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

Ángel del Río 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 , p 2 ) 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 , p f ) , 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 ( Z G ) 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 , p f ) with p a prime number and f 2 .

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 ζ n denotes a complex primitive n-th root of unity, Φ n ( X ) denotes the n-th cyclotomic polynomial, i.e., the minimal polynomial of ζ n over , and for a prime integer p, let v p ( n ) denote the valuation of n at p, i.e., the maximum non-negative integer m with p m n . If F / K is a finite field extension, then Tr F / K : F K 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 ( ζ n ) / ( ζ d ) = 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 = p n p and n p = p v p ( n ) .

If moreover x , then we set

( : x n ) = representative of the class of x modulo n in the interval ( - n 2 , n 2 ] ,
| : x n | = the absolute value of ( : x n ) ,
γ n ( x ) = p n | : x n p | < n p 2 p p ,
γ ¯ n ( x ) = p n | : x n p | n p 2 p p = { 2 γ n ( x ) if | : x n 2 | = n 2 4 , γ n ( 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 , y Z . Then the following conditions hold.

1. (1)

If p γ ¯ n ( x ) , then ( : x n p p ) x mod n p .

2. (2)

Let d n such that x y mod n d . If d divides both γ ¯ n ( x ) and γ ¯ n ( y ) , then x y mod n .

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

x n y x ± y mod n .

We denote by Γ n the set of these equivalence classes.

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

δ x , y ( n ) = { 1 if x n y , 0 otherwise , and κ x ( n ) = { 2 if x 0 mod n or x n 2 mod n , 1 otherwise .

For an integer x (or x Γ n ), we set α x ( n ) = ζ n x + ζ n - x . Observe that ( α 1 ( n ) ) is the maximal real subfield of ( ζ n ) and [ α 1 ( n ) ] is the ring of integers of ( α 1 ( n ) ) . If n n 2 , then let p 0 denote the smallest odd prime dividing n. Then let

𝔹 n = { x n : for every p n , either | : x n p | > n p 2 p or p = 2 , n n 2 , | : x n 2 | = n 2 4 , n p 0 x and ( x : n 2 ) ( x : n p 0 ) > 0 }

and

n = { { α b ( n ) : b 𝔹 n } if n n 2 , { 1 } { α b ( n ) : b 𝔹 n } otherwise .

For b 𝔹 n and x , let

β b , x ( n ) = { - 1 if n n 2 , | : x n 2 | = n 2 4 and ( : x n 2 ) ( : b n p 0 ) < 0 , 1 otherwise .

The following proposition extends [20, Proposition 3.5]. The first statement implies that n is a -basis of ( α 1 ( n ) ) . For x ( α 1 ( n ) ) and b 𝔹 n , we use C b ( x ) to denote the coefficient of α b ( n ) in the expression of x in the basis n .

## Proposition 2.2.

Let n be a positive integer.

1. (1)

Then n is a -basis of [ α 1 ( n ) ] .

2. (2)

If b 𝔹 n and i , then C b ( α i ( n ) ) = κ i ( n ) μ ( γ ( i ) ) β b , i ( n ) δ b , i ( n / γ ¯ ( i ) ) .

## Proof.

We only prove the proposition in the case n n 2 , as the proof in the case n = n 2 is similar (actually simpler). It is easy to see that

| n | φ ( n ) 2 = [ ( α 1 ( n ) ) : ] .

Thus it is enough to prove the equality

α i ( n ) = κ i ( n ) μ ( γ ( i ) ) b 𝔹 n b n / γ ¯ ( i ) i β b , i ( n ) α b ( n ) .

Actually, we will show

ζ n i = μ ( γ ( i ) ) b i mod n / γ ¯ ( i ) b 𝔹 n β b , i ( n ) ζ n b ,

which easily implies the desired expression of α i ( n ) .

Indeed, for every p n , let ζ n p denote the p-th part of ζ n , i.e., ζ n p is a primitive n p -th root of unity and ζ n = p n ζ n p . Let J be the set of tuples ( j p ) p γ ¯ ( i ) satisfying the following conditions:

• If p γ ( i ) , then j p { 1 , , p - 1 } .

• If p = 2 and | : i n 2 | = n 2 4 , then

j 2 = { 1 if ( : i n 2 ) ( i + : j p 0 n p 0 p 0 n p 0 ) < 0 , 0 otherwise .

For every j J , let b j n be given by

b j { i + j p n p p mod n p if p γ ¯ ( i ) , i mod n p otherwise .

Then { b j : j J } is the set of elements in 𝔹 n satisfying i b mod n γ ¯ ( i ) . From

0 = ζ n p i ( 1 + ζ n p n p p + ζ n p 2 n p p + + ζ n p ( p - 1 ) n p p ) ,

we obtain

ζ n p i = - j p = 1 p - 1 ζ n p i + j p n p p .

Therefore, if | : i n 2 | n 2 4 , then γ ( i ) = γ ¯ ( i ) , β b , i ( n ) = 1 for every b 𝔹 n and

ζ n i = p n p γ ( i ) ζ n p i p n p γ ( i ) ( - j p = 1 p - 1 ζ n p i + j p n p p ) = μ ( γ ( i ) ) j J ζ n b j = μ ( γ ( i ) ) b i mod n / γ ¯ ( i ) b 𝔹 n ζ n b .

This gives the desired equality in this case.

Suppose that | : i n 2 | = n 2 4 . Then ζ n 2 i = β b j , i ( n ) ζ n 2 b j for every j J . Then a small modification of the argument in the previous paragraph gives

ζ n i = ζ n 2 i p n p γ ( i ) ζ n p i p n p γ ( i ) ( - j p = 1 p - 1 ζ n p i + j p n p p ) = μ ( γ ( i ) ) j J β b , i ( n ) ζ n b j = μ ( γ ( i ) ) b i mod n / γ ¯ ( i ) b 𝔹 n β b , i ( n ) ζ n b .

## 3 Group theoretical preliminaries

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

ε g ( α ) = h g G α h .

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

1. (A)

If g Z ( G ) and u g , then ε g ( u ) = 0 (Berman–Higman theorem, [15, Proposition 1.5.1]).

2. (B)

If g G and ε g ( u ) 0 , then | g | divides n [13, Theorem 2.3].

3. (C)

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

4. (D)

See [16, 13]. Let F be a field of characteristic t 0 with t n . Let ρ be an F-representation of G. If t 0 , then let ξ n be a primitive n-th root of unity in F, so that if t = 0 , then ξ n = ζ n . 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 ξ n to ζ n ). Then, for every integer , the multiplicity of ξ n as eigenvalue of ρ ( u ) is

1 n x T d n ε x ( u d ) Tr ( ζ n d ) / ( χ ( x ) ζ n - 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 : G G ¯ denote the natural projection, which we extend by linearity to a ring homomorphism π : G G ¯ .

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

1. (E)

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 2 p , 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. (F)

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 C 1 and C 2 of G with C 2 = J C 1 . Furthermore, if n is a multiple of 4, then the elements of C 1 and C 2 have order 2 n , while if n is not a multiple of 4, then one of the classes C 1 or C 2 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 ( Z G ) . Set G ¯ = PSL ( 2 , q ) and n = | u | . Then the following statements hold.

1. (1)

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

2. (2)

| π ( u ) | = n gcd ( 2 , n ) .

3. (3)

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

4. (4)

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

5. (5)

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

6. (6)

Suppose that q = p f with p prime, f 2 and p n . Then u is rationally conjugate to an element of G.

7. (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 u 2 = 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 ε g ( u ) 0 , then | g | is odd by B, and hence ε g ( u ) = ε π ( g ) ( π ( u ) ) 0 , by F. Thus u is rationally conjugate to an element of G.

4 Let q = p f , where p is an odd prime number and p n . 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 ε g ( u ) = 0 for every g C , and hence u is rationally conjugate to an element of G by C.

If 4 n , then | π ( u ) | is coprime with 2 q 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 2 p 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 ) S 3 and SL ( 2 , 4 ) A 5 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 t n , and let ξ n 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 Θ m of G of degree 1 + m such that

(3.1) Θ m ( g ) = { diag ( 1 , ξ n 2 , ξ n - 2 , , ξ n m , ξ n - m ) if  2 m , diag ( ξ n , ξ n - 1 , ξ n 3 , ξ n - 3 , , ξ n m , ξ n - m ) if  2 m .

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

χ m ( g i ) = j = - m j m mod 2 m ζ n i j .

## 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 ( Z G ) . 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 | = 2 r with r 3 . We argue by induction on r. So we assume that units of order 2 k with 1 k r - 1 are rationally conjugate to an element of G. By Proposition 3.15 and E, G has an element g 0 of order 2 r such that { g 0 k : k = 0 , 1 , 2 , , 2 r - 1 } is a set of representatives of the conjugacy classes of G with order a divisor of 2 r . By B, the only possible non-zero partial augmentations of u are the integers ε k = ε g 0 k ( u ) , with k = 1 , , 2 r - 1 - 1 . By the induction hypothesis, if 1 i r and g G , then ε g ( u 2 i ) 0 . Hence, by C, it suffices to prove that ε k = 0 for all but one k = 0 , 1 , , 2 r - 1 .

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

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

A ( χ , ) = k = 1 2 r - 1 - 1 ε k Tr ( ζ 2 r ) / ( χ ( g 0 k ) ζ 2 r - ) ,
B ( χ , ) = k = 0 r - 1 Tr ( ζ 2 k ) / ( χ ( g 0 2 r - k ) ζ 2 k - ) .

Then, by D, we have

(4.1) 1 2 r ( A ( χ , ) + B ( χ , ) ) 0 .

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

(4.2) if B ( χ , ) = 0 , then A ( χ , ) = 0 ,
(4.3) if B ( χ , ) = 2 r - 1 , then A ( χ , ) = ± 2 r - 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  0 h r - 2 and  2 r - 1 , then B ( χ 2 h , ) = { 2 r - 1 if h 1 , 0 if h = 0 ,

and

(4.5) if  0 h r - 3 ,  2 h and  2 r - 1 , then B ( χ 2 h , ) = { 2 r - 1 if ± 2 h mod 2 r - 1 , 0 otherwise .

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

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

This implies

B ( χ 2 h , ) = B ( χ 2 h - 1 , ) + j = 2 2 j 2 h - 1 k = 0 r - 1 Tr ( ζ 2 k ) / ( ζ 2 k 2 h - 1 + j + ζ 2 k - 2 h - 1 - j ) = 2 r - 1 .

This finishes the proof of (4.4).

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

B ( χ 2 h , ) = j = 2 2 j 2 h - 1 k = 0 r - 1 Tr ( ζ 2 k ) / ( ( ζ 2 k 2 h - 1 + j + ζ 2 k - ( 2 h - 1 + j ) ) ζ 2 k - ) .

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

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

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

B ( χ 2 h , ) = k = 0 h Tr ( ζ 2 k ) / ( ( ζ 2 k 2 h + ζ 2 k - 2 h ) ζ 2 k - ) + ϵ 2 h + 1 + k = h + 3 r - 1 Tr ( ζ 2 k ) / ( ( ζ 2 k 2 h + ζ 2 k - 2 h ) ζ 2 k - ) ,

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

k = 0 h Tr ( ζ 2 k ) / ( ( ζ 2 k 2 h + ζ 2 k - 2 h ) ζ 2 k - ) = 2 h + 1 ,
k = h + 3 r - 1 Tr ( ζ 2 k ) / ( ( ζ 2 k 2 h + ζ 2 k - 2 h ) ζ 2 k - ) = { 0 if  2 h + 1 , 2 r - 1 - 2 h + 2 if  2 h + 1 and ± 2 h mod 2 r - 1 , - 2 h + 2 if  2 h + 1 and ± 2 h mod 2 r - 1 .

This finishes the proof of (4.5).

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

(4.6) k X h ( ε k - ε k + 2 r - h - 1 ) = ± 1 , where X h = { i { 1 , , 2 r - 2 } : i ± 1 mod 2 r - h } ,
(4.7) if i ± j mod 2 r - h - 1 and i 0 , ± 1 mod 2 r - h - 1 , then ε i = ε j ,

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

(4.8) if i 0 mod 2 r - h - 1 , then ε i = 0 .

Observe that X 0 = { 1 } . Fix an integer i. Then, for every integer k, we have

Tr ( ζ 2 r ) / ( ( ζ 2 r k + ζ 2 r - k ) ζ 2 r - i ) = { 2 r - 1 if k i mod 2 r , - 2 r - 1 if k 2 r - 1 - i mod 2 r , 0 otherwise .

Thus A ( χ 1 , i ) = 2 r - 1 ( ε i - ε i + 2 r - 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 ε 2 r - 1 = 0 .

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

A ( χ 2 h , 2 h i ) = k = 1 2 r - 1 - 1 ε k Tr ( ζ 2 r ) / ( ζ 2 r - 2 h i ) + j = 2 2 j 2 h - 2 k = 1 2 r - 1 - 1 ε k Tr ( ζ 2 r ) / ( ( ζ 2 r k j + ζ 2 r - k j ) ζ 2 r - 2 h i ) + k = 1 2 r - 1 - 1 ε k Tr ( ζ 2 r ) / ( ζ 2 r 2 h ( k - i ) + ζ 2 r - 2 h ( k + i ) ) .

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

Tr ( ζ 2 r ) / ( ζ 2 r k j - 2 h i ) = Tr ( ζ 2 r ) / ( ζ 2 r - k j - 2 h i ) = 0 .

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

k = 1 2 r - 1 - 1 ε k Tr ( ζ 2 r ) / ( ( ζ 2 r k j + ζ 2 r - k j ) ζ 2 r - 2 h i )

as

l 2 r - h - 1 ε l Tr ( ζ 2 r ) / ( ζ 2 r l - 2 h i ( a = 0 2 h - 1 ( ζ 2 r 2 r - h - 1 j ) a ) + ζ 2 r - l - 2 h i ( a = 0 2 h - 1 ( ζ 2 r - 2 r - h - 1 j ) a ) ) ,

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

Tr ( ζ 2 r ) / ( ζ 2 r 2 h ( k - i ) + ζ 2 r - 2 h ( k + i ) ) = { 2 r - 1 if k X h , i , - 2 r - 1 if k - 2 r - h - 1 X h , i , 0 otherwise ,

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

A ( χ 2 h , 2 h i ) = 2 r - 1 k X h , i ( ε k - ε k + 2 r - h - 1 ) .

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

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

A ( χ 2 h , 0 ) = k = 1 2 r - 1 - 1 ε k Tr ( ζ 2 r ) / ( 1 + ζ 2 r 2 h k + ζ 2 r - 2 h k ) + k = 1 2 r - 1 - 1 ε k Tr ( ζ 2 r ) / ( j = 1 2 h - 1 - 1 ( ζ 2 r 2 j k + ζ 2 r - 2 j k ) ) = k = 1 , 2 r - h k 2 r - 1 - 1 ε k Tr ( ζ 2 r ) / ( 1 + ζ 2 r 2 h k + ζ 2 r - 2 h k ) .

As

Tr ( ζ 2 r ) / ( 1 + ζ 2 r 2 h k + ζ 2 r - 2 h k ) = { 2 r - 1 if  2 r - h - 1 k , - 2 r - 1 if  2 r - h - 1 k and  2 r - h k ,

we obtain

A ( χ 2 h , 0 ) = 2 r - 1 ( 2 r - h - 1 k ε k - 2 r - h - 1 k ε k ) = 2 r - 1 ( 1 - 2 k X ε k ) = 2 r - 1 ( 1 - 2 | X | ε k ) .

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

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

We claim that if ε i 0 , then i ± 1 mod 2 r - 1 . Otherwise, there are integers 2 v r - 2 and 2 < i < 2 r - 1 - 1 satisfying i ± 1 mod 2 v + 1 and ε i 0 . We choose v minimum with this property for some i. Then

1. (1)

ε k = 0 for every k ± 1 mod 2 v ,

2. (2)

i ± ( k + 2 v ) mod 2 v + 1 for every k X r - v - 1 .

Statement (1) implies

k X r - v - 1 ( ε k + ε k + 2 v ) = 1 .

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

k X r - v - 1 ( ε k - ε k + 2 v ) = ± 1 .

Using | X r - v - 1 | = 2 r - v - 1 and ε i 0 , we deduce that

2 r - v ε i = 2 k X r - v - 1 ε k + 2 r - v = 2 ,

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

Then the only possible non-zero partial augmentations of u are ε 1 and ε 2 r - 1 - 1 . Hence we have ε 1 + ε 2 r - 1 - 1 = 1 , and by applying (4.6) with h = 0 , we deduce that ε 1 - ε 2 r - 1 - 1 = ± 1 . Therefore, either ε 1 = 0 or ε 2 r - 1 - 1 = 0 , i.e., ε i 0 for exactly one i { 1 , , 2 r - 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 = t f and t an odd prime number, G ¯ = PSL ( 2 , q ) , π : G G ¯ 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 n 12 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 = χ 1 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 = γ n , γ ¯ = γ ¯ n , α x = α x ( n ) , κ x = κ x ( n ) , β b , x = β b , x ( n ) , 𝔹 = 𝔹 n , = n .

We argue by induction on n. So we also assume that u d is rationally conjugate to an element of G for every divisor d of n with d 1 .

We will use the representations Θ m and t-Brauer characters χ m introduced in (3.1). Observe that the kernel of Θ m 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 Θ m ( u ) is n 2 if m is even, while if m is odd, then the order of Θ m ( u ) is n. Combining this with Proposition 3.17, we deduce that Θ 1 ( u ) is conjugate to diag ( ζ , ζ - 1 ) for a suitable primitive n-th root of unity ζ. Hence there exists an element g 0 G of order n such that Θ 1 ( g 0 ) and Θ 1 ( u ) are conjugate. The element g 0 G 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

Θ m ( g 0 ) is conjugate to { diag ( 1 , ζ 2 , ζ - 2 , , ζ m , ζ - m ) if  2 m , diag ( ζ , ζ - 1 , ζ 3 , ζ - 3 , , ζ m , ζ - m ) if  2 m ,

and

(5.1) χ m ( g 0 i ) = j = - m j m mod 2 m ζ i j = { 1 + α 2 i + α 4 i + + α m i if  2 m , α i + α 3 i + + α m i if  2 m .

By the induction hypothesis on n, if c is a divisor of n with c 1 , then u c is rationally conjugate to g 0 i for some i, and hence ζ c = ζ ± i . Therefore, c n i , and hence u c is conjugate to g 0 c .

By E, two elements of g 0 are conjugate in G if and only if they are equal or mutually inverse. Moreover, every element g G , with g n = 1 , is conjugate to an element of g 0 . Therefore, x ( g 0 x ) G induces a bijection from Γ n to the set of conjugacy classes of G formed by elements of order dividing n. For an integer x (or x Γ n ), we set

ε x = ε g 0 x ( u ) and λ x = i Γ n ε i α i x .

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 g 0 if and only if λ i = α i for any positive integer i.

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

x y ζ n x and ζ n y are conjugate in ( ζ n ) over F .

## Corollary 5.2.

Let n be a positive integer; let F be a subfield of Q ( ζ n ) , and let R be the ring of integers of F. For every x Γ F , let B x be an integer, and for every integer i, define

ω i = x Γ F B x Tr ( ζ n ) / F ( ζ n i x ) .

Let d be a divisor of n such that ω d q = 0 for every prime power q dividing d with q 1 . Then ω d d R .

By Lemma 5.1, in order to achieve our goal, it is enough to prove that λ i = α i for every positive integer i. We argue by contradiction, so we assume that λ d α d for some positive integer d, which we assume to be minimal with this property. Observe that if λ i = α i and j is an integer such that gcd ( i , n ) = gcd ( j , n ) , then there exists σ Gal ( ( α 1 ) / ) such that σ ( α i ) = α j , and applying σ to the equation λ i = α i , we obtain λ j = α j . This implies that d divides n. Note that α 1 = λ 1 by our choice of g 0 , and hence d 1 . Moreover, d n because λ n = 2 x Γ n ε x = 2 = α n as the augmentation of u is 1.

We claim that

(5.2) λ d = α d + d τ for some τ [ α 1 ] .

Indeed, for any x Γ n , let B x = ε x - 1 if x n 1 , and B x = ε x otherwise. Then, for any integer i, we have λ i - α i = x Γ n B x Tr ( ζ ) / ( α 1 ) ( ζ i x ) . The claim then follows by applying Corollary 5.2 with F = ( α 1 ) , R = [ α 1 ] and ω i = λ i - α i . Observe that, in the notation of that corollary, Γ n = Γ F .

By (5.1), we have

(5.3)

χ d ( g 0 ) = i = 0 i d mod 2 d α i ,
χ d ( u ) = x Γ n ε x χ d ( g 0 x ) = x Γ n ε x i = 0 i d mod 2 d α i x = i = 0 i d mod 2 d λ i .

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

χ d ( u ) = χ d ( g 0 ) + d τ .

Furthermore, τ 0 , as λ d α d . Therefore,

(5.4) C b ( χ d ( u ) ) C b ( χ d ( g 0 ) ) mod d for every b 𝔹

and

(5.5) d | C b 0 ( χ d ( u ) ) - C b 0 ( χ d ( g 0 ) ) | for some b 0 𝔹 .

The bulk of our argument relies on an analysis of the eigenvalues of Θ d ( 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 Θ d ( g 0 ) and Θ d ( u ) . Of course, we do not know the eigenvalues of the latter. However, we know the eigenvalues of Θ d ( u c ) for every c n with c 1 because we know the eigenvalues of Θ d ( g 0 ) and u c is conjugate to g 0 c . This provides information on the eigenvalues of Θ d ( u ) . For example, recall that if ξ is an eigenvalue of Θ d ( u ) , then ξ and ξ - 1 have the same multiplicity as eigenvalues of Θ d ( u ) . Therefore, if 3 h , then the sum of the multiplicities of the eigenvalues of Θ d ( u ) of order h is even. Moreover, for every t-regular element g of G, the multiplicity of 1 as eigenvalue of Θ d ( g ) is congruent modulo 2 with the degree d + 1 of χ d . As n is not a prime power, there is an odd prime p dividing n. By the induction hypothesis, Θ d ( u p ) is rationally conjugate to Θ d ( g 0 p ) . Thus the multiplicity of -1 as eigenvalue of Θ d ( u p ) is even. As the latter is the sum of the multiplicities as eigenvalues of Θ d ( u ) of -1 and the elements of order 2 p , we deduce that the multiplicity of -1 as eigenvalue of Θ d ( u ) is even. Using this, we can see that Θ d ( u ) is conjugate to diag ( ζ ν - d , ζ ν 2 - d , , ζ ν d - 2 , ζ ν d ) for integers ν - d , ν - d + 2 , , ν d - 2 , ν d such that ν - i = - ν i for every i. Let X d = { i : 1 i d , i d mod 2 } . Then, by (5.3) and Proposition 2.2, for every b 𝔹 , we have

(5.6) C b ( χ d ( u ) ) - C b ( χ d ( g 0 ) ) = i X d ( κ ν i β b , ν i μ ( γ ( ν i ) ) δ b , ν i ( n / γ ¯ ( ν i ) ) - κ i β b , i μ ( γ ( i ) ) δ b , i ( n / γ ¯ ( i ) ) ) .

Moreover, if c n with c 1 , then the lists ( c ν i ) i X d and ( c i ) i X d represent the same elements in Γ n , up to ordering, and hence ( ν i ) i X d and ( i ) i X d represent the same elements of Γ n c , up to ordering, which we express by writing ( ν ( X d ) ) n c ( X d ) . This provides restrictions on d, n and ν i .

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

## Lemma 5.3.

The following statements hold:

1. (1)

Let i X d . If κ i 1 , then n = 2 d and i = d . If κ ν i 1 , then n d is the smallest prime dividing n and κ ν j = 1 for every j X d { i } .

2. (2)

If d > 2 , then n is not divisible by any prime greater than d. In particular, if d is prime, then κ ν i = 1 for every i X d .

## Proof.

Let p denote the smallest prime dividing n.

1 The first statement is clear. Suppose that κ ν i 1 . Then either p = 2 and ν i 0 mod n 2 or ν i 0 mod n . We deduce that k 0 mod n p for some k X d since ( X d ) n p ( ν ( X d ) ) . Therefore, d = k = n p , and for every j X d { i } , we have ν j 0 mod n p . Thus κ ν j = 1 .

2 Suppose that q is a prime divisor of n with d < q . Then n d p , and therefore, by 1, κ i = κ ν i = 1 for every i X d . Thus, by (5.5) and (5.6), it is enough to show that δ b , i ( n / γ ¯ ( i ) ) 0 for at most one i X d and δ b , ν i ( n / γ ¯ ( ν i ) ) 0 for at most one i X d . Observe that if i X d , then q i , and hence n γ ¯ ( i ) is a multiple of q. Moreover, if i and j are different elements of X d , then i and j have the same parity and - q < i - j < i + j < 2 q . Therefore, i q j . Thus either

δ b , i ( n / γ ¯ ( i ) ) = 0 or δ b , j ( n / γ ¯ ( j ) ) = 0 .

As ( X d ) q ( ν ( X d ) ) , this also proves

δ b , ν i ( n / γ ¯ ( ν i ) ) = 0 or δ b , ν j ( n / γ ¯ ( ν j ) ) = 0 .

We obtain an upper bound for | C b ( χ d ( u ) ) - C b ( χ d ( g 0 ) ) | in terms of the number of prime divisors P ( d ) of d.

## Lemma 5.4.

For every b B , we have

| C b ( χ d ( u ) ) - C b ( χ d ( g 0 ) ) | 2 + 2 P ( d ) + 1 .

## Proof.

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

| { i X d : gcd ( d , γ ¯ ( i ) ) = e , δ b , i ( n / γ ¯ ( i ) ) = 1 } | 1 ,
| { i X d : gcd ( d , γ ¯ ( ν i ) ) = e , δ b , ν i ( n / γ ¯ ( ν i ) ) = 1 } | 1 .

We prove the second inequality, only using ( ν ( X d ) ) d ( X d ) . This implies the first inequality by applying the second one to u = g 0 .

Let Y e = { i X d : gcd ( d , γ ¯ ( ν i ) ) = e , δ b , ν i ( n / γ ¯ ( ν i ) ) = 1 } . By changing the sign of some ν i ’s, we may assume without loss of generality that if δ b , ν i ( n / γ ¯ ( ν i ) ) = 1 , then b ν i mod n γ ¯ ( ν i ) . Thus if i Y e , then b ν i mod n γ ¯ ( ν i ) . We claim that if i , j Y e , then ν i ν j mod d . Indeed, let p be prime divisor of d. If n p d p or p e , then clearly ν i ν j mod d p . Otherwise, i.e., if n p = d p and p e , then p divides both γ ¯ ( ν i ) and γ ¯ ( ν j ) and ν i ν j mod d p p . Therefore, by Lemma 2.12, ν i ν j mod n p , as desired. As ( ν ( X d ) ) d ( X d ) and the elements of X d represent different classes in Γ d , we deduce that | Y e | 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 d 2 + 2 P ( d ) + 1 , and using this, it is easy to show that d 6 or d = 10 . Indeed, if P ( d ) 3 , then

2 + 2 P ( d ) + 1 d 2 3 5 2 P ( d ) - 3 > 14 + 2 P ( d ) + 1 ,

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

Suppose that d = 2 . Then ν 2 n p 2 for every prime p. By the assumptions on n and Lemma 5.31, this implies κ 2 = κ ν 2 = 1 , γ ( 2 ) = γ ( ν 2 ) and β b 0 , 2 = β b 0 , ν 2 . Therefore,

| C b 0 ( χ 2 ( u ) ) - C b 0 ( χ 2 ( g 0 ) ) | = | μ ( γ ( 2 ) ) ( δ b 0 , 2 ( n / γ ¯ ( 2 ) ) - δ b 0 , ν 2 ( n / γ ¯ ( ν 2 ) ) ) | 1 ,

Suppose that d = 3 . By the assumptions on n and Lemma 5.3, κ i = κ ν i = 1 for every i X 3 and n = 6 . If 2 4 n or 3 2 n , then

| { i = 1 , 3 : δ b , i ( n / γ ¯ ( i ) ) = 1 } | 1 and | { i = 1 , 3 : δ b , ν i ( n / γ ¯ ( ν i ) ) = 1 } | 1 ,

which implies | C b 0 ( χ 3 ( u ) ) - C b 0 ( χ 3 ( g 0 ) ) | 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 ,

β b , 1 = β b , 3 = 1 and C b ( χ 3 ( g 0 ) ) = - δ b , 1 ( 12 ) - δ b , 3 ( 8 ) for every b 𝔹 .

We may assume that 3 ν 3 and 3 ν 1 because ( ν ( X 3 ) ) 3 ( X 3 ) . Suppose that ν 3 8 3 and ν 1 8 1 . Then γ ¯ ( ν 1 ) = γ ( ν 1 ) = 2 , γ ¯ ( ν 3 ) = γ ( ν 3 ) = 3 , β b 0 , ν 3 = 1 and δ b 0 , 3 ( 8 ) = δ b 0 , ν 3 ( 8 ) , which implies | C b 0 ( χ 3 ( u ) ) - C b 0 ( χ 3 ( g 0 ) ) | 2 , contradicting (5.5). Suppose now that ν 3 8 1 and ν 1 8 3 . This implies ν 1 ± 3 mod 8 and ν 1 ± 1 mod 3 (because 3 ν 3 but 3 ν 1 ). Thus either ν 1 ± 11 mod 24 or ν 1 ± 5 mod 24 . As ( ν ( X 3 ) ) 12 ( X 3 ) , we deduce that the only possibility is ν 1 ± 11 mod 24 . In this case, γ ¯ ( ν 1 ) = γ ( ν 1 ) = 1 and γ ¯ ( ν 3 ) = γ ( ν 3 ) = 6 . Hence

C 11 ( χ 3 ( u ) ) - C 11 ( χ 3 ( g 0 ) ) = δ 11 , ν 1 ( 24 ) + δ 11 , ν 3 ( 4 ) + δ 11 , 1 ( 12 ) + δ 11 , 3 ( 8 ) = 4 ,

Suppose that d = 4 . By the assumptions on n and Lemma 5.3, κ i = κ ν i = 1 for every i X 4 and n = 6 . If 3 3 n or 2 3 n , then

| { i = 2 , 4 : δ b , i ( n / γ ¯ ( i ) ) = 1 } | 1 ,

which implies | C b 0 ( χ 4 ( u ) ) - C b 0 ( χ 4 ( g 0 ) ) | 3 , contradicting (5.5). Thus n = 36 . In this case, we have γ ( 2 ) = 1 = β b 0 , 2 = β b 0 , 4 and γ ( 4 ) = 2 , which implies | C b 0 ( χ 4 ( g 0 ) ) | 1 , and hence

| C b 0 ( χ 4 ( u ) ) - C b 0 ( χ 4 ( g 0 ) ) | 3 ,

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

| { i = 1 , 3 : δ b , i ( n / γ ¯ ( i ) ) = 1 } | 1 and | { i = 1 , 3 : δ b , ν i ( n / γ ¯ ( ν i ) ) = 1 } | 1 .

On the other hand, since n 10 , we deduce that κ i = 1 for every i X 5 by Lemma 5.31. Therefore, using (5.5) and (5.6), we deduce that κ ν 5 = 2 , in contradiction with Lemma 5.31.

Suppose that d = 6 . By Lemma 5.3, we have n 30 and κ i = κ ν i = 1 for every i X 6 because n 12 . If 25 n , or 9 n or 8 n , then

| { i = 2 , 4 , 6 : δ b , i ( n / γ ¯ ( i ) ) = 1 } | 2 and | { i = 2 , 4 , 6 : δ b , ν i ( n / γ ¯ ( ν i ) ) = 1 } | 2 .

This implies | C b 0 ( χ 6 ( u ) ) - C b 0 ( χ 6 ( g 0 ) ) | 4 , yielding a contradiction with (5.5). Therefore, n = 60 , and hence β b , 2 = β b , 4 = β b , 6 = 1 , γ ¯ ( 2 ) = 1 , γ ¯ ( 4 ) = γ ( 4 ) = 2 and γ ¯ ( 6 ) = γ ( 6 ) = 3 . This implies | C b 0 ( χ 6 ( g 0 ) ) | 2 , and hence

| C b 0 ( χ 6 ( u ) ) - C b 0 ( χ 6 ( g 0 ) ) | 5 ,

Suppose that d = 10 . If 5 n γ ¯ ( i ) for some i X 10 , then n 5 = ( γ ¯ ( i ) ) 5 = 5 , and hence 5 i . The same also holds for ν i . Therefore, if 5 i , then 5 n γ ¯ ( i ) , and if 5 ν i , then 5 n γ ¯ ( ν i ) . Thus

| { i X 10 : 5 i , δ b , i ( n / γ ¯ ( i ) ) = 1 } | 2 ,
| { i X 10 : 5 ν i , δ b , ν i ( n / γ ¯ ( ν i ) ) = 1 } | 2 .

This implies | C b 0 ( χ 10 ( u ) ) - C b 0 ( χ 10 ( g 0 ) ) | 8 , contradicting (5.5).

Funding source: Ministerio de Economía y Competitividad

Award Identifier / Grant number: MTM2012-35240

Award Identifier / Grant number: MTM2016-77445-P

Funding source: Fundación Séneca

Award Identifier / Grant number: 19880/GERM/15

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