is a difficult problem to find the minimal dimension of M for a givenholonomy group G.
In the rest of this section we show that axe-saw conjecture is related to a form of this minimal dimension problem. We consider the case where M is a permutation module. Recallthat a module is a permutation module if it is a direct sum of modules of the form IndG H Z. Notice that the Z-rank of M G is equal to k, the number of Question 6.3. Let G and M be as above. If there is a special class in H2(G, M ), does itfollow that saw(G) ≤ k ? It is known that the answer is affirmative when G is abelian (see [1], [14]). We shall show below that an affirmative answer to this question implies the axe-saw conjecture.
Let G be a finite group and H a subgroup of G. Let Y ∈ Hom(H, C×) be a 1-dimensional representation of H and let X = IndG H Y . Recall that there is an isomorphism which maps each 1-dimensional representation to its first Chern class (see page of 67 of [12]).
Let γ denote the first Chern class of Y . Note that the above isomorphism commutes withrestrictions. Hence, ResH C ch1(Y ) = ch1(ResH C (Y )) for any subgroup C of H . In fact, this last identity holds more generally for higher dimensional representations as well.
We want to define a class in H2(G, IndG H Z) associated to a given γ in H 2(H, Z). This can be done using Shapiro’s Lemma, which states that for every subgroup H in G. Let α be the image of γ under Shapiro’s isomorphism.
Lemma 6.4. If g is a Cauchy element which acts freely on X, then ResG Proof. Let us denote the cyclic subgroup generated by g by C. By Lemma 3.1, the element glies in the core of H, so C ∩ Hx = C for every x ∈ G. Now, consider the following double cosetformula, where E denotes the set of coset representatives: C α is nonzero if one of the components is zero. By considering the component corresponding to x = 1, we see that it suffices to show ResH Since C acts freely on Y , the restricted module ResG C Y is non-trivial, hence ch1(ResH non-trivial. By the naturality of Chern isomorphism, we conclude that C [ch1(Y )] = ch1[ResH Observe that the converse of the lemma does not hold in general. To see this, observe that, for a prime order cyclic subgroup C, the restriction ResG of the terms in the double coset formula is non-zero. The terms in the double coset formulacorrespond to first Chern classes of summands of ResG For C to act freely on X, all these summands must be non-trivial. So, in general ResG does not imply that C acts freely on X.
Now, let X = {X1, . . . , Xk} be a set of monomial representations and {H1, . . . , Hk} the set of subgroups such that Xi = IndG i for some Yi ∈ Hom(G, C×). For each i, let γi denote the first Chern class of Yi in H2(Hi, Z), and let αi ∈ H2(G, IndG Shapiro’s isomorphism. Putting these together we get an element αX = (α1, . . . , αk) ∈ H2(G, We now come to the main result of this section: Proposition 6.5. If G acts freely on the set X , then αX is a special class. Proof. Given a cyclic subgroup C in G, we can pick a Cauchy element g in C. Since G actsfreely on X , there is at least one Xi such that g acts freely on Xi. By Lemma 6.4, we haveResG <g>αi = 0. So, α restricts to < g > non-trivially. Since ResG Corollary 6.6. If the answer to Question 6.3 is affirmative, then the axe-saw conjecture holds. Let us note some other consequences of Proposition 6.5 for Bieberbach groups.
Corollary 6.7. Let G be a supersolvable group with saw(G) = s. Then there is a Bieberbachgroup Γ which fits into an extension of the form 0 → M → Γ → G → 0 such that the Z-rankof the M G is s. Proof. By Lemma 2.2, we can find a free linear action on a set X with |X | = s. Associatedwith this set there is a group extension 0 → M → Γ → G → 1, where M is a permutationmodule with s summands such that the extension class is special.
Notice that in the corollary, the permutation module M need not be faithful. In other words, the holonomy group of Γ need not be G. However, replacing M with M = M ⊕ IndG and Γ with Γ , where Γ is the extension group of M and G with the extension class α = (α, 0)in H2(G, M ), we obtain a Bieberbach group Γ with holonomy group G, and where the Z-rankof the center Z(Γ ) = M is s + 1. From this it follows: Corollary 6.8. Let G be a supersolvable group with saw(G) = s. Then there is a flat Rieman-nian manifold X with holonomy group G such that the holonomy representation is a permuta-tion module and the first Betti number in rational coefficients is equal to s + 1. The converses of Proposition 6.5 and Corollary 6.6 may fail as did the converse of Lemma 6.4. As an example, let G = D8, the dihedral group of order 8, and let X = {XC | |C| = 2}where XC = IndG C YC and YC is the non-trivial one dimensional representation of C. It is clear that αX is a special class. But G does not act freely on X because if g is a non-central Cauchyelement of G, then it will fix a point on all X ∈ X .
One can try to get a converse to these results by assuming that each X in X is induced from a normal subgroup. But then, writing X = IndG H Y and K = ker Y , we see that a Cauchy element acts freely on X if g ∈ H − Kx for all x, whereas the restriction of the correspondingcohomology class is non-zero if g ∈ H − Kx for some x. So, the converse still fails. However,if we assume that both H and K are normal, then the two conditions are equivalent. Wecall an induced representation X = IndG H Y a dinormal representation if both H and ker Y are normal subgroups of G. Similarly, we call a class α ∈ H2(G, IndG provided H is normal and the kernel of the associated class γ ∈ H2(H, Z) is normal.
Proposition 6.9. The following are equivalent:(i) G acts freely on X = {X1, . . . , Xk} where each Xi is a dinormal representation.
(ii) There is a special class α = (α1, . . . , αk) in H2(G, mal class.
(iii) There exist normal subgroups Hi, Ki for i = 1, . . . , k such that the quotients Hi/Ki are non-trivial and cyclic, and every Cauchy element is in Proof. It follows from the discussion above.
When G is a p-group of exponent p, part (iii) says that G is covered by index p sections where each Hi and Ki are normal. It is an interesting group theoretical question as to whetherthe non-trivial elements of a p-group of order ps can be covered using less than s sectionsHi − Ki such that |Hi : Ki| = p (no longer assuming that the Hi and Ki are normal). Moregenerally one can ask: Question 6.10. Let G be a p-group of order ps. For i = 1, . . . , k, let Hi and Ki be subgroups such that |Hi : Ki| = p. If (Hi − Ki) = G − {1}, then does it follow that where t is the minimum of logp |G : Ki| over all i? This question is related to earlier questions and conjectures only by its form. When the Hi and Ki are not normal, there seem to be no implications between possible answers to thesequestions.
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