Accessible Published by De Gruyter May 7, 2019

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 ,

contradicting (5.5).

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 ,

contradicting (5.4).

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 ,

contradicting again (5.5).

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 ,

yielding a contradiction with (5.5).

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

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