On double overs of the generalized alternating group
Z
d oA
m
as Galois groups over algebrai number elds
by
MartinEpkenhans (Paderborn)
Let Z
d oA
m
be the generalized alternating group. We prove that all
double overs ofZ
d oA
m
o ur asGaloisgroupsoveranyalgebrai number
eld. We further realize some of these double overs as the Galois groups
of regular extensions of Q(T). If d is odd and m > 7, then every entral
extension of Z
d oA
m
o urs as the Galois group of a regular extension of
Q(T). We further improve some of our earlier results on erning double
overs ofthe generalizedsymmetri groupZ
d oS
m .
1. Introdu tion and notations. Serre'sformula on tra e forms[15℄,
[16℄ relates theobstru tionto ertainembeddingproblems
1!Z
2
! e
G!G!1
ofa nitegroupG toinvariantsofthetra eformofa eldextension. Using
this,N.Vila[19℄ realizedtheunique overinggroup e
A
m ofA
m
;m8;m
0;1mod8astheGaloisgroupofaregularextensionoftherationalfun tion
eld Q(T). J. F. Mestre [11℄ extended this result to all m 4. Following
Mestre'sideas,J.Sonn[18℄improvedoneofhispreviousresultson overing
groups of the symmetri group S
m
. We an summarize these results as
follows.
Every nite entral extension of S
m
and of A
m
, m4, is realizable as
the Galois group of a regular extension of Q(T).
Vila, Sonn and S ha her [19℄, [20℄, [17℄, [13℄ used trinomials f(X) =
X m
+aX l
+bwithGaloisgroupS
m
,resp. A
m
. We knowthetra eformof
atrinomial[15℄,[3℄. The tra eformofatrinomialwithsquare dis riminant
depends onlyon l. It is not always possible to hoose l <n su h that the
obstru tion vanishes. This explainswhyVila'sresultsare not omplete.
1991 Mathemati sSubje tClassi ation: Primary12F12.
Mestre gave a one-parameter deformation of a polynomial of odd de-
gree to an irredu ible polynomialwith the same tra e form. Volklein [21℄
obtainedMestre'sresult on e
A
m
withouttra e form onsiderations.
In a previous paper [4℄ we realized some of the double overs of the
generalizedsymmetri group Z
d oS
m
asGaloisgroupsoverK(T),whereK
isan algebrai number eldwhi h ontainsthe dthrootsof unity.
In thispaperweinvestigate double oversof thegeneralizedalternating
groupasaGaloisgroupovernumber eldsandoverrationalfun tion elds.
UsingIshanov'stheorem we prove that all double overs of Z
d oA
m
and of
Z
d oS
m
o ur as the Galois groups over any algebrai number eld. If d
is odd, then the unique non-trivial double over of Z
d oA
m
o urs as the
Galoisgroup ofa regularextension ofQ(T).
Kotlar,S ha herandSonn[8, Theorem6℄redu edthequestionwhether
a entral extension of S
m
is a Galois group over K to ertain pull-ba ks
of stem overs of S
m
with y li groups. Following their arguments we
showthat all entralextensions ofZ
d oA
m
arethe Galoisgroupsofregular
extensions ofQ(T) ifd is odd and m>7.
Let us x some notation. Let G be a nite group. Then G 0
denotes
the ommutator subgroup ofG and M(G) is the S hurmultiplierof G. Let
1 :H
1
!G;
2 :H
2
!G behomomorphisms of groups. Then H
1
G H
2
isthe asso iated pull-ba k.
Let K be a eld. Then K denotesan algebrai losure of K.
d
K is
the group of dth roots of unity. Let f(X) 2K[X℄ be a polynomial. Then
dis(f) is the dis riminantof f(X), and Gal(f) stands forits Galois group.
LetT;U;V;X;Y denote indeterminates.
2. The embedding problem. Let K be a eld of har(K)6=2,K
s a
separable losureofK,and
K
:=G(K
s
=K)theabsoluteGaloisgroupofK.
Let L=K be a separable eld extension of nite degree n, N L a normal
losure of L=K insideK
s
,and G =G(N=K) theGaloisgroup of N=K. By
Galoistheory we have homomorphisms%:
K
!G and e:
K
!S
n . Let
0!A
!E!G !0
be a group extension of G with abelian kernel A. We say the embedding
problem withabeliankernelde ned bythe diagram
K
1
A E
G 1
%
// // // //
hasa(proper)solutioni thereisa(surje tive)homomorphism':
K
!E
we allita entral embedding problem. An abelianembeddingproblemover
an algebrai number eldhas apropersolutionifithas asolution(Ikeda's
Theorem [6℄). If the order jAj of A is a prime and if E is a non-trivial
extension of G, then every solution of the embeddingproblem is a proper
solution.
Let H m
(G;A), m 2 Z, denote the mth ohomology group of the G-
module A. The group extension1 !A! E !G ! 1 withabeliankernel
A de nes an element "2 H 2
(G;A). Let Br(K) be theBrauer group of K
andletinf:H 2
(G;A)!Br(K)bethein ationmapindu edby%:
K
!G.
Hoe hsmann'sTheorem. Theembedding problem asso iated with"2
H 2
(G;A) has a solution ifand only if inf(")=02Br(K).
With the help of Serre's formula we are able to al ulate the obstru -
tion inf(") for some embedding problems. By Kummer theory we know
H 1
(
K
;Z
2
) ' Hom (
K
;Z
2 ) ' K
?
=K
?2
. For a;b 2 K
?
, (a;b)
K
denotes
the generalized quaternion algebra generated over K by i;j and satisfying
i 2
=a, j 2
= b,ij = ji. The lass of (a;b)
K
inBr(K) is also denoted by
(a;b)
K
. Let be a (non-degenerate) quadrati form over K. The Hasse
invariant (se ondStiefel{Whitney lass) isde nedby
w
2 :=
O
1i<jn (a
i
;a
j )
K
2Br(K);
where '
K ha
1
;:::;a
n
iisadiagonalizationof . Here'
K
denotestheiso-
metry ofquadrati formsde nedoverK. The determinant of is denoted
bydet
K .
Now were all ade nitionof two overinggroupsof S
n . S
n
hasa stan-
dardpresentationwithgenerators t
1
;:::;t
n 1 (t
i
=(i;i+1))and relations
t 2
i
=1; (t
i t
i+1 )
3
=1; t
i t
j
=t
j t
i
if ji jj2:
LetS
n
be thegroupgenerated by !;
e
t
1
;:::; e
t
n 1
withrelations
e
t 2
i
=1=! 2
; ! e
t
i
= e
t
i
!; ( e
t
i e
t
i+1 )
3
=1;
e
t
i e
t
j
=! e
t
j e
t
i
if ji jj2:
LetS +
n
be thegroupgenerated by!;
e
t
1
;:::; e
t
n 1
withrelations
e
t 2
i
=!; ! 2
=1; ! e
t
i
= e
t
i
!; ( e
t
i e
t
i+1 )
3
=1;
e
t
i e
t
j
=! e
t
j e
t
i
if ji jj2:
Denotebys
n
;s +
n 2H
2
(S
n
;Z
2
)the ohomology lassesasso iatedwiththese
groupextensions. Thesignaturehomomorphism"
n :S
n
!Z
2
istheunique
non-zeroelementofH 1
(S
n
;Z
2
) ifn2. WeknowH 2
(S
n
;Z
2 )'Z
2
Z
2
=
f0;s +
n
;s
n
;"
n ["
n
g if n 4. Here [ denotes the usual up produ t of
ohomology lasses.
The tra e map tr
L=K
:L! K de nes a quadrati form over K on the
K-ve tor spa e L by x 7! tr
L=K (x
2
). We denote the asso iated quadrati
e :
K
! S
n
de nes a homomorphisme
?
:H 2
(S
n
;Z
2
) ! Br
2
(K), where
Br
2
(K) is the subgroup of elements x 2 Br(K) with 2x =0. Now Serre's
formulaasserts:
Proposition 1(Serre [15℄). 1: e
?
(s +
n
)=( 2;det
K hLi)
K w
2 hLi:
2: e
?
(s
n
)=(2;det
K hLi)
K w
2 hLi:
3: e
?
("
n ["
n
)=(det
K
hLi; 1)
K :
Let
inf:H 2
(G;Z
2 )!H
2
(
K
;Z
2 )
bethein ation homomorphismindu edby% and let
res:H 2
(S
n
;Z
2 )!H
2
(G;Z
2 )
be therestri tion homomorphismindu edbythe inje tionG ,! S
n
. Then
e
?
=infÆres . CombiningSerre'sformula withHoe hsmann'sresult we get
Proposition 2. The embedding problem asso iated with the group ex-
tension res(s +
n
) (resp. res(s
n
)) has a solution i
w
2
hLi=( 2;det
K hLi)
K
(resp:w
2
hLi=(2;det
K hLi)
K ):
3. The wreath produ t. The generalized alternating group Z
d oA
m
is the wreath produ t of Z
d
and A
m
. We now re all the de nition of the
wreath produ tof groups.
Definition 1.Let G be a permutation group on a nite set . Let H
be a nite group and set H
= ff : ! H g. Then f 7!
f = f Æ 1
,
2G, de nes an a tion of G on H
. Now the wreath produ t HoG of H
and G isthe semidire tprodu tof H
and G indu edbythe a tionabove.
In thesequel we needthe ommutator subgroupofa wreath produ t.
Lemma 1.Let G bea permutation group of degree m, and H ;H
1
;H
2 be
groups.
1: If G a ts transitively,then
(HoG) 0
=f(h
1
;:::;h
m
;) jh
1 :::h
m 2H
0
; 2G 0
g
and
(HoG)=(HoG) 0
'H =H 0
G=G 0
:
3: If G 0
a ts doubly transitively, then
(HoG) 00
=f(h
1
;:::;h
m
;) jh
1 :::h
m 2H
00
; 2G 00
g:
3: (H
1 oG)
G (H
2
oG)'(H
1
H
2 )oG:
Proof. 1. Let
1
: H ! H =H 0
and
2
: G ! G=G 0
be the anoni al
1 1
f(h
1
;:::;h
m
;)jh
1 :::h
m 2H
0
; 2G 0
g. Sin e H =H 0
is abelian,
HoG!H =H 0
G=G 0
:(h
1
;:::;h
m
;)7!(
1 (h
1 :::h
m );
2 ())
is a homomorphism with kernel K . Hen e (H oG) 0
K . De ne f
i;a :
! H , a 2 H , by f
i;a
(i) = a, f
i;a
(k) = 1 if k 6= i. Let i 6= j. Sin e
G a ts transitively, there is a permutation 2 G with (i) = j. Then
[(f
i;a
;id);(1;)℄ = (f
i;a
f
j;a
1;id). Hen e f(h
1
;:::;h
m
;id) j h
1 :::h
m
=
1g (HoG) 0
: If h
m 2 H
0
, then (1;:::;1;h
m
;id) 2 (HoG) 0
. We get the
assertionfrom (h;id)(1;)=(h;).
2. If G 0
a ts doublytransitively on , then we an hoose 2G 0
with
(i)=i, (j) =k,wherei6=j;k. Then
[(f
i;a
f
j;a
1;id);(1;)℄=(f
k;a
f
j;a
1;id)2(HoG) 00
:
3. De ne
':(H
1 oG)
G (H
2
oG)!(H
1
H
2 )oG
by ((h;);(g;)) 7! ((h
1
;h
2
);), where (h
1
;h
2
) : ! H
1
H
2
: j 7!
(h
1 (j);h
2
(j)): Then 'is anisomorphism.
In thefollowingtwo lemmaswestudyin ation maps.
Lemma2.LetGbeapermutationgroupofdegreem,andletHbea nite
group. Let A be a nite abelian group, onsidered as a trivial G-module.
Then the in ation map
inf:H 2
(G;A)!H 2
(HoG;A)
indu ed by the anoni al proje tion %:HoG!G isinje tive.
Proof. An element "2H 2
(G;A) orresponds to a entralextension of
G with kernel A. The image of "under thein ation map orresponds to a
pull-ba k,i.e. thereis a ommutative diagram
1
A
E
G
(HoG) HoG
1
1
A E
G 1
// //
// //
%
'
xx
q q
q q
q q
q q
q q
// // // //
Weknowinf(")=0ifandonlyiftheuppersequen esplits. Bytheuniversal
propertyof the pull-ba kthisis equivalent to the existen e of a homomor-
phism ' : HoG ! E making the above diagram ommutative. We know
: G ! HoG : 7! (0;) is a monomorphism. Now '((0;)) = e gives
=%((0;))=id. Hen eG'' Æ (G)isasubgroupofE. Letx2A \ ' Æ (G).
Then x = '((0;)) and (x) =id= Æ'((0;)) = gives x = e. Hen e
E 'AG.
The next lemmaredu es ourapproa hto double overs of Z
d
oG,where
f
Lemma 3.Let G bea permutation group of degree m. Let :H
1
!H
2
bean epimorphism of nite groups H
1
;H
2
. Let A be an abelian group with
order relatively prime tothe order of ker (). Then the in ation map
inf:H 2
(H
2
oG;A)!H 2
(H
1 oG;A)
indu ed by %:H
1
oG !H
2
oG :(h;)7!(Æh;) is an isomorphism.
Proof. The sequen e
1!ker () m
!H
1
oG !H
2
oG !1
is exa t. Sin e theorder of A isrelativelyprime to theorder of ker (), we
getH 1
(ker () m
;A)=H 2
(ker() m
;A)=0 (see [1,II.10.2℄). Hen e
0!H 2
(H
2 oG;A)
inf
!H 2
(H
1 oG;A)
res
!H 2
(ker () m
;A)=0
isan exa t sequen e (see [14, VII,x 7, Proposition5℄).
4. The restri tion map res : H 2
(S
md
;Z
2 ) ! H
2
(Z
d oA
m
;Z
2 ). The
image of the restri tion map determines the double overs whi h an be
shown to be Galoisgroupsbytheuseof Serre'sformula.
We know
H 2
(G;A)'((G=G 0
)A)(M(G)A);
with an abeliangroup A (see [7, 2.1.20℄). In [7, Theorem 6.3.13℄ we found
a listof therelevant S hurmultipliers. Together withLemma1 we get
H 2
(Z
d oA
m
;Z
2 )'
Z
2
ifd 1mod2,
Z
2
Z
2
Z
2
ifd 0mod2;
and m4. We furtherknow
Z
d oA
m
=hs
1
;:::;s
m 2
;w
1
;:::;w
m js
3
1
=s 2
j
=(s
j 1 s
j )
3
=1;
1<jm 2; (s
i s
j )
2
=1; 1i<j 1; jm 2; w d
j
=1;
w
i w
j
=w
j w
i
; s
i w
j
=w
j s
i
; j 6=1;2;i+1;i+2;
s
i w
i+1
=w
i+2 s
i
; i=2;:::;m 2;
s
1 w
3
=w
1 s
1
; s
i w
1
=w
2 s
i
; i=1;:::;m 2i:
Let 1 ! f1;!g ! E ! Z
d oA
m
! 1 be an exa t sequen e. Then eg 2 E
denotesa preimage ofg 2Z
d oA
m
inE. We an hoose a set ofgenerators
s
1
;:::;s
m 2
;w
1
;:::;w
m ofZ
d oA
m
su hthat
E =h!;es
1
;:::;es
m 2
;we
1
;:::;we
m j!
2
=1; !se
i
=es
i
!; !we
j
=we
j
!;
e s 3
1
=1; es 2
j
=
3
; ( es
j 1 e s
j )
3
=1; j =2;:::;m 2; ( es
i e s
j )
2
=
3
;
1i<j 1; jm 2; we d
j
=
2
; we
i e w
j
=
4 e w
j e w
i
; i6=j;
e s
i e w
j
=we
j e s
i
; j6=1;2;i+1;i+2; se
i e w
i+1
=we
i+2 e s
i
; i6=1;
e
s we =we es ; eswe =we esi;
