Journal of Algebra 240, 836
858 Ž2001.
doi:10.1006
jabr.2000.8698, available online at http:

www.idealibrary.com on
A Connection between Units of Burnside Rings and the
Exterior Algebra of Elementary Abelian 2-Groups
Michael A. Alawode
Department of Mathematics, Uniersity of Ibadan, Ibadan, Nigeria
Communicated by Walter Feit
Received September 18, 2000
INTRODUCTION
Let G be a finite group, ŽG. the Burnside ring of G, that is, the
Grothendieck ring obtained from the semi-ring of G-isomorphism classes
of finite G-sets under addition and multiplication induced respectively by
the disjoint union and the Cartesian product. The goal of this paper is to
give the connection between the structure of the group ŽG.
 of units of
ŽG. and the associated Exterior Algebra, where
G 

2   
2
n-times
is an elementary abelian 2-group of order 2 n.
In Section 1 we discuss the condition ŽUB. and show how an element of
ŽG.
 can be identified. In Section 2, we show that the map
 i : G
i   ŽG.
i1
 Ž


G. i
is multilinear and that
 i Ž g1 , . . . , gi .  0
if

 g1 , . . . , gi4  i .
Finally, we show that  i induces an isomorphism between 
i ŽG. and
ŽG.
 
i1
ŽG.i .
836
0021-8693
01 $35.00
Copyright  2001 by Academic Press
All rights of reproduction in any form reserved.
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 837
1. CONDITION ŽUB.
1.1. Let
G 

2   
2
n-times
be an elementary abelian 2-group of order 2 n and let SubŽG. denote its
subgroup lattice. It is well known that the group ŽG.
 of units in the
Burnside ring ŽG. of G is canonically isomorphic to the group of maps
e : SubŽG.  14
satisfying the following condition:
ŽUB. For all U,U1,U2 ,U3,V  SubŽG. with ŽV :U .  4 such that
U1 ,U2 ,U34  W  SubŽG.  U  W  V 4 ,
we have
eŽU .  eŽU1 .  eŽU2 .  eŽU3 .  1.
1.2. THEOREM. For eery H  G, the map
eH : SubŽG.  14
U  1  2G , HU  ½ 1, if H  U  G1, if H  U  G
satisfies ŽUB. and hence represents an element in ŽG.
.
Proof. Assume that U,V,U1,U2 ,U3 are as in ŽUB.. We distinguish the
following cases:
Case 1. If
eH ŽU .  eH ŽU1 .  eH ŽU2 .  eH ŽU3 .  1
there is nothing to prove.
Case 2. If
eH ŽU .  1, that is, H  U  G ,
then
G 	 H  Ui 	 H  U  G ,
so H  Ui  G for i  1, 2, 3 and therefore eH ŽUi.  1 for i  1, 2, 3. So
also in this case
eH ŽU . e ŽU
4
 H 1 .  eH ŽU2 .  eH ŽU3 .  Ž1.  1.
Case 3. If
eH ŽU .  1, eH ŽUi .  1
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838 MICHAEL A. ALAWODE
for at least one i  1, 2, 34, say i  1, then we argue as follows. We have
U  H  G
and
U1H  G.
We have to show that we can neither have
U2  H  U3  H  G
nor
U2  H  G , U3  H  G.
To this end we prove first the following.
1.3. LEMMA. If G is a group, H a normal subgroup, and W1,W2 are
subgroups of G with W1  W2 , then
ŽW2 :W1 .
.
Ž H  W2 : H  W1 .
Proof. Given that W1  W2 and H
 G, then we have
HW1  HW2  G , W1  H
 W1 , and W2  H
 W2 .
Consider
W1  H .
Let  be an arbitrary element of W1  H. Then   H and   W1. But
W1  W2; this implies   W2 . Now   W2 and   H; this implies
  W2  H.
So we have
  W1  H    W2  H .
Hence,
W1  H  W2  H
and as both intersection are subgroups
W1  H  W2  H .
In particular,
H  W2 
H  W1 
is a positive integer. We now consider
H  W2  Ž H   W2 .
H  W2  W2   H  W1 
 
H  W1  Ž H   W1 .
H  W1  W1   H  W2 
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 839
which implies that
ŽW2 :W1 .  Ž H  W2 : H  W1 . Ž H  W2 : HW1 . .
This implies that ŽHW2 : HW1. divides ŽW2 :W1., since ŽH  W2 : H  W1.
is a positive integer.
Hence
ŽW2 :W1 .
.
Ž H  W2 : H  W1 .
It also follows from this result that if
ŽW2 :W1 .  2, then Ž H  W2 : H  W1 .  2.
Next, we show that with G, H,U,U1,U2 ,U3,V as above.
1.4. LEMMA. If ŽG :U  H .  2, then one has
Uj  H  G  Uj  U  H for j  1, 2, 3.
Proof. Assume first that Uj  H  G. Then since U  H  G, we obtain
that U  H  Uj  H, and this implies that Uj  H  U  H, so we have that
Uj  U  H, since H  U  H. Thus, Uj  H  G  Uj  U  H.
Conversely, suppose that Uj  U  H. Then Uj  H  U  H. It also follows
that U  H  Uj  H, since U  Uj and therefore U  H  Uj  H but U  H 
Uj  H. Now, as U  H  G, we get that U  H  G. We know also that
Uj  H  G. So it follows that
U  H  Uj  H  G.
Hence
2  ŽG :U  H .  ŽG :Uj  H .  ŽUj  H :U  H .
together with ŽUj  H :U  H .  1 and therefore Žby Lemma 1.3.
ŽUj  H :U  H .  2.
This implies
2  ŽG :Uj  H .  2
or
ŽG :Uj  H .  1.
So we must have that
G  Uj  H .
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840 MICHAEL A. ALAWODE
FIG. 1. Step 1.
Thus,
Uj  U  H  Uj  H  G ,
and so
Uj  H  G  U1  U  H .
Now we continue with the proof of Case 3.
Consider the stepwise diagrams shown in Figs. 1
3 with the motive of
getting a final result for Case 3.
Step 1. Consider H  U1  G Žsee Fig. 1..
Step 2. See Fig. 2.
Žiii. To show that U  H  V  U1, assume U  H  G. We must
show that
Žii. H  U1  U by first showing that
Ži. H  U  U1  U.
Proof of Ži.. In the first place, it is evident that
U  U  H  U1 ,
FIG. 2. Step 2.
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 841
FIG. 3. Step 3.
since
U  U  H , U  U1.
To prove
U  H  U1  U,
consider
U  H  U1.
By definition of intersection,
U  H  U1  U1.
So,
U  U  H  U1  U1.
But since ŽU1 :U .  2, it then follows that either
U  U  H  U1
or
U  H  U1  U1.
But then, by our assumption that
U  H  G ; U1  H  G ,
which also implies
U1  U  H Žsee Lemma 1.4.
we obtain that
U  H  U1  U1
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842 MICHAEL A. ALAWODE
because U  H  U1  U1 would imply that U1  U  H which in turn gives
a contradiction to our assumption. Hence
U  H  U1  U.
Step 3 Žsee Fig. 3.. Since we obtain from Step 2 that
U  U  H  U1  U1 ,
this implies U1  U  H and therefore
U  H  V  U1.
Observe that
U  U  H  V  V .
Step 4. See Fig. 4.
Step 5. See Fig. 5.
Without loss of generality, say, U  H  V  U2 and U  H  V  U3.
We must show first that
U  H  V  U2 ,U34 .
Proof. As
U  H  U  H  V  G
and
V 
 U  H  V ,
we obtain
U  H  V 
 U  H ,
U  H  V   G   2 n .
FIG. 4. Step 4.
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 843
FIG. 5. Step 5.
We also have that
U  H   2 n ,
since U  H  G, and Ui  H  G for any i together implies
1  ŽG :U  H .  2.
This implies
ŽG :U  H .  2
and hence that
U  H   2 n1.
Next, we consider the equation
U  H   V 
U  H  V  
ŽU  H .  V
which implies that
2 n1  V 
2 n   2 n1 V : U  H  V .
Ž .  Ž Ž . .U  H V
This implies
2 n
ŽV : ŽU  H .  V . 
2 n1
 2,
and since U  U  H and U  V implies U  U  H  V, because
ŽV : ŽU  H .  V .  2  ŽV : V . , ŽV :U .
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844 MICHAEL A. ALAWODE
we obtain
U  H  V  V ,U
and also by Step 3,
U  H  V  U1.
Hence
U  H  V  U2 ,U34 .
We shall finally prove Step 5.
Since
U  H  V  U2 ,U34
by Step 4, we may then assume without loss of generality that
U  H  V  U2 ; U  H  V  U3 .
This implies
U2  U  H ; U3  U  H , since U3  V .
This implies
U2  H  UH  G ; U3  U  H .
Hence,
U2 H  G ; U3  H  G Žby Lemma 1.4. .
Therefore the proof of Case 3 is complete.
So we conclude by Case 1, Case 2, and Case 3 that the map
eH : SubŽG.  14
satisfies condition ŽUB..
2. MULTILINEARITY CONDITION
2.1. Now for each i  0, 1, 2, . . . , we define
Ž 
 
 G. i  e   ŽG.  eŽU .  1 for all U  G with U   2 ni4
and observe that


eH   ŽG. i if H   2 i1 , H  G.
2.2. THEOREM. Define the map
: i  Ž .
 
 Ž .
 l G  G i1  G i
by
: Ž g1 , . . . ,


gi .  e² g1 , . . . , g : ŽG. .i i
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 845
Then
2.2.1.  i is multilinear, and
2.2.2.  iŽ g1, . . . , gi.  0 if 
 g1, . . . , gi4  i.
Before we prove Statement 2.2.1, we shall first state and prove the
following useful lemmata:
2.3. LEMMA. Let G be a group of order 2 n and K , H
 G, such that
K   2 i and H   2 n i. Then K  H  G  K  H  1.
Proof. As K  G and H
 G imply
K  H  ²K , H:  G ,
we consider the equation
K   H 
K  H  
K  H 
2 i  2 n i

K  H 
2 ini

K  H 
2 n
 .
K  H 
Hence
K  H   G   2 n  K  H   1.
Therefore
G  K  H  K  H  1.
2.4. LEMMA. Let
G 

2  
2   
2 , A  G.
n-times
Then
² A:  2
A .
Proof. Assume 
A  i. Label the elements in A as, say, a1, . . . , ai, so
that
A   a1 , . . . , ai4 .
Then
² A:  a11 , . . . , a ii  1 , . . . , i  0, 14 4 .
To see this, let H be the set on the righthand side above. Since ² A: is
closed under multiplication and the forming of inverses, H  ² A:. But
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846 MICHAEL A. ALAWODE
also, by definition, ² A: is the unique smallest subgroup of G containing
A in the sense that, for all U  G, whenever A  U  G, then ² A:  U.
Obviously, since all the elements of A are used up in the construction of
an element of H, A  H. Now, let h1, h2  H. Then h  a1 . . . a i1 1 i for
all choices of 1, . . . , i  0, 14, and h2  a1 . . . ai1 i for every choice of
1, . . . ,i  0, 14. Next we consider
h  h11 2  Ža1  a 2 . . . a i11 2 i1  a ii . Ža1  a2 . . . a i 111 2 i1  a ii . ,
 a1  a2 . . . a i1  a i  a i  a i1 . . . a 2  a 11 2 i1 i i i1 2 1
 a1  a 1  a2 . . . a i1  a i  a i  a i1 . . . a 21 1 2 i1 i i i1 2
 a1  a 1  a2  a 2 . . . a i1  a i  a i  a i11 1 2 2 i1 i i i1 . . . .
Continuing in this way, we get
h  h1  a1 1  a2 2 . . . a i1 i1  i i1 2 1 2 i1  ai
 a1  a2 . . . a i11 2 i1  a ii ,
where in view of the special structure of G,  i is determined by i and i
according to the scheme
   0 1
0 0 1
1 1 0
Hence h  h11 2  H, and this implies that H  G. So we have
² A:  H , and therefore ² A:  H .
Hence, we obtain ² A:   2
A, since
2 i  
 Ž 1 , . . . , i .  i  0, 14 4 .
Proof of Statement 2.2.1. Let r be such that 1  r  i. For every r and
gr , hr  G. Consider
 i Ž g1 , . . . , gr1 , gr hr , gr1 , . . . , gi .
  i Ž g1 , . . . , gr , . . . , gi .   i Ž g1 , . . . , hr , . . . , gi . .
To see this, we must prove that
e² g1 , . . . , g r  hr , . . . , g :Ž H . ei ² g1 , . . . , g r , . . . , g :Ž H . e² g Ž H .  1i 1 , . . . , hr , . . . , g i:
for all H  G with H   2 n i.
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Without loss of generality we may assume that r  i, put a  gi, b  hi,
c  gihi so that a  b  c  1. Then we can define
A  ² g1 , . . . , gi1 , a:
B  ² g1 , . . . , gi1 , b:
C  ² g1 , . . . , gi1 , c:.
Note that since
 A   2 i ,
B   2 i ,
C   2 i ,
we have by the above result that
eA   Ž .


G i1,
eB   ŽG.


i1,
and
eC 


 ŽG. i1,
respectively, that is, the following case is obvious. For any H  G with
H   2 n i, we get that
eAŽ H .  eB Ž H .  eC Ž H .  1.
So, we consider the only non-trivial case H   2 n i.
Next we shall discuss under this case some of the useful consequences
derived for members in the set
 A , B ,C4
and with respect to distinguished cases as
Ži. Assume  A   B   C   2 i.
Žii. Assume  A   2 i, B   C   2 i.
Žiii. Assume  A   2 i, B   2 i, C   2 i.
Živ. Assume  A   2 i, B   2 i, C   2 i.
First, we discuss case Živ. as follows: As
 A   2 i1 , B   2 i1 , C   2 i1 ,
we obtain
H  A   2 n i  2 i1  2 n1  G ,
H  B   2 n i  2 i1  2 n1  G ,
H  C   2 n i  2 i1  2 n1  G ,
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848 MICHAEL A. ALAWODE
and it follows by definition that
eAŽ H .  eB Ž H .  eC Ž H .  1.
Second, we discuss cases Žii. and Žiii. by proving the following lemma:
2.5. LEMMA. The following are equialent
Ži. A  ² g1, . . . , gi1:
Žii. a  ² g1, . . . , gi1:
Žiii. B  C.
Proof. Ži.  Žii., i.e., A  ² g1, . . . , gi1:  a  ² g1, . . . , gi1:. As-
sume
A  ² g1 , . . . , gi1:.
Since
a  A  ² g1 , . . . , gi1 , a: and A  ² g1 , . . . , gi1:
it follows that a  ² g1, . . . , gi1:.
Žii.  Žiii., i.e.,
a  ² g1 , . . . , gi1:  B  C.
Assume a  ² g1, . . . , gi1:. Then we have
a  g 11 , . . . , g  i1i1 for some choices 1 , . . . , i1  0, 14 .
In view of g1, . . . , gi1  C by definition, it is enough to observe that
b  ac  g 11 . . . g  i1i1  c
 ² g1 , . . . , gi1 , c:  C ,
hence, B  C.
Similarly, on the other hand, in view of g1, . . . , gi1  B we observe that
c  ab  g 1 . . . g  i11 i1  b
 ² g1 , . . . , gi1 , b:  B.
Hence, C  B; therefore, B  C.
Žiii.  Žii., i.e., B  C  a  ² g1, . . . , gi1:. Assume B  C. Then
there exist  , . . . ,  ,  and  , . . . , ,  with b  g 1  i1 1 i1 1 i1 1 . . . gi1  c and
c  g1 . . . g i1  b1 i1 . Now if   1, then
a  bc  g 1 . . . g  i11 i1  c  c  g 11 . . . g
 i1
  ²i 1 g1 , . . . , gi1:.
Similarly, if   1, then
a  bc  cb  g1 . . . g i11 i1  b  b  g11 . . . g
i1
i ²1  g1 , . . . , gi1:
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 849
and if     0 then
b  ² g1 , . . . , gi1:, c  ² g1 , . . . , gi1:.
This implies
a  bc  ² g1 , . . . , gi1:
hence,
a  ² g1 , . . . , gi1:.
Žii.  Ži., i.e., a  ² g1, . . . , gi1:  A  ² g1, . . . , gi1:. Assume
a  ² g1 , . . . , gi1:.
Then we have
² g1 , . . . , gi1:  A ,
since
a  A  ² g1 , . . . , gi1 , a:
also
A  ² g1 , . . . , gi1:,
since by assumption
a  ² g1 , . . . , gi1:,
hence
A  ² g1 , . . . , gi1:,
and the proof of the lemma is complete.
Continuation of the Proof of Statement 2.2.1. By Lemma 2.4, we get that
² g1 , . . . , gi1:  2 i1.
Hence, in this case,
 A   2 i   A   2 i1  A  ² g1 , . . . , gi1:.
In view of the above considerations we conclude that Case Žii. is possible
and our formula
eAŽ H .  eB Ž H .  eC Ž H .  1
is almost trivially satisfied, since
H  A   2 n i  2 i1  2 n1  G , implies eAŽ H .  1,
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850 MICHAEL A. ALAWODE
and either H  B  H  C  G and then we have
H  B   2 n i  2 i  2 n  G , implies eB Ž H .  1,
and
H  C   2 n i  2 i  2 n  G , implies eC Ž H .  1,
or
H  B  H  C  G.
Then we obtain
H  B   2 n i  2 i1  2 n1  G , implies eB Ž H .  1,
H  C   2 n i  2 i1  2 n1  G , implies eC Ž H .  1.
But case Žiii. is not possible. So we are left to discuss case Ži. as follows.
Assume
 A   B   C   2 i
and consider
A  ² g1 , . . . , gi1:²a:,
B  ² g1 , . . . , gi1:  ²b:,
C  ² g1 , . . . , gi1:  ²c:,
²a:
² : ² :  Ž A : ² g1 , . . . , g :.  2.g1 , . . . , g i1i1  a
Since
² g1 , . . . , gi1:  2 i1 ,
this implies
²a:  2, ² g1 , . . . , gi1:  ²a:  1.
Similarly, we obtain
ŽB : ² g1 , . . . , gi1:.  2,
ŽC : ² g1 , . . . , gi1:.  2.
Also, since it is clear that g1, . . . , gi1  A and g1, . . . , gi1  B implies
g1, . . . , gi1  A  B, this implies
² g1 , . . . , gi1:  A  B  A , B
and A  B  A or B because A  B by Lemma 2.5.
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Similarly, we obtain
² g1 , . . . , gi1:  B  C  B ,C , B  C
² g1 , . . . , gi1:  A  C  A ,C , A  C
so we must have
Ž A : A  B.  2, Ž B : A  B.  2,
ŽC : B  C .  2, Ž B : B  C .  2,
ŽC : A  C .  2, Ž A : A  C .  2.
Next we consider
Ž A : A  B.  Ž A  B : ² g1 , . . . , gi1:.  Ž A : ² g1 , . . . , gi1:. .
Then we have
2  Ž A  B : ² g1 , . . . , gi1:.  2.
This implies
Ž A  B : ² g1 , . . . , gi1:.  1
hence,
A  B  ² g1 , . . . , gi1:.
Similarly, we obtain
B  C  ² g1 , . . . , gi1:,
A  C  ² g1 , . . . , gi1:.
It also follows that
 A  B   B  C    A  C   2 i1.
Also, since
g1 , . . . , gi1 , a, b  ² g1 , . . . , gi1 , a:  ² g1 , . . . , gi1 , b:,
g1 , . . . , gi1 , b , c  ² g1 , . . . , gi1 , b:  ² g1 , . . . , gi1 , c:,
g1 , . . . , gi1 , a, c  ² g1 , . . . , gi1 , a:  ² g1 , . . . , gi1 , c:,
and a  b  c  1, we get that
D  A  B , because c  a  b ,
D  B  C , because a  b  c,
D  A  C , because b  a  c,
where
D  ² g1 , . . . , gi1 , a, b:,
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852 MICHAEL A. ALAWODE
and since
A  D , B  D implies A  B  D ,
B  D ,C  D implies B  C  D ,
A  D ,C  D implies A  C  D ,
it follows that
D  A  B  B  C  A  C.
Now we compute
 A  B  B  C   A  C 
D      2 i1.
 A  B  B  C   A  C 
2.6. We are now set to give the proof of non-trivial case: That is, we
must prove that if H  G, H   2 n i and H  A  G then either H  B 
G and H  C  G or vice-versa.
Note that H  A  G implies H  D  G.
Proof. Since
H  D 
H  D   ,
H  D 
we have
2 n i  2 i1
2 n  ,
H  D 
and hence
2 n i  2 i1
H  D    2 nii1n 1n  2  2.2
Similarly, we obtain
H  ² A  B:  H  ²B  C:  H  ² A  C:  2,
since
D  A  B  B  C  A  C ,
Now as H  ² A  C:   2, there exists precisely one element, say u  G,
such that u  1 and ²u:   2 with ²u:  H and ²u:  ² A  C:, since
H  ² A  C:  ²u:. Similarly, we get H  ² A  B:  H  ²B  C:  H
 D  ²u:. But then by our hypothesis H  A  G implies H  A  1,
and it follows that ²u:  A, as ²u:  H.
Now we know the following:
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 853
H  A  G and H  A  1, and this implies u  A,
H  B  G  H  B  1,
H  C  G  H  C  1,
H  D  ²u: 	 H  B, H  C 	 1, and this implies
H  B  1  u  B and H  C  1  u  C
or, in other words,
H  B  1  u  B and H  C  1  u  C.
We must show either
u  B and u  C
or
u  B and u  C
To see this, we have to show that neither
u  B and u  C
nor
u  B and u  C
can hold.
Hence, assume first that on the contrary u  B and u  C. Then we
consider B  C, and use the fact that
Ž A : ² g1 , . . . , gi1:.  2, B  C  ² g1 , . . . , gi1:.
We obtain
B  C  A ,
and it follows that u  A, a contradiction. So we can’t have
u  B and u  C.
Assume second that
u  B and u  C.
Then we have
u  B  C  D  ² g1 , . . . , gi1 , a, b:.
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854 MICHAEL A. ALAWODE
This implies
u  g 1 . . . g  i1  a i1 i1  b i1
for some choices of 1, . . . , i1  0, 14.
Now, by hypotheses
	 	 	
u  A implies u  g 11 . . . g  i1  a ii1 for every choice of 
	 , . . . ,  	1 i 
0, 14,

 
 

u  B implies u  g 11 . . . g  i1  b ii1 for every choice of 


1 , . . . , 


i 
0, 14, and
  
u  C implies u  g 1 . . . g  i1  i1 i1  c for every choice of 

1 , . . . , 

i 
0, 14.
Now if i1  0, then we shall have
u  g 11 . . . g  i1i1  a i  A  ² g1 , . . . , gi1 , a:,
hence
i1  0  i1  1.
If i  0, then we get that
u  g 11 . . . g  i1  b i1i1  B  ² g1 , . . . , gi1 , b:,
hence
i  0  i  1,
and so we have
i  1, i1  1.
Hence we obtain
u  g 1 . . . g  i11 i1  Ž ab.
1  g 11 . . . g  i1i1  c1 since abc  1
 C  ² g1 , . . . , gi1 , c:, a contradiction.
So we cannot have
u  B and u  C.
Therefore, the proof of the non-trivial case is complete.
Hence
e² g1 , . . . , g r  hr , . . . , g i:Ž H .  e² g1 , . . . , g :Ž H .  ei ² g1 , . . . , hr , . . . , g i:Ž H .  1
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 855
for all H  G with H   2 n i, and by definition, we obtain
 i Ž g1 , . . . , gr hr , . . . , gi .
  i Ž g1 , . . . , gi .   i Ž g1 , . . . , hr , . . . , gi . .
Thus,  i is a multilinear map.
Now, consider again the map
 ii :G   ŽG.


i1
 Ž .


G i
: Ž g1 , . . . ,


gi .  e² g1 , . . . , g : ŽG. .i i
If we impose on this map the condition that 
 g1, . . . , gi4  i then we
obtain as follows. As 
 g1, . . . , gi4  i  1, then we obtain by Lemma 2.4
that
² g , . . . , g :  2 i11 i
and this implies


e² g1 , . . . , g :   ŽG.i i
or


e² g , . . . , g : ŽG. i   Ž


G. .
1 i i
So by definition, we get
 i Ž g1 , . . . , gi .  0.
Hence,  iŽ g1, . . . , gi.  0 whenever 
 g1, . . . , gi4  i, and the proof of
Theorem 2.2 is complete.
3. ISOMORPHISM BETWEEN 
i ŽG. AND ŽG.
i1
ŽG.


i
3.1.  i induces a canonical map
i
ˆ 
 
 i : 
 ŽG.   ŽG. i1
 ŽG. i
which maps
i
g1    gi  
 ŽG. onto


e² g1 , . . . , g : ŽG.i i .
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856 MICHAEL A. ALAWODE
3.2. CLAIM. ˆi is an isomorphism.
Proof. First, we note that ˆi is a well defined linear map, because of
the universal properties of 
i ŽG. and the particular properties established
above of the map  i.
Second,
i

 Ž 
 
G. and  ŽG. i1
 ŽG. i
are both of the same order, because
Ž Ž .
 
 n G :  ŽG. .  2Ž .i1 i i
and by standard results, we know that as dimŽG
2 .  n, we have
dim
i ŽG.  Žn.i , and this implies
i

 Ž .  nG 2Ž .i .
Third, we establish injectivity of ˆi in the following way. Now assume
that 	  
i ŽG. satisfies
ˆi Ž 	.  0, that is,
Ž ˆ Ž . .  1, . . . , n4
T  i 	 1 for all T  ž n  i / ,
where 
T : Ž


 G. i1  14 .
This implies that for every
 ž 1, 2, . . . , n4T n  i /
we have
ˆi Ž 	. Ž²ei  i  T:.  1,
where ei  Ž1, . . . , 1,1, 1, . . . , 1.  G
ith position
for all i  1, . . . , n.
As G  ²e1, . . . , en: and 	  
i ŽG. there are unique coefficients
Ck , . . . , k  2 Ž1  k1    ki  n. such that1 i
	  Ý Ck1 . . . k ei k    e .1 k i
1k1k2  k in
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BURNSIDE RINGS AND EXTERIOR ALGEBRAS 857
Hence, for any
 ž 1, . . . , n4T ,n  i /
we have

T Žˆi Ž 	. .  ˆi Ž 	. Ž²ei  i  T:.
 Ł e² e , . . . , e :Ž²ei  i  T:.Ck1 . . . k i  1,k k
1k1k   k n
1 i
2 i
where
e 
T
k1 , . . . , k i4
² ek , . . . , e :Ž²ei  i  T:.  Ž1. ,1 k i
we define with
½ 0, if  k1 , . . . , ki4  T   Tk1 , . . . , k i4  1, if  k1 , . . . , ki4  T  .
This means that
 Tk , . . . , k 4  1 if and only if T  1, . . . , n4
 k1 , . . . , ki4 .1 i
Applying this definition on individual factors of the above products rela-
tion, we obtain for a fixed k 01 , . . . , k
0
i 4 with T  1, . . . , n4
k 01 , . . . , k 0i 4
that
 T Tk 01 , . . . , k 04  1 and i k1 , . . . , k 4  0i
for
 k 0 01 , . . . , ki4  k1 , . . . , ki 4 .
This implies
e² e , . . . , e :Ž²ei  i  T:0 0 .  1, ek k ² ek , . . . , e :Ž²ei  i  T:.  11 i 1 k i
and substituting this in the above products relation, we get that
T Ck . . . k i
1  
T Žˆi Ž 	. .  Ł žŽ1.  1k1 , . . . , k i4 /
1k1  k in
 Ž1. Ck 01 . . . k 0i ,
and therefore,
Ck 01 . . . k 0  0.i
IBADAN UNIVERSITY LIBRARY
858 MICHAEL A. ALAWODE
Hence, for every k 01 , . . . , k
0
i 4 we must have
Ck 01 . . . k 0  0,i
which implies that
	  Ý Ck . . . k ek    e  0,1 i 1 k i
1k1k2  k in
hence, ˆi is injective.
So it is clear from the above considerations that ˆi is surjective and that
the vectors
e² ek , . . . , e : ,  k1 , . . . , ki4  1, 2, . . . , n41 k i
generate
Ž .
 
 
 G i1  ŽG. i .
Therefore, ˆi is an isomorphism.
ACKNOWLEDGMENTS
I am sincerely grateful to my Ph.D. thesis supervisors Professor Andreas Dress and
Professor Aderemi O. Kuku for their guidance, patience, and generosity.
REFERENCES
1. M. A. Alawode, ‘‘The Group of Units of Burnside Rings of Various Finite Groups,’’ Ph.D.
thesis, University of Ibadan, Nigeria, 1999.
2. H. Bender, On groups with abelian sylow 2-subgroups, Math. Z. 117 Ž1970., 164
176.
3. A. Dress, A characterization of solvable groups, Math. Z. 110 Ž1969., 213
217.
4. A. Dress, Notes on the theory of representation of finite groups, Bielefeld Notes, 1971.
5. W. Feit and J. Thompson, Solvability of groups of odd order, Pacific J. Math. 13 Ž1963.,
775
1029.
6. W. H. Greub, ‘‘Multilinear Algebra,’’ Springer-Verlag, Berlin
Heidelberg
New York,
1967.
7. A. O. Kuku, Axiomatic theory of induced representations of finite groups, in ‘‘Group
Representation and Its Applications: Les cours du C.I.M.P.A.’’ ŽA. O. Kuku, Ed.., 1985.
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