PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 5, May 1998, Pages 1257–1265 S 0002-9939(98)04340-8

FREE PRODUCTS OF ABELIAN GROUPS IN THE UNIT GROUP OF INTEGRAL GROUP RINGS

ERIC JESPERS AND GUILHERME LEAL

(Communicated by Ronald M. Solomon)

Abstract. We classify finite groups G which are such that the unit group of the integral group ring ZG has a subgroup of finite index which is a non-trivial free product of abelian groups.

For a finite commutative group G it is well known, and first discovered by Higman, that the unit group U(ZG) of the integral group ring ZG is the direct product of the trivial units and a finitely generated free abelian group. In particular, the structure of the unit group is known. For a nonabelian finite group G the structure of the unit group U(ZG) is much more complicated. The few known structure theorems state that U(ZG) seldom belongs to a well studied class of groups, such as solvable and nilpotent groups (see for example [9, 10]). The reason for this is that U(ZG) contains a free subgroup of rank 2 except when G is abelian or G is a Hamiltonian 2-group, a result due to Hartley and Pickel. Recently, Marciniak and Sehgal explicitly described generators for such a free subgroup.

The study of free subgroups in U(ZG) was continued in [4]; it is shown that there are only four finite groups G which are such that U(ZG) contains a noncyclic free subgroup of finite index. In [5] and [6] a classification is given of the finite groups G which are such that U(ZG) has a subgroup of finite index which is the direct product of noncyclic free subgroups.

In this paper we answer the following question posed to the authors by Z. Marciniak: describe the finite groups G which are such that the unit group U(ZG) contains a subgroup of finite index which is a free product of free abelian groups. As in [4, 5, 6] it turns out that such groups can be characterised in terms of the Wedderburn decomposition of their rational group algebra. We will actually answer a slightly more general question.

Throughout the paper the following notation is used. The centraliser of an element g of a group G is denoted by C_{G}(g). The dihedral group of order 2n is denoted D_{2n}, the quaternion group of order 2n is denoted Q_{2n} and the cyclic group of order n is denoted C_{n}. We also need some additional notation for four particular

\begin{matrix}c \\ ◯ \end{matrix}1998 American Mathematical Society

Received by the editors October 7, 1996.

1991 Mathematics Subject Classification. Primary 16U60, 16S34.

The first named author is supported in part by NSERC grant OGP0036631, Canada.

The second named author, partially supported by CNPq, Brazil, wishes to thank the Memorial University of Newfoundland for its support and friendly atmosphere.

groups of order 16 and one of order 32:

P = ⟨a, b | a^{4} = 1 = b^{4}, ba = a^{3}b⟩,
D = ⟨a, b, c | a^{2} = b^{2} = c^{4} = 1, ac = ca, bc = cb, ba = c^{2}ab⟩,
D_{16}^{+} = ⟨a, b | a^{8} = b^{2} = 1, ba = a^{5}b⟩,
D_{8} Y Q_{8} = the central product of D_{8} and Q_{8},

with their respective centres amalgamated

By M_{n}(R) we denote the n-by-n matrix ring over a ring R, and by H^{t}(K) we denote a totally definite quaternion algebra over a real field K. By H(K) we denote the classical quaternion algebra over a field K. The unit group of a ring R is denoted by U(R). For a subgroup H of a finite group G we denote by Ĥ the idempotent \frac{1}{|H|} ∑_{h∈H} h of QG.

We are now able to state the theorem.

Theorem 1. Let G be a finite nonabelian group which is not a Hamiltonian 2group. Consider the following properties:

(1) U(ZG) contains a subgroup of finite index which is a non-trivial free product of abelian groups.

(2) For any non-torsion unit u ∈ U(ZG) there exists a positive integer n such that C_{U(ZG)}(u^{n}) is abelian-by-finite.

(3) The simple Wedderburn components of QG are of one of the following six types: Q, K, H^{t}(Q), M_{2}(Q), M_{2}(K) or M_{2}(H^{t}(Q)), where K is a quadratic imaginary extension of Q. Moreover, exactly one matrix component occurs.

(4) The group G is isomorphic to one of the following: D_{6}, D_{8}, Q_{12}, P, D, D_{16}^{+}, Q_{8} × C_{3}, Q_{8} × C_{4} and D_{8} Y Q_{8}.

Then (1) ⇒ (2) ⇒ (3) ⇒ (4).

Proof. (1) ⇒ (2). Let H be a subgroup of finite index, say n, in U(ZG) which is a free product of abelian groups H_{1}, H_{2}, ⋯, H_{l}. Then, for any non-torsion unit u of ZG, 1 ≠ u^{n} ∈ H. The description of subgroups of free products, given by Kurosh (see for example [8, Theorem 6.3.1]), implies that C_{H}(u^{n}) is a free product of a free group and conjugates of subgroups of the abelian groups H_{i}, 1 ≤ i ≤ l. Since the centre of a free product of two nontrivial groups is trivial (see for example [8, Proposition 6.2.6]) and because 1 ≠ u^{n} is in the centre of C_{H}(u^{n}), it follows that C_{H}(u^{n}) is abelian. Because H is of finite index in U, it is then clear that C_{H}(u^{n}) = C_{U(ZG)}(u^{n}) ∩ H is of finite index in C_{U(ZG)}(u^{n}). So this shows that (2) holds.

(2) ⇒ (3). Write QG = ⨁ M_{n_{i}}(D_{i}), where m_{i} ≥ 1 and each D_{i} is a division algebra. Let O_{i} be a Z-order in D_{i}; then both ZG and Λ = ⨁ M_{n_{i}}(O_{i}) are Zorders in QG. Hence, there exists a positive integer n(Λ) such that n(Λ)-th power of a unit of Λ is a unit in ZG. Hence, by [10, Proposition 1.9], periodic subgroups of U(ZG) and of U(M_{n_{i}}(O_{i})) are of bounded period. Because such groups are also linear, it follows that periodic subgroups are finite.

If e_{i} is the identity of M_{n_{i}}(O_{i}), we often identify U(M_{n_{i}}(O_{i})) with (1 − e_{i}) +U(M_{n_{i}}(O_{i})) ⊆ QG. So, under this identification, ZG ∩ ((1 − e_{i}) + U(M_{n_{i}}(O_{i}))) will be written as ZG ∩ U(M_{n_{i}}(O_{i})).

First we show that each n_{i} ≤ 2. Assume the contrary; that is, suppose m = n_{i} ≥ 3. Let E_{ij}, 1 ≤ i, j ≤ m, be a set of matrix units for M_{m}(O_{m}). Then there exists a positive integer n(Λ) such that the group H = ⟨1 + n(Λ)E_{ij} | m ≥ j > i > 1⟩ ⊆ U(ZG). Since H is not abelian and has a non-periodic central element, this yields a contradiction.

Second, we show that there exists precisely one index i so that M_{n_{i}}(O_{i}) has infinitely many units, or equivalently the unit group of M_{n_{i}}(O_{i}) is non-periodic. By the Hartley-Pickel result and its proof (see Theorem 5.1 in [10]), and because G is not abelian and not a Hamiltonian 2-group, there exists an index j so that U(M_{n_{j}}(O_{j})) ∩ U(ZG) contains a non-cyclic free group F. Suppose now that i ≠ j and U(M_{n_{i}}(O_{i})) is infinite. So let u ∈ U(M_{n_{i}}(O_{i})) be a non-periodic unit. Thus, there exists a positive integer n(Λ) so that 1 + u^{n(Λ)} ∈ U(M_{n_{i}}(O_{i})) ∩ U(ZG), and clearly F ⊂ C_{U(ZG)}(u^{n(Λ)}), in contradiction with (2). This shows the claim.

Next we show that there exists a unique index i with n_{i} ≠ 1. Suppose the contrary, that is, QG is a direct sum of division rings and fields. Hence G is a Hamiltonian group, and since it is not a two group, G contains H = Q_{8} ×C_{p} with p an odd prime. As QH does not contain non-zero nilpotent elements, p ≥ 7 (Lemma VI.1.18 in [9]). Hence the Wedderburn decomposition of Q(Q_{8} × C_{p}) contains the direct sum Q(ζ_{p}) ⊕ H(Q(ζ_{p})).

Now U(ZH)∩U(Z[ζ_{p}]) contains a non-periodic element u and U(ZH)∩U(H[Zζ_{p}]) contains a non-cyclic free group (Hartley- Pickel) F_{2}. Clearly F_{2} ⊂ C_{U(ZG)}(u), again in contradiction with (3).

So far we have shown that there exists a unique index j with n_{j} = 2, and that all other n_{i} are 1. Furthermore it follows that for every i ≠ j, U(O_{i}) is finite. We now show that also U(O_{j}) is finite. For this notice that for any u ∈ U(O_{j}) ∩ U(ZG) there exists an integer m ≥ 2 such that

u_{1} = [ \begin{matrix}1 & m \\ 0 & 1 \end{matrix} ] ,

u_{2} = [ \begin{matrix}1 & 0 \\ m & 1 \end{matrix} ]

and

v = [ \begin{matrix}u^{m} & 0 \\ 0 & u^{m} \end{matrix} ]

belong to U(ZG); so the group ⟨u_{1}, u_{2}⟩ is free and ⟨u_{1}, u_{2}⟩ ⊂ C_{U(ZG)}(v). It follows from the assumption (2) that v, and thus u, are periodic. Hence U(O_{i}) is periodic and thus finite.

The Dirichlet Unit Theorem and Lemma 21.3 in [10] now yield that O_{i} is either the ring of integers of a totally imaginary quadratic extension of Q, or an order in a totally definite quaternion algebra over Q. In particular, (3) follows.

(3) ⇒ (4) A finite group whose rational group algebra satisfies condition (3) we call admissible. Assume G is admissible. From the group tables of groups of small order (see for example [11]) one can easily deduce that if |G| ≤ 32 then G is one of the groups listed in (4). Hence it is sufficient to establish the following five steps.

(i) If QG does not have a non-commutative division ring as a simple component, then |G| ≤ 32.

(ii) If QG has a non-commutative division ring as a simple component, then Q_{8}

or Q_{12} is a homomorphic image of G. (iii) If Q_{12} is a homomorphic image of G, then G = Q_{12}. (iv) If Q_{8} is a homomorphic image of G and |G| > 32, then G is a 2-group and

either G/Z(G) ≅ C_{2} × C_{2} or G contains a Hamiltonian subgroup of index 2.

(v) If G is a 2-group and Q_{8} is a homomorphic image of G, then |G| ≤ 32.

To prove these statements we need six technical lemmas.

Lemma 2. If N is a subgroup of G, then QN does not have two matrix rings as simple components, nor does it contain a matrix ring over a division ring with infinitely many units in its orders.

Proof. Since orders in matrix rings have infinitely many units, any matrix component of QN is contained in the matrix component of QG. If QN had two matrix components, then this would yield four orthogonal idempotents in a 2-by-2 matrix ring component of QG, which is impossible.

On the other hand if M_{2}(D_{1}), D_{1} a division ring, is a simple component of QH, then it follows that M_{2}(D_{1}) ⊆ M_{2}(D) = (QG)e, where e is the primitive central idempotent defining the matrix component of QG. Since, by assumption, orders in D have a finite unit group, the same holds for orders in D_{1}.

Lemma 3. The possible orders of elements of G are 1, 2, 3, 4, 6, 8 or 12, and the centre of G has exponent at most 6.

Proof. Let g ∈ G be an element of order n. Consider the group algebra Q⟨g⟩ = Q(ζ_{n}) ⊕ M with Q(ζ_{n}) = Q⟨g⟩f, ζ_{n} a primitive n-th root of unity, f a primitive idempotent and M a direct sum of fields. Let e be a primitive central idempotent of QG such that fe ≠ 0. So Q(ζ_{n}) ⊆ (QG)e.

From the assumptions we obtain that

[(QG)e : Q] = [(QG)e : Z((QG)e)] [Z((QG)e) : Q] = 2^{i},

with 0 ≤ i ≤ 4, and therefore φ(n) = 2^{i}, 0 ≤ i ≤ 4. Note that actually i < 4, for otherwise (Q⟨g⟩)e = (QG)e is a field of Q-dimension 16, a contradiction. Clearly we obtain that n = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24 or 30.

In the cases n = 10, 15, 16, 20, 24 or 30, one can write

Q⟨g⟩ = Q⟨g⟩f_{1} ⊕ Q⟨g⟩f_{2} ⊕ M′,

where M′ is a direct sum of fields, and F_{1} = Q⟨g⟩f_{1} and F_{2} = Q⟨g⟩f_{2} are fields each of which have ring of integers with infinite group of units. Since orders in the division ring components of QG have finite unit groups, it follows that f_{1}e ≠ 0, f_{2}e ≠ 0 and QGe = M_{2}(D), D a division ring. Hence F_{i} ⊆ ef_{i}QGef_{i} ≅ D. Thus D = F_{i}, and thus an order in D contains infinitely many units, again a contradiction. So we have shown that the order of g is 1, 2, 3, 4, 5, 6, 8 or 12. From the above method it also follows that the centre of G has exponent at most 6.

So it remains to show that n ≠ 5. Assume the contrary. Then Q⟨g⟩ = Q⟨g⟩ĝ ⊕ ⟨g⟩Q(1 − ĝ), Q⟨g⟩ĝ ≅ Q and Q⟨g⟩(1 − ĝ) ≅ Q(ζ_{5}). Since the unit group of Z[ζ_{5}] is infinite and because G is admissible, (1 − ĝ)e ≠ 0, (QG)e is a matrix ring and (1 − ĝ)f = 0 for any primitive central idempotent f ∈ QG with f ≠ e. Hence 1 − ĝ = (1 − ĝ)e. Therefore the idempotent 1 − ĝ belongs to QGe. Now, 1 − ĝ is central, for otherwise Qĝ(1 − ĝ) ⊆ (1 − ĝ)eQG(1 − ĝ) ≅ D. So D ≅ Q(ζ_{5}) and therefore an order in D has infinitely many units, a contradiction. Hence it follows that 1−ĝ = e. Since e is central this implies that the group ⟨g⟩ is normal in G. Now, since (QG)e is nonabelian, there exists h ∈ G such that he ∉ ⟨g⟩e. Because g has order 5 and the order of h is 2, 3, 4, 5, 6, 8 or 12, we may assume, by replacing h by some power if necessary, that he has order 2, 3 or 5. Furthermore, there must exist such an h with order of he different from 5; for otherwise Ge is a 5-group and thus QG would not have a simple component which is a 2-by-2 matrix ring. If he has order 2, then, since G does not have elements of order 10, ⟨ge, he⟩ ≅ D_{10}. However, because of Lemma 2, this is impossible as QD_{10} has M_{2}(F) as a simple component, where F a quadratic real field extension of Q. If he has order 3, then ⟨ge, he⟩ is a group of order 15, and thus commutative. Again this yields a contradiction, as G does not have elements of order 15.

Lemma 4. If N is a subgroup of G and QN has a simple component which is a matrix ring (that is, N is nonabelian and not Hamiltonian), then N is admissible. Furthermore, if QN has a matrix component of the same Q-dimension as the matrix component of QG, then N = G.

Proof. Let f be a primitive central idempotent of QN and let e be a primitive central idempotent of QG with ef ≠ 0. Since (QN)f is simple, (QN)f ⊆ (QG)e.

We first deal with the case that (QN)f is a matrix ring. Note that, since QG has only one matrix component, e is the only primitive central idempotent of QG with ef ≠ 0. Hence ef = f. First assume (QN)fe = (QG)e = M_{2}(D), where D is a division ring (this occurs precisely when dim_{Q}((QN)fe) = dim_{Q}((QG)e)). Then f = fe = e. Hence Gf = Ge ⊆ QNf. Consequently, G ⊆ supp(Gf) ⊆ supp((QN)f) ⊆ N; and thus G = N.

Next assume (QN)f is properly contained in (QG)e and (QN)f is a matrix ring. Then dim_{Q}((QN)f) = 4 or 8. So either (QN)f = M_{2}(Q) or (QN)f = M_{2}(F), where F is a quadratic field extension of Q. Now, if (QN)f = M_{2}(F), then a set of matrix units of (QN)f is also a set of matrix units of (QG)e; and thus the centraliser of these matrix units in (QN)f is contained its centraliser in (QG)e. Thus F ⊂ D, where D is a division ring such that (QG)e = M_{2}(D). So the ring of integers O of F is contained in an order of D. The latter has finitely many units; therefore the same holds for O. Hence F is quadratic imaginary. So the matrix component of QN is of a good type.

Next assume (QN)f = D, a division ring. Note that the simple matrix component of QN and D cannot belong to the 2-by-2 matrix component of QG. So, up to isomorphism, (QN)f = D is contained in a division ring component of QG. Since an order in the latter has finitely many units, the same holds for an order in D. Hence QNf = QGe is a totally definite quaternion algebra over Q; that is, it is again of a good type. It follows that N is admissible.

Note that the rational group algebra of the Hamiltonian group Q_{8} × C_{3} has a matrix ring as simple component. It follows from the two previous lemmas that the only Hamiltonian subgroups or homomorphic images of G which are such that their rational group algebra is a sum of division rings are Hamiltonian 2-groups. We will use this remark several times.

Lemma 5. If N is a subgroup of G such that QN has a matrix ring and a noncommutative division ring as simple components, then N = G.

Proof. Let e_{1} and e_{2} be central primitive idempotents of QN such that QNe_{1} is a matrix ring and QNe_{2} is a non-commutative division ring. Once again, it is clear that (QN)e_{1} and (QN)e_{2} are not contained in the same matrix simple component of QG. Since noncommutative division ring components of QG have Q-dimension 4, QNe_{2} = QGe is a division ring, for some primitive central idempotent e of QG. Hence G ⊆ supp(Ge_{2}) ⊂ QNe_{2} ⊆ N, and thus G = N.

Lemma 6. Assume N is a nonabelian subgroup of G which is not a Hamiltonian 2-group. If |N| ≥ 32, then N = G.

Proof. Since N is nonabelian, dim_{Q}QN(1 − N̂′) ≥ |N|/2 ≥ 16. We consider two cases. First assume QN(1−N̂′) is simple. In particular, QN has only one noncommutative simple component. If this component were a division ring, then N would be a Hamiltonian 2-group, in contradiction with the assumption. So, by Lemma 4, QN(1 − N̂′) = M_{2}(D), D a division ring of dimension at most 4 over Q. Because dim_{Q}QN(1 − N̂) ≥ 16, we obtain dim_{Q}D = 4. Hence, by Lemma 4, N = G.

Second, assume QN(1 − N̂′) is not simple; then it must contain a matrix ring and a non-commutative division ring. So, by Lemma 5, N = G.

Lemma 7. If N is a proper subgroup of G, then N is abelian or a Hamiltonian 2-group, or |N| ≤ 16.

Proof. By Lemma 5, QN does not contain a matrix ring and a non-commutative division ring as simple components. So either QN is a sum of division rings, and then N is abelian or a Hamiltonian 2-group; or QN has only one non-commutative component and this component is a matrix ring. By Lemma 4, this matrix ring is of the type M_{2}(F) with Q-dimension of F equal to 1 or 2. Then

8 ≥ dim_{Q}QN(1 − N̂′) ≥ N/2.

So |N| ≤ 16.

We now proceed with the proofs of the five earlier mentioned main steps.

Proof of (i). Suppose G does not have a noncommutative division ring as simple component. Since G is admissible, this means that QG(1 − Ĝ) is a matrix ring. Hence dim_{Q}QG(1 − Ĝ) ≤ 16. Now, once again, QG ≅ Q[G/G′] ⊕ QG(1 − Ĝ′). So it follows that |G| ≤ \frac{|G|}{2} + 16. Hence |G| ≤ 32.

Proof of (ii). Assume (QG)e is a noncommutative division ring component of QG, where e is a primitive central idempotent. Because of Lemma 3, the only possible primes dividing |G| are 2 and 3, and G does not contain elements of order 16 nor of order 9. Hence, since Ge is a subgroup of a division ring (and thus is fixed point free), it follows from [7, Theorem 18.1] that the 3-Sylow subgroups of Ge are cyclic, and thus of order 3, and the 2-Sylow subgroups are either cyclic or a generalised quaternion group. Hence they are either C_{2}, C_{4}, C_{8}, Q_{8} or Q_{16}. However, because of Lemma 4, the group Q_{16} is excluded. Hence |Ge| is 24, 12 or 8 (note that Ge is nonabelian and cannot be of order 6). Since the group algebra Q[Ge] has at most one matrix ring as a simple component, it follows from the Wedderburn decompositions of the groups of such orders that: if |Ge| = 8, then Ge = Q_{8}; if |Ge| = 12, then Ge = Q_{12}; if |Ge| = 24, then Ge = Q_{8} × C_{3}. However, H(Q) is the only noncommutative division ring simple component of the rational group algebra of the latter group. Since Q_{8} × C_{3} is not embedded in H(Q), this situation actually does not occur.

Proof of (iii). Let N be a normal subgroup of G with G/N ≅ Q_{12}. Clearly G contains normal subgroups H_{1} and H_{2} so that N ⊆ H_{1} ⊆ H_{2}, G/H_{1} ≅ D_{6} and H_{2}/H_{1} is of order 3. By Lemma 3, G does not contain elements of order 9; hence there exists an element g ∈ G of order 3 such that H_{2}/H_{1} is generated by the natural image of g in this group.

We claim that ⟨g⟩ is a normal subgroup of G. Indeed, note that the subgroup ⟨g, H_{1}⟩ is normal in G and the idempotent ĝ(1 − Ĥ_{1}) belongs to (QG)(1 − Ĥ_{1}), a direct sum of division rings (note that QGĤ_{1} contains a matrix ring as simple component). Hence, ĝĤ_{1} and ĝ(1 − Ĥ_{1}) are central in QG. Hence, ĝ is central and thus ⟨g⟩ is normal in G.

Again by Lemma 3, there exists h ∈ G such that the order of h is 2^{i}, 1 ≤ i ≤ 3, and the natural image of h in G/H_{1} is of order 2. Then, since ⟨g⟩ is normal in G, T = ⟨g, h⟩ = {h^{i}g^{j}}; and similarly as above we obtain that T̂ = ĝĥ is central in QG. Therefore T is a normal nonabelian subgroup of G with |T | = 2^{i}3. Thus |T | = 24, 12 or 6.

If |T | = 6 then T ≅ D_{6}. It follows that G ≅ T × H_{1} ≅ D_{6} × H_{1}. Hence QG has more than one simple component which is a matrix ring, a contradiction.

If |T | = 12, then since D_{6} is a homomorphic image of T, the subgroup T is admissible. Hence T = Q_{12}, and thus, by Lemma 5, T = G.

If |T | = 24, once again as before one shows that T is admissible. Thus T = C_{3} × Q_{8}, and thus it follows again that T = G. However, this is impossible, as Q_{12} is not a homomorphic image of C_{3} × Q_{8}.

So we have shown that G = Q_{12}.

Proof of (iv). Assume G/N ≅ Q_{8} and |G| > 32. Because of Lemma 7, any subgroup containing N and of index 2 in G is either abelian or a Hamiltonian 2-group. In the latter case G itself is a 2-group and we are done. So we may assume that all such subgroups are abelian. Then it follows easily that G/Z(G) ≅ C_{2} × C_{2}. In particular G is nilpotent, and thus G is the direct product of its Sylow 2 and 3 subgroups, which we denote by G_{2} and G_{3} respectively. Because of Lemma 3, G_{3} ⊆ N is an elementary abelian 3-group. Note that G_{2} is nonabelian. If QG_{2} contains a matrix ring as a simple component, then clearly G_{3} is trivial, as G is admissible. So G_{2} is a Hamiltonian 2-group. We claim that G_{3} is trivial. For otherwise, Q_{8} ×C_{3} is a subgroup of G. Since Q[Q_{8} × C_{3}] contains a matrix ring and a noncommutative division ring as simple components, it follows from Lemma 5 that G = Q_{8} × C_{3}, a contradiction as |G| > 32.

Proof of (v). Assume G is a 2-group with normal subgroup N such that G/N = Q_{8}. We have to prove that |G| ≤ 32. Suppose the contrary. By step (iv), either G/Z(G) ≅ C_{2} × C_{2} or G contains a Hamiltonian normal subgroup N of index 2.

First we deal with the case that G/Z(G) ≅ C_{2} × C_{2}. Note that then G′ = {1, c}, with c a central element of order 2, and that Z(G) has exponent 2 or 4 (by Lemma 3). So (QG)\frac{1−c}{2} is the direct sum of all the noncommutative simple components of QG, and all these simple components are generalised quaternion algebras over Q or Q(i). In the latter case such a simple component is a matrix ring. Hence, since G is admissible, there exists at most one matrix component over Q(i). We claim that Z(G) is an elementary abelian 2-group. Suppose the contrary. Since |G| > 32, |Z(G)| > 8 and thus Z(G) contains C_{4} × C_{2}. It follows that Q[C_{4} × C_{2}]\frac{1−c}{2} is contained in the centre of (QG)\frac{1−c}{2} and contains 2 copies of the field Q(i), a contradiction. So indeed Z(G) has exponent 2, and therefore there exist elements a, b ∈ G such that both a and b have order 4 and Q_{8} is an epimorphic image of ⟨a, b⟩. Since c ∈ ⟨a, b⟩ and because c has order 2, the group ⟨a, b⟩ is normal in G and is of order at most 32. Because Z(G) is of exponent 2, it follows that G ≅ ⟨a, b⟩ × E, where E is an elementary abelian 2-group. As G is admissible, the group algebra Q⟨a, b⟩ contains a matrix ring as a simple component. Furthermore, because QG has precisely one matrix ring as simple component, we obtain that E is trivial. This finishes the proof for this type of groups.

Next assume N is a Hamiltonian subgroup of index 2 in G. Therefore Q_{8} = ⟨i, j⟩ ⊆ N and N = Q_{8} × E, with E an elementary abelian 2-group. Note that i^{2} is central in G as N is normal in G and i^{2} is the unique nontrivial commutator in N. Let a ∈ G be such that its natural image in G/N generates this group. We claim that we may assume that Q_{8} is normal in G. This is clear if the cyclic groups ⟨i⟩ and ⟨j⟩ are normal in G. On the other hand, if, say, ⟨i⟩ is not normal, then take for Q_{8} the group ⟨i, aia^{−1}⟩. Next we claim that E is not central. For otherwise, one can write G = ⟨Q_{8}, a⟩ × T, for some subgroup T of E. Since G is admissible, this implies that T is trivial, and thus |G| ≤ 32, a contradiction.

Let c_{1} ∈ E be non-central in G. Because \frac{1−i^{2}}{2} is a central idempotent, one of

the idempotents (\frac{1−i^{2}}{2} ) (\frac{1−c_{1}}{2} ) or f = (\frac{1+i^{2}}{2} ) (\frac{1−c_{1}}{2} ) is central. It is easily seen

that the former one does not commute with a. Hence f = afa^{−1} = (\frac{1+i^{2}}{2} ) (\frac{1−c_{2}}{2} ),

where c_{2} = ac_{1}a^{−1} ≠ c_{1}. Consequently, c_{1} = i^{2}c_{2}.

Write a^{2} = qe with q ∈ Q_{8} and e ∈ E. Then, qe = aa^{2}a^{−1} = aqea^{−1} ∈ Q_{8}aea^{−1}. So, e(aea^{−1}) is an element of order at most 2 in Q_{8}, and thus either ea = ea (that is, e is central in G) or aea^{−1} = i^{2}e.

Assume e is central. Then f_{2} = \frac{1+e}{2} and f_{3} = \frac{1−e}{2} are orthogonal central idempotents. So one of the idempotents f_{2} \frac{1+c_{1}}{2} , f_{3} \frac{1+c_{1}}{2} has to be central. Because ac_{1}a^{−1} = i^{2}c_{1}, this is only possible if e = 1. But then the group ⟨Q_{8}, a, c_{1}⟩ has order 32. Since its group algebra has non-central idempotents, this group is admissible (by Lemma 4). Since not all its elements of order 2 are central, this group is not Hamiltonian. Hence it must be the group D_{8} Y Q_{8}. Lemma 4 then yields that G = D_{8} Y Q_{8}.

Next assume aea^{−1} = i^{2}e. Then the group algebra of the group H = ⟨Q_{8}, a, e⟩ has \frac{1+e}{2} as a non-central idempotent. So QH contains a matrix component, and thus, by Lemma 4, H is admissible. Since |H| = 32 and because H has noncentral elements of order 2, we obtain that H = D_{8} Y Q_{8}. Hence, by Lemma 4, H = G.

As mentioned earlier, from the group tables of groups of small order one obtains that (4) ⇒ (3). Since the groups listed in (4) only have one simple component with infinitely many units, and because the centraliser of a non-central element of GL_{2}(C) is abelian, one obtains that (4) ⇒ (2), except possibly for D_{8} Y Q_{8}.

Finally we remark that finite groups with reduced Q-degrees either 1 or a fixed number k have been investigated in [2, 3]. Among the several results obtained, it is shown that such groups are metabelian. If furthermore these groups have a faithful irreducible representation, then a classification of these groups can be deduced from the methods used in [2, 3]. Note that subgroups of M_{2}(D), D a division ring, also have been classified by Banieqbal in [1]. The use of these papers could potentially simplify some of our proofs. However, we elected to present a comprehensive proof.

References

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Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, Newfoundland, Canada A1C 5S7

  E-mail address: ejespers@albert.math.mun.ca

Instituto de Matemática, Universidade Federal do Rio de Janeiro, Rio de Janeiro RJ, Brazil

  E-mail address: gleal@mat.dme.ufrj.br