Units of some group algebras of groups of order 12 over any finite field of characteristic 3
The structure of the unit groups of the group algebra of the groups D₁₂ and C₃⋊C₄ over any finite field of characteristic 3 is established.
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irk-123456789-1547612019-06-16T01:32:41Z Units of some group algebras of groups of order 12 over any finite field of characteristic 3 Gildea, J. Monaghan, F. The structure of the unit groups of the group algebra of the groups D₁₂ and C₃⋊C₄ over any finite field of characteristic 3 is established. 2011 Article Units of some group algebras of groups of order 12 over any finite field of characteristic 3 / J. Gildea, F. Monaghan // Algebra and Discrete Mathematics. — 2011. — Vol. 11, № 1. — С. 46–58. — Бібліогр.: 11 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:16U60, 16S34 http://dspace.nbuv.gov.ua/handle/123456789/154761 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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The structure of the unit groups of the group algebra of the groups D₁₂ and C₃⋊C₄ over any finite field of characteristic 3 is established. |
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Gildea, J. Monaghan, F. |
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Gildea, J. Monaghan, F. Units of some group algebras of groups of order 12 over any finite field of characteristic 3 Algebra and Discrete Mathematics |
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Gildea, J. Monaghan, F. |
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Gildea, J. |
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Units of some group algebras of groups of order 12 over any finite field of characteristic 3 |
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Units of some group algebras of groups of order 12 over any finite field of characteristic 3 |
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Units of some group algebras of groups of order 12 over any finite field of characteristic 3 |
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Units of some group algebras of groups of order 12 over any finite field of characteristic 3 |
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Units of some group algebras of groups of order 12 over any finite field of characteristic 3 |
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units of some group algebras of groups of order 12 over any finite field of characteristic 3 |
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Інститут прикладної математики і механіки НАН України |
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2011 |
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Units of some group algebras of groups of order 12 over any finite field of characteristic 3 / J. Gildea, F. Monaghan // Algebra and Discrete Mathematics. — 2011. — Vol. 11, № 1. — С. 46–58. — Бібліогр.: 11 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT gildeaj unitsofsomegroupalgebrasofgroupsoforder12overanyfinitefieldofcharacteristic3 AT monaghanf unitsofsomegroupalgebrasofgroupsoforder12overanyfinitefieldofcharacteristic3 |
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2025-07-14T06:52:08Z |
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2025-07-14T06:52:08Z |
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1837604199857127424 |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 11 (2011). Number 1. pp. 46 – 58
c© Journal “Algebra and Discrete Mathematics”
Units of some group algebras of groups of order
12 over any finite field of characteristic 3
Joe Gildea, Faye Monaghan
Communicated by M. Ya. Komarnytskyj
Abstract. The structure of the unit groups of the group
algebra of the groups D12 and C3 ⋊ C4 over any finite field of
characteristic 3 is established.
1. Introduction
Let U(RG) be the group of units of the group ring RG of the group G
over the commutative ring R. It is well known that
U(RG) = V (RG)× U(R),
where V (RG) = {
∑
g∈G αgg ∈ U(RG) |
∑
g∈G αg = 1} is the group of
normalized units of RG and U(R) is the group of units of R. For further
details and background see Polcino Milies and Sehgal [9].
We are interested in the structure of U(KG) when of order apk where
p is a prime, a, k ∈ N0 and (a, p) = 1. It is well known that |V (KG)| =
|K||G|−1 if G is a finite p-group and K is a finite field of characteristic p.
A basis for V (FpG) is determined in [10], where Fp is the Galois field of
p elements and G is an abelian p-group. Note that several results were
obtained in the case where K is a field of characteristic p and G is a
non-abelian p-group, see [1, 2, 3] for further detials.
In [6], the order of U(FpkD2pm) is determined where Fpk is the Galois
field of pk-elements, D2pm = 〈a, b | ap
n
= 1, b2 = 1, ab = a−1〉 is the
2000 Mathematics Subject Classification: 16U60, 16S34.
Key words and phrases: unit group, group ring, group algebra.
J. Gildea, F. Monaghan 47
dihedral group of order 2pm and p is an odd prime. In [4], the structure
of U(F3kD6) was determined in terms of split extensions of elementary
abelian groups. Additionally in [8], it is shown that Z(V1) and V1/Z(V1)
are elementary abelian 3-groups where V1 = 1+J(F3kD6), J(F3kD6) is the
Jacobson Radical of F3kD6 and Z(V1) is the center of V1. The structure
of FA4 is established in [11] where F is any finite field and A4 is the
alternating group of degree 4. Our main result is the following:
Theorem 1. Let V (FG) be the group of normalized units of the group
algebra FG of a group G over a finite field F of 3k elements. The following
conditions hold:
(i) If G = 〈x, y | x6 = y2 = 1, xy = x−1〉 ∼= D12, then
V (FG) ∼= (C3
6k ⋊ C3
2k)⋊ C3
3k−1
.
(ii) If G = 〈x, y | x4 = y3 = 1, yxy = x〉 ∼= C3 ⋊ C4, then
V (FG) ∼=
{
(C3
6k ⋊ C3
2k)⋊ C3
3k−1
if 4 | (3k − 1)
(C3
6k ⋊ C3
2k)⋊ (C3k−1 × C32k−1) if 4 ∤ (3k − 1).
The next two results can be found in [7].
Theorem 2. Let F be an arbitrary field of characteristic p > 0, let G be
a p-solvable group and c be the sum of all p-elements of G including 1,
then
J(FG) = lann(c)
where J(FG) is the Jacobson Radical of FG and lann(c) is the left anni-
hilator of c.
Theorem 3. Let N be a normal subgroup of G such that G/N is p-solvable.
If |G/N | = npa where (n, p) = 1, then
J(FG)p
a
⊆ FG · J(FN) ⊆ J(FG)
where F is a field of characteristic p > 0. In particular, if G is p-solvable
of order npa where (n, p) = 1, then
J(FG)p
a
= 0.
Denote by ĝ =
∑
h∈〈g〉 h ∈ RG.
48 Units of some group algebras
2. Proof of Main Theorem
Case (i). Let G = D12. Consider the ring homomorphism θ : F3kD12 →
F3k(C2 × C2) given by
2∑
i=0
x2i(ai + ai+3x
3 + ai+6y + ai+9x
3y) 7→
2∑
i=0
(ai + ai+3x̄+ ai+6ȳ + ai+9x̄ȳ).
where C2 × C2 = 〈x, y |x2 = y2 = 1, x y = y x〉 and αi ∈ F3k .
We can now construct the group homomorphism θ′ : U(F3kD12) →
U(F3k(C2 × C2)), since units gets mapped to units. We define the group
epimorphism φ : U(F3k(C2 × C2)) → U(F3kD12) given by a+ bx̄+ cȳ +
dx̄ȳ 7→ a+ bx3 + cy + dx3y where a, b, c, d ∈ F3k . Clearly θ′ ◦ φ = 1 and
therefore U(F3kD12) ∼= H ⋊ U(F3k(C2 × C2)) where H = ker(θ′). Let
α =
2∑
i=0
x2i(ai + ai+3x
3 + ai+6y + ai+9x
3y) ∈ F3kD12,
where αi ∈ F3k . Then α ∈ H iff
2∑
i=0
ai = 1,
2∑
j=0
aj+3 =
2∑
k=0
ak+6 =
2∑
l=0
al+9 = 0.
Lemma 1. H has exponent 3.
Proof. D12 is solvable and hence p-solvable. Clearly |D12| = 4.3 and by
Theorem 3, J(F3kD12)
3 = 0 and 1 + J(F3kD12) has exponent 3. Now by
Theorem 2, J(F3kD12) = {α ∈ F3kD12|αx̂2 = 0}. Let
α =
2∑
i=0
x2i
[
αi+1 + αi+4x
3 + αi+7y + αi+10x
3y
]
∈ U(F3kD12),
then
α ∈ J(F3kD12) ⇐⇒
3∑
i=1
αi =
3∑
j=1
αj+3 =
3∑
l=1
αl+6 =
3∑
m=1
αm+9 = 0
where αi ∈ F3k . Therefore H ∼= 1 + J(F3kD12).
J. Gildea, F. Monaghan 49
Lemma 2. Every element of the centralizer of x2 in H (CH(x2)) has the
form
2∑
i=0
x2i(αi+αi+3x
3)+ (α6+α7x
3)x̂2y where
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0
and αi ∈ F3k . Also CH(x2) ∼= C3
6k.
Proof. CH(x2) = {h ∈ H | hx2 = x2h}. Let h =
∑
2
i=0
x2i(αi + αi+3x
3 +
αi+6y + αi+9x
3y) ∈ H where αi ∈ F3k ,
∑
2
i=0
αi = 1 and
∑
2
i=0
αi+3 =∑
2
i=0
αi+6 =
∑
2
i=0
αi+9 = 0.
(hx2 − x2h) =
(
2∑
i=0
αi+6x
2iy +
2∑
i=0
αi+9x
2i+3y
)
x2−
− x2
(
2∑
i=0
αi+6x
2iy +
2∑
i=0
αi+9x
2i+3y
)
=
2∑
i=0
αi+6x
2i+4y +
2∑
i=0
αi+9x
2i+1y −
2∑
i=0
αi+6x
2i+2y −
2∑
i=0
αi+9x
2i+5y
=
2∑
i=0
[
αi+6x
2i+4y − αi+6x
2i+2y
]
+
2∑
i=0
[
αi+9x
2i+1y − αi+9x
2i+5y
]
= (α6 − α7) x̂2y + (α9 − α10)x
3x̂2y.
Therefore h ∈ CH(x2) if and only if α6 = α7 = α8 and α9 = α10 = α11.
Hence every element of CH(x2) has the form
∑
2
i=0
x2i(αi+αi+3x
3)+(α6+
α7x
3)x̂2y where
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0 and αi ∈ F3k . It remains to show
thatCH(x2) is abelian. Let α =
∑
2
i=0
x2i
(
αi + αi+3x
3
)
+(α6+α7x
3)x̂2y ∈
CH(x2) and β =
∑
2
i=0
x2i
(
βi + βi+3x
3
)
+(β6+β7x
3)x̂2y ∈ CH(x2) where
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0,
2∑
i=0
βi = 1,
2∑
i=0
βi+3 = 0 and αi, βj ∈ F3k . Note
that (x̂2)2 = 0. Then
αβ =
2∑
i=0
x2i
(
ai + ai+3x
3
) 2∑
i=0
x2i
(
bi + bi+3x
3
)
+
2∑
i=0
x2i
(
ai + ai+3x
3
)
(b6 + b7x
3)x̂2y
+ (a6 + a7x
3)x̂2y
2∑
i=0
(
bix
2i + bi+3x
2i+3
)
50 Units of some group algebras
+ (a6 + a7x
3)x̂2y(b6 + b7x
3)x̂2y
=
2∑
i=0
x2i
(
ai + ai+3x
3
) 2∑
i=0
x2i
(
bi + bi+3x
3
)
+
2∑
i=0
(
ai + ai+3x
3
)
x̂2(b6 + b7x
3)y
+
2∑
i=0
(
bi + bi+3x
3
)
x̂2(a6 + a7x
3)y
=
2∑
i=0
x2i
(
ai + a3x
2i+3
) 2∑
i=0
x2i
(
bi + bi+3x
3
)
+ x̂2(b6 + b7x
3)y + x̂2(a6 + a7x
3)y
=
2∑
i=0
x2i
(
ai + ai+3x
3
) 2∑
i=0
x2i
(
bix
2i + bi+3x
2i+3
)
+ ((a6 + b6) + (a7 + b7)x
3)x̂2y
and
βα =
2∑
i=0
x2i
(
bi + bi+3x
3
) 2∑
i=0
x2i
(
ai + ai+3x
3
)
+
2∑
i=0
x2i
(
bi + bi+3x
3
)
(a6 + a7x
3)x̂2y
+ (b6 + b7x
3)x̂2y
2∑
i=0
x2i
(
ai + ai+3x
3
)
+ (b6 + b7x
3)x̂2y(a6 + a7x
3)x̂2y
=
2∑
i=0
x2i
(
bi + bi+3x
3
) 2∑
i=0
x2i
(
ai + ai+3x
3
)
+
2∑
i=0
(
bi + bi+3x
3
)
x̂2(a6 + a7x
3)y
+
2∑
i=0
(
ai + ai+3x
3
)
x̂2(b6 + b7x
3)y
=
2∑
i=0
x2i
(
bi + bi+3x
3
) 2∑
i=0
x2i
(
ai + ai+3x
3
)
J. Gildea, F. Monaghan 51
+ x̂2(a6 + a7x
3)y + x̂2(b6 + b7x
3)y
=
2∑
i=0
x2i
(
bi + bi+3x
3
) 2∑
i=0
x2i
(
ai + ai+3x
3
)
+ ((a6 + b6) + (a7 + b7)x
3)x̂2y.
Let γ =
∑
2
i=0
x2i
(
βi + βi+3x
3
)
∈ H and δ =
∑
2
i=0
x2i
(
αi + αi+3x
3
)
∈
H . Clearly γ, δ ∈ V (F3kC6) and γδ = δγ since V (F3kC6) is abelian. There-
fore CH(x2) is abelian.
Lemma 3. Let S be the subgroup of H consisting of elements of the form
1+ x̆(r+ r1x
3)(1 + y) where x̆ =
2∑
i=0
ix2i and r, r1 ∈ F3k . Then S ∼= C3
2k.
Proof. Let s1 = 1+x̆(r+r1x
3)(1+y) ∈ S and s2 = 1+x̆(s+s1x
3)(1+y) ∈
S where r, s, r1, s1 ∈ F3k , then
s1s2 = 1 + x̆(s+ s1x
3)(1 + y) + x̆(r + r1x
3)(1 + y)
+ x̆(r + r1x
3)(1 + y)x̆(s+ s1x
3)(1 + y)
= 1 + x̆
(
(s+ r) + (s1 + r1)x
3
)
(1 + y)
+ x̆(r + r1x
3)(1 + y)(1− y)x̆(s+ s1x
3)
= 1 + x̆
(
(s+ r) + (s1 + r1)x
3
)
(1 + y)
+ x̆(r + r1x
3)(1− 1)x̆(s+ s1x
3)
= 1 + x̆
(
(s+ r) + (s1 + r1)x
3
)
(1 + y).
Therefore S is closed.
Lemma 4. H ∼= CH(x2)⋊ S.
Proof. Clearly H = CH(x2).S since CH(x2) ∩ S = {1}. It remains to
show that S normalizes CH(x2). Let s = 1 + x̆(r + r1x
3)(1 + y) , then
s−1 = 1− x̆(r+ r1x
3)(1 + y) where r, r1 ∈ F3k . Also let c =
∑
2
i=0
x2i(αi +
αi+3x
3) + (α6 + α7x
3)x̂2y ∈ CH(x2) where
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0 and
αi ∈ F3k . Note that x̆2 = x̂2x̆ = 0. Then
cs =
(
1− x̆(r + r1x
3)(1 + y)
)( 2∑
i=0
x2i(αi + αi+3x
3) + (α6 + α7x
3)x̂2y
)
×
(
1 + x̆(r + r1x
3)(1 + y)
)
52 Units of some group algebras
=
(
1− x̆(r + r1x
3)(1 + y)
)
(
2∑
i=0
x2i(αi + αi+3x
3)
+
2∑
i=0
x2i(αi + αi+3x
3)(x̆(r + r1x
3)(1 + y)) + (α6 + α7x
3)x̂2y
)
=
2∑
i=0
x2i(αi + αi+3x
3) +
2∑
i=0
x2i(αi + αi+3x
3)(x̆(r + r1x
3)(1 + y))
+ (α6 + α7x
3)x̂2y − x̆(r + r1x
3)(1 + y)
2∑
i=0
x2i(αi + αi+3x
3)
=
2∑
i=0
x2i(αi + αi+3x
3) +
2∑
i=0
x2i(αi + αi+3x
3)(x̆(r + r1x
3)(y))
+ (α6 + α7x
3)x̂2y − x̆(r + r1x
3)(y)
2∑
i=0
x2i(αi + αi+3x
3)
=
(
2∑
i=0
x2i(αi + αi+3x
3) + ((α6 + δ) + (α7 + δ)x3)x̂2y
)
∈ CH(x2)
where δ = r(α1 − α2) + r1(α4 − α5) and αi, r, r1 ∈ F3k .
Recall that U(F3kD12) ∼= H⋊U(F3k(C2×C2)). Consider F3k(C2×C2).
F3k(C2 × C2) ∼= (F3kC2)C2
∼= (F3k ⊕ F3k)C2
∼= F3kC2 ⊕ F3kC2
∼= F3k ⊕ F3k ⊕ F3k ⊕ F3k .
Thus U(F3k(C2 × C2)) ∼= C3k−1
4. Therefore
U(F3kD12) ∼= [C3
6k × C3
2k]⋊ C3k−1
4
∼= [(C3
6k × C3
2k)⋊ C3k−1
3]× U(F3k)
Case (ii). Let G ∼= C3 ⋊ C4. Define the group epimorphism
τ : U(F3k(C3 ⋊ C4)) −→ U(F3kC4)
given by
J. Gildea, F. Monaghan 53
2∑
i=0
(αi + αi+3x+ αi+6x
2 + αi+9x
3)yi 7→
2∑
i=0
(αi + αi+3x+ αi+6x
2 + αi+9x
3)
where x generates C4.
Define the group homomorphism φ : U(F3kC4) −→ U(F3k(C3 ⋊ C4))
by a+ bx+ cx2 + dx3 7→ a+ bx+ cx2 + dx3 where a, b, c, d ∈ F3k . Then
τ ◦ φ = 1, therefore U(F3k(C3 ⋊ C4)) ∼= L⋊ U(F3kC4) where L ∼= ker(τ).
Let
l =
2∑
i=0
(αi + αi+3x+ αi+6x
2 + αi+9x
3)yi ∈ U(F3k(C3 ⋊ C4)),
then l ∈ L if and only if
∑
2
i=0
αi = 1 and
2∑
j=0
αj+3 =
2∑
l=0
αl+6 =
2∑
m=0
αm+9 = 0 (αi ∈ F3k)
Therefore |L| = 38k.
Lemma 5. L has exponent 3.
Proof. By Theorem 3, J(F3k(C3⋊C4))
3 = 0 and 1+J(F3k(C3⋊C4)) has
exponent 3. By Theorem 2, J(F3k(C3⋊C4)) = {α ∈ F3k(C3⋊C4)|αŷ = 0}.
Therefore L ∼= 1 + J(F3k(C3 ⋊ C4)).
Lemma 6. Every element of the centralizer of y in L (CL(y)) has the
form
2∑
i=0
(αi + αi+4x
2)yi + (α3x+ α7x
3)ŷ where
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0
and αi ∈ F3k . Also CL(y) ∼= C3
6k.
Proof. CL(y) = {l ∈ L | ly = yl}. Let
l =
2∑
i=0
(αi + αi+3x+ αi+6x
2 + αi+9x
3)yi ∈ L.
where
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0,
2∑
i=0
αi+6 = 0,
2∑
i=0
αi+9 = 0 and αi ∈ F3k .
54 Units of some group algebras
Then
ly − yl =
(
2∑
i=0
(αi+3x+ αi+9x
3)yi
)
y − y
(
2∑
i=0
(αi+3x+ αi+9x
3)yi
)
=
2∑
i=0
(αi+3x+ αi+9x
3)yi+1 −
2∑
i=0
(αi+3yx+ αi+9yx
3)yi
= (α5 − α4)xŷ + (α11 − α10)x
3ŷ.
Therefore l ∈ CL(y) if and only if α3 = α4 = α5 and α9 = α10 = α11.
Thus every element of CL(y) is of the form
2∑
i=0
(αi + αi+4x
2)yi + (α3x+ α7x
3)ŷ
where
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0 and αi ∈ F3k . Let α =
∑
2
i=0
(αi+αi+4x
2)yi+
(α3x+ α7x
3)ŷ ∈ CL(y) and β =
∑
2
i=0
(βi + βi+4x
2)yi + (β3x+ β7x
3)ŷ ∈
CL(y)
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0,
2∑
i=0
βi = 1,
2∑
i=0
βi+3 = 0 and αi, βj ∈ F3k .
Then
αβ =
2∑
i=0
(αi + αi+4x
2)yi
2∑
i=0
(βi + βi+4x
2)yi
+
2∑
i=0
(αi + αi+4x
2)yi(β3x+ β7x
3)ŷ
+ (α3x+ α7x
3)ŷ
2∑
i=0
(βi + βi+4x
2)yi + (α3x+ α7x
3)ŷ(β3x+ β7x
3)ŷ
=
2∑
i=0
(αi + αi+4x
2)yi
2∑
i=0
(βi + βi+4x
2)yi
+
2∑
i=0
(αi + αi+4x
2)ŷ(β3x+ β7x
3) + (α3x+ α7x
3)
2∑
i=0
(βi + βi+4x
2)ŷ
=
2∑
i=0
(αi + αi+4x
2)yi
2∑
i=0
(βi + βi+4x
2)yi
+ ŷ(β3x+ β7x
3) + (α3x+ α7x
3)ŷ
J. Gildea, F. Monaghan 55
=
2∑
i=0
(αi + αi+4x
2)yi
2∑
i=0
(βi + βi+4x
2)yi
+ ((α3 + β3)x+ (α7 + β7)x
3)ŷ
βα =
2∑
i=0
(βi + βi+4x
2)yi
2∑
i=0
(αi + αi+4x
2)yi
+
2∑
i=0
(βi + βi+4x
2)yi(α3x+ α7x
3)ŷ
+ (β3x+ β7x
3)ŷ
2∑
i=0
(αi + αi+4x
2)yi + (β3x+ β7x
3)ŷ(α3x+ α7x
3)ŷ
=
2∑
i=0
(βi + βi+4x
2)yi
2∑
i=0
(αi + αi+4x
2)yi
+
2∑
i=0
(βi + βi+4x
2)ŷ(α3x+ α7x
3)
+ (β3x+ β7x
3)
2∑
i=0
(αi + αi+4x
2)ŷ
=
2∑
i=0
(βi + βi+4x
2)yi
2∑
i=0
(αi + αi+4x
2)yi
+ ŷ(α3x+ α7x
3) + (β3x+ β7x
3)ŷ
=
2∑
i=0
(αi + αi+4x
2)yi
2∑
i=0
(βi + βi+4x
2)yi
+ ((α3 + β3)x+ (α7 + β7)x
3)ŷ
Let γ =
∑
2
i=0
(αi + αi+4x
2)yi ∈ L and δ =
∑
2
i=0
(βi + βi+4x
2)yi ∈ L.
Clearly γ, δ ∈ V (F3kC6) since 〈x2, y〉 ∼= C6 and γδ = δγ. Therefore
CL(y) ∼= C3
6k.
Lemma 7. Let T be the subgroup of L consisting of elements of the form
1 + (a + bx2)ŷ + (rx + r1x
3)y̌ where y̌ =
2∑
i=0
iyi and r, r1 ∈ F3k . Then
T ∼= C3
4k.
Proof. Let t1 = 1 + (a1 + b1x
2)ŷ + (rx+ r1x
3)y̌ ∈ T and t2 = 1 + (a2 +
56 Units of some group algebras
b2x
2)ŷ + (sx+ s1x
3)y̌ ∈ T where ai, bj , r, r1, s, s1 ∈ F3k . Then
t1t2 = 1 + (a2 + b2x
2)ŷ + (sx+ s1x
3)y̌ + (a1 + b1x
2)ŷ
+ (a1 + b1x
2)ŷ(a2 + b2x
2)ŷ + (a1 + b1x
2)ŷ(sx+ s1x
3)y̌
+ (rx+ r1x
3)y̌ + (rx+ r1x
3)y̌(a2 + b2x
2)ŷ
+ (rx+ r1x
3)y̌(sx+ s1x
3)y̌
= 1 + ((a1 + a2) + (b1 + b2)x
2)ŷ + ((r + s)x+ (r1 + s1)x
3)y̌
+ (rx+ r1x
3)y̌(sx+ s1x
3)y̌
= 1 + ((a1 + a2 + δ1) + (b1 + b2 + δ2)x
2)ŷ
+ ((r + s)x+ (r1 + s1)x
3)y̌ ∈ T
where δ1 = 2(rs1 + sr1) and δ2 = 2(rs + r1s2). It can easily be shown
that T is abelian. Therefore T ∼= C3
4k.
Lemma 8. L ∼= C3
6k ⋊ C3
2k.
Proof. Let t = 1 + (a + bx2)ŷ + (rx + r1x
3)y̌ ∈ T and c =
2∑
i=0
(αi +
αi+4x
2)yi + (α3x + α7x
3)ŷ ∈ CL(y) where
2∑
i=0
αi = 1,
2∑
i=0
αi+3 = 0 and
αi, a, b, r, r1 ∈ F3k . Then
ct = (1 + (a+ bx2)ŷ + (rx+ r1x
3)y̌)2
×
(
2∑
i=0
(αi + αi+4x
2)yi + (α3x+ α7x
3)ŷ
)
× (1 + (a+ bx2)ŷ + (rx+ r1x
3)y̌)
= (1− (a+ bx2)ŷ − (rx+ r1x
3)y̌ − (rx+ r1x
3)2ŷ)
×
(
2∑
i=0
(αi + αi+4x
2)yi + (a+ bx2)ŷ
+
2∑
i=0
(αi + αi+4x
2)yi(rx+ r1x
3)y̌ + (α3x+ α7x
3)ŷ
)
=
2∑
i=0
(αi + αi+4x
2)yi +
2∑
i=0
(αi + αi+4x
2)yi(rx+ r1x
3)y̌
+ (α3x+ α7x
3)ŷ − (rx+ r1x
3)y̌
2∑
i=0
(αi + αi+4x
2)yi
J. Gildea, F. Monaghan 57
=
2∑
i=0
(αi + αi+4x
2)yi + ((α3 + δ1)x+ (α7 + δ2)x
3)ŷ ∈ CL(y)
where δ1 = r(α2 − α1) + r1(α6 − α5), δ2 = r(α6 − α5) + r1(α2 − α1).
Clearly ct ∈ CL(y) and T normalizes CL(y). Let
U = CL(y) ∩ T =
{
1 + (a+ bx2)ŷ | a, b ∈ F3k
}
.
Clearly |CL(y) ∩ T | = 32k and by the second Isomorphism Theorem
|CL(y)T | = 38k. Therefore L = CL(y)T . T is an elementary abelian 3-
group (Y ∼= U × V ). Clearly V ∩ CL(y) = {1} and V normalizes CL(y).
Therefore L = CL(y)V ∼= C3
6k ⋊ C3
2k.
Recall that U(F3k(C3 ⋊ C4)) ∼= L⋊ U(F3kC4). Now
F3kC4
∼=
{
F3k
4 if 4 | (3k − 1)
F3k
2 ⊕ F32k if 4 ∤ (3k − 1).
F3kC4
∼= F3k
2 ⊕ F32k . Thus U(F3kC4) ∼= C3k−1
4 if 4 | (3k − 1) or
U(F3kC4) ∼= C3k−1
2 × C32k−1 if 4 ∤ (3k − 1). Therefore
U(F3k(C3 ⋊ C4)) ∼=
∼=
{
[(C3
6k ⋊ C3
2k)⋊ C3
3k−1
]× U(F3k) if 4 | (3k − 1)
[(C3
6k ⋊ C3
2k)⋊ (C3k−1 × C32k−1)]× U(F3k) if 4 ∤ (3k − 1).
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Contact information
J. Gildea,
F. Monaghan
School of Engineering, Institute of Technology
Sligo, Ireland
E-Mail: gildea.joe@itsligo.ie,
fayemonaghan@hotmail.com
URL: http://www.itsligo.ie/staff/jgildea
Received by the editors: 16.02.2011
and in final form 07.05.2011.
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