aa r X i v : . [ m a t h . G R ] F e b MANY FROBENIUS COMPLEMENTSHAVE EVEN ORDER
RON BROWN
Abstract.
The theory of Frobenius groups with Frobenius com-plements of even order largely reduces to tractable algebraic num-ber theory. If we consider only Frobenius complements with anupper bound s on the number of distinct primes dividing the orderof their commutator subgroups, then the proportion of these withodd order is less than 1 / s . A positive lower bound is also given. Introduction and the Main Theorem
Frobenius groups play a significant role in the theory of finite groups,and, in particular, in the study of simple groups. For example it is notunusual in finite group theory that a proof analyzing a minimal coun-terexample will end up considering Frobenius groups. Every Frobeniusgroup G is a semidirect product K ⋊ H where K is a canonical normalsubgroup of G (the Frobenius kernel of G ) and H is a subgroup of G (a Frobenius complement of G ) [1, 35.25(1)]. The theory of Frobeniusgroups with abelian Frobenius kernel largely reduces to algebraic num-ber theory and indeed to a tractable part of algebraic number theory:the study of unramified primes in abelian extensions of the field of ra-tional numbers [2]. For example it is fairly easy to count the exactnumber of isomorphism classes of such groups of order less than 10 ;there are 569 ,
342 of them [2, p. 85]. It is natural then to ask if it iscommon for Frobenius groups to have abelian kernel.QUESTION. Do almost all (or even a positive proportion of) Frobe-nius groups have abelian Frobenius kernel?When the complement of a Frobenius group has even order, then itis known that the kernel is abelian [3, Theorem 3.4A]. About 88 .
8% ofFrobenius complements of order at most 10 have even order [2, p. 54]. Date : February 9, 2021.2010
Mathematics Subject Classification.
Primary: 20D60; Secondary: 11N13,20E99.
Key words and phrases.
Frobenius group, Frobenius complement, Sylow-cyclicgroup, Frobenius triple, distribution of primes. ONJECTURE. Almost all (isomorphism classes of) Frobenius com-plements have even order.In fact, all Frobenius complements have even order except for someof those which are Z -groups, i.e., groups all of whose Sylow subgroupsare cyclic [2, Theorem 1.4]. We follow Lam in calling these groups Sylow-cyclic groups [5]. Our Main Theorem below would appear tosupport the stronger conjecture that almost all Sylow-cyclic Frobeniuscomplements have even order.By the breadth of a Sylow-cyclic Frobenius complement we meanthe number of distinct primes dividing the order of its commutatorsubgroup. In this paper we will show that most of the members ofthe family of Sylow-cyclic Frobenius complements whose breadth isbounded above by some large number have even order. More preciselywe prove the1.1.
Main Theorem.
Suppose that s ≥ . Then for all sufficientlylarge N , the proportion of all Sylow-cyclic Frobenius complements oforder at most N and breadth at most s which have odd order is lessthan / s . The required size of N will of course depend on the choice of s .The above “proportion” really refers to the proportion of isomorphismclasses of Sylow-cyclic Frobenius complements. We will often omit thephrase “isomorphism classes of” below, just as we did in the statementof the Main Theorem.In the next section we establish some formulas for the number T s ( N )of Sylow-cyclic Frobenius complements of breadth s and order at most N and the number O s ( N ) of these of odd order. After establishingsome preliminary lemmas in Section 3 we give a lower bound for T s ( N )(Section 4) and an upper bound for O s ( N ) (Section 5). These are usedin Section 6 to show that ρ s ( N ) := X r ≤ s O r ( N ) . X r ≤ s T r ( N )is less than 1 / s for sufficiently large N . In Section 7 we sketch acomputation of the exact proportion of Sylow-cyclic Frobenius com-plements with breadth at most s which have odd order and use thisresult to provide a lower bound and an improved upper bound for thisproportion.Some conventions with our notation will prove convenient. Lowercase Roman letters (with the exception of r and x ), with or withoutsubscripts, will always denote positive integers; r will always denote anonnegative integer. The lower case p , with or without subscripts, will lways denote a prime. The notation t wll denote the product of thedistinct primes dividing t (so, for example, 12 = 6 and 1 = 1). Theleast common multiple of a , · · · , a s will be denoted by [ a , · · · , a s ]. |B| denotes the number of elements in the finite set B . φ of course denotesEuler’s φ -function. Throughout the paper s will denote a fixed positive integer. Theletters
A, B, C, D, E, F, F ′ , G, H, H ′ , I, I ′ , J, J ′ will denote various con-stants, some depending on the choice of s . Finally, N will always beassumed to be a positive integer large enough that log log log N >
N > (log log N ) s +6 .The prerequisites from [2] for reading this paper are modest; specif-ically we use 5.1 (A), (B) and (C) (definitions), 5.2 (A), 5.3 (A) (inthe statement of which the numbers 2 and 3 were unfortunately trans-posed), and 11.1 (A) and (B1); all of this is largely self-contained.Sylow-cyclic Frobenius complements are also called “1-complements”in [2]. 2. Two counting formulas
The isomorphism classes of Sylow-cyclic Frobenius complements oforder at most N and breadth s correspond bijectively to the “properFrobenius triples of order at most N and breadth s ”, i.e., to triples( m, n, h r + m Z i ) (where h r + m Z i is the multiplicative group generatedby a unit r + m Z of the ring Z /m Z ) such that n > m is relativelyprime to n and to r − ∤ m ), r n/n ≡ m ), mn ≤ N and m is divisible by exactly s distinct primes [2, Theorem 5.2 A and Lemma5.3 A]. Each such triple uniquely defines an ordered pair( k, { ( p a , t ) , · · · , ( p a s s , t s ) } ) (2.1)where p a · · · p a s s is the prime factorization of m , each t i is the mul-tiplicative order of r + p i Z in the ring Z /p i Z , and k = n/tt where t = [ t , · · · , t s ]; by [2, Proposition 11.1 (A)] these ordered pairs satisfyfor all i ≤ s and j ≤ s :(1) t i > p i ≡ t i );(2) p j ∤ t i and if i = j, then p i = p j ;(3) p a · · · p a s s [ t , · · · , t s ][ t , · · · , t s ] ≤ N ; and(4) p j ∤ k and p a · · · p a s s [ t , · · · , t s ][ t , · · · , t s ] k ≤ N .Note that k above is an integer since r n/n ≡ m ).On the other hand each ordered pair of the form (2.1) satisfying theabove four conditions arises as above from exactly φ ([ t , ··· ,t s ]) Q ≤ i ≤ s φ ( t i )proper Frobenius triples of order at most N and breadth s [2, Propo-sition 11.1 (B1)]. Let D denote the set of such ordered pairs. ach ordered pair (2.1) in D gives rise to exactly s ! ordered pairs( k, (( p a , t ) , · · · , ( p a s s , t s ))) (2.2)satisfying the four conditions above. For any t > D t denote theset of ordered pairs (2.2) with t = [ t , · · · , t s ] and also satisfying thefour conditions above.For any t > U N ( t ) denote the set of s -tuples τ = (( p a , t ) , · · · , ( p a s s , t s )) (2.3)with t = [ t , · · · , t s ] and also satisfying conditions (1), (2), and (3)above; for any such τ ∈ U N ( t ) set φ ( τ ) = 1 φ ( t ) Y ≤ i ≤ s φ ( t i )and K ( τ ) = { k : ( k, τ ) ∈ D t } . If τ ∗ is the set of all s coordinates of τ ∈ U N ( t ), then we also set φ ( τ ∗ ) = φ ( τ ).Note that if U N ( t ) is nonempty, so that it contains some τ as in (2.3),then t ≤ tt · · · t s ≤ tt p a · · · p a s s ≤ N, so t ≤ √ N .Combining these several observations we can give formulas for T s ( N )and O s ( N ).2.1. Proposition. s ! T s ( N ) = X ≤ t ≤√ N X τ ∈U N ( t ) | K ( τ ) | φ ( τ ) . Proof. s ! T s ( N ) = s ! X ( k,τ ) ∈D φ ( τ ) = X ≤ t ≤√ N X ( k,τ ) ∈D t φ ( τ )= X ≤ t ≤√ N X τ ∈U N ( t ) | K ( τ ) | φ ( τ ) . (cid:3) A similar argument yields the next2.2.
Proposition. s ! O s ( N ) = X ≤ t ≤√ N ∤ t X τ ∈U N ( t ) |{ k ∈ K ( τ ) : 2 ∤ k }| φ ( τ ) . . Some preliminary lemmas
We continue to fix s ≥
1. For any t > t ) = { ( a , · · · , a s ) : [ a , · · · , a s ] = t } and Γ ∗ ( t ) = { ( a , · · · , a s ) ∈ Γ( t ) : a i > i ≤ s } . We also set γ ( t ) = | Γ( t ) | , γ ∗ ( t ) = | Γ ∗ ( t ) | , and γ (1) = 1.3.1. Lemma.
For all t > and r, p, m, and n we have:(A) if m and n are relatively prime, then γ ( mn ) = γ ( m ) γ ( n ) ;(B) if ∤ t , then γ ∗ (2 r t ) ≥ γ (2 r ) γ ∗ ( t ) ;(C) γ ( p r ) = ( r + 1) s − r s ; and(D) γ ∗ ( p m ) = m s − ( m − s .Proof. Parts (B) and (C) follow trivially from the definition of γ (1)if r = 0 so suppose r >
0. Similarly we may suppose that m > n >
1. We have inverse maps α : Γ( m ) × Γ( n ) −→ Γ( mn ) and β : Γ( mn ) −→ Γ( m ) × Γ( n ) with α (( a , · · · , a s ) , ( b , · · · , b s )) = ( a b , · · · , a s b s )and β ( a , · · · , a s ) = ((( a , m ) , · · · , ( a s , m )) , (( a , n ) , · · · , ( a s , n )))where in the above display we have let “( a, b )” denote the greatestcommon divisor of a and b . This implies part (A) above; part (B)follows since we can restrict the map α to an injective map Γ(2 r ) × Γ ∗ ( t ) −→ Γ ∗ (2 r t ) when m = 2 r and n = t .Next, an element of Γ( p r ) will have the entry p r in some coordinate;for any 1 ≤ j ≤ s there are (cid:0) sj (cid:1) ways of filling j coordinates withthe power p r ; the remaining s − j coordinates can be filled with anycombination of the powers p k − where 1 ≤ k ≤ r . Therefore the numberof elements of Γ( p r ) is X ≤ j ≤ s (cid:18) sj (cid:19) r s − j j = ( r + 1) s − r s . Similarly, an element of Γ ∗ ( p m ) must have entry p m in j coordinateswhere 1 ≤ j ≤ s and the remaining s − j coordinates can be filledwith any p i where 1 ≤ i < m , so γ ∗ ( p m ) = P ≤ j ≤ s (cid:0) sj (cid:1) ( m − s − j j = m s − ( m − s . (cid:3) . .2. Remark.
Assertions (A) and (C) of the preceding lemma give away of computing γ ( t ) from the prime factorization of t . Let us in thisremark write γ s instead of γ to indicate the dependence of the function γ on the choice of s . Then we can also compute the values of γ ∗ ( t )using the formula γ ∗ = s − X r =0 ( − r (cid:18) sr (cid:19) γ s − r . We will set f ( t ) := 1 / ( tt φ ( t )). Note that f ( t ) = 1 / ( t φ ( t )).3.3. Lemma.
The series X ≤ t< ∞ f ( t ) γ ∗ ( t )(log t ) s converges.Proof. There is a constant
A > s ) such that thenumber d ( m ) of positive divisors of any positive integer m is at most Am / (10 s ) [4, Theorem 315]. Then ∞ X t =2 f ( t ) γ ∗ ( t )(log t ) s ≤ ∞ X t =2 f ( t ) d ( t ) s (log t ) s ≤ ∞ X t =2 A s t / (log t ) s t φ ( t ) < ∞ . (cid:3) Corollary.
The series P ≤ t< ∞ ∤ t f ( t ) γ ∗ ( t ) and P ≤ t< ∞ f ( t ) γ ∗ ( t ) both converge. In Theorem 7.1 below we give the exact proportion of Sylow-cyclicFrobenius complements of breadth at most s which have odd order interms of the convergent series of the above corollary.3.5. Lemma.
For integers t ≥ and u ≥ we have φ ( t ) X p ≡ t ) p | u p ≤ u/ log 2) . Proof.
Let q c · · · q c r r be the prime factorization of u , so that u ≥ r and hence r ≤ log u/ log 2. For each j ≥ p j denote the j -th primecongruent to 1 modulo t , so p j ≥ jt and so j ≤ p j /t . By theBrun-Titchmarsh theorem [6] if j >
1, then the number j of primescongruent to 1 modulo t and at most equal to p j satisfies j ≤ p j φ ( t ) log( p j /t ) ≤ p j φ ( t ) log j , o p j ≥ φ ( t ) j log j, (3.1)which is of course also true if j = 1. The lemma is trivally true if r < u we have log log(log u/ log 2) ≥
0. Hencesuppose that r ≥
4. Then the inequality (3.1) implies that φ ( t ) X p ≡ t ) p | u p ≤ φ ( t ) (cid:18) p + 1 p + 1 p (cid:19) + 2 Z r dxx log x ≤ r ≤ u/ log 2) . (cid:3) Note. If u ≤ √ N , then by the above theorem φ ( t ) (cid:0) X p ≡ t ) p | u p (cid:1) ≤ √ N / log 2) ≤ N. The next lemma is simply a variant of a special case of a lemma ofK. K. Norton.3.6.
Lemma.
Suppose t > . Then there is a constant C > (inde-pendent of N and t ) with (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ ( t ) X p ≡ t ) p ≤ N p − log log N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C log t. Proof.
By [7, Lemma 6.3] there is a constant B (independent of N and t ) such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X p ≡ t ) p ≤ N p − log log Nφ ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ B log(3 t ) φ ( t )(in Norton’s lemma set L = { } , k = t , and x = N ). The lemmafollows immediately since t ≥ (cid:3) Lower bounds for T s ( N )It will be convenient to use in this section some abbreviations in-volving summation notation. In the expression X = X t X h t i i X h p i i X p s e let P t sum over all t with 1 < t ≤ log log N ; P h t i i will abbreviate P t · · · P t s where for each i < s , P t i sums over all t i > t i | t and P t s sums over all t s > t , · · · , t s ] = t . (Thus if s = 1 we have P t = P t P h t i i .) Also P h p i i abbreviates P p · · · P p s − where each P p i sums over all primes p i ≡ t i ) with p i ∤ t , p i ∤ p · · · p i − , and p i ≤ N / (3 s ) ; and, finally, where P p s sums over all p s ≡ t s ) with p s ∤ t , p s ∤ p · · · p s − and tt p · · · p s ≤ √ N . We are interpreting emptyproducts (such as p · · · p s − if s = 1) as equal to 1. Thus, for example,if s = 1 then P p s sums over all p s ≡ t s ) (so consequently p s ∤ t )and tt p s ≤ √ N , and P = P t P p s . Note that for all s , γ ∗ ( t ) = X h t i i t > . It must be kept in mind that this notation depends heavily on con-text. Thus, for example, the precise meaning of P t s depends on theprevious choices of the parameters t, t , · · · , t s − , and similarly P p i de-pends on the previous choices of t, t i , p , · · · , p i − (as well as N ). Forsome choices of t, t , · · · , t s , p , · · · , p s − it could be the case that P p s isan empty sum. If not, and t, t , · · · , t s , p , · · · , p s is a set of parametersfor a term of the summation P , then 1 < t ≤ log log N < √ N and(( p , t ) , · · · , ( p s , t s )) ∈ U N ( t ). It therefore follows from Proposition 2.1that s ! T s ( N ) ≥ X Q ≤ i ≤ s φ ( t i ) φ ( t ) | K ((( p , t ) , · · · , ( p s , t s ))) | . (4.1)4.1. Lemma.
Suppose that t , N and the s -tuple τ = (( p , t ) , · · · , ( p s , t s )) appear in the inequality (4.1) (with t = [ t , · · · , t s ] ). Set ∆ = N/ ( tt p · · · p s ) .Then | K ( τ ) | ≥ ∆ Q ≤ i ≤ s (1 − p i ) − s .Proof. For any subset α of { , , · · · , s } let P α = Q i ∈ α p i (so P ∅ = 1).For any r ≤ s let A r denote the set of subsets of { , · · · , s } of order r .An inclusion/exclusion argument [4, Theorem 261] shows that | K ( τ ) | = s X r =0 ( − r X α ∈A r (cid:20) ∆ P α (cid:21) = s X r =0 ( − r X α ∈A r ∆ P α − s X r =0 ( − r X α ∈A r (cid:0) ∆ P α − (cid:20) ∆ P α (cid:21) (cid:1) ≥ ∆ s Y i =1 (1 − p i ) − s X r =0 X α ∈A r ≥ ∆ s Y i =1 (1 − p i ) − s . (cid:3) Lemma.
Let i ≤ s . For all choices of t, t , · · · , t i , p , · · · , p i − (orjust of t and t if i = 1 ) in display (4.1) we have X p i φ ( t i ) (cid:18) p i − p i (cid:19) ≥ log log N − E log log log N (4.2) for a positive constant E (depending only on s ).Proof. By our choice of N in § √ N / ( tt p · · · p s − ) ≥ N / (3 s ) .Thus X p i φ ( t i ) (cid:18) p i − p i (cid:19) ≥ X p i ≤ N / (3 s ) φ ( t i ) p i − X p i | tp ··· p s − φ ( t i ) p i − X p i ≤ N φ ( t i ) p i (4.3)where each sum above is only over primes p i ≡ t i ).First, X p i ≤ Np i ≡ t i ) φ ( t i ) p i ≤ ∞ X k =1 t i (1 + kt i ) ≤ ∞ X k =1 k = π . Next, if the number of prime divisors of tp · · · p s − is less than 4.then X p i | tp ··· p s − p i ≡ t i ) φ ( t i ) p i ≤ N. (4.4)If the number is greater than 3, then2 ≤ tp · · · p s − ≤ (log log N )( N / (3 s ) ) s − ≤ √ N so by Lemma 3.5 (and the Note following it), the inequality (4.4) stillholds.Finally, by Lemma 3.6 and the choice of t we have X p i ≤ N / (3 s ) p i ≡ t i ) φ ( t i ) p i ≥ log log N / (3 s ) − C log t i ≥ log log N − log 3 s − C log t ≥ log log N − D log log log N for a constant D depending only on s .Applying the inequalities of the last three paragraphs to the inequal-ity (4.3) we deduce that X p i φ ( t i ) (cid:18) p i − p i (cid:19) ≥ log log N − D log log log N − π − − N log log N − E log log log N where the positive constant E depends only on s. (cid:3) Observe that since N ≥ (log log N ) s +6 , therefore X Y ≤ i ≤ s φ ( t i ) ! /φ ( t ) ≤ X t X h t i i X h p i i √ Ntt p · · · p s − Y ≤ i ≤ s φ ( t i ) ! /φ ( t ) ≤ √ N X t X h t i i X h p i i ≤ √ N (log log N ) s +1 ( N / (3 s ) ) s − ≤ N. We can now give our lower bound for T s ( N ).4.3. Theorem.
There is a constant I depending only on s such that s ! T s ( N ) is greater than or equal to N X t f ( t ) γ ∗ ( t ) ! (log log N ) s − IN (log log N ) s − log log log N. Proof.
Applying the last paragraph and the previous two lemmas tothe terms of display (4.1) we have s ! T s ( N ) ≥ X φ ( t ) Y ≤ i ≤ s φ ( t i ) ! − s + Ntt p · · · p s Y ≤ i ≤ s (cid:18) − p i (cid:19)! ≥ − s N + N X t f ( t ) X h t i i X h p i i X p s Y ≤ i ≤ s ( φ ( t i )( 1 p i − p i )) ≥ − s N + N X t f ( t ) γ ∗ ( t ) ! (log log N − E log log log N ) s ≥ N X t f ( t ) γ ∗ ( t ) ! (log log N ) s − IN (log log N ) s − log log log N for a constant I > s . (Recall that P t f ( t ) γ ∗ ( t )converges by Corollary 3.4.) (cid:3) Upper bounds for O s ( N )We introduce abbreviations for summation notation to be used onlyin this section. We will consider expressions X = X t X h t i i X h p i ,a i i X p s X a s here P t sums over all odd t with 3 ≤ t ≤ √ N ; P h t i i has thesame meaning as in § P h t i i γ ∗ ( t )); P h p i ,a i i abbrevi-ates P p P a · · · P p s − P a s − where for each i ≤ s , P p i is the sumover all p i ≡ t i ) with p i ≤ N and p i ∤ tp · · · p i − ; and for i < s , P a i is the sum over all a i with p a i i ≤ N , while P a s is the sum over all a s with tt p a · · · p a s s ≤ N .Then arguing as in the proof of the inequality (4.1) we can see thatProposition 2.2 implies that s ! O s ( N ) ≤ X (cid:18) Q ≤ i ≤ s φ ( t i ) φ ( t ) (cid:19) |{ k ∈ K ((( p a , t ) , · · · , ( p a s s , t s ))) : 2 ∤ k }| . (5.1)After all, any sequence t , · · · , t s , p , a , · · · , p s , a s which satisfies con-ditions (1), (2) and (3) in Section 2 and has 2 ∤ t i for all i ≤ s will, with t = [ t , · · · , t s ], be a sequence of parameters in the summation P .In this section we set p = 2 so p = 2. Consider an s -tuple τ =(( p a , t ) , · · · , ( p a s s , t s )) appearing in the inequality (5.1). For any r ≤ s + 1 let B r denote the set of subsets of { , , · · · , s } of order r ; for anysubset β of { , , · · · , s } let P β = Q r ∈ β p r . Also set ∆ = N/ ( tt p a · · · p a s s ).Arguing much as in § |{ k ∈ K ( τ ) : 2 ∤ k }| = X ≤ r ≤ s +1 X β ∈B r ( − r (cid:20) ∆ P β (cid:21) ≤ X ≤ r ≤ s +1 ( − r X β ∈B r ∆ P β + X ≤ r ≤ s +12 ∤ r X β ∈B r (cid:18) ∆ P β − (cid:20) ∆ P β (cid:21)(cid:19) ≤ ∆ Y ≤ r ≤ s (cid:18) − p r (cid:19) + X ≤ r ≤ s +12 ∤ r |B r | = ∆2 Y ≤ r ≤ s (cid:18) − p r (cid:19) + 2 s since 2 s is the number of subsets of { , , · · · s } of odd order. Hence s ! O s ( N ) ≤ N X f ( t ) Y ≤ i ≤ s φ ( t i ) (cid:18) p a i i (1 − p i ) (cid:19) + 2 s X Q ≤ i ≤ s φ ( t i ) φ ( t ) . (5.2)5.1. Lemma.
For some constant G depending only on s we have X Y ≤ i ≤ s φ ( t i ) ! /φ ( t ) ≤ GN (log log N ) s − . roof. Given t, t , · · · , t s , p , a , · · · , p s , a s as in the summation nota-tion P for this section we have p a s s ≡ t s ), so X p s X a s φ ( t s ) ≤ φ ( t s ) (cid:12)(cid:12) { k : 1 + kt s ≤ N/ ( tt p a · · · p a s − s − ) } (cid:12)(cid:12) ≤ ( φ ( t s ) /t s ) N/ ( tt p a · · · p a s − s − ) ≤ N/ ( tt p a · · · p a s − s − ) . (Here, if s = 1, we regard the expression p a · · · p a s − s − as an emptyproduct, so it is equal to 1.)Also for all i < s , φ ( t i ) X p i X a i p a i i ≤ φ ( t i ) X p i ∞ X k =1 p ki ≤ φ ( t i ) X p i p i ≤ φ ( t i ) X p i ≤ Np i ≡ t i ) p i ≤ N + 2 C log t i by Lemma 3.6. Therefore since 2 log t > t > X Q ≤ i ≤ s φ ( t i ) φ ( t ) ≤ X t φ ( t ) X h t i i X p X a φ ( t ) p · · · X p s − X a s − φ ( t s − ) p s − X p s X a s φ ( t s ) p · · · p s − ! ≤ X t φ ( t ) X h t i i (2 log log N + 2 C log t ) s − Ntt ≤ N X t f ( t ) γ ∗ ( t ) ! (2 log log N + 2 C log t ) s − ≤ N X ≤ r ≤ s − X t f ( t ) γ ∗ ( t ) (cid:18) s − r (cid:19) (2 log log N ) r (2 C log t ) s − − r ≤ N X ≤ r ≤ s − (2 C ) s (cid:18) s − r (cid:19) X t f ( t ) γ ∗ ( t )(log t ) s ! (log log N ) s − ≤ N G (log log N ) s − for some positive constant G depending only on s (cf., Lemma 3.3) (cid:3) We now give our upper bound for O s ( N ).5.2. Theorem.
There is a constant F depending only on s with s ! O s ( N ) ≤ N X t f ( t ) γ ∗ ( t ))(log log N ) s + N F (log log N ) s − . roof. For any i ≤ s we have φ ( t i ) X p i X a i (cid:18) p a i i − p a i i (cid:19) ≤ φ ( t i ) X p i ∞ X j =1 (cid:18) p ji − p j +1 i (cid:19) ≤ log log N + C log t by Lemma 3.6. Applying this inequality and that of the previous lemmato the inequality (5.2) we deduce that s ! O s ( N ) is at most N X t f ( t ) X h t i i X h p i ,a i i X p s X a s s Y i =1 φ ( t i ) (cid:18) p a i i − p a i i (cid:19) +2 s GN (log log N ) s − ≤ N X t f ( t ) γ ∗ ( t )(log log N + C log t ) s + 2 s GN (log log N ) s − ≤ N X t f ( t ) γ ∗ ( t ))(log log N ) s + N X t f ( t ) γ ∗ ( t ) s X r =1 (cid:18) sr (cid:19) (log log N ) s − r ( C log t ) r +2 s GN (log log N ) s − ≤ N X t f ( t ) γ ∗ ( t ))(log log N ) s + N X t f ( t ) γ ∗ ( t )(log t ) s )( s X r =1 (cid:18) sr (cid:19) C r )(log log N ) s − +2 s GN (log log N ) s − ≤ N X t f ( t ) γ ∗ ( t ))(log log N ) s + F N (log log N ) s − for a constant F > s (using again Lemma 3.3). (cid:3) Proof of the main theorem
We will prove that for sufficiently large N (depending only on thechoice of s ) the ratio ρ s ( N ) := X r ≤ s O r ( N ) . X r ≤ s T r ( N )is less than 1 / s .We define σ ( x ) = P f ( t ) γ ∗ ( t ), where the summation is over all odd t with 3 ≤ t ≤ x (for any x > σ ( x ) is bounded above(c.f., Corollary 3.4).The Sylow-cyclic Frobenius complements of breadth 0 are cyclicgroups and correspond to proper Frobenius triples ( m, n, h r + m Z i ) ith m = 1, and hence T ( N ) ≤ N and O ( N ) ≤ ( N + 1) /
2. This,together with the theorems of the previous two sections, show that for s > s ! s X r =0 O r ( N ) ≤ N N ) s σ ( √ N ) + J N (log log N ) s − , and that s ! P sr =0 T r ( N ) is at least N X ≤ t ≤ log log N f ( t ) γ ∗ ( t ) ! (log log N ) s − HN (log log N ) s − log log log N for constants J and H depending only on s .One checks that X r =1 r s r ≥ s s ≤
4. Indeed the equation (6.1) holds for all s since if s >
4, then X r =1 r s r ≥ s + 3 s ≥ s ( 116 + (3 /
64 ) ≥ s . Thus by Lemma 3.1 (C) we have X r =0 γ (2 r )4 r = X r =0 ( r + 1) s − r s r = 5 s
256 + 3 X r =1 r s r ≥
160 + 2 s − . Consequently using Lemma 3.1 (B) we deduce that X ≤ t ≤ log log N f ( t ) γ ∗ ( t ) ≥ X r =0 X ≤ t ≤ (log log N ) / ∤ t f (2 r t ) γ ∗ (2 r t ) ≥ X r =0 X ≤ t ≤ (log log N ) / ∤ t r t ) φ ( t ) γ (2 r ) γ ∗ ( t )= X r =0 γ (2 r )4 r σ ((log log N ) / ≥ ( 160 + 2 s − ) σ ((log log N ) / . Since σ ( √ N ) > σ (3) therefore the ratio ρ s ( N ) is at most( N/ N ) s σ ( √ N ) + J N (log log N ) s − N ( + 2 s − ) σ ((log log N ) / N ) s − HN (log log N ) s − log log log N Jσ (3) log log N ( + 2 s ) σ ((log log N ) / σ ( √ N ) − H log log log Nσ (3) log log N . Now let ǫ = s )+31 . We can pick N sufficiently large (where “suffi-ciently large” depends only on s ) that the numerator in the last displayis at most 1 + ǫ , that H log log log Nσ (3) log log N is less than ǫ and that σ ((log log N ) / σ ( √ N ) is greater than 1 − ǫ . (This is possible since σ is nondecreasing andbounded above.) Then one checks that ρ s ( N ) < ǫ ( + 2 s )(1 − ǫ ) − ǫ = 12 s , which was to be proven.7. Upper and lower bounds for T s ( N ) , O s ( N ) and ρ s ( N )We adapt the arguments of Sections 4 and 5 to give upper bounds for T s ( N ) and lower bounds for O s ( N ). We will then be able to give theexact proportion of Sylow-cyclic Frobenius complements of width atmost s which have odd order in terms of the infinite sums of Corollary3.4. Rather crude estimates of these infinite sums will show that theabove proportion is between 1 / (6Υ) and 1 / ((21 + 6(2 s ))Υ) where Υ = P ∞ i =1 i s i .Using Proposition 2.2 and the arguments of Section 4 for a lowerbound for T s ( N ), we easily obtain a lower bound for O s ( N ): s ! O s ( N ) ≥ N σ (log log N )(log log N ) s − N I ′ (log log N ) s − log log log N for some constant I ′ depending only on s . Of course this implies that s ! s X r =0 O r ( N ) ≥ N σ (log log N )(log log N ) s − N J ′ (log log N ) s − log log log N for a constant J ′ depending only on s . Combining our upper and lowerbounds on O s ( N ) (and also Corollary 3.4) we can deduce that s ! s X r =0 O r ( N ) ∼ N σ ( N )(log log N ) s (7.1)(i.e., the ratio of the two sides approaches 1).Similarly Proposition 2.1 and the arguments of Section 5 for an up-per bound for O s ( N ) easily adapt to give an upper bound for T s ( N ), amely, s ! T s ( N ) ≤ N (log log N ) s ( X ≤ t ≤√ N f ( t ) γ ∗ ( t )) + N F ′ (log log N ) s − for a constant F ′ depending only on s , so that s ! s X r =0 T r ( N ) ≤ N (log log N ) s ( X ≤ t ≤√ N f ( t ) γ ∗ ( t )) + N H ′ (log log N ) s − for a constant H ′ depending only on s , and hence s ! s X r =0 T r ( N ) ∼ N (log log N ) s ( X ≤ t ≤ N f ( t ) γ ∗ ( t )) . (7.2)Combining the above formulas (7.1) and (7.2) with the convergenceof the infinite sums in Corollary 3.4, we have the7.1. Theorem. lim ρ s ( N ) = P ≤ t< ∞ ∤ t f ( t ) γ ∗ ( t )2 P ≤ t< ∞ f ( t ) γ ∗ ( t ) . Recall that Υ = P ∞ r =1 r s / r .7.2. Theorem. ≥ lim ρ s ( N ) ≥ s )) . Proof.
The left-hand inequality of our theorem follows from the previ-ous theorem and the calculation (using Lemma 3.1 (B) and (C)): ∞ X t =2 f ( t ) γ ∗ ( t ) ≥ ∞ X r =0 ∞ X t =32 ∤ t f (2 r t ) γ ∗ (2 r t ) ≥ ∞ X r =0 ∞ X t =32 ∤ t r t φ ( t ) γ (2 r ) γ ∗ ( t )= ∞ X r =0 ( r + 1) s − r s r ∞ X t =32 ∤ t f ( t ) γ ∗ ( t ) = 3Υ ∞ X t =32 ∤ t f ( t ) γ ∗ ( t ) . Each element of Γ( t ) has an entry 1 in, say, r coordinates where0 ≤ r < s , and for each such element the remaining s − r coordinatescan be filled with positive integers in at most γ ∗ ( t ) ways. (Replacingthe 1’s in the s -tuple by t ’s would give an element of Γ ∗ ( t ).) Thus γ ( t ) ≤ X ≤ r ≤ s − (cid:18) sr (cid:19) γ ∗ ( t ) ≤ (2 s − γ ∗ ( t ) . (7.3) he right-hand inequality of our theorem follows from the followingcalculation using the above inequality (7.3) and Lemma 3.1 (A) and(D): ∞ X t =2 f ( t ) γ ∗ ( t ) ≤ ∞ X r =0 ∞ X t =32 ∤ t f (2 r t ) γ ∗ (2 r t ) + ∞ X r =1 f (2 r ) γ ∗ (2 r ) ≤ ∞ X r =0 ∞ X t =32 ∤ t r t φ ( t ) γ (2 r ) γ ( t ) + ∞ X r =1 r s − ( r − s r ≤ s − ∞ X t =32 ∤ t f ( t ) γ ∗ ( t ) + 34 Υ ≤ (3(2 s −
1) + 3 / /
18 )Υ ∞ X t =32 ∤ t f ( t ) γ ∗ ( t ) , since P ∞ t =32 ∤ t f ( t ) γ ∗ ( t ) > f (3) γ ∗ (3) = 1 / (cid:3) Remark.
For small s the left-hand inequality of the last theoremgives a modest improvement of the Main Theorem, but the improve-ment can be substantial for large s . For example, when s = 10 then 6Υis about 99 , . Indeed if s = 20 then6Υ exceeds 25 × while 2 s is of course a bit more than a million.(The calculations of Υ were made using WolframAlpha.) References [1] M. Aschbacher,
Finite Group Theory, 2nd ed. . (Cambridge University Press,Cambridge, 2000).[2] R. Brown, Frobenius groups and classical maximal orders,
Mem. Amer. Math.Soc. (2001).[3] J. Dixon and B. Mortimer,
Permutation Groups . (Springer-Verlag, New York,1996).[4] G. H. Hardy and E. M. Wright,
An Introduction to the Theory of Numbers,5th ed. . (Oxford University Press, Oxford, 1979).[5] T. Y. Lam, Finite groups embeddable in division rings,
Proc. Amer. Math.Soc. (2001), 3161-3166.[6] H. L. Montgomery and R. C. Vaughn, The large sieve,
Mathematika (1973),119-134.[7] K. K. Norton, On the number of restricted prime factors of an integer, IllinoisJ. Math. (1976), 681-705. epartment of Mathematics, University of Hawaii, 2565 McCarthyMall, Honolulu, HI 96822 Email address : [email protected]@math.hawaii.edu