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Publicly Available Published by De Gruyter June 11, 2016

Subgroup Isomorphism Problem for units of integral group rings

Leo Margolis EMAIL logo
From the journal Journal of Group Theory

Abstract

The Subgroup Isomorphism Problem for Integral Group Rings asks for which finite groups U it is true that if U is isomorphic to a subgroup of V(G), the group of normalized units of the integral group ring of the finite group G, it must be isomorphic to a subgroup of G. The smallest groups known not to satisfy this property are the counterexamples to the Isomorphism Problem constructed by M. Hertweck. However, the only groups known to satisfy it are cyclic groups of prime power order and elementary-abelian p-groups of rank 2. We give a positive solution to the Subgroup Isomorphism Problem for C4×C2. Moreover, we prove that if the Sylow 2-subgroup of G is a dihedral group, any 2-subgroup of V(G) is isomorphic to a subgroup of G.

1 Introduction

The first to study the unit group of an integral group ring G of a finite group G was G. Higman in his PhD thesis [20]. There, in Section 5, he states it to be “plausible” that any finite subgroup of units U is isomorphic to a group of trivial units, i.e. units of the form ±g with gG. See [30] for details on the thesis. Denote by V(G) the normalized units of G, i.e. the units whose coefficients sum up to 1. In view of the fact that ±V(G) are all the units of G it is enough to study normalized units.

After Higman’s question remained open for quite a long time, a negative answer to it was given by M. Hertweck’s counterexample to the isomorphism problem [16]: There are finite groups G and H such that G is isomorphic to H, but G and H are not isomorphic. In particular, V(G) contains a subgroup isomorphic to H, but G does not, and thus provides a counterexample to the question of Higman. However Hertweck’s counterexample is a rather complicated group of or order 2219728 and derived length 4 and leaves Higman’s question totally open for smaller and/or “easier” groups. This leads to the following question.

Subgroup Isomorphism Problem (SIP)

For which finite groups U does the following statement hold: If V(G) contains a subgroup isomorphic to U for some finite group G, then G contains a subgroup isomorphic to U?

Not explicitly studied, this was only known to hold for cyclic groups of prime power order, following from a result of Cohn and Livingstone that the exponents of G and V(G) coincide [9, Corollary 4.1]. Z. Marciniak raised the question for the Klein four group at an ICM Satellite Conference in 2006 and W. Kimmerle immediately gave a positive answer [25], followed by a positive answer for elementary abelian p-groups of rank 2 for odd primes p by M. Hertweck [18]. Since then no further progress has been made on the question. The name “Subgroup Isomorphism Problem” was coined by W. Kimmerle in several talks and is also recorded in [26]. The problem appeared for the first time in the literature in [24, Problem 19]. Denote by Cn the cyclic group of order n. The purpose of this article is the proof of the following result:

Theorem 1

Let G be a finite group. If V(ZG) contains a subgroup isomorphic to C4×C2, then so does G, i.e. the Subgroup Isomorphism Problem has a positive solution for C4×C2.

Doing just a little more we obtain also the following results, being of independent interest. The investigation of the following Theorem was proposed in [18, Comment before Example 7].

Theorem 2

Let G be a finite group possessing a dihedral Sylow 2-subgroup. Then any 2-subgroup of V(ZG) is isomorphic to a subgroup of G.

Moreover, we obtain the following generalisation of [3, Proposition 3.4] and [26, Proposition 4.7].

Proposition 3

Let G be a finite group whose Sylow 2-subgroup has at most order 8 and assume moreover that G is not isomorphic to the alternating group of degree 7. Then any 2-subgroup U of V(ZG) is rationally conjugate to a subgroup of G, i.e. there exists a unit x in the rational group algebra QG and a subgroup P of G such that x-1Ux=P.

2 Preliminaries

The following lemma from elementary group theory is the key observation to allow the proof of Theorem 1.

Lemma 1

Let P be a finite 2-group such that P contains no subgroup isomorphic to C4×C2. Then P is one of the following: Elementary-abelian, cyclic, a generalized quaternion group, a dihedral group or a semidihedral group.

Remark

Note that semidihedral groups are called “quasidihedral” groups by many authors. A presentation of a semidihedral group of order 2n+1 is given by

h,ah2n=a2=1,ha=h-1+2n-1.

For definitions of all groups mentioned in the lemma see [27, Section 5.3].

Proof.

If P is abelian, the statement is clear and if P is not abelian and contains only one involution, P is a quaternion group [27, 5.3.7]. So assume P is not abelian and contains at least two involutions. Since P then contains an element of order 4, say g, we have C2Z(P)=g2. Assume P possesses a non-cyclic abelian normal subgroup M, which must be elementary-abelian, and let a be an element of M which is not central in P. Then CP(M)=N is also normal, elementary-abelian and contains g2. If the order of N is 4, we have M=N=CP(a)C2×C2 and P is a dihedral or semidihedral group by [27, 5.3.10]. So assume |N|8 and let b be an element of N which is not central in P such that ab is also not central in P. Since aga commutes with g, we have ag=ag2 and in the same way bg=bg2. But then (ab)g=ab, contradicting the existence of N and M. Hence there is no non-cyclic abelian normal subgroup in P and P is dihedral or semidihedral by [27, 5.3.9]. ∎

Today an important open question concerning torsion units in G is the Zassenhaus Conjecture. It states that for every finite group G and every torsion unit uV(G) there exist a unit x in the rational group algebra G and an element gG such that x-1ux=g. In this case one says that u and g are rationally conjugate. The main notion to study torsion subgroups of G are the so-called partial augmentations. For a conjugacy class xG in G and an element u=gGzgg in G the sum

εx(u)=gxGzg

is called the partial augmentation of u with respect to x. The connection between partial augmentations and the Zassenhaus Conjecture is established by [29, Theorem 2.5]: A torsion unit uV(G) of order n is rationally conjugate to an element of G if and only if εx(ud)0 for all xG and divisors d of n. If u is a torsion unit and x is an element of G of order not dividing the order of u, then εx(u)=0 (see [17, Theorem 2.3]). Moreover, εz(u)=0 for every central element z of G, unless u=z by the Berman–Higman Theorem [23, Proposition 1.5.1]. These facts will be used in the rest of the paper without further mention.

A method to study the torsion units in integral group rings was introduced in [28] for ordinary characters and later generalized in [17] for Brauer characters and has, in recent years, became well known as the HeLP-method. Though developed to study finite cyclic subgroups of V(G) its idea does also work for non-cyclic groups and has been applied for ordinary characters a few times (namely in [21], [19], [18] and [5]). Since for non-cyclic groups Brauer characters have not been used so far, the following lemma seems necessary. Note that for a prime p and a p-Brauer character φ of a finite group G with corresponding representation D we can extend the domain of φ to the torsion units of V(G) of order prime to p by linearly extending D to G, see [17, Section 3].

Lemma 2

Let G be a finite group, U a finite subgroup of V(ZG) of order n and p a prime not dividing n. Let χ be the extension of a p-Brauer character of G to the p-regular elements of V(ZG) and let ψ be an ordinary character of U.

  1. [17, Theorem 3.2] For every uU we have

    χ(u)=xGxp-regularεx(u)χ(x).
  2. 1nuUχ(u)ψ(u-1) is a non-negative integer.

Proof.

Since p does not divide n, there is an ordinary character χ of U such that χ(u)=χ(u) for every uU by [11, Corollary 18.11]. So

1nuUχ(u)ψ(u-1)=χ,ψU

equals the scalar product of two ordinary characters and thus it is a non-negative integer. ∎

The proof of Theorem 1 will be achieved by checking all possible cases for Sylow 2-subgroups given by Lemma 1. The knowledge so far is recorded in [26].

Proposition 3

Proposition 3 ([26, Theorem 4.1, Proposition 4.8])

Let G be a finite group whose Sylow 2-subgroup is either abelian or a (generalized) quaternion group. Then any 2-subgroup of V(ZG) is isomorphic to a subgroup of G.

Groups possessing dihedral or semidihedral Sylow 2-subgroups were classified in a series of articles [14] (see also [6] for an alternative proof) and [2]. These will be the basis for the rest of the proofs. We will denote by O2(G) the biggest normal subgroup of odd order of a group G. The following Lemma follows from [12, Lemma 2.1, Theorem 2.2].

Lemma 4

Let G be a finite group, N a normal subgroup of G and U a finite subgroup of V(ZG) such that the orders of N and U are coprime. Then U is rationally conjugate to a subgroup of G if and only if the image of U in V(Z(G/N)) is rationally conjugate a subgroup of G/N. Moreover, V(Z(G/N)) possesses a subgroup isomorphic to U.

In the proofs of the main results we will need a lot of information about finite classical groups. We are not going to cite the original literature here of Dickson, Schur, Dieudonné, Brauer, just to mention a few, but literature which is more accessible and easier to understand nowadays. We will also mostly avoid the use of any character tables accessible in the literature, using natural representations of the groups involved instead.

3 Proof of main results

The first lemma concerns a big class of the groups involved in the results.

Lemma 1

Let G be a subgroup of PΓL(2,q)=Aut(PSL(2,q)) containing PSL(2,q) or PGL(2,q) as a normal subgroup of odd index, where q is an odd prime power. Then units of order 2 and 4 in V(ZG) are rationally conjugate to elements of G. In particular, squares of units of order 4 from V(ZG) are rationally conjugate to involutions in PSL(2,q).

Proof.

The following facts are given in [14, Lemma 3.1]. The Sylow 2-subgroup of G is a dihedral group or an elementary abelian-group of rank 2. The group PSL(2,q) possesses exactly one conjugacy class of involutions, say 2a, while the group PGL(2,q) possesses exactly two conjugacy classes of involutions, the one not lying in PSL(2,q) will be called 2b. If PGL(2,q) is not a normal subgroup of G, the conjugacy class 2b can be ignored in the following computations. If G possesses no elements of order 4, then PGL(2,q) is not a subgroup of G and there exists exactly one conjugacy class of involutions in G. Since for uV(G) of order 2 and xG of order different from 2 we have εx(u)=0, this immediately yields the desired result. So assume that G contains elements of order 4 and denote by 4a the conjugacy class of these elements. There is exactly one such class, since this is the case in a dihedral group. The class 4a may lie in PSL(2,q) or outside of it and these two cases will be partly separated in the following computations.

If PGL(2,q) is a subgroup of G, then G/PSL(2,q) has a Sylow 2-subgroup of order 2. Hence it possesses a normal 2-complement by [27, 7.2.2]. Thus the group G maps onto a cyclic group of order 2 such that PSL(2,q) is in the kernel of this map and elements of PGL(2,q) outside of PSL(2,q) are not. Hence the group G possesses a 1-dimensional representation mapping PSL(2,q) to 1 and elements of PGL(2,q) outside of PSL(2,q) to -1. The character corresponding to this representation will be called χ. If PGL(2,q) is not a subgroup of G, the character χ can be ignored.

Moreover, GL(2,q) acts via conjugation on the 2×2-matrices over 𝔽q with trace 0. The kernel of this action is exactly the centre of GL(2,q), thus inducing a 3-dimensional representation of PGL(2,q). The character corresponding to the induced representation on G and its twist with χ will be called ψ+ and ψ-. Note that these characters take only integral values on the classes we are interested in. Let N be the normal subgroup of G which is PSL(2,q) or PGL(2,q) such that [G:N]=m is odd. The characters described above take the values given in Table 1 and Table 2 on the classes of interest.

Table 1

Some characters of G, if elements of order 4 are in PSL(2,q).

1a2a4a2b
χ111-1
ψ+3m-mm-m
ψ-3m-mmm

Table 2

Some characters of G, if elements of order 4 are not in PSL(2,q).

1a2a4a2b
χ11-1-1
ψ+3m-mm-m
ψ-3m-m-mm

Let uV(G) be of order 2. We have ε2a(u)+ε2b(u)=1. Moreover, from χ we obtain

χ(u)=ε2a(u)-ε2b(u){±1}.

Thus (ε2a(u),ε2b(u)){(1,0),(0,1)}, proving the claim for units of order 2.

Now let uV(G) be of order 4. Note that since χ(u) is integral, χ(u){±1} and thus χ(u2)=1 meaning that u2 is rationally conjugate to elements in 2a. We have

(3.1)ε2a(u)+ε2b(u)+ε4a(u)=1.

Let D be a representation corresponding to ψ+ or ψ-. Since ψ-(u2)=ψ+(u2)=ψ+(2a)=-m the eigenvalues of D(u2), with multiplicities, are 2m times -1 and m times 1. Thus 2m of the eigenvalues of D(u) are i or -i, where i denotes a primitive 4th root of unity in 𝔽q2. But since ψ+ and ψ- do only take real values, m eigenvalues of D(u) must be i and m must be -i. In particular, we have -mψ+(u)m and -mψ-(u)m.

Assume firstly that 4a is lies in PSL(2,q). Then using χ we obtain

χ(u)=ε2a(u)-ε2b(u)+ε4a(u){±1}.

Subtracting (3.1) this gives ε2b(u){0,1}. Using ψ+ and ψ- we obtain

-m-mε2a(u)+mε4a(u)-mε2b(u)m

and

-m-mε2a(u)+mε4a(u)+mε2b(u)m

respectively. Adding or subtracting (3.1) gives ε4a(u){0,1} and ε2a(u){0,1} respectively. Since we have ε4a(u)0mod2 and ε2a(u)ε2b(u)0mod2 by [9, Theorem 4.1], this implies the equality (ε2a(u),ε2b(u),ε4a(u))=(0,0,1).

Now assume 4a is not in PSL(2,q). Doing analogous computations to the above χ gives ε2a(u){0,1}, while ψ+ and ψ- give ε2b(u){0,1} and ε4a(u){0,1} respectively. By [9, Theorem 4.1] however ε4a(u)0mod2 and thus we have (ε2a(u),ε2b(u),ε4a(u))=(0,0,1), proving that units of order 4 are rationally conjugate to elements of G. ∎

We are now ready to prove the main results.

Proof of Theorem 2.

By Lemma 4 we may assume that O2(G)=1. So by [6, Theorem of Gorenstein and Walter] one of the following three cases may be assumed:

  1. G is a subgroup of PΓL(2,q) containing PSL(2,q) or PGL(2,q) as a normal subgroup of odd index m. Here q denotes an odd prime power.

  2. G is the alternating group of degree 7.

  3. G is a 2-group.

In case (iii) the result follows from [32, Theorem 1] and case (ii) has been handled in [18, Example 7]. So (i) is the only remaining case and we are exactly in the situation of Lemma 1. We may moreover assume that G contains elements of order 4, since otherwise the result follows from Proposition 3. We will frequently use Lemma 1 and the characters from Tables 1 and 2.

Assume V(G) contains a subgroup U isomorphic to C4×C2. Note that U contains three involutions and four elements of order 4 and ψ+ takes the value -m on involutions of U while taking the value m on all elements of order 4 in U. Let 𝟏 be the trivial character of U. Then by Lemma 2

18uUψ+(u)𝟏(u-1)=18(3m-3m+4m)=m2

is a non-negative integer, contradicting the existence of U since m is odd.

Now in view of Lemma 1 and the fact that the exponents of G and V(G) coincide, it only remains to show that there are no elementary-abelian groups of rank bigger than 2, quaternion or semidihedral groups in V(G). Since every quaternion and semidihedral group contains a quaternion group of order 8, it will be enough to prove that V(G) does not contain neither a subgroup isomorphic to C2×C2×C2 nor Q8. Let U be an elementary-abelian group of rank 3. Then ψ+(u)=-m for every u1 in U and by Lemma 2

18uUψ+(u)𝟏(u-1)=18(3m-7m)=-m2

is a non-negative integer, contradicting the existence of U.

Finally, let UV(G) be a quaternion group of order 8. Note that U contains exactly one involution and six elements of order 4. Since the involution in U is a square of an element of order 4, it is rationally conjugate to elements in 2a. We claim that 4a is in PSL(2,q). Indeed, otherwise

18uUχ(u)𝟏(u-1)=18(2-6)=-12,

contradicting Lemma 2. So assume that 4a lies in PSL(2,q). Note that since the 2-Sylow subgroup of PSL(2,q) is dihedral and |PSL(2,q)|=(q-1)q(q+1)2, this is equivalent to q±1mod8. Any irreducible ordinary character of degree q±1 of PSL(2,q) may be extended to a character of PGL(2,q) by [33, Lemma 4.5], meaning it has the same values on the elements lying in PSL(2,q). If PGL(2,q) is not a normal subgroup of G, this fact may just be ignored. In any case every ordinary irreducible character η of degree q±1 of PSL(2,q) can be induced to a character η of G such that η(1a)=mη(1a),η(2a)=mη(2a) and η(4a)=mη(4a). Since in PGL(2,q) every element is conjugate to its inverse, the extension of η to PGL(2,q) is real-valued. Moreover, the Schur index of this extension is 1 by [15, Theorem 2 (a)] and hence it is affordable by a real representation and then so is η.

We will apply the same arguments as in the last paragraph of the proof of [19, Theorem 2.1]. Let ε{±1} such that qεmod4. The ordinary character table of PSL(2,q) was first computed independently by Schur and Jordan and is also given in [19, Table 1]. Let η be an irreducible ordinary character of PSL(2,q) of degree q+ε such that η(2a)=-2ε and η(4a)=0. Let η be the character of G obtained from η the way described in the preceding paragraph. Then η(1a)=m(q+ε), η(2a)=-2mε and η(4a)=0. Let λ be the 2-dimensional irreducible character of U, so λ takes the value -2 on the involution in U and the value 0 on elements of order 4. Taking the scalar product of η and λ with respect to U we obtain

λ,ηU=18(2m(q+ε)+4mε)=m(q-ε)4+mε.

Since q±1mod8, the number q-ε4 is even, thus λ,ηU is odd. By the preceding paragraph η is affordable by a real representation. Linearly extending this representation we obtain a real representation of U with character η|U. Since the Frobenius–Schur indicator of λ is -1, it is not affordable by a real representation [22, Chapter XI, Theorem 8.3]. So by [22, Chapter XI, Theorem 8.9] the integer λ,ηU must be even, a contradiction. Thus V(G) does not contain a quaternion group of order 8 and this finishes the proof of the theorem. ∎

Proof of Proposition 3.

By [26, Proposition 4.7] we may assume that the Sylow 2-subgroup of G is not abelian, thus it is a dihedral or quaternion group of order 8. First assume that the Sylow 2-subgroup is dihedral. By Lemma 4 we may assume that O2(G)=1 and so again by [6, Theorem of Gorenstein and Walter] one of the three cases given in the beginning of the proof of Theorem 2 may be assumed. If G is a 2-group, the result is well known and follows also from [32, Theorem 1]. Since by assumption G is not isomorphic to an alternating group of degree 7, it is a subgroup of PΓL(2,q) containing PSL(2,q) or PGL(2,q) as a normal subgroup of odd index with some odd prime power q. Let U be a 2-subgroup of V(G). Then U is isomorphic to a subgroup of G by Theorem 2. Let the conjugacy classes of elements of order 2 and 4 be named as in the proof of Lemma 1. If PGL(2,q) is not a subgroup of G, Lemma 1 implies that any isomorphism between U and a subgroup S of G preserves values of all ordinary irreducible characters of G and then U and S are rationally conjugate by [31, Lemma 4].

So assume that PGL(2,q) is a subgroup of G. Note that in this case elements of order 4 in G do not lie in PSL(2,q) since otherwise the Sylow 2-subgroup of G would have more then eight elements. We thus may assume that G has characters as given in Table 2. If U is cyclic, then it is rationally conjugate to a subgroup of G by Lemma 1. If U is elementary-abelian of rank 2, either all elements of U are rationally conjugate to elements in 2a or exactly two elements are rationally conjugate to elements of 2b, since otherwise 14uUχ(u) is not a non-negative integer, contradicting Lemma 2. Thus U is rationally conjugate to a subgroup of G by [31, Lemma 4], since in both cases we can find a subgroup S of G and an isomorphism between U and S preserving values on all ordinary characters of G.

So finally assume that U is isomorphic to a dihedral group of order 8. The central involution of U is rationally conjugate to elements in 2a by Lemma 1 since it is a square of an unit of order 4. Since 18uUχ(u) is an integer, one of the non-central involutions in U is rationally conjugate to elements of 2b and the character values of χ then determine for every involution in U, whether it is rationally conjugate to an element of 2a or 2b. This allows again to construct an isomorphism between U and a Sylow 2-subgroup of G preserving character values and thus U is rationally conjugate to Sylow 2-subgroups of G by [31, Lemma 4].

It remains to consider the case where the Sylow 2-subgroup of G is a quaternion group of order 8. By [26, Proposition 4.5] it is enough to show that units of order 2 and 4 are rationally conjugate to elements of G. By the famous theorem of Brauer and Suzuki [8], G contains a central involution, say g, and by the Berman–Higman Theorem, g is the only involution in V(G). So assume uV(G) has order 4. Then εg(u)=0 by the Berman–Higman Theorem. If G contains only one conjugacy class of elements of order 4, this implies that u is rationally conjugate to an element of G. So assume that G has more then one conjugacy class of elements of order 4 and that u is not rationally conjugate to an element of G. Different conjugacy classes of elements of order 4 in G map to different conjugacy classes of involutions in G/g. Thus, if u has non-trivial partial augmentations, the image of u in V((G/g)) is an involution with non-trivial partial augmentations. Such an involution does not exist by [3, Proposition 3.4] and hence u is rationally conjugate to an element of G and the proposition is proved. ∎

Remark

As already remarked in [18, Example 7] it is not known whether units of order 4 in V(A7) are rationally conjugate to elements of the group base. If one could prove that this is the case, then the exception of A7 from Proposition 3 would not be necessary.

Proof of Theorem 1.

Let G be a group not containing a subgroup isomorphic to C4×C2 such that V(G) contains a subgroup U=t×sC4×C2. By Lemma 1 we may assume that the Sylow 2-subgroup P of G is abelian, (generalized) quaternion, dihedral or semidihedral. By Proposition 3 and Theorem 2 only the case where P is semidihedral remains and by Lemma 4 we may assume O2(G)=1. The groups of interest are classified in [2], but are not given in a single theorem there. By [2, Chapter II, Section 1, Proposition 1] four different cases may appear. These cases get their names in [2, Chapter II, Section 2, Definition 1]. So G is one of the following:

  1. A QD-group. G possesses exactly one conjugacy class of involutions and one conjugacy class of elements of order 4.

  2. A Q-group. G possesses exactly two conjugacy classes of involutions and one conjugacy class of elements of order 4.

  3. A D-group. G possesses exactly one conjugacy class of involutions and two conjugacy classes of elements of order 4.

  4. A 2-group.

In case (iv) [32, Theorem 1] gives the result, so we have to consider the first three cases.

Case (i): G is a QD-group. Denote by 2a the conjugacy class of involutions in G and by 4a the class of elements of order 4.

By [2, Chapter II, Section 2, Proposition 2], G possesses a simple normal subgroup N of odd index m and by [2, Third Main Theorem] the normal subgroup N is either PSL(3,q) with q-1mod4 or PSU(3,q) with q1mod4, for some prime power q, or the Mathieu group M11. Note that PSU(3,q) is actually defined over 𝔽q2, the field with q2 elements. The groups SL(3,q) and SU(3,q) act on the homogeneous polynomials of degree 3 in three commuting variables x,y,z over the field 𝔽q and 𝔽q2 respectively. See e.g. [1, pp. 14–16] for a discussion of the analogous action of SL(2,q). The centres of the groups are in the kernel of the action and thus this action supplies a 10-dimensional representation of PSL(3,q) and PSU(3,q) respectively. Let χ be the character of this representation. Elements of SL(3,q) and SU(3,q) projecting to elements in the classes 2a and 4a in G are given by

A=(1000-1000-1)andB=(1000010-10)

where the Gram matrix of the underlying unitary form is taken to be the identity matrix. I.e. we understand SU(3,q) to be those matrices of GL(3,q2) having determinant 1 and leaving the unitary form

α:𝔽q23×𝔽q23𝔽q2,(x,y)σ(x)t(100010001)y

invariant. Here σ denotes the automorphism of 𝔽q2 of order 2.

The eigenspace of A for the eigenvalue 1 is spanned by x3,xy2,xz2,xyz while the eigenspace for the eigenvalue -1 is spanned by y3,z3,x2y,x2z,y2z,yz2. Denote by i a primitive 4th root of unity in 𝔽q2. The eigenspaces of B are spanned by:

  1. for the eigenvalue i by y3+iz3,yz2+iy2z,x2y+ix2z,

  2. for the eigenvalue -i by y3-iz3,yz2-iy2z,x2y-ix2z,

  3. for the eigenvalue -1 by xy2-xz2,xyz,

  4. for the eigenvalue 1 by x3,xy2+xz2.

One thus obtains χ(1)=10, χ(2a)=-2 and χ(4a)=0. The Mathieu group M11 does also possess an (ordinary) character with the same values, see e.g. [10]. Inducing this character of N to G we get the character given in Table 3.

Table 3

A character of a QD-group G.

1a2a4a
χ10m-2m0

Unlike in Lemma 1 it is not possible to show using only the HeLP-method that units of order 4 in V(G) are rationally conjugate to elements of G. This may be checked e.g. using a GAP-package implementing the method [4]. See also [7] concerning M11. Let uV(G) be of order 4. If D is a representation affording χ, then χ(u) is integral and the eigenvalues of D(u) are 4th roots of unity and thus χ(u)=χ(u3). Recall that C4×C2U=t×s is the group we want to study. There are three involutions in U taking the value -2m for χ. Moreover, the elements of order 4 in U are t,t3,st,st3 and for them we have χ(t)=χ(t3) and χ(st)=χ(st3). So by Lemma 2

18uUχ(u)=18(10m-6m+2χ(t)+2χ(st))

is a non-negative integer. Note that since χ(4a)=0, we have χ(t)=-2mε2a(t) and χ(st)=-2mε2a(st). As ε2a(t)ε2a(st)0mod2 by [9, Theorem 4.1], this implies χ(t)χ(st)0mod4. Thus 2χ(t)2χ(st)0mod8, implying 84m, a contradiction.

Case (ii): G is a Q-group. Denote by 2a and 2b the conjugacy classes of involutions in G and by 4a the class of elements of order 4.

By [2, Chapter II, Section 3, Proposition 2], G possesses a normal subgroup isomorphic to SL(2,q) for some odd prime power q and hence by [2, Chapter II, Section 3, Proposition 3], G possesses a normal subgroup N isomorphic to a group denoted by Alperin, Brauer and Gorenstein as SLk(2,q) or SUk(2,q) and [G:N]=m is odd. The groups appearing for N are defined in [2, p. 17, above Lemma 1]. The only relevant fact for us will be that these are subgroups of GL(2,q) and GU(2,q) respectively and that they contain all matrices of these groups having determinant 1 or -1. Representatives of the conjugacy classes of interest in N are given by (-100-1), (100-1) and (0-110) where again the Gram matrix of the underlying unitary form is taken to be the identity matrix. Inducing the character of N given by the determinant map to a character of G turns out to be already enough. The values of this character χ are given in Table 4.

Table 4

A character of a Q-group G.

1a2a2b4a
χmm-mm

Claim

Elements of order 2 and 4 in V(ZG) are rationally conjugate to elements of G. Hence if uV(ZG) is of order 4, then u2 is rationally conjugate to elements in 2a.

Let first uV(G) be of order 2. If u is not in 2a, then ε2a(u)=0 by the Berman-Higman Theorem, since 2a consists of a single element which is central in G. Thus u is rationally conjugate to an element of 2b in this case. If u is of order 4, again ε2a(u)=0. So from ε2b(u)+ε4a(u)=1 and the inequalities

-m-mε2b(u)+mε4a(u)m

coming from χ together with the fact that ε4a0mod2 (see [9, Theorem 4.1]) we conclude (ε2a(u),ε2b(u),ε4a(u))=(0,0,1).

Now let again C4×C2U=t×s be a subgroup of V(G). As t2 is a square of an element of order 4, it must be the element of 2a by the above claim. The other involutions in U must thus be rational conjugates of elements in 2b since the element of 2a, being central, has no conjugates. Then

18uUχ(u)=18(6m-2m)=m2

is a non-negative integer by Lemma 2, a contradiction.

Case (iii): G is a D-group. Denote by 2a the conjugacy class of involutions in G and by 4a and 4b the classes of elements of order 4.

By [2, Chapter II, Section 3, Proposition 4] we may assume that G is a PGLn(2,3), defined in [2, p. 20, before Lemma 5], or A7 or it is a subgroup of PΓL(2,q)=Aut(PSL(2,q)), for some prime power q, and contains one of the groups PSL(2,q), PGL(2,q) or PGL*(2,q) as a normal subgroup N such that [G:N]=m and q are both odd. In case G is a PGLn(2,3) its Sylow 2-subgroup is not a semidihedral group by [2, Chapter II, Section 2, Lemma 5]. In case G is A7 or N is PSL(2,q) or PGL(2,q), the Sylow 2-subgroup is also not semidihedral by [2, Chapter II, Section 2, Lemma 3]. So only the case N=PGL*(2,q) remains, this group is defined in [2, p. 20, above Lemma 4], and we are going to describe it now.

The group N=PGL*(2,q) is a non-split extension of PSL(2,q) of degree 2 and does only exist, if q is a square, say q=r2. The groups PGL*(2,q) are Zassenhaus groups and were introduced, to my knowledge, in [34] where they are denoted by Mq. They are also discussed in [22, Chapter XI, Example 1.3 c)]. Let σ be the unique automorphism of 𝔽q of order 2. So in particular, the elements of 𝔽r are fixed by σ. By acting entrywise on GL(2,q) the automorphism σ induces an automorphism, also called σ, of GL(2,q) and after projection an automorphism σ¯ of PSL(2,q). Denote by ¯ the natural homomorphism from GL(2,q) to PGL(2,q). Then via conjugation every AGL(2,q) induces an automorphism A¯ of PSL(2,q). We understand PSL(2,q) to be its own group of inner automorphisms. Let α be an element of maximal 2-power order in 𝔽q and i a primitive 4th root of unity in 𝔽q. Set

g=(100α)if r-1mod4,
g=(0α-10)if r1mod4.

Note that σ(α)=-α-1, if r-1mod4, and σ(α)=-α, if r1mod4. Then g¯σ¯ is an automorphism of order 4 of PSL(2,q) such that (g¯σ¯)2=(i00-i)¯ is an inner automorphism of PSL(2,q). We set PGL*(2,q)=PSL(2,q),g¯σ¯, understanding it as a subgroup of the automorphism group of PSL(2,q). The concrete choice of g is inspired by the proof of [13, Lemma 2.3] where the structure of the Sylow 2-subgroup of PGL*(2,q) is analysed.

Since G/PSL(2,q) has a Sylow 2-subgroup of order 2, it contains a normal 2-complement by [27, 7.2.2]. Thus G maps onto a cyclic group of order 2 and PSL(2,q) is in the kernel of this map while elements of N outside of PSL(2,q) are not. Hence G has a 1-dimensional representation containing PSL(2,q) in its kernel and mapping elements of N outside of PSL(2,q) to -1. Call the corresponding character χ. Moreover, SL(2,q) acts via conjugation on the 2×2-matrices over 𝔽q having trace 0 giving a 3-dimensional representation of PSL(2,q). This then induces a 6m-dimensional representation of G. Call the corresponding character ψ.

Furthermore GL(2,q) acts on the 4-dimensional 𝔽r-vector space

={(acσ(c)b)|a,b𝔽r,c𝔽q}

by A*H=σ(A)tHA for AGL(2,q) and H, where Xt denotes the transpose of a matrix X. The kernel of this operation is {(x00x)x𝔽q,xσ(x)=1}. Since the centre of SL(2,q) is contained in this kernel, we obtain a 4-dimensional 𝔽r-representation of PSL(2,q) which we want to extend to a representation of N. Also σ acts on by entry-wise application. Let H=(acσ(c)b) be an arbitrary element of . In case r-1mod4 the element gσ, as an element of the semilinearities ΓL(2,q), has order 4 and the action of gσ on has trivial kernel. Thus setting

g¯σ¯*H=(aασ(c)σ(α)c-b)

extends the action of PSL(2,q) to an action of N providing a 4-dimensional 𝔽r-representation of N. In case r1mod4 the order of the semilinearity gσ is twice the order of α while the order of g¯σ¯ is 4. We have

gσ*H=(b-ασ(c)-σ(α)c-α2a),

so (gσ)4*H=α4H. Hence viewing a basis of as a basis of the 4-dimensional 𝔽q-vector space 𝔽q𝔽r and setting

g¯σ¯*H=α-1(b-ασ(c)-σ(α)c-α2a)=(α-1b-σ(c)c-αa),

we obtain a 4-dimensional 𝔽q-representation of N. Let η be the character corresponding to the induced representation of G. Computing the eigenvalues of these actions one obtains the character values given in Table 5.

Table 5

Some characters of a D-group G.

1a2a4a4b
χ111-1
ψ6m-2m2m0
η4m0-2m0

Claim

Elements of order 2 and 4 in V(ZG) are rationally conjugate to elements of G. Hence especially an element of order 4 is rationally conjugate to its inverse.

This is clear for elements of order 2, so let uV(G) be of order 4. From χ we obtain the equation

ε2a(u)+ε4a(u)-ε4b(u){±1},

and adding ε2a(u)+ε4a(u)+ε4b(u)=1 onto this gives ε4b(u){0,1}. Since u2 is rationally conjugate to an element in 2a under a representation affording η, the unit u2 has 2m times the eigenvalue 1 and 2m times the eigenvalue -1. This implies

-2mη(u)=-2mε4a(u)2m

and thus ε4a(u){-1,0,1}. Under a representation affording ψ the involution u2 has 4m times the eigenvalue -1 and 2m times the eigenvalue 1. Thus

-2mψ(u)=-2mε2a(u)+2mε4a(u)2m,

implying

-ε2a(u)+ε4a(u){-1,0,1}.

Together with the fact that ε2a(u)0mod2 and ε4a(u)+ε4b(u)0mod2 by [9, Theorem 4.1] these equations prove the claim. The fact that u is rationally conjugate to its inverse follows from the character values of χ.

So let C4×C2U=t×s. Then by Lemma 2

18uUχ(u)=18(1+3+2χ(t)+2χ(st))

is a non-negative integer. Since χ(t),χ(st){±1}, this implies χ(t)=χ(st) and thus t and st are rationally conjugate. Now

18uUη(u)=18(4m+30+4η(t))

is also a non-negative integer. Since η vanishes on the conjugacy classes 2a and 4b, we obtain that η(t)=η(4a)ε4a(t)=-2mε4a(t). So 4η(t)0mod8 meaning that m2 has to be an integer. This contradicts the existence of U and finalizes the proof of the theorem. ∎


Communicated by Evgenii I. Khukhro


Funding source: DFG

Award Identifier / Grant number: Sonderforschungsbereich 701

Funding statement: This research was partly supported by the Sonderforschungsbereich 701 at the University of Bielefeld.

Acknowledgements

I am thankful to Markus Stroppel for many interesting conversations about actions of classical groups. I am moreover thankful to the referee for his corrections and suggestions improving the readability of the paper.

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Received: 2015-12-20
Revised: 2016-5-13
Published Online: 2016-6-11
Published in Print: 2017-3-1

© 2017 by De Gruyter

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