HOUSTON
JOURNAL
@
FRATTINI
LIKE
PROPER,TIES
OF MATHEMATICS
1998 University
of Houston
Volume
4, 1998
24, No.
OF KERNELS
OF SOME
CHARACTERS
YAKOV
COMMUNICATED
BY
Let K be minimal
ABSTRACT.
BERKOVICH
BERNHARD
H.
NEUMANN
among the kernels of the irreducible
ters of a finite group G, and let Q/K be a normal subgroup
Q is r-closed,
where T is a set of primes,
sult improves
Theorem
Academic
if and only if Q/K
12.24 in I.M. lsaacs,
Characters
Press, New York, 1976. In particular,
charac-
of G/K.
of Finite
Groups,
G is rr-closed if and only if
G/K is. The analog of the result above holds for minimal quasikernels
Some related
Throughout
sidered.
results
follows,
is n-closed
if its maximal
Obviously,
subgroups
too.
are proved.
this note only finite groups
In what
Then
is. This re-
and their complex
7r, I? are complementary
normal
7r-subgroup
and epimorphic
O,(G)
images
characters
sets of primes.
is a T-Hall
of n-closed
groups
are con-
A group
subgroup
G
of G.
are r-closed.
A
group G is p-&potent
(G( and Irr(G),
if it is {p}‘-closed. Let r(G) be the set of prime divisors
nonlinear irreducible characters
Irrl(G) th e sets of irreducible,
of
of
G, respectively.
It is well known
[G] that
the Frattini
subgroup
Q(G) of G has the following
properties:
(al) If N is normal
(Q2) If N is normal
in G, then N is 7r-closed if and only if N@(G)/@(G)
in G, then Q(N) 5 Q(G).
is.
(@3) r(G/@(G))
= r(G).
(+4) If N is normal in G and N 5 @(A) for some A 5 G, then N 5 a(G).
(al)
and (ip2) are also true if N is subnormal
1991 Mathematics
Subject
mary 2OC15.
Key words and phrases.
ernel, Hall subgroup,
Frattini
The author was supported
Class$cation.
p-nilpotent
subgroup,
in G.
1991 Mathematics
Subject
Classification.
Pri-
group, n-closed group, minimal kernel, minimal quasiksubnormal
subgroup.
in part by the Ministry
639
of Absorption
of Israel.
YAKOV
640
In this note we will show that
similar
the kernels
of some characters
have properties
to (@1))(@3).
Let K be minimal
that
BERKOVICH
among
the kernels of irreducible
case, we will call K a minimal
nilpotent.
Using that
result,
of a group G (in
12.20, K is
By [I], Corollary
we will prove the following
Theorem 1. Let K = ker(x)
subgroup of G/K.
kernel of G).
characters
be a minimal
kernel
of G and Q/K
a normal
Then Q is r-closed if and only if Q/K is.
By [I], Corollary 12.20, K is nilpotent.
If Q is rr-closed, then, obviously, Q/K is. Therefore,
PROOF.
it remains
to show that,
if Q/K is rr-closed, then Q is. In what follows we assume, therefore, that Q/K
is rr-closed. Let Qo/K be a (normal) rr-Hall subgroup of Q/K. Since Qo/K is
characteristic
in Q/K, it follows that
is a 7r-number.
Since K is nilpotent,
that
Qc contains
a T-Hall
subgroup
Qo is normal
in G.
Obviously,
it follows by Schur-Zassenhaus
H (indeed,
Qo contains
a normal
IQ : Qe]
Theorem
nilpotent
r/-Hall subgroup that coincides with the r/-Hall subgroup of K). Obviously,
is a T-Hall subgroup
of Q. Suppose that we have proved that H is normal
H
in
Qo. Since H is characteristic
in Qo, it is normal in Q, and so Q is 7r-closed.
Therefore we may assume, without loss of generality, that Qe = Q. Since, in that
case, Q/K is a 7r-group, we get Q = HK. We have to prove that H is normal
in G. Assume that this is false. Since all T-Hall subgroups
of Q are conjugate
by the Schur-Zassenhaus
Theorem, G = NG(H)Q = NG(H)HK = NG(H)K, by
Frattini’s argument;
NG(H) < G since H is not normal in G. If M is a maximal
subgroup of G containing
NG(H), then G = MK, and so x~ = X E Irr(M).
Since ker(XG) < ker(X) < M, it follows that K $ ker(XG).
Therefore,
there
exists 29 E Irr(XG)
Assume
K $ L = ker(d).
such that
L < M. Then,
that
L = M n L = M n ker(d)
contrary
to the
choice
G = LM.
In that
reciprocity,
and
= ker(8M)
5 ker(X) = ker(xM)
of K as a minimal
case, r9~ E Irr(M).
M n L = M
Put ker(X) = N. Note that
Since K
by reciprocity,
kernel.
Then
Thus
= M II K < K,
L $ M, and
n ker(d) = ker(tiM) = ker(X) = K f? M.
N = ker(xM)
hence
29~~ = X since X E Irr(bM)
= K rl M.
n L n M = K n (K n M) = K n M we get K n M 5 L and
by
FRATTINI
and hence KnL
Thus,
< M. Therefore,
LIKE
KnL
PROPERTIES
< KnM
641
5 KnL,
and so KnL
= KnM.
LnM=KnM=KnL=N.
(I)
Now, by what has been proved
and (l),
IQ:QnMI=IQM:MI=IG:MI=IML:MI=IL:LnMI
(2)
=IL:KnLI=(KL:KI.
Since H < QflM and H is a s--Hall subgroup of Q, it follows from (2) that IKL/KI
is a 7r’-number. Since K H/ K = Q/K IS a n-group, we obtain KL/K n KH/K =
{l},
and so KL/K
KL
are normal
of G.
[L, KH]
In particular,
[L, NH]
and KH/K
5 [L, KH]
< K.
5 N 5 NH.
KH = Q, MQ = G = MK
NH = (MnK)H
= MnKH
Therefore,
element
[L, KH]
Therefore,
Thus
KH
by element:
in G as kernels
5 K
L < NG(NH).
n L = N, so that
n K,
Since N = M
> (M, L) = G and NH is normal
in G. Since H is a
of KH = Q, it follows that H is a T-Hall subgroup
since NH
= Q and
of characters
and H < M, it follows, by the modular law, that
= MnQ, and hence NH is normal in M since Q is
in G. Then NG(NH)
normal
T-Hall subgroup
commute
in G since K and L are normal
as a subgroup
of Q has a normal
r/-Hall
of NH < KH.
subgroup,
T-Hall
subgroups
of NH are conjugate
by the Schur-Zassenhaus
Theorem,
an so, by
Frattini’s
argument,
G = NG(H)HN
= NG(H)N
5 MN = M, contrary to
assumption.
Thus H is normal in G, as claimed.
0
The theorem
is also true if Q is subnormal
in G.
Remark. If K is as in the theorem and Q/K is a nilpotent
normal
G, then Q is nilpotent
as well. It follows that, if a minimal kernel
with
the Fitting
If F/ker(x),
not nilpotent,
coincides
subgroup
of a nonidentity
group
G, then
subgroup
of
K coincides
G/K is semisimple.
where x E Irr(G), is the Fitting subgroup of G/ ker(>o and F is
then ker(x) is not a minimal kernel, by the theorem.
This result
with [I], Theorem
12.24.
Corollary 2. If K = ker(x)
(x E Ix(G))
zs a minimal
kernel
ofG,
then x(G) =
4GIW.
PROOF. Set n = n(G/K)
subgroup
theorem
Then
and let K1 be a rr’-Hall
K (see [I], Corollary
12.20); obviously,
G = Q x K1, where Q is a n-Hall
Ki > (1).
subgroup
x ker(p)
of the nilpotent
in G. Then by the
of G. Assume
We have x = T x I*, where r E Irr(Q),
K = ker(>o = ker(r)
subgroup
KI is normal
P E Irr(Ki).
that x(G) # r.
Since
= (K n Q) x (K il KI) = (K n Q) x KI,
642
YAKOV
it follows that ker(p)
Ki.
= Ki,
i.e., p = 1~~. Let ,ui be a nonprincipal
Set xi = r x pl. Then
Hence K1 = (1)
We see that
minimal
they are nilpotent.
(P(G)
(take G = E(p”),
the example
Let x E Irr(G).
Z(x)/ ker(x)
kernels
But,
of x.
behave
similarly
the elementary
The
subgroup
the Frattini
minimal
abelian
Z(x)
It is known that
= Z(G/ ker()o)
quasikernel
If Z(T)
Z(r) ker(x),
= {z
Z(x)
kernels
subgroup
and, as
are not contained
in
group of order p” > p; see also
quasikernel
th e result
we see that H/ ker(x)
to a minimal
= x(l)}
in G, ker(x)
is called
5 Z(x)
see [I], Lemma
and
2.27.
A
kernel.
If T E
of G, where x E Irr(G).
then Z(T) = Z(x).
< ker(x),
5 Z(x),
E G 1 Ix(x)l
is normal
and the last group is cyclic;
is defined similarly
is such that ker(r)
PROOF.
to the choice of K.
cl
in general,
Lemma 3. Let Z(x) be a minimal
Irr(G)
= K, contrary
< ker(x)
of
below).
the quasikernel
minimal
ker(xi)
character
= rr = n(G/K).
and r(G)
(P(G),
BERKOVICH
holds since Z(x)
< Z(G/ ker(>o)
is minimal.
Setting
= Z(x)/ ker(x),
H =
and so Z(r)
<
0
H I Z(x).
Corollary 4. If2 = Z(x) (x E Irr(G))
zs a minimal
quasikernel
of G, then Z is
nilpotent.
PROOF.
By the lemma,
a minimal
kernel of G.
we may assume,
without
In that case, Z/ ker(x)
loss of generality,
= Z(G/ ker(>o)
that ker(x)
is nilpotent,
the result follows from the theorem.
Corollary
Q/Z(x)
5.
of G.
generality,
z/ker(x)
0
(x E Irr(G))
be a minimal
Then Q/Z is x-closed
In view of the lemma,
PROOF.
kernel
Let Z = Z(x)
be normal in G/Z(x).
we may assume
As in the proof of the theorem,
that
Q/Z
1s
. a r-group.
= Z(G/ker(x))
by
In that
a n-group
that
quasikernel
K = ker(x)
we may assume,
case,
and let
Q/ ker(x)
Q/Z, is r-closed.
is a minimal
without
loss of
as an extension
By the theorem,
of
Q is
0
Let Z be as in Corollary
But the analog of Corollary
abelian
ofG
if and only if Q is.
7r-closed, as desired.
nonidentity
is
and
5. Then G is nilpotent
if and only if G/Z is nilpotent.
1 does not hold for minimal
quasikernels
(let G be a
group).
Suppose that x E Irr(G) is such that
Corollary 6. Let n be a set of primes.
1ker(x)ln is minimal with respect to divisibility. Then ker(x) = A x B, where A
is a nilpotent T-Hall subgroup of ker(>o.
FRATTINI
PROOF.
Set K = ker(X).
kernel with L 5 K.
choice of K.
ILIT divides
Since L is nilpotent
of L (and of K),
7r’-group.
Therefore,
We will show that
kernels.
Let H = PI
x
.
P,,
x
Let & be an irreducible
and X = $1 x ...
is a minimal
is normal
is the only restriction
such that
of
= (~1,. . ,p,}.
group of order pa, i = 1, . . , n.
of Pi of degree pi, i = 1, . . , n. Set G = H x K
Then
X E Irr(G)
and ker(X)
= K.
We claim that K
let N be a proper normal divisor of K; then N
ker(r)
IS
. not cyclic.
of G, Z(G/N)
= N ([I], Lemma
with the same pi as above.
2.27(d),
Therefore,
(f)).
This
there
means
let G = C x K, where C is cyclic of
kernel of G. Similarly,
X of G is minimal.
on the structure
group and rr(K)
where Pi is a nonabelian
character
Question 1. Let G be a nonabelian
character
is a
subgroup
0
K, C are minimal
Then
in this case, K is the kernel of a linear character
Moreover,
K/L
Obviously,
K = A x B, where B is a +-Hall
in G. By the construction
order pl . . .p,
it follows that A, a K-Hall
12.20),
let K be a nilpotent
x & x 1~.
K is a minimal
is a minimal
IKIT = IL/,, by the
and A 5 L, the result follows.
kernel of G. Indeed,
is no r E Irr(G)
that
L = ker(r)
be such that
in G and so in K.
is normal
the nilpotency
Indeed,
643
lK(ls, it follows that
([I], Corollary
by the theorem,
Since L is nilpotent
minimal
PROPERTIES
Let r E Irr(G)
Since
subgroup
of K.
LIKE
p-group
such that the kernel
Study the embedding
kernels of G.
of G.
of ker(X)
of some linear
in G.
Question 2. Let G be a nonabelian p-group such that some minimal kernel K =
(X E Irr(G) as nonlinear) has index p3 in G. Study the embedding of K in
ker(X)
G.
Question 3.
elementary
Classify minimal
abelian p-groups,
Proposition
ker(x)]
Then D = nxEx
ker(X)
Let X E X,
H < G such that
K II H
x
of:
(a) a standard wreath product
5 Q(H).
I x E Irr(G)},
X = {X ] X E Irr(G),
I ker(X)] = m}.
5 G(G).
K = ker(X).
G = KH.
Obviously,
Suppose
We assume
that
that
K $ a(G).
Then
H
n ker(r)
XH = 0 E Irr(H).
= ker(rH)
< ker(0)
there exists
IHI is as small as possible.
Let r E Irr(BG).
reciprocity,
(3)
of two
B in terms of A and B.
7. Let
m = mini]
PROOF.
kernels
(b) A
= ker(XH)
= K f~ H.
Then
Then,
by
644
YAKOV
ker(r)
<_ H.
Assume
that
minimal
choice of K. Since K f H, we obtain
that HL = G = HK,
Then,
BERKOVICH
by (3), ker(r)
where L = ker(r).
5 K, and so ker(r)
a contradiction.
= K by the
Hence it follows
5 IKI,
Then by (3), (LI = IKI. $$$#
so that (L( = (K( by the choice of K and T E X, (H CTL( = (H n K(. Hence by
(3), Hnker(r)
= Hn K for every 7 E Irr(OG). S’mce ker(BG) < H, it follows that
ker(BG) = H
it follows that
n K. Since ker(OG) is normal in G and ker(BG) = H n K 5 Q(H),
that Irr(eG)
proposition.
c X.
ker(BG) < @(G) by (Q4). S’mce r E Irr(eG)
Therefore,
D = n,,,
ker($)
It is easy to prove that the intersection
such that x(1) = max{~(l)
( T E Irr(G)}
Let p be a prime. Set
Irrr(G,p’)
= {x E Irri(G)
is arbitrary,
we obtain
5 ker(BG) < (P(G), proving
of the kernels of characters
is also contained in (P(G).
I ~‘i x(l)), G(d) =
0
the
0
x E Irr(G)
k-(x),
xEWG,P’)
WG,p)
=
lx
E WG)
I P I x(l)),
G(P) =
n
Wx).
XEWG,P)
Remark. We claim that G(p) is p-closed and its Sylow p-subgroup
is abelian. By
the Michler-Ito
Theorem
(see [I], Corollary
12.34 and [MI), it suffices to show
that all characters in Irri(G(p))
have p’-degrees. Assume that p E Irri(G(p))
has
degree divisible by p. Take x E Irr(p’);
then G(p) $ ker(>o. Then, by Clifford’s
Theorem,
~(1) divides
G(p) is characteristic
and so G(p) < ker(x), a contradiction.
in G. It is known that G(p’) has a normal
x(l),
The subgroup
p-complement
and solvable; see [B], Proposition
9 and Remark 1 following it. Next, we will prove
that G(p’) < G’ unless p divides the degrees of all nonlinear irreducible characters
of G (and then G is pnilpotent
by [I], Corollary 12.2). Indeed, let X be a nonlinear
irreducible
character
of G of p/-degree,
X a linear character
of G, z E G(p’). Then
Xx E Irr(G,p’),
and so x(1) = x(z) = (XX))(Z) = X(z)x(s) = X(x)x(l).
It follows
that X(z) = 1 so z E G’ since X is an arbitrary
linear character of G. Similarly,
G(p) 5 G’ unless all characters in Irr(G) h ave p’-degrees (and then G is p-closed
with abelian
classification
Sylow p-subgroup
by [Ml; note that the proof in [M] depends
on finite simple groups).
Let N be a normal
general,
N(p’)
subgroup
of G.
Then
Q(N)
5 (P(G), by (Q2).
on the
But,
in
$ G(p’) (take G = S4, N = A4 and p = 3: in that case, N = N(3’)
is of order 12 and ]G(3’)] = 4), and N(p) $ G(p) (take the same G, N and p = 2).
However, the following result holds:
FRATTINI
Proposition
particular,
PROOF.
sition.
LIKE
PROPERTIES
8. If N is a normal subgroup
if N(p’)
f G(p’),
Let x E Irrl(G,p’)
IfpI
645
of G, then N(p’)
fl N’ < G(p’).
In
then N is p-nilpotent.
and let XN = e(pl+.
= 1, then N(p’)nN’
..+pLt) be the Clifford decompo-
5 N’ 5 ker(pl+...+pt)
= ker(XN)
5 ker(x).
If ~1( 1) > 1, then p does not divide pi(l) for all i since pi( 1) divides x(1) by
Clifford’s Theorem, and so N(p’) n N’ < N(p’) 5 ker(pl +.
+ ,ut) = ker(x(ni) <
ker(x).
N(p’)
Therefore,
n N’ < nXEIrrlCG,p,j ker(x)
Th en,
G(P’).
N(P’) $
Irrl(N,p’)
12.2.
is empty
by what
has just
by the Remark;
been
= G(p’).
$
in that case, N is p-nilpotent
N’,
that
and
so
by [I], Corollary
0
For a normal subgroup N of G, set Irrl(G 1 N) = Irrl(G)
The following proposition
was inspired by [IK], Theorem
Proposition
9. Let K 5 N be normal subgroups
(a) If NG(p)/G(p)
and all characters
is p-closed,
in Irq(G
(b) If NG(p’)/G(p’)
nilpotent
(moreover,
compare to [IK],
then N is.
is p-nilpotent,
in Irrl(G
1K)
if N/K
is solvable,
Therefore,
of minimal
Without
of N/G(p).
we may assume
is p-
then N is also solvable;
order.
loss of generality,
we may assume
by induction,
applied to N/O,(G(p))
< G/O,(G(p)),
= {l}, i.e., G(p) is a p’-subgroup
by the Remark. Let P/G(p)
Sylow p-subgroup
that
P is p-closed
Obviously,
P = N, i.e., N/G(p)
(1) < G(p) < N. Assume
p-closed,
In particular,
is.
< N and,
get O,(G(p))
(normal)
is p-closed
h ave degrees divisible by p, then N is
N/K
(a) Let us prove the first assertion.
G(p)
if N/K
0).
Let G be a counterexample
PROOF.
of G.
In particular,
then N
if, in addition,
Theorem
- Irr(G/N).
D.
1K) h awe p/-degrees, then N is p-closed.
and all characters
p-nilpotent
that
Suppose
N(p’)
proved,
that
O,(N)
if and only if N is.
is a p-group.
> (1).
we
be a
Since N is not
Set c = G/O,(N).
Take
x E Irr(G,p)
(‘1f such a x does not exist, ?? is p-closed, by Ito-Michler Theorem;
in that case, G and so N are p-closed). We see that O,(N)G(p)
5 ker(x) (here
we consider
NH/H,
x as a character
as an epimorphic
it follows by induction
is p-closed
of G).
NH
that
G(p)
It follows that
image of N/G(p),
is p-closed.
is p-closed.
Since
so N is, and G is not a counterexample.
L/G(p)
be a minimal
L/G(p)
is a (nonidentity)
elementary
proved,
L is not p-closed.
Therefore,
normal
subgroup
of G/G(p)
abelian
O,(N)
Thus
contained
p-group.
by Ito’s Theorem
5 G(p)
Since NH/H
= E. Then
-E NH/H,
is a p-group,
O,(N)
= (1).
in N/G(p).
By what
NH
Let
Then
has been just
(see [I], Corollary
12.34
646
YAKOV
BERKOVICH
for solvable groups, but the same proof works for p-solvable groups as well), L
has an irreducible
character
X of degree divisible by p.
Let x E Irr(XG).
By
Clifford’s Theorem,
p divides x(1) so x E Irr(G,p).
Then G(p) < ker(x), and
which
is
not
the case since L/G(p) is
so G(p) 5 L TI ker(x) = ker(xL) < ker(X),
abelian and X is nonlinear. The first assertion in (a) is proved.
Suppose that N/K is p-closed and all characters in Irri (G ( K) have p/-degrees
(here we do not assume that G(p) is contained in N). Then K 5 G(p). In
that case, NG(p)/G(p) E N/N n G(p) as an epimorphic image of the p-closed
group N/K is also pclosed. Therefore, by what has been proved in the previous
paragraph, NG(p) is p-closed so is N. The proof of (a) is completed.
(b) Let us prove the first assertion.
Without loss of generality, we may assume
that G(p’) < N and, by induction, O,! (G(p'))
= {l}, i.e., G(p’) is apsubgroup by
the Remark. Let H/G($) b e a ( normal) p/-Hall subgroup of N/G($). Obviously,
N is p-nilpotent
if and only if H is. Therefore,
we may assume, without loss of
generality, that H = N; in that case N/G@‘) is a p’-subgroup and G(p’) is a psubgroup. Since G is a counterexample, N > G(p’) > (1). Let P E Syl,(G); then
G(p’) < P. We claim that G(p’) f Q(P). A ssume that this is false. Then G(p’) <
Q(G) by (+4), and so N is pnilpotent
since N/G(p’) is, which is a contradiction.
Since P’ < Q(P), it follows that G(p’) $ P’, and so P possesses a linear character
X such that G(p’) $ ker(X). S’
mce XG(l) = IG : P( $ 0 (mod p), there exists
x E Irr(XG) wh ose degree is not divisible by p. By reciprocity,
By the Remark,
G(p’) 5 G’ (otherwise, G is p-nilpotent,
case, G’ $ ker(x), i.e., x is nonlinear and x E Irri(G,p’).
contrary to what has just been proved.
Now suppose that N/K
Irri(G
is p-nilpotent
G(p’) $ ker(>o.
and so N is). In that
Then G(p’) < ker()o,
and the degrees of all characters
in
1K) are divisible by p (here we do not assume that G(p’) is contained in
N). In that case, K 5 G(p’).
Then NG(p’)/G(p’)
E N/N fl G(p’) is p-nilpotent
as an epimorphic image of the p-nilpotent group N/K. By what has been proved
in the previous paragraph, NG(p’) is p-nilpotent so is N. Since K is solvable, the
0
last assertion of (b) is obvious.
In general, G(p’), G(p) are not necessarily subgroups of Q(G). If G = S4, then
G(2) = G(3’) $ Q(G) = (1) since IG(2)I = 4. If W = ASL(3,2) = SL(3, 2).E(23),
where E(23) is elementary abelian of order 8, and G/E(23) is a subgroup of order
21 in W/E(23), then G(3) = G(7’) = E(23) $ a(G) = (1).
Let N be subnormal in G (i.e., N is a member of some composition series of
G), let NG(p)/G(p)
K/G(p)
be p-closed and K/G(p)
its normal p-Sylow subgroup. Then
is a subnormal p-subgroup of G/G(p).
It is easy to prove by induction
FRATTINI
that
K/G(p)
5 H/G(p)
LIKE PROPERTIES
= O,(G/G(p)).
647
By Proposition
7(a),
H is p-closed.
K is p-closed and subnormal
in N. Since IN : KI is a $-number,
a
P of K is a Sylow subgroup of N. Since P is subnormal
in G,
Sylow p-subgroup
it is subnormal
in N. Then P is normal in N, and hence N is p-closed. Similarly,
Therefore,
a subnormal
subgroup N is p-nilpotent
if and only if NG(p’)/G(p’)
I am indebted to the referee for remarks and comments.
is.
REFERENCES
[B] Y. Berkovich, Finite groups with small sums of degrees of some characters,
J Algebra 171
(1995), 426443.
[G] W. Gaschiitz,
[I]
Uber die a-Untergruppe
I.M. Isaacs, Character
endlicher Gruppen, Math. 2. 58 (1953), 160-170.
Theory of Finite Groups, Academic
[IK] I.M. Isaacs and Greg Knutson,
Irreducible
character
Press, New York, 1976.
degrees and normal subgroups,
J.
Algebra 199 (1998), 302-326.
[M] G.O. Michler, Modular representation
Lect. Notes in Math. 1178
theory and the classification
of finite simple groups,
(1986), 129-142.
Received
Revised
version
August 8, 1997
received
May 5, 1998
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF HAIFA, MOUNT CARMEL, HAIFA 31905,
RAEL
Is-