H. J. Zassenhaus conjectured that any unit of finite order and augmentation 1 in the integral group ring of a finite group G is conjugate in the rational group algebra to an element of G. We prove the Zassenhaus conjecture for the groups and 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 , with f arbitrary and u is a torsion unit of with augmentation 1 and order coprime with p, then u is conjugate in 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 of order a multiple of p and augmentation 1 has order actually equal to p.
For a finite group G, let denote the group of units of augmentation 1 in . We say that two elements of are rationally conjugate if they are conjugate in the units of . The following conjecture stated by H. J. Zassenhaus  (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 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 provided it holds for G. Recently, Eisele and Margolis  announced a metabelian counterexample to the Zassenhaus conjecture. Nevertheless, the Zassenhaus conjecture holds for large classes of solvable groups, e.g. for nilpotent groups , groups possessing a normal Sylow subgroup with abelian complement  or cyclic-by-abelian groups . 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 (see the proof of Theorem C in  and ).
The goal of this paper is proving the following theorem.
Let with q an odd prime power, and let u be a torsion element of 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.
The Zassenhaus conjecture holds for with p a prime number and .
In Section 2, we prove a number theoretical result relevant for our arguments. Known results on and properties of 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, denotes the set of non-negative integers. Let n be a positive integer. Then , and denotes a complex primitive n-th root of unity, denotes the n-th cyclotomic polynomial, i.e., the minimal polynomial of over , and for a prime integer p, let denote the valuation of n at p, i.e., the maximum non-negative integer m with . If is a finite field extension, then denotes the standard trace map. We will frequently use the following formula for d a divisor of n [18, Lemma 2.1]:
We reserve the letter p to denote a positive prime integer, and for every positive integer n, we set
If moreover , then we set
The next lemma collects two facts which follow easily from the definitions.
Let p be a prime dividing n, and let . Then the following conditions hold.
If , then .
Let such that . If d divides both and , then .
For integers x and y, we define the following equivalence relation on :
We denote by the set of these equivalence classes.
If and n are integers with , then let
For an integer x (or ), we set . Observe that is the maximal real subfield of and is the ring of integers of . If , then let denote the smallest odd prime dividing n. Then let
For and , let
The following proposition extends [20, Proposition 3.5]. The first statement implies that is a -basis of . For and , we use to denote the coefficient of in the expression of x in the basis .
Let n be a positive integer.
Then is a -basis of .
If and , then .
We only prove the proposition in the case , as the proof in the case is similar (actually simpler). It is easy to see that
Thus it is enough to prove the equality
Actually, we will show
which easily implies the desired expression of .
Indeed, for every , let denote the p-th part of , i.e., is a primitive -th root of unity and . Let J be the set of tuples satisfying the following conditions:
If , then .
If and , then
For every , let be given by
Then is the set of elements in satisfying . From
Therefore, if , then , for every and
This gives the desired equality in this case.
Suppose that . Then for every . Then a small modification of the argument in the previous paragraph gives
3 Group theoretical preliminaries
Let G be a finite group. We denote by the center of G. If , then denotes the order of g, denotes the cyclic group generated by g, and 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 is an element of a group ring RG, with each , then the partial augmentation of α at g is defined as
We collect here some known results on partial augmentations of an element u of order n in .
If and , then (Berman–Higman theorem, [15, Proposition 1.5.1]).
If and , then divides n [13, Theorem 2.3].
u is rationally conjugate to an element of G if and only if for all and all divisors d of n [17, Theorem 2.5].
See [16, 13]. Let F be a field of characteristic with . Let ρ be an F-representation of G. If , then let be a primitive n-th root of unity in F, so that if , then . Let T be a set of representatives of the conjugacy classes of t-regular elements of G (all the conjugacy classes if ). Let χ denote the character afforded by ρ if , and the t-Brauer character of G afforded by ρ if (using a group isomorphism associating to ). Then, for every integer , the multiplicity of as eigenvalue of is
In the remainder of the paper, fix an odd prime power q, and let , . Let denote the natural projection, which we extend by linearity to a ring homomorphism .
We collect some group-theoretical properties of G and (see, e.g., [10, Theorem 38.1]).
G has a unique element J of order 2 and 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 , and q classes are formed by elements of order dividing or . Furthermore, if g and h are p-regular elements of G and divides , then h is conjugate in G to an element of and two elements of are conjugate in G if and only if they are equal or mutually inverse.
Let C be a conjugacy class of formed by elements of order n. If , then is the only conjugacy class of G formed by elements of order 4. Otherwise, is the union of two conjugacy classes and of G with . Furthermore, if n is a multiple of 4, then the elements of and have order , while if n is not a multiple of 4, then one of the classes or is formed by elements of order n.
The following proposition collects some consequences of these facts.
Let , and let u be a torsion element of . Set and . Then the following statements hold.
J is the unique element of order 2 in .
If and is rationally conjugate to an element of , then u is rationally conjugate to an element of G.
If and either or , then u is rationally conjugate to an element of G.
If , then G has an element of order n.
Suppose that with p prime, and . Then u is rationally conjugate to an element of G.
If ρ is a representation of G and ζ is a root of unity of order dividing n, then ζ and have the same multiplicity as eigenvalues of .
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 , then is odd by B, and hence , by F. Thus u is rationally conjugate to an element of G.
4 Let , where p is an odd prime number and . 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 , then for every , and hence u is rationally conjugate to an element of G by C.
If , then is coprime with by 2, and hence is rationally conjugate to an element of 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].
Observe that, for q odd, Theorem 1.2 follows at once from Theorem 1.1 and Proposition 3.1 6. On the other hand, and 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 , and let 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 of G of degree such that
In particular, the restriction to of the t-Brauer character associated to is given by
4 Prime power order
In this section, we prove the following particular case of Theorem 1.1.
Let with q an odd prime power, and let u be a torsion element of . 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.
By Proposition 3.1 4, we may assume that with . We argue by induction on r. So we assume that units of order with are rationally conjugate to an element of G. By Proposition 3.1 5 and E, G has an element of order such that is a set of representatives of the conjugacy classes of G with order a divisor of . By B, the only possible non-zero partial augmentations of u are the integers , with . By the induction hypothesis, if and , then . Hence, by C, it suffices to prove that for all but one .
For a t-Brauer character χ of G and an integer , define
Then, by D, we have
Observe that and . Therefore, from (4.1) it follows that
We will calculate and for several t-Brauer characters χ and several integers , and for that, we will use (2.1) without further mention. We start proving that
In both cases, we argue by induction on h with the cases and being straightforward. Suppose that , and . If j is even, then straightforward calculations show that
This finishes the proof of (4.4).
Suppose that , and . In this case, the induction hypothesis implies . Arguing as in the previous paragraph, we get
However, if j is even and smaller than , then
Therefore, having in mind that , we have
where if , and otherwise. Then the claim follows using the following equalities that can be proved by straightforward calculations:
This finishes the proof of (4.5).
We now prove by induction on h that the following two statements hold for any integer :
and that the next one holds for every :
Observe that . Fix an integer i. Then, for every integer k, we have
We now prove that the first two summands are 0. This is clear for the first one because . To prove that the second summand is 0, let and . Observe that . Thus if k is odd, then the order of is a multiple of , and as , we deduce that
Thus we only have to consider the summands with k even. Actually, we can exclude also the summands with because, by the induction hypothesis on (4.8), for such k, we have . For the remaining values of k (i.e., k even and not a multiple of ), we have if by the induction hypothesis on (4.7). So we can rewrite
which is 0 because is a root of unity different from 1 and of order dividing , as j is even but not a multiple of . This finishes the proof that the first two summands are 0. To finish the calculation of , we compute
where . So we have proved the following:
Then (4.6) follows from (4.3), (4.5) and the previous formula. Using (4.2), we also obtain if . However, in this case, the induction hypothesis for (4.7) means that the with are all equal and the with are all equal. Hence (4.7) follows.
In order to deal with (4.8), assume that . By the induction hypothesis on (4.8), we have if , and by the induction hypothesis on (4.7), we have that is constant on the set X formed by integers such that . We will use these two facts without specific mention. Arguing as before, we have
To finish the proof of the proposition, it is enough to show that for exactly one . If i is even, then by (4.8) with .
We claim that if , then . Otherwise, there are integers and satisfying and . We choose v minimum with this property for some i. Then
for every ,
for every .
Statement (1) implies
Using and , we deduce that
in contradiction with . This finishes the proof of the claim.
Then the only possible non-zero partial augmentations of u are and . Hence we have , and by applying (4.6) with , we deduce that . Therefore, either or , i.e., for exactly one , as desired. ∎
5 Proof of Theorem 1.1
In this section, we prove Theorem 1.1. Recall that with and t an odd prime number, , is the natural projection and u is an element of order n in with . We have to show that u is rationally conjugate to an element of G. By Proposition 3.1 4, 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 because this case follows easily from known results and the HeLP method. Indeed, if , then has order 6 by Proposition 3.1 2, and hence is rationally conjugate to an element of 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 and , it easily follows that all the partial augmentations of u are non-negative. This can be also obtained using the GAP package HeLP .
In the remainder, we follow the strategy of the proof of the main result of . 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 , where n was odd. On the other hand, for , we have more Brauer characters than for , 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
We argue by induction on n. So we also assume that is rationally conjugate to an element of G for every divisor d of n with .
We will use the representations and t-Brauer characters introduced in (3.1). Observe that the kernel of 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 is if m is even, while if m is odd, then the order of is n. Combining this with Proposition 3.1 7, we deduce that is conjugate to for a suitable primitive n-th root of unity ζ. Hence there exists an element of order n such that and are conjugate. The element 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
By the induction hypothesis on n, if c is a divisor of n with , then is rationally conjugate to for some i, and hence . Therefore, , and hence is conjugate to .
By E, two elements of are conjugate in G if and only if they are equal or mutually inverse. Moreover, every element , with , is conjugate to an element of . Therefore, induces a bijection from to the set of conjugacy classes of G formed by elements of order dividing n. For an integer x (or ), we set
u is rationally conjugate to if and only if for any positive integer i.
For a positive integer n and a subfield F of , let denote a set of representatives of equivalence classes of the following equivalence relation defined on :
Let n be a positive integer; let F be a subfield of , and let R be the ring of integers of F. For every , let be an integer, and for every integer i, define
Let d be a divisor of n such that for every prime power q dividing d with . Then .
By Lemma 5.1, in order to achieve our goal, it is enough to prove that for every positive integer i. We argue by contradiction, so we assume that for some positive integer d, which we assume to be minimal with this property. Observe that if and j is an integer such that , then there exists such that , and applying σ to the equation , we obtain . This implies that d divides n. Note that by our choice of , and hence . Moreover, because as the augmentation of u is 1.
We claim that
Indeed, for any , let if , and otherwise. Then, for any integer i, we have . The claim then follows by applying Corollary 5.2 with , and . Observe that, in the notation of that corollary, .
By (5.1), we have
Combining this with (5.2) and the minimality of d, we deduce that
Furthermore, , as . Therefore,
The bulk of our argument relies on an analysis of the eigenvalues of 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 and . Of course, we do not know the eigenvalues of the latter. However, we know the eigenvalues of for every with because we know the eigenvalues of and is conjugate to . This provides information on the eigenvalues of . For example, recall that if ξ is an eigenvalue of , then ξ and have the same multiplicity as eigenvalues of . Therefore, if , then the sum of the multiplicities of the eigenvalues of of order h is even. Moreover, for every t-regular element g of G, the multiplicity of 1 as eigenvalue of is congruent modulo 2 with the degree of . As n is not a prime power, there is an odd prime p dividing n. By the induction hypothesis, is rationally conjugate to . Thus the multiplicity of -1 as eigenvalue of is even. As the latter is the sum of the multiplicities as eigenvalues of of -1 and the elements of order , we deduce that the multiplicity of -1 as eigenvalue of is even. Using this, we can see that is conjugate to for integers such that for every i. Let . Then, by (5.3) and Proposition 2.2, for every , we have
Moreover, if with , then the lists and represent the same elements in , up to ordering, and hence and represent the same elements of , up to ordering, which we express by writing . This provides restrictions on d, n and .
The following two lemmas are variants of [20, Lemmas 4.2 and 4.3].
The following statements hold:
Let . If , then and . If , then is the smallest prime dividing n and for every .
If , then n is not divisible by any prime greater than d. In particular, if d is prime, then for every .
Let p denote the smallest prime dividing n.
1 The first statement is clear. Suppose that . Then either and or . We deduce that for some since . Therefore, , and for every , we have . Thus .
2 Suppose that q is a prime divisor of n with . Then , and therefore, by 1, for every . Thus, by (5.5) and (5.6), it is enough to show that for at most one and for at most one . Observe that if , then , and hence is a multiple of q. Moreover, if i and j are different elements of , then i and j have the same parity and . Therefore, . Thus either
As , this also proves
We obtain an upper bound for in terms of the number of prime divisors of d.
For every , we have
We prove the second inequality, only using . This implies the first inequality by applying the second one to .
Let . By changing the sign of some ’s, we may assume without loss of generality that if , then . Thus if , then . We claim that if , then . Indeed, let p be prime divisor of d. If or , then clearly . Otherwise, i.e., if and , then p divides both and and . Therefore, by Lemma 2.1 2, , as desired. As and the elements of represent different classes in , we deduce that . 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.
a contradiction. Thus and , or and . Hence d is either or 10. We deal with these cases separately.
Suppose that . By the assumptions on n and Lemma 5.3, for every and . If or , then
which implies , contradicting (5.5). Thus since n is neither 12 nor a prime power and it is a multiple of 4. In this case, we have , ,
We may assume that and because . Suppose that and . Then , , and , which implies , contradicting (5.5). Suppose now that and . This implies and (because but ). Thus either or . As , we deduce that the only possibility is . In this case, and . Hence
Suppose that . By the assumptions on n and Lemma 5.3, for every and . If or , then
which implies , contradicting (5.5). Thus . In this case, we have and , which implies , and hence
contradicting again (5.5).
Suppose that . Since , there is exactly one which is divisible by 5, say, . In particular, for , we have and . Moreover, if j is an integer not a multiple of 5, then . This implies
Suppose that . By Lemma 5.3, we have and for every because . If , or or , then
This implies , yielding a contradiction with (5.5). Therefore, , and hence , , and . This implies , and hence
yielding a contradiction with (5.5).
Suppose that . If for some , then , and hence . The same also holds for . Therefore, if , then , and if , then . Thus
This implies , 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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