UNITS IN FINITE DIHEDRAL AND QUATERNION GROUP ALGEBRAS
NEHA MAKHIJANI*, R. K. SHARMA AND J. B. SRIVASTAVA
Department Of Mathematics, Indian Institute of Technology Delhi,
New Delhi, India
email : [email protected], [email protected], [email protected]
Abstract. Let Fq G be the group algebra of a finite group G over Fq = GF (q). Using the Wedderburn
decomposition of F2k D2n /J(F2k D2n ), we establish the structure of the unit group of F2k G when G is either
D4n , the dihedral group of order 4n or Q4n , the generalized quaternion group of order 4n, n odd.
1. Introduction
Let F G be the group algebra of a finite group G over a field F and U(F G) be its unit group. The study of the
group of units is one of the classical topics in group ring theory. Results obtained in this direction are useful for
the investigation of Lie properties of group rings, isomorphism problem and other open questions in this area [1].
In [2], Bovdi gave a comprehensive survey of results concerning the group of units of a modular group algebra
of characteristic p. There is a long tradition on the study of the unit group of finite group algebras [3–12]. In
general, the structure of U(F G) is elusive if |G| = 0 in F .
Let us introduce the background of our investigation. The structure of U(F2 D2p ) was determined by Kaur and
Khan in [13] for an odd prime p. Recently, the authors generalized this result and computed the structure of the
unit group of F2k D2n when n is odd. In this note, we use the Wedderburn decomposition of F2k D2n /J(F2k D2n )
obtained in [14] to study the unit group of F2k D4n and F2k Q4n when n is odd.
In what follows, q = 2k , ordl (m) denotes the multiplicative order of m modulo l when (l, m) = 1 and ϕ(n)
denotes the Euler’s phi function on a positive integer n.
2. Main Results
In this section, we begin by considering the lemmas that are essential for developing the proof of main results.
Lemma 2.1. Let F be a perfect field, G be a finite group and J(F G) be the Jacobson radical of F G. Then
FG
∼
U(F G) = (1 + J(F G)) o U
J(F G)
Proof. Observe that
1
1 + J(F G)
inc
U(F G)
ψ
U
FG
J(F G)
1
MSC(2010): Primary: 16U60; Secondary: 16S34, 20C05.
Keywords: Group Algebra, Unit Group, Wedderburn decomposition, Jacobson radical.
∗ Corresponding
author.
1
2
NEHA MAKHIJANI*, R. K. SHARMA AND J. B. SRIVASTAVA
is a short exact sequence of groups, where ψ(x) = x + J(F G) ∀ x ∈ U(F G).
By Wedderburn-Malcev theorem [15, Thm. 6.2.1], it follows that there exists a semisimple subalgebra B of F G
such that
F G = B ⊕ J(F G)
FG
and thus for each x + J(F G) ∈
, there exists a unique xB ∈ B such that
J(F G)
x + J(F G) = xB + J(F G)
Define θ : U
FG
J(F G)
→ U(F G) as
θ (x + J(F G)) = xB
∀ x + J(F G) ∈ U
FG
J(F G)
Then θ is a group homomorphism such that ψ o θ = id | U (F G/J(F G)) and hence
FG
∼
U(F G) = (1 + J(F G)) o U
J(F G)
For a normal subgroup H of G, the natural homomorphism εH : G → G/H can be extended to an F -algebra
epimorphism ε∗H : F G → F (G/H). The kernel of ε∗H is denoted by ∆(G, H) and ∆(G) = ∆(G, G).
Lemma 2.2. [16, Lemma 1.17] Let G be a locally finite p-group and F be a field of characteristic p. Then
J(F G) = ∆(G).
Lemma 2.3. [17, Ch. 1, Prop. 6.16] Let f : R1 → R2 be a surjective homomorphism of rings. Then
f (J(R1 )) ⊆ J(R2 )
with equality if ker f ⊆ J(R1 ).
Lemma 2.4. [18, Theorem 7.2.7] Let H be a normal subgroup of G with [G : H] = n < ∞. Then J(F G)n ⊆
J(F H)F G ⊆ J(F G). If in addition n 6= 0 in F , then J(F G) = J(F H)F G.
Lemma 2.5. Let N = 2t n such that 2 - n. Then
Fq Q4N /J(Fq Q4N ) ∼
= Fq D2N /J(Fq D2N )
∼
= Fq ⊕
⊕
m|n, m>1
ϕ(m)
M (2, Fqem ) 2em
where
(
em =
dm /2
if dm is even and q dm /2 ≡ −1 mod m
dm
otherwise
and dm = ordm (q).
Proof. To distinguish the elements of D2N from those of D2n , let D2N be presented by
h A, B | AN , B 2 , B −1 AB = A−1 i
and D2n by
h a, b | an , b2 , b−1 ab = a−1 i
FINITE DIHEDRAL AND QUATERNION GROUP ALGEBRAS
3
From [14], it is known that
Fq D2n /J(Fq D2n ) ∼
=
Fq ⊕
⊕
m|n, m>1
ϕ(m)
M (2, Fqem ) 2em
Now
∆(D2N , h An i)
=
∆(h An i) Fq D2N
= J(Fq h An i) Fq D2N
⊆ J(Fq D2N )
showing that dimFq J (Fq D2N ) ≥ 2N − 2n.
Since D2N /h An i ∼
= D2n , there exists an onto Fq -algebra homomorphism
φ : Fq D2N → Fq D2n /J (Fq D2n )
given by the assignment A 7→ a + J (Fq D2N ) , β 7→ b + J (Fq D2N ) whence J (Fq D2N ) ⊆ ker φ and
dimFq J (Fq D2N ) ≤ 2N − (2n − 1) = 2N − 2n + 1
But there is only one 1-dimensional representation of D2N over Fq . This proves that dimFq J (Fq D2N ) =
2N − 2n + 1 and J (Fq D2N ) = ker φ giving
Fq D2N /J (Fq D2N ) ∼
= Fq D2n /J (Fq D2n )
The decomposition of Fq Q4N /J (Fq Q4N ) can be obtained by working on parallel lines.
Theorem 2.6. If n is odd, then


(2n+1)k
U(Fq D4n ) ∼
o Cq−1 ×
= C2
Y
ϕ(m)
GL(2, Fqem ) 2em 
m|n, m>1
where
(
em =
dm /2
if dm is even and q dm /2 ≡ −1 mod m
dm
otherwise
and dm = ordm (q).
Proof. Let D4n = h α, β | α2n , β 2 , β −1 αβ = α−1 i and X = 1+αn . Then { X, αX, · · · , αn−1 X, βX, βαX, · · · ,
βαn−1 X } is a basis of ∆(D4n , h αn i).
Observe that any W ∈ ∆(D4n , h αn i) is expressible as
W = A1 + A2 α + · · · + An αn−1 + An+1 β + An+2 βα + · · · + A2n βαn−1 X
for some Ai ∈ Fq so that
W2
=
=
A1 + A2 α + · · · + An αn−1 + An+1 β + An+2 βα + · · · + A2n βαn−1
0
That is, 1 + ∆(D4n , h αn i) ∼
= C22nk .
The Fq -algebra homomorphism
θ : Fq D4n → Fq D2n /J(Fq D2n )
2
(1 + αn )2
4
NEHA MAKHIJANI*, R. K. SHARMA AND J. B. SRIVASTAVA
given by the assignment
α 7→ a + J(Fq D2n ), β 7→ b + J(Fq D2n )
n−1
d
is onto and it is known that D
)(1+β), then θ(B) = 0+J(Fq D2n )
2n ∈ J(Fq D2n ). Thus if B = (1+α+· · ·+α
showing that B ∈ ker θ = J(Fq D4n ). In fact, J(Fq D4n ) = ∆(D4n , h αn i) ⊕ Fq B as a vector space over Fq .
Since
B2
=
((1 + β)(1 + α + · · · + αn−1 ))2
=
(1 + β)(1 + βαn+1 ) 1 + α + · · · + αn−1
=
(1 + β + αn+1 + βαn+1 ) 1 + α2 + · · · + α2n−2
=
1 + α2 + · · · + α2n−2 + β + βα2 + · · · + βα2n−2 +
2
αn+1 + αn+3 + · · · + α3n−1 + βαn+1 + βαn+3 + · · · + βα3n−1
0 because n is odd and α2n = 1.
=
and
XB
(1 + αn )(1 + α + · · · + αn−1 )(1 + β) = α
b(1 + β),
=
we find that W B = BW so that
V
=
1 + J(Fq D4n )
=
1 + ∆(D4n , h αn i) × { 1 + ηB | η ∈ Fq }
∼
=
C2
(2n+1)k
This completes the proof.
A group G is said to be the general product of its subgroups L and M if
G = LM, L ∩ M = {1}
In this case, we write G = L o M .
In the subsequent theorem, it is established that 1+J(Fq Q4n ) is a general product of two of its proper subgroups.
As a consequence, the structure of U(Fq Q4n ) is obtained.
i
i
i
Lemma 2.7. [3, Lemma 1.1] Let G be a finite abelian p-group, Gp = { xp | x ∈ G } and pmi = |Gp |. If
k
∼ Q C nii , then
G =
p
i=1
ni = mi−1 − 2mi + mi+1 ∀ 1 ≤ i ≤ k
Theorem 2.8. If n is odd, then

U(Fq Q4n ) ∼
=
(2n−2)k
C2
o
C2k × C4k
o  Cq−1 ×

Y
m|n, m>1
where em as in Theorem 2.6.
Proof. Let Q4n be presented by
h C, D | C 2n , D2 = C n , D−1 CD = C −1 i
GL(2, Fqem )
ϕ(m)
2em

FINITE DIHEDRAL AND QUATERNION GROUP ALGEBRAS
5
Let U = (1 + D)(1 + C + · · · + C n−1 ). Then via similar arguments as in the previous theorem, U ∈ J(Fq Q4n )
and
U2
=
=
b + DC
b
(1 + D) C
b
(1 + D) (1 + C n ) + DC
1 + C + · · · + C n−1
b
= C
b (1 + D)C,
b U } is a basis
Notice that if Y = 1 + C n , then { Y, CY, · · · , C n−2 Y, DY, DCY, · · · , DC n−2 Y, C,
of J(Fq Q4n ) over Fq .
Since Y ∈ Z(Fq Q4n ) and Y 2 = 0, it is therefore evident that
(
!
n−2
n−2
X
X
i
i
H= 1 +
Ai C +
Bi DC
Y
i=0
and H ∼
=
i=0
)
Ai , Bi ∈ Fq ≤ 1 + J(Fq Q4n )
(2n−2)k
C2
.
Also K =
n
o
b + A3 (1 + D)C
b Ai ∈ Fq ≤ 1 + J(Fq Q4n ) and by Lemma 2.7, we find K ∼
1 + A1 C + A2 C
=
C2k × C4k .
Since H ∩ K = { 1 } and |1 + J(Fq Q4n )| = |HK|, it follows that 1 + J(Fq Q4n ) = H o K. This completes the
proof.
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