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Publicly Available Published by De Gruyter May 7, 2019

Zassenhaus conjecture on torsion units holds for SL(2, 𝑝) and SL(2, 𝑝2)

Ángel del Río EMAIL logo and Mariano Serrano
From the journal Journal of Group Theory

Abstract

H. J. Zassenhaus conjectured that any unit of finite order and augmentation 1 in the integral group ring G of a finite group G is conjugate in the rational group algebra G to an element of G. We prove the Zassenhaus conjecture for the groups SL(2,p) and SL(2,p2) with p a prime number. This is the first infinite family of non-solvable groups for which the Zassenhaus conjecture has been proved. We also prove that if G=SL(2,pf), with f arbitrary and u is a torsion unit of G with augmentation 1 and order coprime with p, then u is conjugate in G to an element of G. By known results, this reduces the proof of the Zassenhaus conjecture for these groups to proving that every unit of G of order a multiple of p and augmentation 1 has order actually equal to p.

1 Introduction

For a finite group G, let V(G) denote the group of units of augmentation 1 in G. We say that two elements of G are rationally conjugate if they are conjugate in the units of G. The following conjecture stated by H. J. Zassenhaus [23] (see also [21, Section 37]) has centered the research on torsion units of integral group rings during the last decades.

Conjecture (Zassenhaus conjecture).

If G is a finite group, then every torsion element of V(G) is rationally conjugate to an element of G.

The relevance of the Zassenhaus conjecture is that it describes the torsion units of the integral group ring of G provided it holds for G. Recently, Eisele and Margolis [11] announced a metabelian counterexample to the Zassenhaus conjecture. Nevertheless, the Zassenhaus conjecture holds for large classes of solvable groups, e.g. for nilpotent groups [22], groups possessing a normal Sylow subgroup with abelian complement [12] or cyclic-by-abelian groups [7]. In contrast with these results, the list of non-solvable groups for which the Zassenhaus conjecture has been proved is very limited [16, 9, 13, 14, 5, 6, 3, 8]. For example, the Zassenhaus conjecture has only been proved for sixty-two simple groups, all of them of the form PSL(2,q) (see the proof of Theorem C in [4] and [20]).

The goal of this paper is proving the following theorem.

Theorem 1.1.

Let G=SL(2,q) with q an odd prime power, and let u be a torsion element of V(ZG) of order coprime with q. Then u is rationally conjugate to an element of G.

As a consequence of Theorem 1.1 and known results we will obtain the following theorem which provides the first positive result on the Zassenhaus conjecture for an infinite series of non-solvable groups.

Theorem 1.2.

The Zassenhaus conjecture holds for SL(2,pf) with p a prime number and f2.

In Section 2, we prove a number theoretical result relevant for our arguments. Known results on V(G) and properties of V(SL(2,q)) are collected in Section 3. A particular case of Theorem 1.1 is proved in Section 4. Finally, in Section 5, we prove Theorem 1.1.

2 Number theoretical preliminaries

We use the standard notation for the Euler totient function φ and the Möbius function μ. Moreover, 0 denotes the set of non-negative integers. Let n be a positive integer. Then n=/n, and ζn denotes a complex primitive n-th root of unity, Φn(X) denotes the n-th cyclotomic polynomial, i.e., the minimal polynomial of ζn over , and for a prime integer p, let vp(n) denote the valuation of n at p, i.e., the maximum non-negative integer m with pmn. If F/K is a finite field extension, then TrF/K:FK denotes the standard trace map. We will frequently use the following formula for d a divisor of n [18, Lemma 2.1]:

(2.1)Tr(ζn)/(ζd)=μ(d)φ(n)φ(d).

We reserve the letter p to denote a positive prime integer, and for every positive integer n, we set

n=pnpandnp=pvp(n).

If moreover x, then we set

(:xn)=representative of the class ofxmodulonin the interval(-n2,n2],
|:xn|=the absolute value of(:xn),
γn(x)=pn|:xnp|<np2pp,
γ¯n(x)=pn|:xnp|np2pp={2γn(x)if|:xn2|=n24,γn(x)otherwise.

The next lemma collects two facts which follow easily from the definitions.

Lemma 2.1.

Let p be a prime dividing n, and let x,yZ. Then the following conditions hold.

  1. If pγ¯n(x), then (:xnpp)xmodnp.

  2. Let dn such that xymodnd. If d divides both γ¯n(x) and γ¯n(y), then xymodn.

For integers x and y, we define the following equivalence relation on :

xnyx±ymodn.

We denote by Γn the set of these equivalence classes.

If x,y and n are integers with n>0, then let

δx,y(n)={1ifxny,0otherwise,andκx(n)={2ifx0modnorxn2modn,1otherwise.

For an integer x (or xΓn), we set αx(n)=ζnx+ζn-x. Observe that (α1(n)) is the maximal real subfield of (ζn) and [α1(n)] is the ring of integers of (α1(n)). If nn2, then let p0 denote the smallest odd prime dividing n. Then let

𝔹n={xn:for everypn,either|:xnp|>np2porp=2,nn2,|:xn2|=n24,np0xand(x:n2)(x:np0)>0}

and

n={{αb(n):b𝔹n}ifnn2,{1}{αb(n):b𝔹n}otherwise.

For b𝔹n and x, let

βb,x(n)={-1ifnn2,|:xn2|=n24and(:xn2)(:bnp0)<0,1otherwise.

The following proposition extends [20, Proposition 3.5]. The first statement implies that n is a -basis of (α1(n)). For x(α1(n)) and b𝔹n, we use Cb(x) to denote the coefficient of αb(n) in the expression of x in the basis n.

Proposition 2.2.

Let n be a positive integer.

  1. Then n is a -basis of [α1(n)].

  2. If b𝔹n and i, then Cb(αi(n))=κi(n)μ(γ(i))βb,i(n)δb,i(n/γ¯(i)).

Proof.

We only prove the proposition in the case nn2, as the proof in the case n=n2 is similar (actually simpler). It is easy to see that

|n|φ(n)2=[(α1(n)):].

Thus it is enough to prove the equality

αi(n)=κi(n)μ(γ(i))b𝔹nbn/γ¯(i)iβb,i(n)αb(n).

Actually, we will show

ζni=μ(γ(i))bimodn/γ¯(i)b𝔹nβb,i(n)ζnb,

which easily implies the desired expression of αi(n).

Indeed, for every pn, let ζnp denote the p-th part of ζn, i.e., ζnp is a primitive np-th root of unity and ζn=pnζnp. Let J be the set of tuples (jp)pγ¯(i) satisfying the following conditions:

  1. If pγ(i), then jp{1,,p-1}.

  2. If p=2 and |:in2|=n24, then

    j2={1if(:in2)(i+:jp0np0p0np0)<0,0otherwise.

For every jJ, let bjn be given by

bj{i+jpnppmodnpifpγ¯(i),imodnpotherwise.

Then {bj:jJ} is the set of elements in 𝔹n satisfying ibmodnγ¯(i). From

0=ζnpi(1+ζnpnpp+ζnp2npp++ζnp(p-1)npp),

we obtain

ζnpi=-jp=1p-1ζnpi+jpnpp.

Therefore, if |:in2|n24, then γ(i)=γ¯(i), βb,i(n)=1 for every b𝔹n and

ζni=pnpγ(i)ζnpipnpγ(i)(-jp=1p-1ζnpi+jpnpp)=μ(γ(i))jJζnbj=μ(γ(i))bimodn/γ¯(i)b𝔹nζnb.

This gives the desired equality in this case.

Suppose that |:in2|=n24. Then ζn2i=βbj,i(n)ζn2bj for every jJ. Then a small modification of the argument in the previous paragraph gives

ζni=ζn2ipnpγ(i)ζnpipnpγ(i)(-jp=1p-1ζnpi+jpnpp)=μ(γ(i))jJβb,i(n)ζnbj=μ(γ(i))bimodn/γ¯(i)b𝔹nβb,i(n)ζnb.

3 Group theoretical preliminaries

Let G be a finite group. We denote by Z(G) the center of G. If gG, then |g| denotes the order of g, g denotes the cyclic group generated by g, and gG denotes the conjugacy class of g in G. If R is a ring, then RG denotes the group ring of G with coefficients in R. If α=gGαgg is an element of a group ring RG, with each αgR, then the partial augmentation of α at g is defined as

εg(α)=hgGαh.

We collect here some known results on partial augmentations of an element u of order n in V(G).

  1. If gZ(G) and ug, then εg(u)=0 (Berman–Higman theorem, [15, Proposition 1.5.1]).

  2. If gG and εg(u)0, then |g| divides n [13, Theorem 2.3].

  3. u is rationally conjugate to an element of G if and only if εg(ud)0 for all gG and all divisors d of n [17, Theorem 2.5].

  4. See [16, 13]. Let F be a field of characteristic t0 with tn. Let ρ be an F-representation of G. If t0, then let ξn be a primitive n-th root of unity in F, so that if t=0, then ξn=ζn. Let T be a set of representatives of the conjugacy classes of t-regular elements of G (all the conjugacy classes if t=0). Let χ denote the character afforded by ρ if t=0, and the t-Brauer character of G afforded by ρ if t>0 (using a group isomorphism associating ξn to ζn). Then, for every integer , the multiplicity of ξn as eigenvalue of ρ(u) is

    1nxTdnεx(ud)Tr(ζnd)/(χ(x)ζn-d).

In the remainder of the paper, fix an odd prime power q, and let G=SL(2,q), G¯=PSL(2,q). Let π:GG¯ denote the natural projection, which we extend by linearity to a ring homomorphism π:GG¯.

We collect some group-theoretical properties of G and G¯ (see, e.g., [10, Theorem 38.1]).

  1. G has a unique element J of order 2 and q+4 conjugacy classes. More precisely, if p is the prime dividing q, then two of the classes are formed by elements of order p, another two are formed by elements of order 2p, and q classes are formed by elements of order dividing q+1 or q-1. Furthermore, if g and h are p-regular elements of G and |h| divides |g|, then h is conjugate in G to an element of g and two elements of g are conjugate in G if and only if they are equal or mutually inverse.

  2. Let C be a conjugacy class of G¯ formed by elements of order n. If n=2, then π-1(C) is the only conjugacy class of G formed by elements of order 4. Otherwise, π-1(C) is the union of two conjugacy classes C1 and C2 of G with C2=JC1. Furthermore, if n is a multiple of 4, then the elements of C1 and C2 have order 2n, while if n is not a multiple of 4, then one of the classes C1 or C2 is formed by elements of order n.

The following proposition collects some consequences of these facts.

Proposition 3.1.

Let G=SL(2,q), and let u be a torsion element of V(ZG). Set G¯=PSL(2,q) and n=|u|. Then the following statements hold.

  1. J is the unique element of order 2 in V(G).

  2. |π(u)|=ngcd(2,n).

  3. If 4n and π(u) is rationally conjugate to an element of G¯, then u is rationally conjugate to an element of G.

  4. If gcd(n,q)=1 and either n=4 or 4n, then u is rationally conjugate to an element of G.

  5. If gcd(n,q)=1, then G has an element of order n.

  6. Suppose that q=pf with p prime, f2 and pn. Then u is rationally conjugate to an element of G.

  7. If ρ is a representation of G and ζ is a root of unity of order dividing n, then ζ and ζ-1 have the same multiplicity as eigenvalues of ρ(u).

Proof.

1 This is a direct consequence of A and E.

2 By the main result of [19], if π(u)=1 then u2=1 and hence either u=1 or u=J, by 1. Then 2 follows.

3 Suppose that n is not a multiple of 4. If n is even, then the order of Ju is odd by 1. Thus we may assume without loss of generality that the order of u is odd. If εg(u)0, then |g| is odd by B, and hence εg(u)=επ(g)(π(u))0, by F. Thus u is rationally conjugate to an element of G.

4 Let q=pf, where p is an odd prime number and pn. By E, G has a unique conjugacy class C formed by elements of order 4 and a unique element of order 2. Thus, by A and B, if n=4, then εg(u)=0 for every gC, and hence u is rationally conjugate to an element of G by C.

If 4n, then |π(u)| is coprime with 2q by 2, and hence π(u) is rationally conjugate to an element of G¯ by [20, Theorem 1.1]. Then u is rationally conjugate to an element of G by 3. Further, 5 is a consequence of 2 and [13, Proposition 6.7].

6 In this case, |π(u)|=p by 2 and [2, Theorem A]. Then n is either p or 2p by 2, and π(u) is rationally conjugate to an element of G¯ by [13, Proposition 6.1]. Thus u is rationally conjugate to an element of G by 3.

7 This is a consequence of E and the formula in D. ∎

Observe that, for q odd, Theorem 1.2 follows at once from Theorem 1.1 and Proposition 3.16. On the other hand, SL(2,2)S3 and SL(2,4)A5 for which the Zassenhaus conjecture is well known. So, in the remainder of the paper, we concentrate on proving Theorem 1.1. For that, from now on, t denotes the prime dividing q (we want to use freely the letter p to denote an arbitrary prime), and we introduce some t-Brauer characters of G.

Let g be an element of G of order n with tn, and let ξn denote a primitive n-th root of unity in a field F of characteristic t. Adapting the proof of [18, Lemma 1.2], we deduce that, for every positive integer m, there is an F-representation Θm of G of degree 1+m such that

(3.1)Θm(g)={diag(1,ξn2,ξn-2,,ξnm,ξn-m)if 2m,diag(ξn,ξn-1,ξn3,ξn-3,,ξnm,ξn-m)if 2m.

In particular, the restriction to g of the t-Brauer character associated to Θm is given by

χm(gi)=j=-mjmmod2mζnij.

4 Prime power order

In this section, we prove the following particular case of Theorem 1.1.

Proposition 4.1.

Let G=SL(2,q) with q an odd prime power, and let u be a torsion element of V(ZG). If the order of u is a prime power and it is coprime with q, then u is rationally conjugate to an element of G.

Proof.

By Proposition 3.14, we may assume that |u|=2r with r3. We argue by induction on r. So we assume that units of order 2k with 1kr-1 are rationally conjugate to an element of G. By Proposition 3.15 and E, G has an element g0 of order 2r such that {g0k:k=0,1,2,,2r-1} is a set of representatives of the conjugacy classes of G with order a divisor of 2r. By B, the only possible non-zero partial augmentations of u are the integers εk=εg0k(u), with k=1,,2r-1-1. By the induction hypothesis, if 1ir and gG, then εg(u2i)0. Hence, by C, it suffices to prove that εk=0 for all but one k=0,1,,2r-1.

By [18, Theorem 2] and Proposition 3.12, π(u) is rationally conjugate to an element of order 2r-1 in G¯, and hence ε2r-2=επ(g0)2r-2(π(u))=0 by F.

For a t-Brauer character χ of G and an integer , define

A(χ,)=k=12r-1-1εkTr(ζ2r)/(χ(g0k)ζ2r-),
B(χ,)=k=0r-1Tr(ζ2k)/(χ(g02r-k)ζ2k-).

Then, by D, we have

(4.1)12r(A(χ,)+B(χ,))0.

Observe that B(χ,+2r-1)=B(χ,) and A(χ,+2r-1)=-A(χ,). Therefore, from (4.1) it follows that

(4.2)ifB(χ,)=0,thenA(χ,)=0,
(4.3)ifB(χ,)=2r-1,thenA(χ,)=±2r-1.

We will calculate B(χ,) and A(χ,) for several t-Brauer characters χ and several integers , and for that, we will use (2.1) without further mention. We start proving that

(4.4)if 0hr-2and 2r-1,thenB(χ2h,)={2r-1ifh1,0ifh=0,

and

(4.5)if 0hr-3, 2hand 2r-1,thenB(χ2h,)={2r-1if±2hmod2r-1,0otherwise.

In both cases, we argue by induction on h with the cases h=0 and h=1 being straightforward. Suppose that 1<hr-2, 2r-1 and B(χ2h-1,)=2r-1. If j is even, then straightforward calculations show that

k=0r-1Tr(ζ2k)/(ζ2k2h-1+j+ζ2k-2h-1-j)=0.

This implies

B(χ2h,)=B(χ2h-1,)+j=22j2h-1k=0r-1Tr(ζ2k)/(ζ2k2h-1+j+ζ2k-2h-1-j)=2r-1.

This finishes the proof of (4.4).

Suppose that 1<hr-3, 2h and 2r-1. In this case, the induction hypothesis implies B(χ2h-1,)=0. Arguing as in the previous paragraph, we get

B(χ2h,)=j=22j2h-1k=0r-1Tr(ζ2k)/((ζ2k2h-1+j+ζ2k-(2h-1+j))ζ2k-).

However, if j is even and smaller than 2h-1, then

k=0r-1Tr(ζ2k)/((ζ2k2h-1+j+ζ2k-(2h-1+j))ζ2k-)=0.

Therefore, having in mind that ζ2h+22h+ζ2h+2-2h=0, we have

B(χ2h,)=k=0hTr(ζ2k)/((ζ2k2h+ζ2k-2h)ζ2k-)+ϵ2h+1+k=h+3r-1Tr(ζ2k)/((ζ2k2h+ζ2k-2h)ζ2k-),

where ϵ=1 if 2h+1, and ϵ=-1 otherwise. Then the claim follows using the following equalities that can be proved by straightforward calculations:

k=0hTr(ζ2k)/((ζ2k2h+ζ2k-2h)ζ2k-)=2h+1,
k=h+3r-1Tr(ζ2k)/((ζ2k2h+ζ2k-2h)ζ2k-)={0if 2h+1,2r-1-2h+2if 2h+1and±2hmod2r-1,-2h+2if 2h+1and±2hmod2r-1.

This finishes the proof of (4.5).

We now prove by induction on h that the following two statements hold for any integer 0hr-3:

(4.6)kXh(εk-εk+2r-h-1)=±1,whereXh={i{1,,2r-2}:i±1mod2r-h},
(4.7)ifi±jmod2r-h-1andi0,±1mod2r-h-1,thenεi=εj,

and that the next one holds for every 0hr-2:

(4.8)ifi0mod2r-h-1,thenεi=0.

Observe that X0={1}. Fix an integer i. Then, for every integer k, we have

Tr(ζ2r)/((ζ2rk+ζ2r-k)ζ2r-i)={2r-1ifkimod2r,-2r-1ifk2r-1-imod2r,0otherwise.

Thus A(χ1,i)=2r-1(εi-εi+2r-1), and hence, for h=0, (4.6) and (4.7) follows at once from (4.2), (4.3) and (4.5). Moreover, for h=0, (4.8) is clear because ε2r-1=0.

Suppose 0<hr-3 and (4.6), (4.7) and (4.8) hold for h replaced by h-1. Suppose also that i0mod2r-h-1. To prove (4.6) and (4.7), we first compute A(χ2h,2hi), which we split in three summands as follows:

A(χ2h,2hi)=k=12r-1-1εkTr(ζ2r)/(ζ2r-2hi)+j=22j2h-2k=12r-1-1εkTr(ζ2r)/((ζ2rkj+ζ2r-kj)ζ2r-2hi)+k=12r-1-1εkTr(ζ2r)/(ζ2r2h(k-i)+ζ2r-2h(k+i)).

We now prove that the first two summands are 0. This is clear for the first one because 2r-12hi. To prove that the second summand is 0, let 2j2h-2 and 2j. Observe that 2hj. Thus if k is odd, then the order of ζ2r±kj-2hi is a multiple of 2r-h-1, and as hr-3, we deduce that

Tr(ζ2r)/(ζ2rkj-2hi)=Tr(ζ2r)/(ζ2r-kj-2hi)=0.

Thus we only have to consider the summands with k even. Actually, we can exclude also the summands with 2r-hk because, by the induction hypothesis on (4.8), for such k, we have εk=0. For the remaining values of k (i.e., k even and not a multiple of 2r-h), we have εk=εl if klmod2r-h-1 by the induction hypothesis on (4.7). So we can rewrite

k=12r-1-1εkTr(ζ2r)/((ζ2rkj+ζ2r-kj)ζ2r-2hi)

as

l2r-h-1εlTr(ζ2r)/(ζ2rl-2hi(a=02h-1(ζ2r2r-h-1j)a)+ζ2r-l-2hi(a=02h-1(ζ2r-2r-h-1j)a)),

which is 0 because ζ2r2r-h-1j is a root of unity different from 1 and of order dividing 2h, as j is even but not a multiple of 2h. This finishes the proof that the first two summands are 0. To finish the calculation of A(χ2h,2hi), we compute

Tr(ζ2r)/(ζ2r2h(k-i)+ζ2r-2h(k+i))={2r-1ifkXh,i,-2r-1ifk-2r-h-1Xh,i,0otherwise,

where Xh,i={k{1,,2r-2}:k±imod2r-h}. So we have proved the following:

A(χ2h,2hi)=2r-1kXh,i(εk-εk+2r-h-1).

Then (4.6) follows from (4.3), (4.5) and the previous formula. Using (4.2), we also obtain kXh,iεk=kXh,iεk+2r-h-1 if i±1mod2r-h-1. However, in this case, the induction hypothesis for (4.7) means that the εk with kXh,i are all equal and the εk+2r-h-1 with kXh,i are all equal. Hence (4.7) follows.

In order to deal with (4.8), assume that 0<hr-2. By the induction hypothesis on (4.8), we have εk=0 if 2r-hk, and by the induction hypothesis on (4.7), we have that εk is constant on the set X formed by integers 1k2r-1 such that k2r-h-1mod2r-h. We will use these two facts without specific mention. Arguing as before, we have

A(χ2h,0)=k=12r-1-1εkTr(ζ2r)/(1+ζ2r2hk+ζ2r-2hk)+k=12r-1-1εkTr(ζ2r)/(j=12h-1-1(ζ2r2jk+ζ2r-2jk))=k=1,2r-hk2r-1-1εkTr(ζ2r)/(1+ζ2r2hk+ζ2r-2hk).

As

Tr(ζ2r)/(1+ζ2r2hk+ζ2r-2hk)={2r-1if 2r-h-1k,-2r-1if 2r-h-1kand 2r-hk,

we obtain

A(χ2h,0)=2r-1(2r-h-1kεk-2r-h-1kεk)=2r-1(1-2kXεk)=2r-1(1-2|X|εk).

From (4.3) and (4.4), we deduce that if kX, then 1-2|X|εk=±1, and hence εk=0 since |X|=2r-h-12, as hr-2. This finishes the proof of (4.8).

To finish the proof of the proposition, it is enough to show that εi0 for exactly one i{1,,2r-1-1}. If i is even, then εi=0 by (4.8) with h=r-2.

We claim that if εi0, then i±1mod2r-1. Otherwise, there are integers 2vr-2 and 2<i<2r-1-1 satisfying i±1mod2v+1 and εi0. We choose v minimum with this property for some i. Then

  1. εk=0 for every k±1mod2v,

  2. i±(k+2v)mod2v+1 for every kXr-v-1.

Statement (1) implies

kXr-v-1(εk+εk+2v)=1.

On the other hand, 1r-v-1r-3, and hence, applying (4.6) and (4.7) with h=r-v-1, we deduce from (2) that εi=εk+2v for every kXr-v-1 and

kXr-v-1(εk-εk+2v)=±1.

Using |Xr-v-1|=2r-v-1 and εi0, we deduce that

2r-vεi=2kXr-v-1εk+2r-v=2,

in contradiction with 2r-v. This finishes the proof of the claim.

Then the only possible non-zero partial augmentations of u are ε1 and ε2r-1-1. Hence we have ε1+ε2r-1-1=1, and by applying (4.6) with h=0, we deduce that ε1-ε2r-1-1=±1. Therefore, either ε1=0 or ε2r-1-1=0, i.e., εi0 for exactly one i{1,,2r-1-1}, as desired. ∎

5 Proof of Theorem 1.1

In this section, we prove Theorem 1.1. Recall that G=SL(2,q) with q=tf and t an odd prime number, G¯=PSL(2,q), π:GG¯ is the natural projection and u is an element of order n in V(G) with gcd(n,q)=1. We have to show that u is rationally conjugate to an element of G. By Proposition 3.14, we may assume that n is a multiple of 4 and, by Proposition 4.1, that n is not a prime power. Moreover, we may also assume that n12 because this case follows easily from known results and the HeLP method. Indeed, if n=12, then π(u) has order 6 by Proposition 3.12, and hence π(u) is rationally conjugate to an element of G¯ by [13, Proposition 6.6]. Using this and the fact that G has a unique conjugacy class for each of the orders 3, 4 or 6 and two conjugacy classes of elements of order 12, and applying D with χ=χ1 and =1,5, it easily follows that all the partial augmentations of u are non-negative. This can be also obtained using the GAP package HeLP [1].

In the remainder, we follow the strategy of the proof of the main result of [20]. The difference with the arguments of that paper is twofold: On the one hand, in our case, n is even (actually, a multiple of 4), and this introduces some difficulties not appearing in [20], where n was odd. On the other hand, for SL(2,q), we have more Brauer characters than for PSL(2,q), and this will help to reduce some cases.

As the order n of u is fixed throughout, we simplify the notation of Section 2 by setting

γ=γn,γ¯=γ¯n,αx=αx(n),κx=κx(n),βb,x=βb,x(n),𝔹=𝔹n,=n.

We argue by induction on n. So we also assume that ud is rationally conjugate to an element of G for every divisor d of n with d1.

We will use the representations Θm and t-Brauer characters χm introduced in (3.1). Observe that the kernel of Θm is trivial if m is odd, and otherwise it is the center of G. Using this and the induction hypothesis on n, it easily follows that the order of Θm(u) is n2 if m is even, while if m is odd, then the order of Θm(u) is n. Combining this with Proposition 3.17, we deduce that Θ1(u) is conjugate to diag(ζ,ζ-1) for a suitable primitive n-th root of unity ζ. Hence there exists an element g0G of order n such that Θ1(g0) and Θ1(u) are conjugate. The element g0G and the primitive n-th root of unity ζ will be fixed throughout, and from now on, we abuse the notation and consider ζ both as a primitive n-th root of unity in a field of characteristic t and as a complex primitive n-th root of unity. Then

Θm(g0)is conjugate to{diag(1,ζ2,ζ-2,,ζm,ζ-m)if 2m,diag(ζ,ζ-1,ζ3,ζ-3,,ζm,ζ-m)if 2m,

and

(5.1)χm(g0i)=j=-mjmmod2mζij={1+α2i+α4i++αmiif 2m,αi+α3i++αmiif 2m.

By the induction hypothesis on n, if c is a divisor of n with c1, then uc is rationally conjugate to g0i for some i, and hence ζc=ζ±i. Therefore, cni, and hence uc is conjugate to g0c.

By E, two elements of g0 are conjugate in G if and only if they are equal or mutually inverse. Moreover, every element gG, with gn=1, is conjugate to an element of g0. Therefore, x(g0x)G induces a bijection from Γn to the set of conjugacy classes of G formed by elements of order dividing n. For an integer x (or xΓn), we set

εx=εg0x(u)andλx=iΓnεiαix.

Our main tool is the following lemma whose proof is exactly the same as the one of [20, Lemma 4.1]. We also collect [20, Corollary 3.3].

Lemma 5.1.

u is rationally conjugate to g0 if and only if λi=αi for any positive integer i.

For a positive integer n and a subfield F of (ζn), let ΓF denote a set of representatives of equivalence classes of the following equivalence relation defined on :

xyζnxandζnyare conjugate in(ζn)overF.

Corollary 5.2.

Let n be a positive integer; let F be a subfield of Q(ζn), and let R be the ring of integers of F. For every xΓF, let Bx be an integer, and for every integer i, define

ωi=xΓFBxTr(ζn)/F(ζnix).

Let d be a divisor of n such that ωdq=0 for every prime power q dividing d with q1. Then ωddR.

By Lemma 5.1, in order to achieve our goal, it is enough to prove that λi=αi for every positive integer i. We argue by contradiction, so we assume that λdαd for some positive integer d, which we assume to be minimal with this property. Observe that if λi=αi and j is an integer such that gcd(i,n)=gcd(j,n), then there exists σGal((α1)/) such that σ(αi)=αj, and applying σ to the equation λi=αi, we obtain λj=αj. This implies that d divides n. Note that α1=λ1 by our choice of g0, and hence d1. Moreover, dn because λn=2xΓnεx=2=αn as the augmentation of u is 1.

We claim that

(5.2)λd=αd+dτfor someτ[α1].

Indeed, for any xΓn, let Bx=εx-1 if xn1, and Bx=εx otherwise. Then, for any integer i, we have λi-αi=xΓnBxTr(ζ)/(α1)(ζix). The claim then follows by applying Corollary 5.2 with F=(α1), R=[α1] and ωi=λi-αi. Observe that, in the notation of that corollary, Γn=ΓF.

By (5.1), we have

(5.3)

χd(g0)=i=0idmod2dαi,
χd(u)=xΓnεxχd(g0x)=xΓnεxi=0idmod2dαix=i=0idmod2dλi.

Combining this with (5.2) and the minimality of d, we deduce that

χd(u)=χd(g0)+dτ.

Furthermore, τ0, as λdαd. Therefore,

(5.4)Cb(χd(u))Cb(χd(g0))moddfor everyb𝔹

and

(5.5)d|Cb0(χd(u))-Cb0(χd(g0))|for someb0𝔹.

The bulk of our argument relies on an analysis of the eigenvalues of Θd(u) and the induction hypothesis on n and d. More precisely, we will use (5.4) and (5.5) to obtain a contradiction by comparing the eigenvalues of Θd(g0) and Θd(u). Of course, we do not know the eigenvalues of the latter. However, we know the eigenvalues of Θd(uc) for every cn with c1 because we know the eigenvalues of Θd(g0) and uc is conjugate to g0c. This provides information on the eigenvalues of Θd(u). For example, recall that if ξ is an eigenvalue of Θd(u), then ξ and ξ-1 have the same multiplicity as eigenvalues of Θd(u). Therefore, if 3h, then the sum of the multiplicities of the eigenvalues of Θd(u) of order h is even. Moreover, for every t-regular element g of G, the multiplicity of 1 as eigenvalue of Θd(g) is congruent modulo 2 with the degree d+1 of χd. As n is not a prime power, there is an odd prime p dividing n. By the induction hypothesis, Θd(up) is rationally conjugate to Θd(g0p). Thus the multiplicity of -1 as eigenvalue of Θd(up) is even. As the latter is the sum of the multiplicities as eigenvalues of Θd(u) of -1 and the elements of order 2p, we deduce that the multiplicity of -1 as eigenvalue of Θd(u) is even. Using this, we can see that Θd(u) is conjugate to diag(ζν-d,ζν2-d,,ζνd-2,ζνd) for integers ν-d,ν-d+2,,νd-2,νd such that ν-i=-νi for every i. Let Xd={i:1id,idmod2}. Then, by (5.3) and Proposition 2.2, for every b𝔹, we have

(5.6)Cb(χd(u))-Cb(χd(g0))=iXd(κνiβb,νiμ(γ(νi))δb,νi(n/γ¯(νi))-κiβb,iμ(γ(i))δb,i(n/γ¯(i))).

Moreover, if cn with c1, then the lists (cνi)iXd and (ci)iXd represent the same elements in Γn, up to ordering, and hence (νi)iXd and (i)iXd represent the same elements of Γnc, up to ordering, which we express by writing (ν(Xd))nc(Xd). This provides restrictions on d, n and νi.

The following two lemmas are variants of [20, Lemmas 4.2 and 4.3].

Lemma 5.3.

The following statements hold:

  1. Let iXd. If κi1, then n=2d and i=d. If κνi1, then nd is the smallest prime dividing n and κνj=1 for every jXd{i}.

  2. If d>2, then n is not divisible by any prime greater than d. In particular, if d is prime, then κνi=1 for every iXd.

Proof.

Let p denote the smallest prime dividing n.

1 The first statement is clear. Suppose that κνi1. Then either p=2 and νi0modn2 or νi0modn. We deduce that k0modnp for some kXd since (Xd)np(ν(Xd)). Therefore, d=k=np, and for every jXd{i}, we have νj0modnp. Thus κνj=1.

2 Suppose that q is a prime divisor of n with d<q. Then ndp, and therefore, by 1, κi=κνi=1 for every iXd. Thus, by (5.5) and (5.6), it is enough to show that δb,i(n/γ¯(i))0 for at most one iXd and δb,νi(n/γ¯(νi))0 for at most one iXd. Observe that if iXd, then qi, and hence nγ¯(i) is a multiple of q. Moreover, if i and j are different elements of Xd, then i and j have the same parity and -q<i-j<i+j<2q. Therefore, iqj. Thus either

δb,i(n/γ¯(i))=0orδb,j(n/γ¯(j))=0.

As (Xd)q(ν(Xd)), this also proves

δb,νi(n/γ¯(νi))=0orδb,νj(n/γ¯(νj))=0.

We obtain an upper bound for |Cb(χd(u))-Cb(χd(g0))| in terms of the number of prime divisors P(d) of d.

Lemma 5.4.

For every bB, we have

|Cb(χd(u))-Cb(χd(g0))|2+2P(d)+1.

Proof.

Using (5.6), it is enough to prove that iXdκiδb,i(n/γ¯(i))1+2P(d) and iXdκνiδb,νi(n/γ¯(νi))1+2P(d). This is a consequence of Lemma 5.31 and the following inequalities for every e dividing d:

|{iXd:gcd(d,γ¯(i))=e,δb,i(n/γ¯(i))=1}|1,
|{iXd:gcd(d,γ¯(νi))=e,δb,νi(n/γ¯(νi))=1}|1.

We prove the second inequality, only using (ν(Xd))d(Xd). This implies the first inequality by applying the second one to u=g0.

Let Ye={iXd:gcd(d,γ¯(νi))=e,δb,νi(n/γ¯(νi))=1}. By changing the sign of some νi’s, we may assume without loss of generality that if δb,νi(n/γ¯(νi))=1, then bνimodnγ¯(νi). Thus if iYe, then bνimodnγ¯(νi). We claim that if i,jYe, then νiνjmodd. Indeed, let p be prime divisor of d. If npdp or pe, then clearly νiνjmoddp. Otherwise, i.e., if np=dp and pe, then p divides both γ¯(νi) and γ¯(νj) and νiνjmoddpp. Therefore, by Lemma 2.12, νiνjmodnp, as desired. As (ν(Xd))d(Xd) and the elements of Xd represent different classes in Γd, we deduce that |Ye|1. This finishes the proof of the lemma. ∎

We are ready to finish the proof of Theorem 1.1. Recall that we are arguing by contradiction.

By (5.5) and Lemma 5.4, we have d2+2P(d)+1, and using this, it is easy to show that d6 or d=10. Indeed, if P(d)3, then

2+2P(d)+1d2352P(d)-3>14+2P(d)+1,

a contradiction. Thus P(d)=2 and d10, or P(d)=1 and d5. Hence d is either 2,3,4,5,6 or 10. We deal with these cases separately.

Suppose that d=2. Then ν2np2 for every prime p. By the assumptions on n and Lemma 5.31, this implies κ2=κν2=1, γ(2)=γ(ν2) and βb0,2=βb0,ν2. Therefore,

|Cb0(χ2(u))-Cb0(χ2(g0))|=|μ(γ(2))(δb0,2(n/γ¯(2))-δb0,ν2(n/γ¯(ν2)))|1,

contradicting (5.5).

Suppose that d=3. By the assumptions on n and Lemma 5.3, κi=κνi=1 for every iX3 and n=6. If 24n or 32n, then

|{i=1,3:δb,i(n/γ¯(i))=1}|1and|{i=1,3:δb,νi(n/γ¯(νi))=1}|1,

which implies |Cb0(χ3(u))-Cb0(χ3(g0))|2, contradicting (5.5). Thus n=24 since n is neither 12 nor a prime power and it is a multiple of 4. In this case, we have γ¯(1)=γ(1)=2, γ¯(3)=γ(3)=3,

βb,1=βb,3=1andCb(χ3(g0))=-δb,1(12)-δb,3(8)for everyb𝔹.

We may assume that 3ν3 and 3ν1 because (ν(X3))3(X3). Suppose that ν383 and ν181. Then γ¯(ν1)=γ(ν1)=2, γ¯(ν3)=γ(ν3)=3, βb0,ν3=1 and δb0,3(8)=δb0,ν3(8), which implies |Cb0(χ3(u))-Cb0(χ3(g0))|2, contradicting (5.5). Suppose now that ν381 and ν183. This implies ν1±3mod8 and ν1±1mod3 (because 3ν3 but 3ν1). Thus either ν1±11mod24 or ν1±5mod24. As (ν(X3))12(X3), we deduce that the only possibility is ν1±11mod24. In this case, γ¯(ν1)=γ(ν1)=1 and γ¯(ν3)=γ(ν3)=6. Hence

C11(χ3(u))-C11(χ3(g0))=δ11,ν1(24)+δ11,ν3(4)+δ11,1(12)+δ11,3(8)=4,

contradicting (5.4).

Suppose that d=4. By the assumptions on n and Lemma 5.3, κi=κνi=1 for every iX4 and n=6. If 33n or 23n, then

|{i=2,4:δb,i(n/γ¯(i))=1}|1,

which implies |Cb0(χ4(u))-Cb0(χ4(g0))|3, contradicting (5.5). Thus n=36. In this case, we have γ(2)=1=βb0,2=βb0,4 and γ(4)=2, which implies |Cb0(χ4(g0))|1, and hence

|Cb0(χ4(u))-Cb0(χ4(g0))|3,

contradicting again (5.5).

Suppose that d=5. Since (ν(X5))5(X5), there is exactly one νi which is divisible by 5, say, ν5. In particular, for i5, we have 5nγ¯(νi) and 5nγ¯(i). Moreover, if j is an integer not a multiple of 5, then |{i=1,3:νi5j}|1. This implies

|{i=1,3:δb,i(n/γ¯(i))=1}|1and|{i=1,3:δb,νi(n/γ¯(νi))=1}|1.

On the other hand, since n10, we deduce that κi=1 for every iX5 by Lemma 5.31. Therefore, using (5.5) and (5.6), we deduce that κν5=2, in contradiction with Lemma 5.31.

Suppose that d=6. By Lemma 5.3, we have n30 and κi=κνi=1 for every iX6 because n12. If 25n, or 9n or 8n, then

|{i=2,4,6:δb,i(n/γ¯(i))=1}|2and|{i=2,4,6:δb,νi(n/γ¯(νi))=1}|2.

This implies |Cb0(χ6(u))-Cb0(χ6(g0))|4, yielding a contradiction with (5.5). Therefore, n=60, and hence βb,2=βb,4=βb,6=1, γ¯(2)=1, γ¯(4)=γ(4)=2 and γ¯(6)=γ(6)=3. This implies |Cb0(χ6(g0))|2, and hence

|Cb0(χ6(u))-Cb0(χ6(g0))|5,

yielding a contradiction with (5.5).

Suppose that d=10. If 5nγ¯(i) for some iX10, then n5=(γ¯(i))5=5, and hence 5i. The same also holds for νi. Therefore, if 5i, then 5nγ¯(i), and if 5νi, then 5nγ¯(νi). Thus

|{iX10:5i,δb,i(n/γ¯(i))=1}|2,
|{iX10:5νi,δb,νi(n/γ¯(νi))=1}|2.

This implies |Cb0(χ10(u))-Cb0(χ10(g0))|8, contradicting (5.5).


Communicated by Pavel A. Zalesskii


Award Identifier / Grant number: MTM2012-35240

Award Identifier / Grant number: MTM2016-77445-P

Funding source: Fundación Séneca

Award Identifier / Grant number: 19880/GERM/15

Funding statement: Partially supported by Ministerio de Economía y Competitividad projects MTM2012-35240 and MTM2016-77445-P and Fondos FEDER and Fundación Séneca of Murcia 19880/GERM/15.

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Received: 2018-05-04
Revised: 2019-03-25
Published Online: 2019-05-07
Published in Print: 2019-09-01

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