Para **compra cialis** puede ser visto como un desafío. Aumenta Smomenta, y todos los que se poco a poco abrumado, como es lógico, cada vez más hombres están diagnosticados con disfunción eréctil.

## Fen.bilkent.edu.tr

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 that

*Gk − 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 C

*G*-module

*X *provided no non-zero element of

*X *is fixed by

*g*. Given a set

*X *of C

*G*-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

*G*action 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 to

*A*5 or

*A*6. The solvable group

*A*4 is not good since its involutions do not act freely on anyC

*A*4-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 C

*C*-module. The non-trivialelements of

*C *act freely on the induced C

*G*-module Ind

*G*
*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 C

*G*-module.

We define axe(

*G*), called the axe rank of

*G*, to be the minimum size of a set of C

*G*-
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 =

*G*0

*✁ . ✁ 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, when

*G *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

*p*3 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 regular

*p*-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∗ *for

*G *cuts the set of non-trivial elements neatly into slices

*G*1

*− G*0

*, ., Gt − Gt−*1. A Cauchy factor

*Gk/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 of

*p*-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 of

*k *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

*X*1

*, . . . , Xk*, the diagonal

*G*-action on

*X*1

*×· · · × Xk *is called a product action. Notice that free linear actions are product actions where

*G *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 C

*Gk*-module such that ker

*Yk *=

*Gk−*1. Let

*Xk *= Ind

*G*
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 *)

*and*saw(

*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 *and

*G *act freely on

*{X*1

*, ., Xα} *and

*{X*1

*, ., Xα}*, then

*G *acts freely on

*{X*1

*⊗ X*1

*, ., 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

*{X*1

*, ., Xa}*, and let

*χ*1

*, ., χa *be the corresponding characters. Thefunction

*g → *(

*χ*1(

*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 C

*G*-module has index

*p*in

*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

*{ *log

*p |K| *:

*K ✂ G, *[

*K, K*]

*≤ Z*(

*K*)

*, *exp(

*K*) =

*p }.*
Since log

*p |K| *= saw(

*K*) when

*K *has exponent

*p*, and saw(

*K*)

*≤ *saw(

*G*) for all subgroups

*K ≤ 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

*G*1 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 of

*G*. 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∗*. Writing

*A *=

*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 *of

*A*, and also stabilizes the subgroup

*Q*0

*≤ Q *generated by the Cauchy elements. Notice thatsince

*Q *is cyclic,

*Q*0 is cyclic of order

*q*. If

*q *=

*p*, we obtain a contradiction by observing that

*g *and

*Q*0 generate an elementary abelian

*p*-group of rank 2. Supposing now that

*q *=

*p*, thehypothesis implies that

*g Q*0 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 C

*G*-module. Given subgroups

*K *and

*H *of

*G*such that

*K ✂ H ✂ G *and

*H/K *is cyclic, the subset
is called a swath of

*G*. Given a C

*G*-module

*X*, we write

*C*(

*X*) for the set of Cauchy elementsthat act freely on

*X*.

Recall that a C

*G*-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 *C

*H-module with kernel K,*
*and let X be the monomial *C

*G-module *Ind

*G*
*H *(

*Y *)

*. Then C*(

*X *)

*is the set of Cauchy elements*
*Proof. *As a C

*H*-module by restriction,

*X *is the sum of the 1-dimensional modules of the form

*Y ⊗ 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 *C

*G-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 C

*H *-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 *. Replacing

*H *with its core and

*X *with its restriction to the core, we can assume that

*X *is a monomialC

*G *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 C

*G*-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 arbitraryC

*G*-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 *C

*G-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

*C*3

*C*3, let

*H *be the normal subgroup

*C*3

*× C*3

*× C*3,and put

*X *= Ind

*G*
*H *(

*Y *) where

*Y *has kernel

*C*3

*× C*3

*× *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, z*are 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

*K*are cyclic, and 1

*< K < H*, we have

*|K| *=

*p *and

*|H| *=

*p*2.

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

*|H*1 :

*K*1

*| *=

*p *=

*|H*2 :

*K*2

*|*.

First, let us assume that

*p *= 2. If

*|K*1

*| *= 1 =

*|K*2

*|*, then

*H*1 and

*H*2 are normal cyclic
groups of order

*p*, and

*C ∪ {*1

*} *=

*H*1

*∪ H*2. Hence

*G *is cyclic and saw(

*G*) = 1. Suppose that

*|K*1

*| *= 1 =

*|K*2

*|*. Then

*|H*1

*| *=

*p*. All the Cauchy elements of

*K*2 belong to

*H*1, so

*H*1 is theunique subgroup of

*K*2 with order

*p*. So

*K*2 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

*✁ H*1

*✁ H*2

*✂ G*
refines to a chief series with rank 2. So saw(

*G*)

*≤ *2.

Now consider the case where both

*K*1 and

*K*2 are non-trivial. The intersection

*C ∩ K*1

*∩ K*2
is empty, so

*K*1

*∩ K*2 is trivial. The subgroup

*H*1

*∩ K*2 owns all the Cauchy elements of

*K*2 andis isomorphic to a subgroup of the cyclic group

*H*1

*/K*1. Therefore

*K*2 has a unique subgroupof order

*p*. Again

*K*2 is cyclic. Similarly,

*K*1 is cyclic. As before, we may assume that

*H*1and

*H*2 are isomorphic to

*Cp × Cp*. But

*K*1 and

*K*2 are both contained in

*H*1 and are bothcontained in

*H*2, so

*H*1 =

*K*1

*K*2 =

*H*2. The normal series 1

*✁ Z*(

*G*)

*∩ H*1

*✁ H*1

*✂ 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

*K*1 and

*K*2 are normal in

*G*. When both

*K*1 and

*K*2 are trivial, the argument is the same for odd

*p*. Supposing thatonly

*K*1 is trivial, then

*H*1 is the unique subgroup of order 2 in

*K*2. It follows that the normalseries 1

*✁ H*1

*✂ K*1

*✁ H*2

*✂ G *refines to a chief series with rank 2.

Finally, supposing that both

*K*1 and

*K*2 are non-trivial, then they own central involutions

*c*1 and

*c*2, respectively. Furthermore,

*c*1 and

*c*2 are distinct because

*K*1 and

*K*2 intersecttrivially. By Lemma 2.1,

*G *does not contain an elementary abelian subgroup of rank 3. So

*C ⊆ c*1

*, c*2 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 *=

*D*8, 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

*p*6 such that each subgroup of order

*p*4 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

*p*3 and exponent

*p*. For groups of order

*p*5,this observation disposes of the exceptional cases given in [4, Theorem 4.2]. For

*p*-group

*G *withsaw rank 2 of order less than

*p*5, [4, Theorem 3.2] implies that if

*G *is not meta-cyclic, then

*G*is 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 *=

*Q*8, 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

*✁G*1

*✁· · · 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 D*8

*with Q*8

*. This is the quotient group ofD*8

*× Q*8

*with kernel c*1

*c*2

*where c*1

*and c*2

*are central elements of order *2

*in D*8

*and Q*8

*.*

Let a and b be involutions generating D8

*, and let c and d be generators of Q*8

*. 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

*→ *0

*which 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 exponent

*p *case. In Theorem 5.9, we show that the conjecture holds for a certain class of non-regular

*p*-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 *irreducibleC

*G*-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 C

*G*-modules one of which, say

*X*, 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 C

*G*-modules allof dimension greater than unity, we have

*|X | ≥ p *+ 2. Consider an element

*X ∈ X *, and let

*H − 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)

*≥ p*2, 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 *=

*pα*, 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*). Hence

*s ≤ *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

*Sp*2, we see that exp(

*Cp Cp*) =

*p*2. 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 *C

*G-module of dimension greater thanunity is induced from an irreducible *C

*E-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 C

*G *module ofdimension greater than unity, and let

*Y *be a 1-dimensional summand of Res

*G*
nius Reciprocity,

*X *must be summand of Ind

*G*
*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 *F

*p of order p, and hence regard E as an *F

*pC-module. Then G has exponent p ifand only if no direct summand of the *F

*pC-module E is free.*
*Proof. *We may assume that

*E *is indecomposable as an F

*pC*-module. It is well-known thatthe free F

*pC*-module of rank unity has a unique composition series 0

*< M*1

*< . < Mp *= F

*pC*where dim(

*Md*) =

*d*. Furthermore, each

*Md *is indecomposable, and every indecomposableF

*pC*-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 (

*x*1

*, ., xp*)

*gα *with

*xk ∈ *F

*p*; the group
(

*x*1

*, ., xp*)

*gα*(

*y*1

*, ., yp*)

*gβ *= (

*x*1 +

*y*1+

*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 (

*x*1

*, ., xp*) such that

*x*1 +

*. *+

*xp *= 0.

We have
((

*x*1

*, ., xp*)

*gα*)

*p *= (

*x, ., x*)

*,*
where

*x *=

*x*1 +

*. *+

*xp*. So

*MpC − Mp−*1

*C *is precisely the set of elements of order

*p*2. 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 that

*G − 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 C

*G*-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 F

*pC*-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

*p*2.

Using the above formula for the group operation, an easy calculation shows that the conjugacyclass of

*f *is the coset

*Mp−*1

*f*. So every element of

*Mp−*1

*f − Mp−*1 has order

*p*2, and
rk(

*M*1

*✁ . ✁ Mp−*1

*✁ Mp−*1

*f ✁ 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
F

*pC*-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 equality

*s *=

*a *=

*e*.

Special classes associated to free linear actions
Let

*G *be a finite group and

*M *a Z

*G*-lattice (a Z-free Z

*G*-module). Let

*H∗*(

*G, M *) denote thecohomology of

*G *in twisted coefficients

*M *. In particular,

*H*2(

*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 Res

*G*
*H *:

*H ∗*(

*G, M *)

*→ H ∗*(

*H, M *).

Definition 6.1.

*A cohomology class α ∈ H*2(

*G, M *)

*is called as a special class if *Res

*G*
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 ∈ H*2(

*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 Z

*G*-lattice. In fact, given an arbitrary Z

*G*-lattice

*M *with kernel

*K ≤ G*and 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 Z

*G*-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 *Z

*G-lattice. Then the following are equiv-alent:*(i)

*There is a special class α ∈ H*2(

*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 H*1(

*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 R

*n *where

*n *= dim

*M *(see [3]). Since

*M *acts as translations, R

*n/M *=

*T n*, and the group

*G *= Γ

*/M *actsfreely on R

*n/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

*H*2(

*G, M *)has a special class. In fact, if we take

*M *as the direct sum of all induced modules Ind

*G*
all cyclic subgroups

*C ≤ G*, we have

*H*1(

*C, *Q

*/*Z)

*∼*
So, by picking nontrivial homomorphisms for each cyclic subgroup, we can form a special classin

*H*2(

*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 Ind

*G*
*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 H*2(

*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 *= Ind

*G*
*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, Res

*H*
*C *ch1(

*Y *) = ch1(Res

*H*
*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

*H*2(

*G, *Ind

*G*
*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 *Res

*G*
*Proof. *Let us denote the cyclic subgroup generated by

*g *by

*C*. By Lemma 3.1, the element

*g*lies 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 Res

*H*
Since

*C *acts freely on

*Y *, the restricted module Res

*G*
*C Y *is non-trivial, hence ch1(Res

*H*
non-trivial. By the naturality of Chern isomorphism, we conclude that

*C *[ch1(

*Y *)] = ch1[Res

*H*
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 Res

*G*
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 Res

*G*
For

*C *to act freely on

*X*, all these summands must be non-trivial. So, in general Res

*G*
does not imply that

*C *acts freely on

*X*.

Now, let

*X *=

*{X*1

*, . . . , Xk} *be a set of monomial representations and

*{H*1

*, . . . , Hk} *the
set of subgroups such that

*Xi *= Ind

*G*
*i *for some

*Yi ∈ *Hom(

*G, *C

*×*). For each

*i*, let

*γi *denote
the first Chern class of

*Yi *in

*H*2(

*Hi, *Z), and let

*αi ∈ H*2(

*G, *Ind

*G*
Shapiro’s isomorphism. Putting these together we get an element

*αX *= (

*α*1

*, . . . , αk*)

*∈ H*2(

*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 haveRes

*G*
*<g>αi *= 0. So,

*α *restricts to

*< g > *non-trivially. Since Res

*G*
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 ⊕ *Ind

*G*
and Γ with Γ , where Γ is the extension group of

*M *and

*G *with the extension class

*α *= (

*α, *0)in

*H*2(

*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 *=

*D*8, the dihedral group of order 8, and let

*X *=

*{XC | |C| *= 2

*}*where

*XC *= Ind

*G*
*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 *= Ind

*G*
*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 *= Ind

*G*
*H Y *a dinormal representation if both

*H *and ker

*Y*
are normal subgroups of

*G*. Similarly, we call a class

*α ∈ H*2(

*G, *Ind

*G*
provided

*H *is normal and the kernel of the associated class

*γ ∈ H*2(

*H, *Z) is normal.

Proposition 6.9.

*The following are equivalent:*(i)

*G acts freely on X *=

*{X*1

*, . . . , Xk} where each Xi is a dinormal representation.*

(ii)

*There is a special class α *= (

*α*1

*, . . . , αk*)

*in H*2(

*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 *sections

*Hi − 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 *log

*p |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.

A. Adem and D. J. Benson. Abelian groups acting on products of spheres.

*Math. Z.*

228 (1998), 705-712.

M. Ashbacher.

*Finite group Theory *(Cambridge University Press, 1986).

L. Auslander and M. Kuranishi. On the Holonomy Group of Locally Euclidean Spaces.

*Ann. Math. *65 (1957), 411-415.

N. Blackburn. Generalizations of Certain Elementary Theorems on

*p*-Groups.

*Proc.*

Londan Math. Soc. (3) 11 (1961), 1-22.

L. Charlap. Compact Flat Riemannian Manifolds I.

*Ann. Math. *81 (1965), 15-30.

R. Dotzel and G. Hamrick.

*p*-Group Actions on Homology Spheres.

*Invent. Math. *62(1981), 437-442.

M. Hall.

*Theory of groups *(Chelsea, New York, 1976).

P. Landrock.

*Finite group algebras and their modules, *London Mathematical SocietyLecture Note Series, 84 (Cambridge University Press, 1983).

T. Laffey. The Minimum Number of Generators of a Finite

*p*-Group.

*Bull. LondanMath. Soc. *5 (1973), 288-290.

U. Ray. Free linear actions of finite groups on products of spheres.

*J. Algebra *147(1992), 456-490.

J. P. Serre.

*Linear Representations of Finite Groups *(Springer-Verlag GTM 42, Hei-delberg, 1977).

C. B. Thomas.

*Characteristic Classes and the Cohomology of Finite Groups *(Cam-bridge University Press, 1986).

J. A. Wolf.

*Spheres of Constant Curvature *(Publish or Perish, Delaware, 1984).

E. Yal¸cın. Group Actions and Group Extensions.

*Trans. A.M.S. *352 (2000), no. 6,2689-2700.

L. Barker, E. Yal¸cın, Dept. of Mathematics, Bilkent University, Ankara, 06533, Turkey.

E-mail addresses: barker@fen.bilkent.edu.tr, yalcine@fen.bilkent.edu.tr

Source: http://www.fen.bilkent.edu.tr/~yalcine/papers/ranks.pdf

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