Table Of ContentINTEGRAL GROUP RINGS OF SOLVABLE GROUPS WITH
TRIVIAL CENTRAL UNITS
ANDREASBA¨CHLE
Abstract. Theintegral group ringZG of agroup G has only trivialcentral
7 units,iftheonlycentralunitsofZGare±zforz inthecenterofG. Weshow
1 thattheorderofafinitesolvablegroupGwiththisproperty,canonlyhave2,
0 3,5and7asprimedivisors,bylinkingthistoinversesemi-rationalgroupsand
2 extending one result on this class of groups. We also classify the Frobenius
groupswhoseintegralgroupringshaveonlytrivialcentralunits.
b
e
F
3
1. Introduction
]
R By G we always denote a finite group. The integral group ring of G is denoted
G by ZG and its group of units by U(ZG). Clearly, all elements of G are units in
±
ZG, the so-called trivial units. Already G. Higman determined in his PhD thesis
.
h in 1940 the groups such that all units in ZG are trivial: all of them are Dedekind
t
a groups of small exponent. If π(G) denotes the prime spectrum of G, i.e. the set of
m prime divisors of the order of G, then π(G) 2,3 in all these cases and clearly
⊆ { }
[ all of these groups are solvable.
It remains a fundamental problem to describe generators for U(ZG) if not all
3
units are trivial. It is well-known that the the units of reduced norm 1 together
v
7 with the central units generate a subgroup of finite index in U(ZG). For many
4 finite groups it has been shown by E. Jespers, G. Leal, J. Ritter and S. Sehgal
3 thattheso-calledbicyclic unitsgenerateasubgroupoffinite index inthe subgroup
4
of reduced norm 1 elements. These together with the so-called Bass units then
0
generateasubgroupoffiniteindexinthecentralunits,cf.e.g.therecentmonograph
.
1 of E. Jespers and A´. del R´ıo [JdR16], Chapter 11, for a detailed treatment. It is
0
relevant to know when the Bass units can be omitted, in other words when all
7
central units are trivial, i.e. belong to Z(G). Ritter and Sehgal studied groups
1
±
: G with this property [RS90]. They obtained a description of the centers of the
v
Wedderburn components of the rational group algebra of such groups together
i
X withagrouptheoreticcharacterizationintermsofconditionsonconjugacyclasses,
r cf. Proposition 2.2 below. This class of groups has recently again been extensively
a
studied by G.K. Bakshi, S. Maheshwary and I.B.S. Passi [BMP16, Mah16]. We
adopt their notation for these groups.
Definition 1.1. AgroupGissaidto be acut group (“centralunits trivial”)ifZG
only contains trivial centralunits, that is, if u U(ZG) is central, then u= z for
∈ ±
some z Z(G), the center of G.
∈
In[Mah16,Theorems1and2]Maheshwaryshowsthatπ(G) 2,3,5,7 forGa
cutgroupsofoddorderorasolvablecutgroupsofevenordersu⊆ch{thatall}elements
are of prime power order. We show that this is true for all solvable groups:
2010 Mathematics Subject Classification. 16U60,20C15,20E45.
Key words and phrases. integral group ring, central units, inverse semi-rational, solvable
groups,primespectrum,Frobeniusgroups.
Theauthorisapostdoctoral researcheroftheFWO(ResearchFoundationFlanders).
1
2 ANDREASBA¨CHLE
Theorem 1.2. Let G be a solvable cut group. Then π(G) 2,3,5,7 .
⊆{ }
This is best possible in the sense that none of the primes can be removedas can
beseenfromthegroupsinTheorem1.3. Notethatfornon-solvablecutgroupssuch
a restriction on the prime spectrum cannot exists as all Weyl groups, in particular
allsymmetricgroupsS ,arecut groups. Thereisalsoagrouptheoreticalinterpre-
n
tationofandinterestin cut groups. By Proposition2.2, a cut groupis nothing but
aninversesemi-rationalgroupsasintroducedby D.Chillag andS. Dolfiin[CD10].
And this is also the path we will take to prove Theorem 1.2: we will improve one
of the results of Chillag and Dolfi on the prime spectrum of such groups that are
solvable.
Asasecondresult,weclassifytheFrobeniusgroupswithonlytrivialcentralunits
in their integral group rings.
Theorem 1.3. Let K be a Frobenius complement.
(1) If K is even and K is the complement of a cut Frobenius group G, then
| |
G is isomorphic to a group in the series in (1a) – (1f) with b,c,d Z or
≥1
∈
G is one of the groups in (1A) – (1D).
(a) C2b⋊C (A) C2⋊Q
3 2 5 8
(b) C2b⋊C (B) C2⋊(C ⋊C )
3 4 5 3 4
(c) C2b⋊Q (C) C2⋊SL(2,3)
3 8 5
(d) Cc⋊C (D) C2⋊SL(2,3)
5 4 7
(e) Cd⋊C
7 6
(f) C2d⋊(Q C )
Conver7sely, for8×each3oftheabovestructuredescriptions thereisacutFrobe-
nius group of that form and it is unique up to isomorphism.
(2) If K is odd, thenthere is acut Frobenius groupG withcomplement K and
| |
Frobenius kernel F if and only if K C and one of the following holds
3
(a) F is a cut 2-group admitting a fi≃xed point free automorphism of order
3.
In particular, F has order 2a for some a 2Z and is an extension
≥1
∈
of an abelian group of exponent a divisor of 4 by an an abelian group
of exponent a divisor of 4.
(b) F is an extension of an elementary abelian 7-group by an elementary
abelian 7-group, expF = 7 and F admits a fixed point free automor-
phism of order 3 fixing each cyclic subgroup of F.
Theorem1.2isprovedinSection3,Theorem1.3inSection4. Wegivecorollaries
describing cut Frobenius groups with abelian Frobenius kernel (Corollary 4.6) and
cut Camina groups (Corollary 4.8), see end of Section 4.
The notation is mainly standard. C denotes a cyclic group of order n, Q the
n 8
quaternion group of order 8. F denotes a finite field with q elements. Euler’s
q
totient function is indicated by φ. Standard references are [Hup67, Sco87].
2. Preliminaries
We need the following group theoretical notion of (inverse) semi-rationalgroups
that was coined by Chillag and Dolfi [CD10]. By we denote conjugacy (in a
∼
group that should always be clear from the context).
Definition 2.1. Let G be a group and x G.
∈
(1) x is rational in G if xj x for all j Z with (j,o(x))=1.
∼ ∈
(2) The element x is called inverse semi-rational in G if xj is conjugate to x
or x−1 for all j Z with (j,o(x))=1.
∈
INTEGRAL GROUP RINGS WITH TRIVIAL CENTRAL UNITS 3
(3) x is semi-rational in G if each if xj is conjugate to x or xm for all j Z
∈
with (j,o(x))=1 for some fixed m.
Agroupiscalledrational (respectivelyinversesemi-rational,semi-rational)ifevery
of its elements is respectively of that type.
Foracharacterψ(ordinaryorBrauer),Q(ψ)denotesthefieldofcharactervalues
of ψ, that is, the field extensionof Q generatedby ψ(g): g G . It is well-known
{ ∈ }
that agroupGis rationalifand onlyif for eachχ Irr(G), Q(χ)=Q,whence the
∈
name.
The following characterization of cut groups is essentially due to Ritter and
Sehgal. We link it to inverse semi-rational groups.
Proposition 2.2. Let G be a group. The following are equivalent.
(1) G is a cut group.
(2) G is inverse semi-rational.
(3) For every x G and k Z with (k, G)=1, xk x or xk x−1.
m∈ ∈ | | ∼ ∼
(4) If QG ≃ Lk=1Mnk(Dk) is the Wedderburn decomposition (m,nk ∈ Z≥1,
D rational division algebras), then for each k,
k
Z(D ) Q(√ d)
k
≃ −
for some d Z , where Z denotes the center.
≥0
∈
(5) For each χ Irr(G), Q(χ) = Q(√ d) for some d Z , i.e. the field of
≥0
∈ − ∈
character values of χ is Q or an imaginary quadratic extension of Q for
each absolutely irreducible character of G.
Proof. The equivalence of (1), (3), (4), and (5) is proved in [RS90]. Clearly, (2)
implies (3). To see that (3) (2), let x G and j Z with (j,o(x)) = 1 be
⇒ ∈ ∈
given. We can choose k = o(x) s + j for some s Z such that (k, G) = 1
· ∈ | |
(this is possible, for instance, by Dirichlet’s Theoremonthe infinitude ofprimes in
arithmetic progressions). Then xj =xk is conjugate to x or x−1, by (3). (cid:3)
We willuse these equivalences freely without further mention. From(4) it easily
followsthat eachquotientof a cut groupis againa cut group. Note thatQ(G), the
field extension of the rationals generated by the values of all irreducible characters
of the cut group G might have degree larger than 2 over Q as can already be seen
by the non-abelian group of order 21. We have
G: G is rational G: G is cut G: G is semi-rational .
{ } ⊆ { } ⊆ { }
In case of groups of odd order, the first class only contains the trivial group, and
the last two classes coincide, cf. [CD10, Remark 13].
We willuse the followingnotationintroducedby ChillagandDolfitogetherwith
their observations we state here.
Definition 2.3. Let G be a group, x G. Set B (x) = N ( x )/C ( x ), the
G G G
∈ h i h i
automorphisms of x induced by elements of G. We identify B (x) with the
G
h i
corresponding subgroup of Aut( x ).
h i
Lemma 2.4 ([CD10, Lemma 5 (2), (3)]). Let G be a finite group.
(2) An element x is inverse semi-rational in G if and only if Aut( x ) =
h i
B (x) τ , where τ Aut( x ) is the inversion automorphism of x de-
G
h i ∈ h i h i
fined by τ(y) = y−1. In particular for an inverse semi-rational element x
in G, we have [Aut( x ):B (x)] 2.
G
h i ≤
(3) If p is a prime of the form p 1mod4, then a p-element x is inverse
≡
semi-rational in G if and only if x is rational in G.
4 ANDREASBA¨CHLE
3. Solvable Groups
It is a classical result of R. Gow [Gow76] that the orders of solvable rational
groups are divisible only by the primes 2, 3 and 5. In such results, the treatment
of the smallest primes usually requires the most effort; Gow used here local Schur
indices to show that 7 cannot divide the order of a solvable rational group. The
primes spectra of semi-rational groups were studied by Chillag and Dolfi [CD10].
Theyshowedthatπ(G) 2,3,5,7,13,17 forasolvablesemi-rationalgroupG. It
⊆{ }
seemstobestillunknownwhether17candividetheorderofasolvablesemi-rational
group. For solvable groups G that are inverse semi-rational (i.e. cut), they prove
π(G) 2,3,5,7,13 , however it remained open whether 13 can be omitted from
⊆ { }
the list. In [Mah16, Theorems 1 and 2] Maheshwary shows that π(G) 2,3,5,7
for G a (solvable) cut groups of odd order or a solvable cut groups o⊆f e{ven orde}r
such that each element has order a prime power (with different techniques than
we use here). To prove that this is best possible even for all solvable cut groups,
we improve the above mentioned result of Chillag and Dolfi [CD10, Theorem2] on
solvable inverse semi-rational groups, and show that 13 can be removed from the
list.
In the investigationof the prime spectra of classessolvable groups,the notion of
the k-eigenvalue property, as introduced in [FeS86, (2.8) Definition] by E. Farias e
Soares, has proved to be very useful.
Definition 3.1. Let a group G act on a finite dimensional vectorspace V over a
finitefieldF. WesaythatV,ortheactionofGonV,hasthek-eigenvalueproperty,
for some k Z provided that (a) k divides F× and (b) for every v V, there
>0
∈ | | ∈
exists g G such that vg =λv, where λ is some fixed element of order k in F×.
∈
A key point in the proof of Theorem 1.2 is the following result due to Farias e
Soares, so we will cite it here. By ζ we denote a primitive nth root of unity.
n
Theorem 3.2 ([FeS86, (2.9) Theorem B (c)]). Let H be a solvable group acting
on a finite-dimensional vector space V over F , where p ∤ H . Assume that the
p
| |
action has the k-eigenvalue property. Let ϕ be a Brauer character afforded by V
and K =Q(ϕ). If k= K(ζ ):K and K :Q 2, then p 2,3,5,7 .
p
| | | |≤ ∈{ }
Proof of Theorem 1.2. By [CD10, Theorem 2] we have π(G) 2,3,5,7,13 for
solvablecutgroupsG. AssumethatGisasolvablecutgroupof⊆mi{nimalorder}such
that 13 π(G). Let
∈
1=G EG E...EG EG =G
n+1 n 1 0
beachiefseriesofG. NotethateachfactorgroupG/G isagainacutgroup. Hence
j
G is the only chief factor of G divisible by 13 and, as minimal normal subgroup,
n
elementary abelian. Thus V =G Syl (G) and G= V ⋊ H, where H G/V
is also a cut group of order not dniv∈isible13by 13. Now, the gαroup homomo≃rphism
α: H Aut(V) GL(d,13) turns V into a F H-module, which is irreducible as
13
→ ≃
V is a minimal normal subgroup of G. We will write V additively. Let ϕ be the
Brauer character of H afforded by V. Then ϕ is actually an ordinary character as
13 ∤ H . By [Ten12, Lemma 3] and the remark following it, the field of character
| |
values of ϕ is contained in the field of character values of each of its irreducible
constituents. But then [Q(ϕ) : Q] 2 by Proposition 2.2 (1) (2). All elements
≤ ⇒
oforder13inGarerationalinGbyLemma2.4,soforx V wehavethatj x x
∈ · ∼
for each j Z with (j,13) = 1, hence there exists h H such that xh = j x. In
∈ ∈ ·
particular,theH-moduleV hasthe12-eigenvalueproperty. Butthenweareinthe
situation of Theorem 3.2: H is solvable, K = Q(ϕ) has degree at most 2 over Q
and k = [K(ζ ) : K] = 12, as K Q(ζ ) = Q (the unique quadratic subfield of
13 13
∩
Q(ζ ) is real) and H (and hence G) cannot exist. (cid:3)
13
INTEGRAL GROUP RINGS WITH TRIVIAL CENTRAL UNITS 5
4. Frobenius Groups
Frobenius groups that are rational have been determined in [DS04] (there are
two infinite series together with “Markel’s group”), the structure of semi-rational
Frobeniusgroupshasbeenstudiedin[ADD16],thoughacompleteclassificationhas
not been obtained. We will now prove the classification of cut Frobenius groups
given in Theorem 1.3 in the introduction. To this end we will first show that
there are only eight isomorphism types of potential Frobenius complements in cut
Frobenius groups (Proposition 4.2) and use this to determine the structure of all
cut Frobenius groups. The general structure of Frobenius groups is fairly well
understood, standard references include [Hup67, Pas68, Sco87].
Inthefollowingpropositionwestrengthentheconditionsof[ADD16,Lemma2.5]
for cut groups. Note that we can in particular prove that cut Frobenius groups are
solvable,whichisnotthecaseforsemi-rationalFrobeniusgroups(e.g.theFrobenius
groupC2 ⋊SL(2,5)is indeedsemi-rationalbutnotcut). Smallpartsofthe proofs
11
are similar to those in [ADD16], however we include the complete proofs to make
the exposition self-contained.
Proposition 4.1. Let G be a cut Frobenius group. Then G is solvable. If K is a
Frobenius complement of G, then the Sylow p-subgroups of K are cyclic of order at
most p if p is odd and cyclic of order at most 4 or quaternion of order 8 for p=2.
Proof. Assume by way of conradiction that the cut Frobenius group G is not solv-
able. Then a Frobenius complement K of G is not solvable and, by a theorem of
Zassenhaus[Pas68, 18.6.], K has a normalsubgroupof index atmost 2 isomorphic
toSL(2,5) M,whereM isofordercoprimeto30. By[JdR16,Theorem11.4.10]G
×
has a quotient isomorphic to SL(2,5) or to the covering group of S having Small-
5
GroupID[240, 89]inGAP[GAP16]. Itisreadilycheckedthatthelastmentioned
group has fields of character values isomorphic to Q(√2) and Q(√3) and SL(2,5)
has a character field Q(√5), so these groups cannot be cut groups. As one of the
two necessarily appears as a quotient of G, G is also not cut, implying that cut
Frobenius groups are necessarily solvable.
Let p be an odd prime and P Syl (K) with P = 1. Then P = x C by
∈ p 6 h i ≃ pf
[Pas68, 18.1] and Aut(P) = (p 1)pf−1. As P is abelian, p ∤ B (x). But then
K
| | − | |
[Aut(P):B (x)] 2 (cf. Lemma 2.4) implies that f 1.
K
≤ ≤
For p = 2 the Sylow 2-subgroups of K are cyclic or generalized quaternion by
[Pas68, 18.1]. If P Syl (K) is cyclic, then K has a normal 2-complement by
∈ 2
Cayley’s normal 2-complement theorem, i.e. K M ⋊P with M a group of odd
order. Then P is a cyclic 2-group that is a cut g≃roup as a quotient of a cut group
and hence of order at most 4. Now assume that
P = x,y x2n−1 =1, y2 =x2n−2, xy =x−1
h | i
is a (generalized) quaternion Sylow 2-subgroup of K of order 2n, n 3. Then
≥
x EP is a cyclic normal subgroup of order 2n−1. Clearly B (x) = τ , where
P
h i h i
τ denotes the inversion-automorphism of x . But as P Syl (K), the 2-part of
h i ∈ 2
the order ofB (x) and B (x) coincide. As Aut( x ) =φ(2n−1)=2n−2, a power
K P
| h i |
of 2, we have by Lemma 2.4, that x is inverse semi-rational in K if and only if
2n−2 =2, hence n=3 and P is the quaternion group of order 8. (cid:3)
Fromthe factthat the FrobeniuscomplementK is asolvablecut group,we infer
with Theorem 1.2 that π(K) 2,3,5,7 , so in particular there are only finitely
⊆ { }
many isomorphism types of Frobenius complements in Frobenius groups that are
cut.
6 ANDREASBA¨CHLE
Proposition4.2. IfK isaFrobeniuscomplementthatiscut,thenK isisomorphic
to one of the following groups:
C , C , C , C , Q , C ⋊C , SL(2,3), Q C .
2 3 4 6 8 3 4 8 3
×
Proof. Using a direct implementation in GAP [GAP16], it can be easily seen that
the remaining Frobenius complements of cut Forbenius groups are those given in
the proposition.
Tofilterthecorrectcomplements,weuseπ(K) 2,3,5,7 ,thestructureofthe
⊆{ }
SylowsubgroupstogetherwiththefollowingadditionalresultsonFrobeniusgroups
which eliminate another 7 possibilities1:
(1) If the Frobenius complement is of even order, then it has a unique (hence
central) involution [Pas68, Theorem 18.1].
(2) AFrobeniuscomplementcannotcontainasubgroupwhichisitselfaFrobe-
nius group [Sco87, 12.6.11] (cid:3)
In a large part of the proof of Theorem 1.3 we will have to deal with Frobenius
group G that have an elementary abelian p-group as Frobenius kernel F. If, in
this situation, K denotes a Frobenius complement of G, then we can consider F
as an F K-module, which is semi-simple. Hence F V ... V for irreducible
p 1 s
≃ ⊕ ⊕
F K-modules V , which are necessarily faithful. Each V then corresponds to an
p j j
ordinary character of K. If V is an F K-module, such that V ⋊K is a Frobenius
p
group with complement K, then we can define a new group (a subdirect product)
by taking Vs ⋊K with “diagonal action”, i.e. K acts on all s copies of V in the
sameway,thatisif(x ,...,x ) Vs,then(x ,...,x )y =(xy,...,xy)fory K (and
xy is defined in V ⋊K1). Clesar∈ly, this is agai1n a Frsobenius1group.s ∈
j
Definition 4.3. Let V be a F K-module such that G = V ⋊K is a Frobenius
p
group. Let β: K GLF (V) be the corresponding homomorphism. For x
→ p ∈
V denote by Z (x) those automorphisms of x that are induced by elements of
G
h i
β(K) Z(GLF (V)).
∩ p
Thus the elements of Z (x) are exactly those automorphisms of x induced by
G
h i
thosemultiplesoftheidentityofV thatcomefromelementsofK. Actually,Z (x)
G
isequalforeachx V 0 . NotethatifinthepreviousdefinitiontheF K-module
p
∈ \{ }
V is absolutely irreducible, then Z (x) are those automorphisms of x that are
G
h i
inducedby elements ofβ(Z(K)). For lateruse werecordthe followingobservation.
Lemma 4.4. Let K be a group, p an odd prime and let V be an F K-module such
p
that G=V ⋊K is a cut Frobenius group. Then the follwoing are equivalent:
(1) Vs⋊K (with diagonal action) for all s 2 is a cut Frobenius group.
(2) V2⋊K (with diagonal action) is a cut F≥robenius group.
(3) Z (x) τ = Aut( x ) for each x V, where τ is the inversion automor-
G
h i h i ∈
phism of x .
h i
Proof. We will only give a proof in case p 1mod4, the proofin the other case is
≡
similar. Hence an element x of order p is inverse semi-rationalin a group X if and
only if B (x)=Aut( x ), by Lemma 2.4.
X
h i
Clearly, (1) implies (2). Now assume that (2) holds and we want to show (3).
Hence assume that H = V2 ⋊ K is a cut Frobenius group. We have to show
that all automorphisms of x for x V G are induced by scalar multiples
h i ∈ ≤
of the identity that come from an element of K. Let x V 0 and λ F×.
∈ \{ } ∈ p
Then there is a unique k K such that xk = λx, because G is a cut Frobenius
∈
1NamelyS3,C5⋊C4withfaithfulaction,C7⋊C3,C7⋊C6withfaithfulaction,C2×(C7⋊C3),
C15⋊C4 withfaithfulactionandC2×(C7⋊C3).
INTEGRAL GROUP RINGS WITH TRIVIAL CENTRAL UNITS 7
group. Now for any y V, there is an ℓ K such that (x,y) V2 H gets
conjugate by ℓ to λ(v,w∈), as V2⋊K is a cut∈group. Now as the ac∈tion is≤diagonal,
(xℓ,yℓ) = (x,y)ℓ = λ(x,y) = (λx,λy) = (xk,λy). By uniqueness, k = ℓ and the
action of k multiplies all elements of V by the same scalar λ, hence multiplication
by λ is an element of Z (x) for every x V.
G
∈
For the implication (3) (1), assume that Z (x) = Aut( x ) for each x V
G
⇒ h i ∈
and set L = Vs⋊K for some s 2 with diagonal action. Then L is a Frobenius
≥
group. For each y Vs, B (y) = Aut( y ) and y is (inverse semi) rational in L.
L
As L is the union o∈f conjugates of the Frhobienius complement K (which is cut as a
quotientofthe cut groupG)andthe Frobenius kernelVs, it followsthatL is a cut
group. (cid:3)
Proof of Theorem 1.3. Let G always be a Frobenius group with Frobenius kernel
F and Frobenius complement K.
(1) IfK hasevenorder,thentheFrobeniuskernelF isabelian[Hup67,V,8.18].
Let x F 1 . Then C (x)=F andhence B (x)=N ( x )/C ( x ) is
G G G G
∈ \{ } h i h i
isomorphic to a subgroup of K. Note that if x F is an element of order
rn for an odd prime r, then we need that φ(rn)/∈2= (r−1)rn−1 divides K ,
2 | |
by Lemma 2.4; if in this situation the 2-part of (r 1) is at least 4, then
−
we even have φ(rn) divides K , as Aut(C ) is cyclic. From this it follows
rn
| |
that n 1, i.e. all Sylow subgroups of F are elementary abelian.
≤
Goingthroughthepossibilities forevenorderFrobeniuscomplements in
cutgroupsgiveninProposition4.2,weobtainthecandidatesforFrobenius
kernelsstated in Table 1 sortedaccordingto their Frobenius complements.
Table 1. Possible isomorphism types of Frobenius complements
K togetherwiththeirFrobeniuskernelsF of cutFrobeniusgroups.
K Candidates for F
C Cb
2 3
C Cb, Cc, Cb Cc
4 3 5 3× 5
C Cd
6 7
Q Cb, Cc, Cb Cc
8 3 5 3× 5
C ⋊C Cc, Cd, Cc Cd
3 4 5 7 5× 7
SL(2,3) Cc, Cd, Cc Cd
5 7 5× 7
Q C Cc, Cd, Cc Cd
8× 3 5 7 5× 7
For each of these possibilities, we will now prove that either there is a
unique cut Frobenius groupof that form or that no cut Frobenius groupof
that form exists. We will first handle the cases where the Frobenius kernel
F is an elementary abelian p-group, as we have representation theory at
our disposal in these situations. When considering F as an F K-module,
p
it will be written additively. For the convenience of the reader we provide
the GAP SmallGroup ID of the group whenever this is appropriate.
(i) F = Cd, K = C : We provide details in this case. We will prove that
7 6
there is a unique cut Frobenius group with kernel F and complement K
for each d. Note that F is a splitting field for C and there are 2 non-
7 6
isomorphicfaithful irreducible F C -modules W andW . Ageneratort of
7 6 1 2
C actsasmultiplicationby¯3and¯5onW F andW F ,respectively.
6 1 7 2 7
ClearlyW ⋊C andW ⋊C arecutFrobe≃niusgroupsa≃ndsoareWd⋊C
1 6 2 6 1 6
and Wd⋊C for all d 1 by Lemma 4.4. We claim that if
2 6 ≥
(V ... V )⋊C , (1)
1 d 6
⊕ ⊕
8 ANDREASBA¨CHLE
where each V is either isomorphic to W or W , is a cut Frobenius group,
j 1 2
then all V ’s have to be isomorphic. For if not, then for the element x =
j
(¯1,¯1,...,¯1) V1 V2 .. Vd,thesubgroup x containsonlytwoG-conjugates
∈ ⊕ ⊕ ⊕ h i
of x, i.e. B (x) C , and x is not inverse semi-rational in G. Note that
G 2
≃
the isomorphism type of G does not depend on whether all V ’s in (1) are
j
isomorphic to W or to W as t t5 induces an automorphism of C ,
1 2 6
7→
which in turn induces an isomorphism Wd⋊C Wd⋊C . Thus, up to
isomorphism,we get a unique cut Frobeni1us gro6u→pof t2he for6m Cd⋊C for
7 6
each d Z as stated in (1e) of Theorem 1.3.
≥1
∈
(ii) F =Cc, K =C : With the sameargumentas in the previouscase,we
5 4
see that there is a unique cut Frobenius group as given in (1d).
(iii) F =Cb, K =C : There is a unique faithful irreducible F C -module,
3 2 3 2
where C acts by inversion on F . Giving us case (1a) of Theorem 1.3.
2 3
(iv) F = Cb, K =C : There is a unique faithful irreducible F C -module
3 4 3 4
W which is 2-dimensional as F -vector space and for a generator t of C ,
3 4
t2 acts as (cid:0)¯2. ¯2.(cid:1) on W. Hence W ⋊C4 is a cut Frobenius group and, by
Lemma 4.4, we get that Ws⋊C is a cut Frobenius group for each s 1,
4
≥
which is case (1b).
(v) F = Cb, K = Q : Again there is a unique faithful irreducible F Q -
3 8 3 8
moduleW whichis2-dimensionalasF -vectorspaceandtheuniquecentral
3
involution of K acts as (cid:0)¯2 .(cid:1) on F. We can apply the same argument as
. ¯2
above and see that there is a cut Frobenius group of the form Cb ⋊Q if
3 8
and only if b is even. Thus we are in case (1c)
(vi) F =Cc,K =Q : Alsointhiscasethereisauniquefaithfulirreducible
5 8
F Q -module W which is 2-dimensional as F -vectorspace. Now it can be
5 8 5
checked that W ⋊Q is even a rational Frobenius group (also known as
8
Markel’sgroup,GAPSmallGroupID[200, 44]). Notethatonlythecenter
ofQ acts by diagonalmatrices onW. Hence by Lemma 4.4, Ws⋊Q can
8 8
never be a cut Frobenius group if s 2. Thus there is a cut Frobenius
≥
group of the form Cc⋊Q if and only if c=2; case (1A).
5 8
(vii) F =Cc, K =C ⋊C : The groupK hasa unique faithfulirreducible
5 3 4
character,whichisofdegree2andrationalvalued. Hencethereisaunique
faithful F K-module W which is 2-dimensional as F -vectorspace. It can
5 5
be checked that G= W ⋊K (GAP SmallGroup ID [300, 23]) is in fact a
cut Frobenius group. However Z(K) C and hence Ws ⋊K cannot be
2
cut for s 2, by Lemma 4.4, and we j≃ust obtain the group in (1B).
≥
(viii) F = Cd, K = C ⋊C : As in the previous case there is a unique
7 3 4
faithfulF K-moduleW whichis2-dimensionalasF -vectorspace. However
7 7
in this case the group W ⋊K (GAP SmallGroup ID [588, 33]) is not cut
as for three of its irreducible characters the field of character values is the
maximal real subfield of Q(ζ ).
7
(ix) F = Cc, K = SL(2,3): Looking at the ordinary character table, one
5
sees that there are 3 faithful characters of SL(2,3), however only the rep-
resentation corresponding to the rational valued character can be used to
define a Frobenius group. Denote the correspondingF SL(2,3)-module by
5
W. Then W is absolutely irreducible and dimF W = 2. It is easily seen
5
that W ⋊SL(2,3) is a cut group (e.g. because SL(2,3) acts transitively on
W 0 ). AsZ(SL(2,3))iscyclicoforder2,therecannotbeacutFrobenius
\{ }
group of the form Cc ⋊SL(2,3) for c > 2, by Lemma 4.4. Hence there is
5
a unique cut Frobenius group of the form Cc ⋊SL(2,3) and c = 2 in this
5
case (with GAP SmallGroup ID [600, 150]). Case (1C).
INTEGRAL GROUP RINGS WITH TRIVIAL CENTRAL UNITS 9
(x) F = Cd, K = SL(2,3): With the same reasoning as in the previous
7
case,there is a cut Frobenius groupofthe formCd⋊SL(2,3)if andonly if
7
d=2. This group has the GAP SmallGroup ID [1176, 215], case (1D).
(xi) F = Cc, K = Q C : The group K has two faithful irreducible
5 8 × 3
ordinarycharacters,theyarecomplexconjugatesofeachotherandofdegree
2. NotethattheirfieldofvaluesisQ(ζ ),acorrespondingrepresentationin
3
characteristic 5 can be realized over F and a corresponding F K-module
25 5
W has dimension 4. The group W ⋊K has 20 irreducible characters with
field of character values Q(√5). Thus there is no cut Frobenius group of
the form Cc⋊(Q C ).
5 8× 3
(xii) F =Cd, K =Q C : From the previous case we see that there are
7 8× 3
twofaithfulirreducibleF K-modulesW andW whichcanberealizedover
7 1 2
F . Similar as in the case (i), one can argue that in a cut Frobenius group
7
only one of the isomorphism types can be involved. Note that Z(K)
≃
C and hence these elements act in a faithful F -representation as scalar
6 7
multiplesoftheidentity. InparticularW ⋊K isacutFrobeniusgroupand
j
then, by Lemma 4.4, each Ws ⋊K is a cut Frobenius group (j 1,2 ).
j ∈ { }
Again as in the case (i) one can see that both, Ws⋊K and Ws⋊K, lead
1 2
toisomorphicgroups. Hence thereis a(unique)cut Frobeniusgroupofthe
form Cd⋊(Q C ) if and only if d is even, which is case (1f).
7 8× 3
It remains to handle the caseswhere the Frobenius kernelis divisible by
twodifferentprimes. Inwhatfollows,wewillalwaysassumethatb,c,dare
positive. Assume first that G = (Cc Cd)⋊(C ⋊C ) is a cut Frobenius
groupwith kernel F =Cc Cd. Th5e×n G/7O (G)3 Cd4⋊(C ⋊C ) is a cut
5× 7 5 ≃ 7 3 4
Frobeniusgroup,contradictingcase(viii)above. Similarly,the existenceof
(Cc Cd)⋊(Q C ) would contradict (xi).
5× 7 8× 3
Now assume that G is a Frobenius group with kernel Cb Cc and the
3 × 5
complementK iseitherC orQ .
4 8
Let x G be an element of or-
∈ Figure 1. Subgroups
der 15. Then Aut( x ) = α
h i h i× ofAut(C ) C C
β C2 C4. As the Frobe- 15 ≃ 2× 4
nhiuis≃kernel×is abelian, C (x) = hα,βi≃C2×C4
G
F and we have that B (x) =
G
N ( x )/C ( x )isisomorphicto
G G
h i h i
a subgroup of K. But it is also C4≃hβi hα,β2i≃ hαβi≃C4
a subgroup of Aut( x ). As x C2×C2
h i
is inverse semi-rational, we have
B (x) 4, hence B (x) is
G G
| | ≥
cyclicoforder4andB (x)= y¯ ,
G h i hαi hβ2i hαβ2i
whereyactsonxasβorasαβ,cf.
Figure1. Inthefirstcasex5isleft
invariantbyy, hence y C (x5),
G
∈
contradicting the fact that G is
1
Frobenius. In the second case x5
is stabilized by y2 and y2 C (x5), again a contradiction with G being a
G
∈
Frobenius group.
The last remaining case is if G is a cut Frobenius group with kernel
F = Cc Cd and complement K = SL(2,3). If such a group would exist,
5 × 7
consider an element x F of order 35. Then B (x) Aut(C ) has to
G 35
∈ ≤
have order 12 or 24 = φ(35). But SL(2,3) does not contain a subgroup of
order 12 and Aut(C ) is clearly not isomorphic to SL(2,3).
35
10 ANDREASBA¨CHLE
(2) If K is odd, then K C by Proposition 4.2, and the Frobenius kernel
3
| | ≃
F of G is a 2,5,7 -groupby Theorem 1.2. If F would be divisible by 5,
{ } | |
then the nilpotent Frobenius kernel F contains an element z of order 5 in
its center. Then z has to be rationalby Lemma 2.4. But this is impossible
as B (z) C . We infer that F is a 2,7 -group. First assume that F is
G 3
≤ { } | |
divisible by both primes, 2 and 7. Then F has an element z of order 14 in
its center and F C (x). As G is a cut group, B (x) C . But then the
G G 3
≤ ≃
unique involution in x is central in G, a contradiction. Consequently, F
h i
is a p-group for p = 2 or p = 7. By a classical result of Burnside, see e.g.
[Hup67, V, 8.8], the Frobenius kernel F is nilpotent of class at most 2 and
thus metabelian. Let x Z(F). Then C (x) = F and B (x) K C .
G G 3
∈ ≤ ≃
Hence φ(o(x))/2 3 and thus o(x) 4 (in case p = 2) or o(x) 7 (in case
p = 7). Now F/F| ′ ⋊K is an epim|orphic image of G and the|refor a cut
group with F/F′ abelian. Thus for each y F/F′ we have φ(o(y))/2 3
∈ |
andF/F′isofexponent4(ifp=2)orofexponent7(ifp=7),respectively.
AsF ismetabelian,F′ Z(F),andtheclaimonthestructureofthegroup
≤
follows.
Assume now that F is a 2-group and x F is a 2-element, then the
∈
order of Aut( x ) is a power of 2 and hence B (x) =B (x) and F has to
F G
be cut. The ohrdierof F has to be a even power of 2 as 3 (F 1). Onthe
other hand, every cut 2-group admitting a fixed point f|re|e a|u−tomorphism
of order 3 can clearly be used to define a cut Frobenius group.
If F is a 7-group, then clearly expF is a divisor of 72, by the above, we
want to show that expF = 7. Deny and assume that expF = 49, then
there is an element y F of order 49. Then necessarily B (y) C , as
G 21
G is cut. Hence there i∈s an element x F of order k 7,49 no≃rmalizing
∈ ∈{ }
y , but not centralizing it. But then H = x,t C ⋊C , where t is a
k 3
h i h i ≃
generatorofthe Frobeniuscomplement,andB (y)is anepimorphicimage
G
of H. But this is a contradiction since H does not have an epimorphic
image that is cyclic of order 21. Hence F has exponent 7. Each 7-element
x has to be inverse semi-rational in G, thus it has to have 3 conjugates
in the cyclic subgroup generated by it, so the generator t of a Frobenius
complement has to normalize x . Conversely, assume that F is a 7-group
h i
ofexponent7suchthatitadmitsafixedpointfreeautomorphismαoforder
3 mapping each cyclic subgroup to itself. Then, clearly, F ⋊ α defines a
cut Frobenius group. The result is proved. h i (cid:3)
Remark 4.5. As by [CD10, Remark 13], groups of odd order are semi-rational if
andonlyiftheyarecut(=inversesemi-rational),(2b)oftheTheorem1.3alsogives
a more precise description of the Frobenius groups appearing in [CD10, Theorem
3], where the semi-rational groups of odd order are classified.
The arguments from the last part of the proof of (1) can also be used to reduce
the list of potential semi-rational Frobenius groups in [ADD16, Theorem 1.2], e.g.
groupswith complementC orQ ,andFrobenius kernelwithanorderdivisible by
4 8
two different primes can be removed from the list as well as groups with comple-
ment Q and kernel an elementary abelian 17-group. Some of the arguments also
16
apply to show that the isomorphism type of semi-rational Frobenius groups of a
certaintypeareunique. Ingeneral,asemi-rationalFrobeniusgroupisnotuniquely
determined by Frobenius kernel and complement. The two non-isomorphic Frobe-
nius groups (C C )⋊C are both semi-rational (however, only one of them is
5 5 4
cut). ×
We now describe the cut Frobenius groups with abelian kernel.
