Publicacions
Ma emá iques,
Vol
36
(1992),
197-204
.
A
bs ac
UNITARY
SUBGROUP
OF
INTEGRAL
GROUP
RINGS
A
.A
.
BOVDI
AND
S.K
.
SEFIGAL
Le
A
be a
ini e
abelian
g oup
and
G=
A
>G
(b),
b
2
=
1,
ab
=
a
-1
,
da
E
A
.
We
ind
gene a o s
up
o
ini e
index
o
he
uni a y
subg oup
o
7LG
.
In ac ,
he
gene a o s
a e he
bicyclic
uni s
.
Fo
an
a bi a y
g oup
G,
le
B2(7LG)
deno e
he
g oup
gene a ed
by he
bicyclic
uni s
.
We
classi y
g oups
G
such
ha
B
2
(7LG)
is
uni a y
.
Le
7LG
be
he
in eg al
g oup
ing
o
an
a bi a y
g oup
G
and
le
:
G
->
U(7L)
=
{±
1}
be an
o ien a ion
homomo phism
.
Fo
each
x
=

1
:
a
g
g,
we
pu
x
=

a
g
(g)g -1
.
In pa icula ,
i
is
i ial,
gEG
x
coincides
wi h
he
s anda d
x*
.
Le
U(ZG)
be
he
g oup
o
uni s
o
7LG
.
Then
u
E
U(ZG)
is
called
-uni a y
i
u
-l
=
u
o
u
-1
=
-u
.
All
-uni a y
elemen s
o
U(ZG)
o m
a
subg oup
U
(7LG) con aining
G
x
U(7L)
.
We
e e
o
U
(ZZG)
as
he
-uni a y
subg oup
o
U(ZZG)
.
In e es
in
he
g oup
U
(7LG)
a ose
in
algeb aic
opology
and
uni a y
K- heo y
[4]
.
We
a e
in e es ed
in
he cons uc i e
desc ip ion
o
U
(ZLG)
.
I
G
is
ini e
cyclic,
hen
Bo di
[1]
ga e
a
linea ly
independen
se
o
gene a o s
o
a
o sion
ee
subg oup
o
ini e
index
in
U
(7ZG)
.
This
was
ex ended
o
ini e
abelian
g oups
by Hoechsmann-Sehgal
in
[3]
.
We
gi e
gene a o s
up
o
ini e
index
o
U
(7ZG)
i
G
is
a
ini e
dihed al
g oup
.
In
ac ,
he
gene a o s
consis o
he
bicyclic
uni s
.
The
subg oup
B2(7ZG)
o
U(ZG)
gene a ed
by
all
he
bicyclic
uni s o
7LG
plays
an
impo an
ole
in
he
s udy
o
U(ZG)
(see
[5], [6])
.
In
Theo em
2,
we
cha ac e ize
g oups
G
o
which
B
2
(7LG)
is
uni a y
.
This wo k
was suppo ed by
NSERC
G an
A-5300
.
198

A
.A
.
BOVDI
AND
S
.K
.
SEHGAL
We
need
he
2
.
U
(ZG)
o
dihed al
g oups
Fi s ,
we
ecall
some
de ini ions
.
Fo
an
elemen
aE
G
o
ini e
o de
n
w i e
á
=
1
+
a
+
- - -
+
an
-1
.
Deno e by (G)
he
se
o
all
o sion
elemen s
o
G
.
I
a,
b
E
G,

o(a)
<
oo,
hen
ua,b
=
1
+
(1
-
a)bá
has
an
in e se
u~
,
b
=
1
-
(1
-
a)bá
.
Mo eo e ,
u
a
,b
=
1 i
and
only
i
b
no malizes
(a)
.
The
elemen s
u
a
b,
a,
b
E
G
a e called
bicyclic
uni s
o
7LG and
he
g oup
gene a ed
by
hem
is
deno ed
by
B2(7LG)
.
We
ecall
[5]
ha
by Bl
(ZG)
is
unde s ood
he
g oup
gene a ed
by
he
Bass
cyclic
uni s
o
7LG
.
I
is
known
[5]
ha
i
G
is
a
ini e
dihed al
g oup
and
Z
is
he cen e
o
U(ZG)
hen
(Z,B2(7ZG))
(equi alen ly,
(Bl(7ZG),B2(7G)))
is
o
ini e
index
in
U(7LG)
.
We
p o e
Theo em
1
.
Le
G
be
he
dihed al
g oup
D2
a
=(a
-
=1=
b
2
l
a
b=
a_
1
)
.
Suppose
is
an
o ien a ión
homomo phism
o
G
wi h
ke nel
(a)
.
Then
he
index
(U
(7ZG)
:
B2
(ZG»
is
ini e
.
P oposi ion
.
Le
G
be
a g oup
con aining
a subg oup
A
o
index
2
and
an
elemen
b
such
ha
G
=
(A,
b)
and
b-lab
=
a
-1
o
all
a E
A
.
Suppose
ha
A
2
5
E
1
.
I
is
an
o ien a ion
homomo phism
o
G
wi h
ke nel
A,
hen
1)
he
cen e o
U
(7LG)
coincides
wi h 2(A) x
(-1),
whe e
2(A)={aE (A)
:
a2=1}
;
2)
he cen e
o
U(ZG)
is
he
di ec
p oduc
o
2(A)
x
(-1)
and
a
o sion
ee
abelian
g oup
T
such
ha
U(ZA)
=
(-1)
x
A
x
T
and
x=x*
o allxET
.
P oo
..
Le
x
=
xl
+
x2b,

xi
E
7ZA be a
cen al
uni in
7LG
.
Since
G
is
a
subg oup
o
U(7ZG),
x
=
b
-l
xb
=
x*
+
x*b

and

x
=
a
-1
xa
=
xl
+
a-2x2b
o
all
x
E
A
.
Then
xi
=
xi
and
(1)

x2(1
-
a
2 )
=
0
UNITARY
SUBGROUPS

19
9
o
all
a
E
A
.
We
wish
o
p o e
ha
x2
=
0
.
_Le
us
suppose
ha
x2
=,~
0
.
ROM
(1)
we
ob ain
ha
A
2is
ini e
.
Le
A
2
deno e
he
sum
o
all
elemen s
o
A
2
.
I
H
is
a no mal
subg oup
o
G,
hen
deno e
by
0(G,
H)
he
ideal o
7LG
gene a ed
by
elemen s
o
he
o m h
-
1
wi h
h
E
H
.
Clea ly,
7ZG/A(G,
H)
-
7L(G/H)
.
I
X(y)
is
he
sum
o
he
coe icien s
o
y,
hen
he
elemen
x
+
0(G,
A)
=
X(x1)
+
X(x2)b+ 0(G,
A)
is
i ial,
because
IG/AJ
=
2
[7,
p
.
46]
.
This
implies
ha
one
o
he
numbe s
X(xl_)
o
X(X2)
equals
±1
and
he
o he
is
ze o
.
F om
(1)
we
ob ain
x2
=
ZA
2
,
z
E
7LA,
X(x2)
=
X(z)1A
2
1,
and
his
is
possible
only
in
he
case
when
X(x2)
=
0
.
Suppose
A=A
2
.
Then
x2
=
7
Z
:
a
o
some
-y
E
7
.
F om
he
equali y
aEA
X(x2)
=
-
yjAj
=
0
we
ob ain
,y
=
0
and
x2
=
0,
which
leads
o
a
con adic ion
.

Thus
A
=,A
A
2
.
W i e
x2
=
(
aici)
A
2
wi h
al
E
7
i
whe e
ci's
a e
a
ans e sal
o
A
2
in
A
.
Then
x1
+x2b+
A(G,A
2
)
=
XI+
(~a
i
c
i
)A
2
b+0(G,A
2 )
=
x1
+
(¡A
2
1

aici)b+
0(G,
A
2
)
i
is
a
uni
in
7L(G/A
2
)
.
Since
G/A
2 is
an
abelian
g oup
o
exponen
wo,
by
Higman's heo em
[7,
p
.
57],
all
uni s
o
7L(G/A
2
)
a e
i ial
.
Ob iously,
E
al
=
0 and
i
al
7~
0
o
some
i,
hen
n
i
¡
A2
1
=~
±1
.
Thus,
al
=
0
o
all
i
and
he
equali y
x2
=
0
is
con adic o y
.
Hence,
.x
=
xI
E
U(ZLA)
and
x*
=
x
=
xi
=
xi
.
Clea ly,
i
x
E
U(7LA)
and
x*
=
x,
hen
x
is
a
cen al
uni
o
7G
.
I
is
well
known
(see
[2])
ha
U(7 (A))
=
± (A)
x
T
and
U(7A)
_
A
x
T,
whe e
e e y
elemen
u
E
T
sa is ies
he
condi ion
u
=
u*
.
The e o e
he cen e
o
U
(7G)
is
he
di ec
p oduc
o
subg oups
± 2
(A)
and
T
.
This
is
2) o
he
P oposi ion
.
Suppose
ha
x
=
x1
+
x2b
is
a
cen al
uni in
U
(7G)
.
Since
G
is
a
subg oup
o
U
(7G),
x
is
cen al in
U(7LG)
.
I
ollows
ha
x
=
xi
and
xx
=
x1xi
=
xi
=
±1
.
The e o e,
by
Higman's
heo em xI
=
a
whe e
a
E
2(A)
.
This
comple es he
p oo
o
he
P oposi ion
.
20
0

A
.A
.
BOVDI
AND
S
.K
.
SCI
-
IGAL
P oo o
Theo em
1
:
Le
G
be
he
dihed al
o
o de
2n
gi en
by
G=
(a
n
=
1
=
b
2
,
J
=
a
-1
)
.
I
n
=
2,
hen
he
heo em
is
i ial
.
So
we
may
apply
he
las
P oposi ion
.
Le
Z
be
he
cen e
o
U(ZZG)
.
Thenwe know
ha
(U(ZZG)
:
(B2
(7G),
Z))
<
oo
.
We
ha e
seen
in
he
P oposi ion
abo e
ha
Z1,
he
cen e
o
U
(7ZG),
is
ini e
and
Z
1
<
Z
.
I
su ñces
o
p o e,
he e o e,
ha
B2(7LG)
is
uni a y
.
I
u,,
:,
7,
~
A
0,
hen
o(x)
=
2
and u
'
,
y
=
1
+
(1
-
x)y(1
+
x)
.
Now,
y
=
a
i
x',

E
=
0 o
1
.
Since
x(1
+
x)
=
1
+
x,
we
ha e,
in
any
case,
Then
u
,u
=
1+(1
+x)
(ax)
(1-x)
=
1+(1-x)
a
-
Z
(1+x)
.
The e o e,
u=,ay
U
,
,
j
=
1
+
(1
-
x) (a
i
+
a`)
(1
+
x)
=
1
as
(a
l
+
a-
Z)
is
cen al
.
This
comple es
he
p oo
o
he
heo em
.
Rema k
.
The
las
heo em
holds
o
nonabelian
g oups
G
=
(A,
b)
whe e
A
is
ini e
abelian
and b
2
=
1,
a
b
=
a
-,
o
all
a
E
A
.
I
A
is
an
elemen a y
2-g oup,
hen
so
is
G
and
whe e
is
no hing
o
p o e
.
Suppose
A2
:~
1
.
The
nonlinca
i educible
ep esen a ions
p o
G
a e
induced
om
hose
o
A
and
p(ZLG)
=
p(D)
o
some
dihed al
subg oup
D
o
A
.
The
esul
ollows
.
We
need
hc
:
ollowing
.
ux,y=1+(1-x)
a'
(1+
.x)
.
3
.
Uni a i y
o
he
subg oup
B
2
(7G)
Theo em
2
.
Le ,
G
=
(A,
b)
whe e
A
is
he
ke nel
o
he
non i ial
o ien a ion
homomo phism
:
G
-
U(7L)
.
The
subg oup
B2(7LG)
ás
non i ial
and
-uni a y
i
and
only
i
G
is
non-Hamil onian
'in
which
an
elemen
b
=,A
1
o
ini e
o de
can
be
chosen
such
ha
one
o
he
ollowin,g
condi ions
is
ul illed
:
1)
A
is
an
abelian
g oup, he
o de
o
he
elemen
b
di ides
4
and
bab-1
=
a-1
o
all
a
E
A
;
2)
A
is
a,
Hamil onian
2-g oup,
G
is
he
semidi ec
p oduc
o
A
and
(b
1
b 2
=
1),
an,d
e e y subg oup
o
A
is
no mal
in
G
;
3)
A
is
a,
Hamil onian
2-g oup
and
G
is
he
di ec
p oduc
o
a
Hamil-
onian
2-subg oup
o
A
and
a
cyclic
g oup
(b)
o
o de 4
;
4)
(A)
is
an
abelian
g oup,
e e y subg oup o (A)
is
no mal
in
G
and
bab
-1
=
a
-1
b
47
o
all
aE
A,
whe e
he
in ege
i
depends
on
a,
.
UNITARY
SUBGROUPS

20
1
Lemma
.
Suppose
ha
G
has
a subg oup
A
o
index
2
uwi h
G
=
(A,
b)
and
o(b)
<
oo
.
Suppose
u he
ha
A
7¿
NA((b))
and
1)
(A)
is
abelian
and
all
subg oups
o
(A)
a e
no mal
in
A
;
2)
bgb
-1
=
g
-1
o
all
gE
A
NA((b))
.
Then
bab
-1
=
a-1
o
all
aE
A
and
b
4
=
1
.
P oo
:
Le
c
E
N
A
((b))
.
Choose
aE
A
N
A
((b))
.
A
i s ,
suppose
c
has
ini e
o de
.
Then
by
(2)
we
ha e
a
-
'bcb
-1
=
b(ac)b
-
=
c-1a-1
I
aE
(A),
hen
by
(1)
we
ha e
bcb
-1
=
c
-1
.
I
a,
has
in ini e
o de ,
he e
exis s
an
in ege
n
such
ha
anc
--
ca",
sin(,(,
,
(c)
is
no mal
in
A
.
B,y
hypo hesis,
a 'c
~
N((b)) and
ln s
a
-
'Lbcb
-1
=
b(a
n
c)b
-1
=
(
.--l
a-'L
.
I
ollows
ha
bcb
-1
=
c-
as
desi ed
.
Now
i is
enough
o
p o e
ha
c
canno
ha e
in ini e
o de
.
Suppose
ha
o(c)
=
oo
a ld
o(a)
<
oo
.
Then
he e
is
an
n
such
ha
cna
=
ac"
.
Clea ly,
acn
~
N((b))
.
We
ha e
a
-1
bc"b
-1
=
b(ac
n
)b
-1
=
e-na-1
.
I
ollows
ha
bc' b
-1
=
-n
.
Tllis
is
impossible
because
c"
E
N((b))
.
Now
le
o(c)
=
oc,
o(a)
=
oo
.
The e
exis s
an
n
such
ha
bc"
=
c"b
and
a
-1
cn
=
ba
.c"b
-1
=
-n a
-
'
.
I
ollows
ha
[c"
;
a21
=
1
.
Clea ly,
a
2
cn
1
N((b))
and
wc
;
ge
which
implies
c
2
n
=
1,
a
con adic ion
.
Since
b
2
E
A,

bb
2 b
-1
=
b
-2
and
we
ha e
b
4
=
1,
comple ing
he
P oo
o
he
,
lenuna
.
P oo
o
Tlheo em
2
:
"Necessi Y
."
a
-2 c'
=
ba
2
c"b
-
=
a-2c-n
Suppose
ha
B2(ZLG)
is
non i ial
and
-uni a y
.
Le
us
i s
p o e
ha
e e y
ini e
subg oup
(a)
o
A
is
no mal
in
G
.
Le
n
be
he
o de
o
(a)
.
I
gNG((a)),
hen
u,,,
g
=
1
+
(1
-
a)gá
q¿ 1
.
Tilen
o l
he
equali y
u-
=
u
,9
we
ha e
-
dg
-1
(
.g)( 1
-
a-1)
=-
(
1
-
a)gá
.
Mul iplying
by
á
we
ob ain
n(
1-
a)g
-
a
=
0,
which
is
impossible
.
The e-
o e,
e e y
subg oup
o
,
(A)
is
no mal
in
G
.
Because
B
2
(7ZG)
~
1
;

G
A

20
2

A
.A
.
BOVDI
AND
S.K
.
SGI1GAL
con ains
an
elemen
e o
ini e
o de
wi h
(e)
no
no malized
by
A
.
Then
e
2
E
(A)
and c2
is
cen al
in
7LG
.
Clea ly,
U
C
,
9
=
1
+
,(1
-
c)g(1
+
e)c2
and
(e)
=
-1
.
Since
u,,s
is
-úni a y,

u
e
,
9
u,
,9
=
1
and
i
ollows
ha
( 1
)
(g+g-1 (g))(1+c)c2=c(g+g-1 (g))(1+c)c2
.
Choose
b
E
G
A
such
ha
b
is
a
2-elemen
o
leas
o de
and
le
g
E
A
.
In
(1)
aking
c
=
b,
g
=
bg
-1 b
1
+
2=
whene e
g1
NA((b))
.
We
ob ain
bgb-1
=
g-1b
2i'
o
all
g
E
A
NA((b))
and
(bg)
2
=
(g-1b2g)`'+1
.
Clea ly,
bg
is
a
2-elemen
in
G
A
and
i' is
e en,
o he wise
he
o de
o
bg
is
less
han
he
o de
o
b,
which
is
impossible
.
The e o e,
(2)

b
gb-1
=
g-1b
4i
o
all
g
,
e
A
NA((b))
.
a)
Suppose
ha
he
o de o b
di ides
4
.
Then
om
(2)
bgb
-1
=
g-1
o
all
gE
A
NA((b))
.
I
(A)
is
abelian,
hen,
by
he
Lemma,
A
is
abelian
and bab-1
=
a-
o
all
a
E
A
.
This
is
case
1)
o
he
heo em
.
I
(A)
is
nonabelian,
hen
(A)
is
a
Hamil onian g oup
and
(A)
=QxExT
whe e
Q
is
he
qua e nion
g oup
o
o de
8,
E
2
=
1
and
all
elemen s o
T
a e
o
odd
o de
.
We
wisll
o
p o e
ha
A
=
(A)
.
Suppose
ha
g
is
an
elemen
o
in ini e
o de
o
A N((b))
.
Then
g
2
E
CA
(Q)
and
he e
exis s
an
elemen
w
o
o de
4
o
Q
such ha
[b,
u)]
=
1,
because
e e y
subg oup
o
Q
is
no mal
in
G
.
Clea ly,
g
2
w
1
N((b))
and by
(2)
wg
-2
=
bwg
2 b
-1
=
bg
2
wb
-1
=
w-1g-2,
which
is
impossible
.
The e o e,
all
elemen s
o
A
N((b))
ha e
ini e
o de s
.
Le
gbe an
elemen
o
in ini e
o de
om
NA((b))
and
le
an a
E
A N
A
((b))
.
Clea ly
he e
exis s
n
such
ha
[g', a]
=
1,
because
he
ini e
UNITARY
SUBGROUPS

20
3
cyclic
subg oup
(a)
is
no mal
in
G
.
Then
gna
E
A
NA((b))and
gna
is
o
in ini e
o de ,
which
leads
o a
con adic ion
.
The e o e,
(A)
=
A
.
We
claim
ha
T=
1
.
Le
be
an
elemen
o
odd
o de
om
A N((b))
.
Ob iously,
he e
exis s
an
elemen
w
o
o de
4 in
Q
such
ha
[b,
w]
=
1,
as
e e y
subg oup
o
Q
is
no mal
in
G
.
Thus
w
~
N((b))
and by
(2)
which
is
impossible
.
Nex ,
le
be
an
elemen
o
odd
o de
om
NA
«b)
)
.
Because
( )
a
G,
[ ,
b]
=
1
.
Clea ly
he e
is
an
elemen
w
o
o de
4
in
Q
such
ha
b-1
wb
=
w-1
and
w
~
NA((b))
.
Then
which
is
impossible
.
Hence, he
s uc u e
o
G
is
desc ibed
in
case
2)
o
3)
o
he
heo em
.
b)
Suppose
ha
he
o de
o
b
is
2"

(k
>_
3)
.
Then
by
(2)
b
2
belongs
o
he
cen e
o (A),
Because
(A)
is
abelian
o
Hamil onian
.
Hence,
(A)
is
abelian
and
e e y
subg oup
o
(A)
is
no mal
in
G
.
Then
om
(2)
bab-1
=
a
-1
W
o
all
aE
A
NA((b))
.
Deno e by
(b4 )
he
subg oup
gene a ed
by b
4y
=
abab
-1
,
as
a
uns o e
A NA((b))
.
Pu
G
=
G/(b
4
),

A
=
A/(b
4
)
and
b
=
b(b
4
)
.
Then
_
G
_
sa is ies
he
condi io_ns
o
ou
Lemma
and
i
ollows
ha
=
1
and
bab-1
=
a
-1
o
all
a
E
A
.
This
is
case
4) o
he
heo em
.
"Su
ciencg
."
Le
G
sa is y
one
o
he condi ions
1)-4) o
he
heo em
.
Cléa ly,
i
a
ini e
subg oup
(c)
is
no
no mal
in
G,
hen
c
E
bA,
(c
2
)
--
(b
2
)
and
c
2
belongs
o
he
cen e
o
ZG
.
The e o e,
and
and
(g
+
g
-1
)c
2
is
cen al in
7LG
.
-1
w=
b wb
-1
=
w-1 -1,
w
-1
=
bw b
-1
=
-1w-1
u,,
9
=
1
+
(1
-
c)g(1
+
C)2_
uc,g
u
,s
=
1
+
(1
-
c)
(9
+
9
-l
(9))(
1
+
C)2-
.
Suppose
ha
g
E
A
.
Then
(g)
=
1
and
(g
+
g
-1 )c
2
is
a cen al
elemen
.
This
is
ob ious
in cases
1),
2)
and
3)
.
Suppose
ha
G
sa is ies
he
condi ion
4)
o
he
heo em
.
Then
(c
2
)
=
(b
2
),
and bgb-1
=
g-1b4i
and
G/(b
4
)
is
abelian
.
Thus
b(g
+
9
-1
)c
2
b
-1
=
(9
+
9
-1
)P
=
a
-1
(9
+
9-1)72a
204

A
.A
.
BOVDI
AND
S
.K
.
SEIIGAL
2
.
3
.
4
.
I
g
E
bA,
hen
g
=
ba,

(
.g)
=
-1
and
A
.A
.
BOVD1,
Uni a y
subg oup
o
he
mul iplica i e
g oup
o
in-
eg al
g oup
ing
o
a
cyclic
g oup,
Ma h
.
Za ne ki
41
(4)
(1987),
467-474
.
A
.A
.
BOVDI,
The
mul iplica i e
g oup
o
an
in eg al
g oup
ing,
Uzhgo od, 1987
.
K
.
HO cIISMANN
AND
S.K
.
SEIIGAL,
On
a
heo em
o Bo di, o
appea
.
S
.P
.
NOVIKOV,
Algeb aic
cons uc ion
and
p ope ies
o
He mi ian
analogues
o
K- heo y
o e
ings
wi h
in olu ion
om
he
iew-
poin
o
Ha nil onian
o malis n,
Applica ions
o
di e en ial
opol-
ogy
a,nd
i e
heo y o
cha ac e is ic
classes,
II,
Iz
.
Akad
.
Nauk
SSSR
Se
.
Ma
.
34
(1970),
475-500
;
English
ansl
.
i
n
Ma h
.
USSR
Iz
.
4
(1970)
.
5
.

J
.
RITTER
AND
S
.K
.
SEIIGAL,
Gene a o s
o
subg oups
o U(7LG)*,
Con empo a y
Ma h
.
93
(1989),
331-347
.
6
.

J
.
RITTER
AND
S
.K
.
SEIIGAL,
Cons uc ion
o
uni s
in in eg al
g oup
ings,
T aces
.
A
.M
.S
.
324
(1991),
603-621
.
7
.

S.K
.
SE11CAL,
"Topics
in
g oup
ings,"
M
.
Dekke ,
New
Yo k,
1978
.
A
.A
.
Bo di
:
Ma hema ics
Ins i u e
Uz1igo od
S a o
Uni e si y
Uzhgo od
USSR
g-
I
=
a
-
b
-
=
b-Iab4i
.
Clea ly,
g
-l
c
2
=
gc
2and
(g
+
(g)g
-
)c2
=
0
.
The e o e,
Uc,g
~c,g
=
1
and
he
bicyclic
uni s a e
-uni a y
.
Thus
B2(7LG)
is
an
-uni a y
subg oup, p o ing
he
heo em
Re e ences
S .K
.
seligal
:
Depa men
o
Ma hema ics
Uni e si y
o
Albe a
Ed non on, Albe a
CASADA
T6G
2G1
P ime a
e sió
ebudo,
el
24
de
Juliol
de
1991,
Ba e a
e sió
ebudc
el
14
d'Oc ub e
de
1991