TORSION MATRICES OVER COMMUTATIVE INTEGRAL GROUP RINGS

Abstract Let ZA be the integral group ring of a finite abelian group A, and n a positive integer greater than 5. We provide conditions on n and A under which every torsion matrix U , with identity augmentation, in GLn(ZA) is conjugate in GLn(QA) to a diagonal matrix with group elements on the diagonal. When A is infinite, we show that under similar conditions, U has a group trace and is stably conjugate to such a diagonal matrix.

For n = 1, a positive answer for all A is classical (see [9,Corollary 1.6]).Luthar and Passi obtained a positive answer for all A when n = 2, if QA is replaced with CA, in [4].More recently, in [8], Marciniak and Sehgal obtained an affirmative answer for all A and all n ≤ 5. On the other hand, in [3], Cliff and Weiss constructed a counterexample for the case n = 6, A = C 6 × C 6 .Furthermore, they showed that for a given finite abelian (or, indeed, nilpotent) group A, the answer will be yes for all n if and only if A has at most one non-cyclic Sylow subgroup.
Our question, then, is this: if A has two or more non-cyclic Sylow subgroups, by restricting them suitably, can we obtain a positive answer for certain n ≥ 6?We have this result.

Theorem 1.
Let A be a finite abelian group and n ≥ 6. Suppose that either (1) A has at most one non-cyclic Sylow subgroup; or, (2) if q 1 and q 2 are the two smallest (distinct) primes such that the Sylow q 1 -and q 2 -subgroups of A are non-cyclic, then q 1 + q 2 > n 2 +n−8 4 . Then for any torsion matrix U ∈ SGL n (ZA), U is conjugate in GL n (QA) to a diagonal matrix with group elements on the diagonal.
We will follow the same plan of attack as in [8].For any matrix M , let Tr(M ) denote its trace.Also, if α ∈ ZA, we say that α ≥ 0 if every coefficient of α is greater than or equal to zero.
By the proposition in [8], the theorem is equivalent to showing that Tr(U ) ≥ 0. Suppose that our theorem fails.Fix n ≥ 6, and choose an abelian group A of minimal order which provides us with a counterexample U .Note that if A satisfies (1), then [3] tells us that our result holds, hence we may assume that (2) is satisfied.Let us write Tr(U ) = α ∈ ZA, where α ≥ 0. We also write α = α + − α − , where α + , α − ≥ 0, S + = supp α + , S − = supp α − , and S + and S − are disjoint.If h ∈ S − , we may replace U with h −1 U , and therefore assume that 1 ∈ S − .More explicitly, we write α + = g∈S+ α g g, α − = g∈S− α g g.We have Proof: (a) The expression g∈S+ α g − g∈S− α g is the trace of the augmentation of U , namely the trace of the identity matrix.(b) [4,Corollary 2.3].

Matrices over Commutative Integral Group Rings 361
Proof: Suppose this is not the case.Then (α Thus, by Lemma 1, which is a contradiction.
For each prime p, let E p be the set of all subgroups of order p in A. Let E = p E p .We define σ to be |E|.By assumption, we know that A contains a copy of C q1 ×C q1 ×C q2 ×C q2 .Therefore, σ ≥ q 1 +q 2 +2 > n 2 +n 4 .For any x ∈ S − and any H ∈ E, we let T x,H = Hx ∩ S + .We will prove Lemma 3. No T x,H is empty.
Proof: We have the usual projection π : ZA → Z(A/H).Applying this to each element in the matrix, we see that π(U ) is a torsion element in SGL n (Z(A/H)).Now, looking at the restrictions placed upon A in the theorem, we observe that any homomorphic image of an abelian group satisfying (1) will also satisfy (1), and the homomorphic image of an abelian group satisfying (2) must satisfy (1) or (2).Thus, A/H is also a group of the type discussed in the theorem.Since A is a group of minimal order which provides a counterexample, π(α) = Tr(π(U )) ≥ 0. Now, π(x) appears in the support of π(α − ) (since all coefficients are positive).Thus, π(x) must also appear in the support of π(α + ).That is, For each x ∈ S − , we define 4 .Let us examine the intersections of the sets T x .
Lemma 5. Let x and y be distinct elements of S − .If T x ∩ T y is nonempty, then either (i) xy −1 has order pq for distinct primes p and q, and then |T x ∩T y | ≤ 2; or, (ii) xy −1 is a p-element for some prime p, and Proof: If T x ∩ T y is not empty, then we may choose H ∈ E p , K ∈ E q , for (not necessarily distinct) primes p and q, such that T x,H ∩ T y,K is not empty.First, suppose p = q.By Lemma 4, xy −1 ∈ HK\(H ∪ K).But this set is precisely the set of elements of order pq in HK, hence xy −1 has order pq.Also, H and K are uniquely determined as the Sylow subgroups of by Lemma 4.
If p = q, then by Lemma 4, xy −1 ∈ HK, hence it is a p-element.It follows easily that T x,H ∩ T y,K must be empty unless H and K are p-groups.
For the proof of the next lemma, we refer to the proof of [8,Lemma 6], noting only that a homomorphic image of a group satisfying (1) or (2) must also satisfy (1) or (2).
Proof: Suppose that not all coefficients of α − are 1.Then by Lemma 6, (α Just as in the proof of Lemma 2, this is a contradiction. Clearly, since n ≥ 6, this means |S − | > 4. For any distinct x, y ∈ S − , we say that T x and T y have a large intersection if xy −1 is a p-element.Otherwise, the intersection is said to be small.(By Lemma 5, the intersection can contain at most two elements in this case.)Our last lemma is Lemma 9.There exist distinct elements x and y in S − such that T x and T y have small intersection.
Proof: The proof of [8, Lemma 9] carries through verbatim up to the point where they conclude that for n ≥ 6.This is a contradiction.We know that each |T u | ≥ σ.If any two such sets are disjoint, then |S + | ≥ 2σ, and we are done.Thus, we will assume that no two T u 's are disjoint.Suppose that for some pairwise distinct x, y, z ∈ S − , T x and T y have small intersection, and T y and T z have small intersection.We have two cases.First, if T x and T z have small intersection, then

Proof of
since each set has order at least σ, and any pair has at most two elements in common.But then Second, if T x and T z have large intersection, then by Lemma 5, there exists a prime p such that Choose one of {q 1 , q 2 } which is not p (without loss of generality, say q 1 ).Then T x \(T x ∩T z ) ⊇ H∈Eq 1 T x,H , since the T x,H are disjoint, for a fixed x, by Lemma 4. Again, we have Now, |T y | ≥ σ, and since T z and T y have small intersection, T x,H | ≥ q 1 + 1, since the T x,H are nonempty, disjoint, and there are at least q 1 + 1 of them, by choice of q 1 .Now, since q 1 , being a prime, is at least 2. This is what we wanted to know, and therefore ( * ) We may assume that, for any distinct a, b, c ∈ S − , either T a and T b have large intersection, or T b and T c have large intersection.
We know from Lemma 9 that there exist distinct x and z in S − such that T x and T z have small intersection.Since they cannot be disjoint, Lemma 5 tells us that xz −1 has order pq for distinct primes p and q.We know from Lemma 8 that |S − | ≥ 5, so let us say that v, w, x, y and z are distinct elements of S − .By ( * ), T x and T y cannot have small intersection hence, by Lemma 5, xy −1 is an r-element for some prime r.If p = r = q, then yz −1 = (xy −1 ) −1 xz −1 has order divisible by three primes, contradicting Lemma 5. Thus, xy −1 is a p-element or a q-element.Without loss of generality, it is a p-element.Then yz −1 , being the product of an element of order pq and a p-element, must have order q or pq (given the choices afforded by Lemma 5).In the latter case, T y and T z have small intersection, which is disallowed by ( * ), hence yz −1 is a q-element.Again by ( * ), T x and T w have large intersection, hence xw −1 is an r-element for some prime r.If p = r = q, then zw −1 = (xz −1 ) −1 xw −1 has order divisible by p, q, and r, which is impossible.Thus, xw −1 is a p-element or a q-element.Suppose xw −1 is a p-element.Then since xy −1 is a p-element, so is has order pq and xw −1 is a p-element, zw −1 must have order q or pq.Once again, ( * ) disallows the latter, hence zw −1 is a q-element.But wy −1 = (zw −1 ) −1 (yz −1 ) −1 , and both zw −1 and yz −1 are q-elements.Therefore, wy −1 is both a p-element and a q-element, which is impossible.It follows that xw −1 must be a q-element.Thus, wy −1 = (xw −1 ) −1 xy −1 , being the product of a q-element and a p-element, has order pq.
Once again, T x and T v must have large intersection.Thus, xv −1 is an r-element for some prime r, and once again, r = p or q.Suppose xv −1 is a p-element.Then yv −1 = (xy −1 ) −1 xv −1 , being a product of two p-elements, is a p-element.However, zv −1 = (xz −1 ) −1 xv −1 .Since xz −1 has order pq, and xv −1 is a p-element, we again see that zv −1 is a q-element.But yz −1 is also a q-element, hence yv −1 = yz −1 zv −1 is both a p-element and a q-element, giving us a contradiction.Therefore, xv −1 is a q-element.But then yv −1 = (xy −1 ) −1 xv −1 is the product of a p-element and a q-element, hence it has order pq.That is, T y and T v have small intersection, but T w and T y also have small intersection, and this contradicts ( * ).The proof is complete.
Of course, the restriction placed upon the group becomes much harsher as n increases, but for small values of n, it is fairly mild.For instance, if n = 6, we are assuming that q 1 + q 2 ≥ 9.In this case, the theorem reduces to this Corollary.Let A be a finite abelian group.Suppose that at most one of the Sylow p-subgroups, p ≤ 5, is non-cyclic.Then for any torsion matrix U ∈ SGL 6 (ZA), U is conjugate in GL 6 (QA) to a diagonal matrix with group elements on the diagonal.
This improves [5,Theorem 4.6] which requires us to assume that n < p for all primes p dividing the order of A.
Removing the assumption that A is finite, we would also like to know when a torsion matrix U ∈ SGL n (ZA) will have a group trace.A matrix U ∈ GL n (ZA) is said to have a group trace if there exist g 1 , . . ., g n ∈ A such that, for all positive integers m, Tr(U m ) = n i=1 g m i .(Note that this definition applies only to abelian groups.)See [1] and [2] for a more extensive discussion of this property.In [2, Theorem 3.3], it is shown that if n < p for every prime p such that A has p-torsion, then every torsion matrix U ∈ SGL n (ZA) will have a group trace.In a similar vein, we can prove Theorem 2. Let A be an abelian group and n a positive integer.Suppose either that n ≤ 5 or we have (1) every finite subgroup of A has at most one non-cyclic Sylow subgroup; or, (2) if q 1 and q 2 are the two smallest (distinct) primes such that the Sylow q 1 -and q 2 -subgroups of some finite subgroup of A are noncyclic, then q 1 + q 2 > n 2 +n−8
Then every torsion matrix U ∈ SGL n (ZA) has a group trace.
Proof: Since the condition on A is certainly inherited by subgroups, there is no harm in assuming that A is generated by the group elements appearing in the support of one or more entries of U .In particular, A is finitely generated.In [2, pp. 629-630] it is shown that in this case, the elements of infinite order in A do not appear in the support of Tr(U r ) for any r ≥ 1.Let us write A = T × F , where T is finite and F is a free abelian group.Then the support of Tr(U r ) is contained in T for all r ≥ 1.It follows from [1,Proposition 15] that if the image of U in SGL n (Z(A/F )) = SGL n (ZT ) has a group trace, then U has a group trace.In effect, we have reduced the problem to the case in which A is finite.But by the main result of [8] (if n ≤ 5) or Theorem 1 (if n ≥ 6), U is conjugate to a diagonal matrix diag(g 1 , . . ., g n ) in this case.It follows immediately that U has a group trace.
Since we are dealing with abelian groups, the restriction on the Sylow subgroups of finite subgroups of A could be replaced with a restriction on the Sylow subgroups of A. However, requiring such subgroups to be cyclic is too strong a condition.For example, there is no reason to rule out groups which are the direct product of quasicyclic p-groups, Z p ∞ for different primes p.
When dealing with infinite groups, it would be rather optimistic to expect our matrix U to be conjugate to a diagonal matrix, particularly since even (ZC1) fails for infinite nilpotent groups (see [7]).Instead, let us introduce the following notion.Let K be a subfield of the complex numbers and G a group.For any positive integer n, we say that two matrices A, B ∈ GL n (KG) are stably conjugate if there exist roots of unity ξ are conjugate in GL n+k (KG).For the definition of the Bass rank map, to which we will refer in the proof below, we refer the reader to [6, p. 572].Let Q denote the algebraic closure of Q in C. We have Theorem 3. Let A be an abelian group and n a positive integer.Suppose either that n ≤ 5 or else (1) or ( 2) of Theorem 2 holds.Then every torsion matrix U ∈ SGL n (ZA), regarded as a matrix in GL n (QA), is stably conjugate to a diagonal matrix with group elements on the diagonal.
Proof: Once again, we are free to assume that A is finitely generated.Let us write A = T × F , where T is finite and F is a free abelian group.By [6, Theorem 4.1], if K is a splitting field for T in C, then the Bass rank map is injective on K 0 (KA).By Brauer's Theorem, this only requires K to contain a primitive e-th root of unity, where e is the exponent of T .Let m = de, where d is the multiplicative order of U .Then, let us take K = Q(ξ), where ξ is a primitive m-th root of unity.By [1, Proposition 14], U is stably conjugate over KA to a diagonal matrix with group elements on the diagonal if and only if U has a group trace.By Theorem 2, U does indeed have a group trace.Enlarging the field to Q does not harm our conclusion.Therefore, we are done.
Remark.The definition of stable conjugacy in [1] is slightly different from the one we have used.In that paper, the scalars ξ i were not assumed to be roots of unity.However, examining the relevant proofs (to wit, the nonempty. −1 ∈ HK\(H ∪ K) and |T x,H ∩ T y,K | = 1.Proof: See [8,Lemma 4].