volume 73 no. 2 2011, 143-158 the unit groups of fg of ... · and 2008, khan, sharma and srivastava...

International Journal of Pure and Applied Mathematics————————————————————————–Volume 73 No. 2 2011, 143-158

THE UNIT GROUPS OF FG OF GROUPS WITH ORDER 12

Tang Gaohua1 §, Gao Yanyan2

School of Mathematical SciencesGuangxi Teachers Education UniversityNanning, Guangxi 530023, P.R. CHINA

Abstract: Note that there are five mutually non-isomorphic groups of order12: two decomposable Abelian groups G1

∼= C3 × C4 and G2∼= C3×K4; three

indecomposable non-Abelian groups A4, Q12 and D12. For a finite field F , thestructure of U (FA4) was determined by R.K. Sharma, J.B. Srivastava and M.Khan in 2007. In this paper, we determine the structure of the unit groupU (FG) of the group ring FG, where F is a finite field, and G= G1, G2, Q12

or D12.

AMS Subject Classification: 16S34, 16U60, 20C05Key Words: finite field, group ring, unit group

1. Introduction

Let FG be the group ring of a group G over a field F . For a normal subgroupH of G, the natural homomorphism G → G/H : g 7→ gH can be extended toan F -algebra homomorphism from FG to F (G/H) defined by

∑g∈G

agg 7→∑g∈G

ag

gH, for ag ∈ F . The kernel of this homomorphism, denoted by ω(H), is theideal of FG generated by {h − 1 | h ∈ H} in FG. Of course, it is clear thatFG/ω(H) ∼= F (G/H). The ideal ω(G) is known as the augmentation ideal of

Received: June 13, 2011 c© 2011 Academic Publications, Ltd.§Correspondence author

144 T. Gaohua, G. Yanyan

the group ring FG, which is also denoted by ω(FG). It can be seen that ω(H) =ω(FH)FG = FG ω(FH). For convenience, we use ωn(H) to denote (ω(H))n.Also FG/ω(G)∼= F implies that the Jacobson radical, J(FG), is contained inω(FG). It is known that for an ideal I ⊆ J(FG), the natural homomorphismfrom FG to FG/I induces an epimorphism from the unit group of FG, U (FG),to U (FG/I) with kernel 1 + I, so that U (FG)/(1 + I)∼=U (FG/I).

The lower central chain of G is given by

G = γ1(G) ⊇ γ2(G) ⊇ · · · ⊇ γn+1(G) ⊇ · · ·where γm+1(G) = (γm(G),G), form ≥1. For g1, g2 ∈ G, the commutator (g1, g2)= g−1

1 g−12 g1g2. The group G is said to be nilpotent of class n if γn+1(G) = (1)

and γn(G) 6=(1).

Note that there are two Abelian groups of order 12: G1∼= C3 × C4 and

G2∼= C3 × K4, both of them are decomposable. And there are three non-

Abelian groups of order 12, all of them are indecomposable:A4 = 〈a, b | a3 = 1, b2 = 1, (ab)3 = 1〉

= {1, a, a2, b, ba, ba2, ab, aba, aba2, a2b, a2ba, a2ba2},Q12 = 〈a, b | a6 = 1, b2 = a3, ab = ba−1〉

= {1, a, a2, a3, a4, a5, b, ab, a2b, a3b, a4b, a5b}, andD12 = 〈a, b | a6 = 1, b2 = 1, ab = ba−1〉

= {1, a, a2, a3, a4, a5, b, ab, a2b, a3b, a4b, a5b}.

As we know, A4, Q12 and D12 are the alternative group of degree 4, thequaternion group of order 12 and the dihedral group of order 12, respectively.

Given a group ring RG and a finite subset X of the group G, we shalldenote X the following element of RG: X =

∑x∈X

x. In addition, the all distinct

conjugate classes of Q12 and D12 both are C 1 = {1}, C 2 = {a, a5}, C 3 = {a2,a4}, C 4 = {a3}, C 5 = {b, a2b, a4b}, C 6 = {ab, a3b, a5b}. By [6, Theorem 3.6.2],

{C1, C2, C3, C4, C5, C6} forms a basis of the center Z(FQ12) and it also formsa basis of Z(FD12), where Ci denotes the class sum, i =1, 2, . . . , 6.

On the study of unit groups of integral group rings ZG there are manyliteratures, such as [6], [7], [11], [12], et al). Recently, the structure of the unitgroup U (FG) over a finite field F were studied for some little groups. In 2007and 2008, Khan, Sharma and Srivastava determined the structure of the unitgroup U (FG) for groups G = A4 and S4 in [16] and [9], respectively. From2008, Gildea and his cooperators have made many investigations on the unitgroup of U (FG) mainly for the special case that the order of G is dividedby the characteristic of F (for details see [1-5]). In this paper we completelydetermine the structure of the unit group U (FG) over any finite field F andall groups of order 12.

THE UNIT GROUPS OF FG OF GROUPS WITH ORDER 12 145

In what follows F is a finite field, M(n, F ) is the algebra of all n×nmatricesover F , GL(n, F ) is the general linear group of degree n over F , char(F ) is thecharacteristic of F , F ∗ is the multiplicative group of all nonzero elements of F ,Fn is the extension field of F with degree n, Cn is the cyclic group of order nand Ck

n is the direct product of k copies of Cn.

2. Lemmas

Lemma 2.1. (see [8, Lemma 1.17]) Let G be a locally finite p-group, andlet F be a field of characteristic p. Then J(FG) = ω(FG).

Lemma 2.2. (see [8, Proposition 8.1.20]) Let R be a commutative noethe-rian ring and let G be an arbitrary group. Then there exists finitely manyindecomposable rings R1, R2, · · · , Rn such that RG ∼= R1G×R2G×· · ·×RnG.In particular, U (RG) ∼= U (R1G) × U (R2G) × · · · × U (RnG).

Lemma 2.3. (see [10, Lemma 4.11]) If a left (resp., right) ideal I ⊆ R isnil, then I ⊆ J(R).

Lemma 2.4. (see [13, Exercise 4, Page 134]) Let R be a commutative ringand let G, H be groups. Then R(G×H) ∼= (RG)H (the group ring of H overthe ring RG).

Lemma 2.5. (see [13, Theorem 2.6.8]) (Wedderburn-Artin) A ring R issemisimple if and only if it is a direct sum of finite number of matrix algebrasover division rings:

R ∼= M(n1,D1)⊕M(n2,D2)⊕ · · · ⊕M(nr,Dr).

Lemma 2.6. (see [13, Corollary 3.4.8]) Let G be a finite group and let Fbe a field. Then, FG is semisimple if and only if char(F ) ∤ |G|.

Lemma 2.7. (see [14, Exercise 3.6, Page 154]) Let p be an odd prime,and let:

(i) If p ≡ 3(mod 4), then for any integer a we have a2 6≡ −1 (mod p);

(ii) If p ≡ 1(mod 4), then there exists an integer a such that a2 ≡ −1 (modp).

Lemma 2.8. (see [15, Theorem 7.2.7]) Let H be a normal subgroup ofG with [G : H] = n < ∞. Then (J(KG))n ⊆ J(KH)KG ⊆ J(KG). If inaddition n 6= 0 in K, then J(KG) = J(KH)KG.

Lemma 2.9. (see [15, Exercise 9, Page 29]) Let F be a field of character-istic 6= 2 and let K4 be the Klein′s four-group. Then FK4

∼= F ⊕ F ⊕ F ⊕ F .


3. The unit groups of FG1 and FG2

In this section, we give the structures of U (FG1) and U (FG2), where G1∼=

C3 × C4, G2∼= C3 × K4.

Lemma 3.1. Let p be a prime and F an extension field of Zp with ordern. If p ≡ 3 (mod 4) and n is odd, then x2 = −1 has no solution in F .

Proof. (i) If n = 1, by Lemma 2.7, we know that x2 = − 1 has no solutionin Zp.

(ii) Assume that n ≥ 3. If x2 = − 1 has a solution, saying β, then β is notin Zp and β2 +1 = 0. Thus

Zp ⊂ Zp[β] ⊂ F and [Zp[β] : Zp] = 2.which contradicts to [F : Zp] = n and n is odd.This completes the proof of Lemma 3.1.

Lemma 3.2. Let p be a prime and F a finite field of characteristic p with|F | = pn. If p ≡3 (mod 4) and n is odd, then x2 = 1 has exactly eight roots inFC4. Therefore U (FC4) has exactly seven elements of order two.

Proof. Let C4 = 〈τ | τ4 = 1〉 = {1, τ, τ2, τ3}. ∀ x = β0+β1τ+β2τ2+β3τ

3∈FC4, then

x2 = 1 ⇐⇒

β20 + 2β1β3 + β2

2 = 1β21 + 2β0β2 + β2

3 = 0β0β1 + β2β3 = 0β0β3 + β1β2 = 0

⇐⇒

(β0 + β2)2 + (β1 + β3)

2 = 1(β0 − β2)

2 − (β1 − β3)2 = 1

(β0 + β2)(β1 + β3) = 0(β0 − β2)(β1 − β3) = 0

in F

Then βi (i = 0, 1, 2, 3) need to satisfy the following systems of congruenceequations simultaneously:

(1)

{(β0 + β2)

2 + (β1 + β3)2 = 1

(β0 + β2)(β1 + β3) = 0

⇐⇒

{β0 + β2 = 0β1 + β3 = ±1

or

{β0 + β2 = ±1β1 + β3 = 0

and

(2)

{(β0 − β2)

2 − (β1 − β3)2 = 1

(β0 − β2)(β1 − β3) = 0

⇐⇒

{β0 − β2 = 0(β1 − β3)

2 = −1or

{β0 − β2 = ±1β1 − β3 = 0

By Lemma 3.1, we know that (β1 − β3)2 = −1 has no solution in F .


Then the above systems of congruence equations are equivalent to the fol-lowings:

(i)

β1 + β3 = 0β0 + β2 = ±1β1 − β3 = 0β0 − β2 = ±1

(ii)

β0 + β2 = 0β1 + β3 = ±1β1 − β3 = 0β0 − β2 = ±1

By (i), we have the following four solutions:

β0 = 1β1 = 0β2 = 0β3 = 0

β0 = −1β1 = 0β2 = 0β3 = 0

β0 = 0β1 = 0β2 = 1β3 = 0

β0 = 0β1 = 0β2 = −1β3 = 0

By (ii), we have the following four solutions:{β0 = −β2 = ±2−1

β1 = β3 = ±2−1

By verification, x = 1, −1, τ2, −τ2, 2−1(1+τ−τ2+τ3), 2−1(1+τ−τ2+τ3),2−1(1 + τ − τ2 + τ3) and 2−1(1 + τ − τ2 + τ3) are solutions of x2 = 1 in FC4.Hence, x2 = 1 has exactly 8 solutions in FC4.

Lemma 3.3. Let F be a finite field of characteristic p(6= 2) with |F | = pn

and C4 = 〈τ〉 a cyclic group of order 4. Then

FC4∼=

{F ⊕ F ⊕ F ⊕ F, if p ≡1 (mod 4) or n is even;F2 ⊕ F ⊕ F, if p ≡3 (mod 4) and n is odd.

Proof. Suppose x = β0+β1τ+β2 τ2+β3τ3∈ FC4, for βi ∈ F , i = 0, 1, 2, 3.

If p ≡ 1 (mod 4), then pn ≡ 1 (mod 4), for all n. Thus

xpn

= (β0+β1τ+β2τ2+β3τ

3)pn

=βpn

0 +βpn

1 (τ)pn

+βpn

2 (τ2)pn

+βpn

3 (τ3) pn = x.

Also, when p ≡ 3 (mod 4) and n is even, then pn ≡ 1 (mod 4), so that xpn

= x. Hence, FC4∼= F ⊕ F ⊕ F ⊕ F , if p ≡ 1 (mod 4) or n is even.

If p ≡ 3 (mod 4) and n is odd, we get pn ≡ 3 (mod 4). Then p2n ≡ 1(mod 4) which implies xp

2n

= x. Hence, FC4∼= F2 ⊕F ⊕F or FC4

∼= F2 ⊕F2.Clearly, x2 = 1 has 8 roots in F2 ⊕ F ⊕ F and it has 4 roots in F2 ⊕F2. So, byLemma 3.2, we must have FC4

∼= F2 ⊕ F ⊕ F .

This completes the proof of Lemma 3.3.

Theorem 3.4. Let U (FG1) be the unit group of FG1 over a finite fieldF of characteristic p and |F | = pn, where G1

∼= C3 × C4 = C12 = 〈a〉.

(1) If p = 2, then

U (FG1) ∼=

{C3n2 × C3n

4 × C32n−1 if n is even;

C3n2 × C3n

4 × C22n−1 × C2n−1 if n is odd.(2) If p = 3, then


U (FG1) ∼=

{C8n3 × C4

3n−1 if n is even;C8n3 × C32n−1 × C2

3n−1 if n is odd.

(3) If p > 3, then

U (FG1) ∼=

C12pn−1 if p ≡1(mod 12) or n is even;

C4p2n−1 × C4

pn−1 if p ≡5(mod 12) and n is odd.

C3p2n−1 × C6


C5p2n−1 × Cpn−1 if p ≡11(mod 12) and n is odd.

Proof. Let H = 1 + J(FG1), where J(FG1) is the Jacobson radical of thegroup ring FG1. F

∗ is the multiplicative group of F , which is a cyclic group oforder pn − 1, we also denote it by Cpn−1.(1) Let char(F ) = 2 with |F | = 2n. Set C4 = {1, a, a2, a3}, then we haveG1/C4

∼= C3. Thus, by Lemma 2.1 and Lemma 2.8, J(FG1) = J(FC4)FG1 =ω(FC4)FG1 = ω(C4). So, FG1/J(FG1) ∼= F (G1/C4) ∼= FC3, where C3 = 〈σ〉is a cyclic group of order 3. Hence, from the ring epimorphism FG1 → FC3,we get a group epimorphism φ : U (FG1)) → U (FC3) and Kerφ = H =1 + J(FG1) = 1 + ω(C4).

Since C3 is a subgroup of C12, the ring monomorphism FC3 → FG1,α0+α1σ+α2σ

2 → α0+α1a4+α2a

8, reduces a group homomorphism

θ : U (FC3) → U (FG1).

And we can verify that φθ = 1U (FC3). Thus U (FG1) is an extension of U (FC3)by H, so

U (FG1) ∼= H × U (FC3).

Suppose x = α0+α1σ+α2σ2∈ FC3, for αi ∈ F . If n is even, then 3 | (2n−1),

and consequently, x2n

= (α0+α1σ+α2σ2)2

n

=α2n0 +α2n

1 (σ)2n

+α2n2 (σ2)2

n

= x sothat o(x) | (2n − 1), where x ∈ U (FC3). Thus FC3

∼= F ⊕ F ⊕ F , if n is even.When n is odd, 3 ∤ (2n − 1), but 3 | (22n − 1) and as above, x2

2n

= x, for anyx ∈ FC3. Hence, if n is odd, FC3

∼= F2 ⊕ F .

On the other hand,ω(C4) = {

∑11i=0 αia

i|∑3

i=0 α3i+j = 0, j = 0, 1, 2, αi ∈ F}.

By calculation, we have that{α2 | α ∈ ω(C4)} = {

∑5i=0(α

2i + α2

6+i)a2i|

∑3i=0 α3i+j = 0, j = 0, 1, 2, αi ∈ F},

and α4 = 0, for any α ∈ ω(C4).

For every x ∈ H = 1 + J(FG1) = 1 + ω(C4), set x = 1 + α,α ∈ ω(C4), weget x2 = (1 + α)2 = 1 + α2 and x4 = (1 + α)4 = 1 + α4 = 1. Hence, H is anAbelian 2-group and has exponent 4. So H ∼= Ck

2 × C l4 for some non-negative

integers k, l. In addition, FG1/J(FG1) ∼= FC3, we know that dimFJ(FG1) =


9. So |H| = |ω(C4)| = |J(FG1)| = 29n and so 2k × 4l = |H| = 29n. Next, weneed to determine k and l. Set

S = {α ∈ ω(C4) | α2 = 0, and ∃β ∈ ω(C4) such that α = β2}.

By direct computation, we have that

S = {∑2

i=0 α2i(a2i + a6+2i)|, α2i ∈ F}, and therefore |S| = 23n, which

implies that l = 3n and then k = 3n. Thus H ∼= C3n2 × C3n

4 . So, the proof ofstatement (1) is finished.

(2) Let char(F )= 3 and |F | = 3n. Obviously we have G1/C3∼= C4. Then

F (G1/C3) ∼= FC4. Since J(FG1) = J(FC3)FG1 = ω(FC3)FG1 = ω(C3), thenFG1/J(FG1) ∼= FC4, where C4 = 〈τ〉 is a cyclic group of order 4. By Lemma3.3, we know that

U (FC4) ∼=

{F ∗ × F ∗ × F ∗ × F ∗ if n is even;F ∗2 × F ∗ × F ∗ if n is odd.

Let C3 = {1, σ, σ2}. We know that ω(C3) = 〈g − 1 : g ∈ C3〉, ω2(C3) =

〈(g1−1)(g2−1) : g1, g2 ∈ C3〉 = (1+σ+σ2)FG1 and ω3(C3) = 0. For any x ∈ H,set x = 1 + α,α ∈ J(FG1) = ω(C3), we have x

3 = (1+α)3 = 1+3α+3α2+α3

= 1. Hence, H is an elementary Abelian 3-group.

In addition, FG1/J(FG1) ∼= FC4, we know that dimFJ(FG1) = 8. So, |H|= |J(FG1)| = 38n and H ∼= C8n

3 . Since (3, 3k − 1) = 1, it is easy to see thatU (FG1) ∼= H × U (FC4). Hence the statement (2) is hold.

(3) Let char(F ) = p > 3 and |F | = pn.

First, by Lemma 2.4, we have FG1∼= F (C3 × C4) ∼= (FC3)C4. We have

known that

FC3∼=

{F ⊕ F ⊕ F, if p ≡ 1 (mod 3) or n is even;F2 ⊕ F, if p ≡ 2 (mod 3) and n is odd.

Then, we have

FG1∼=

{(F ⊕ F ⊕ F )C4, if p ≡ 1 (mod 3) or n is even;(F2 ⊕ F )C4, if p ≡ 2 (mod 3) and n is odd.

By Lemma 2.2, we have

FG1∼=

{FC4 ⊕ FC4 ⊕ FC4, if p ≡ 1 (mod 3) or n is even;F2C4 ⊕ FC4, if p ≡ 2 (mod 3) and n is odd.

By Lemma 3.3, we get

FC4∼=

{F ⊕ F ⊕ F ⊕ F, if p ≡ 1 (mod 4) or n is even;F2 ⊕ F ⊕ F, if p ≡ 3 (mod 4) and n is odd.

And then, we haveF2C4

∼= F2 ⊕ F2 ⊕ F2 ⊕ F2.

Hence, if p ≡ 1 (mod 12) or n is even, thenFG1

∼= F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ;

if p ≡ 5 (mod 12) and n is odd, thenFG1

∼= F2 ⊕ F2 ⊕ F2 ⊕ F2 ⊕ F ⊕ F ⊕ F ⊕ F ;


if p ≡ 7 (mod 12) and n is odd, thenFG1

∼= F2 ⊕ F2 ⊕ F2 ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F.if p ≡ 11 (mod 12) and n is odd, then

FG1∼= F2 ⊕ F2 ⊕ F2 ⊕ F2 ⊕ F2 ⊕ F ⊕ F.

This completes the proof of the Theorem 3.4.

Theorem 3.5. Let U (FG2) be the unit group of FG2 over a finite fieldF of positive characteristic p and |F | = pn, where G2

∼= C3 × K4.(1) If p = 2, then

U (FG1) ∼=

{C9n2 × C3

2n−1 if n is even;C9n2 × C22n−1 × C2n−1 if n is odd.

(2) If p = 3, thenU (FG2) ∼= C8n

3 ×C43n−1.

(3) If p > 3, then

U (FG1) ∼=

{C12pn−1 if p ≡1(mod 3) or n is even;

C4p2n−1 × C4


Proof. Let H = 1 + J(FG2), where J(FG2) is the Jacobson radical of thegroup ring FG2. F

∗ is the multiplicative group of F , which is a cyclic group oforder pn − 1, we also denote it by Cpn−1.

(1) Let char(F ) = 2 with |F | = 2n. Since G2∼= C3 × K4, we have

G/K4∼= C3. Thus, by Lemma 2.1 and Lemma 2.8, J(FG2) = J(FK4)FG2 =

ω(FK4)FG2 = ω(K4). Hence, FG2/J(FG2) ∼= F (G2/K4) ∼= FC3. From thering epimorphism FG2 → FC3, we get a group epimorphism φ : U (FG2)) →U (FC3) and Kerφ = H = 1 + J(FG2) = 1 + ω(K4).

Since C3 is a subgroup of G2, the ring monomorphism FC3 → FG2,α0+α1σ+α2σ

2 → α0+α1σ+α2σ2, reduces a group homomorphism

θ : U (FC3) → U (FG2).

And we can verify that φθ = 1U (FC3). Thus U (FG2) is an extension of U (FC3)by H, therefore

U (FG2) ∼= H × U (FC3).Set K4 = {a0, a1, a2, a3} and C3 = 〈σ | σ3 = 1〉, then

ω(K4) = {∑3

i=0

∑2j=0 αijaiσ

j |∑3

i=0 αij = 0, j = 0, 1, 2, αij ∈ F}.

By calculation, we have that α2 = 0 for any α ∈ ω(K4).For every x ∈ H = 1+ J(FG1) = 1+ω(K4), set x = 1 + α,α ∈ ω(K4), we

get x2 = (1+α)2 = 1. Hence, H is an elementary Abelian 2-group. In addition,FG2/J(FG2) ∼= FC3, we know that dimFJ(FG2) = 9. So |H| = |ω(K4)| =|J(FG2)| = 29n and so H ∼= C9n

2 .


We have known that

FC3∼=

{F ⊕ F ⊕ F if n is even;F2 ⊕ F if n is odd.

Thus, we have finished the proof of (1).

(2) Let char(F ) = 3 and |F | = 3n. Obviously we know G2/C3∼= K4.

Also, by Lemma 2.1 and Lemma 2.8, J(FG2) = J(FC3)FG2 = ω(FC3)FG2 =ω(C3), then FG2/J(FG2) ∼= F (G2/C3) ∼= FK4. By Lemma 2.9, we haveFK4

∼= F ⊕ F ⊕ F ⊕ F .

The rest part of the proof of (2) is similar to the proof of (1), we omit it.

(3) Let char(F ) = p > 3 and |F | = pn, we can conclude that FG2∼=

F (C3 ×K4) ∼= (FC3)K4, by Lemma 2.4. We know that

FC3∼=

{F ⊕ F ⊕ F, if p ≡ 1(mod 3) or n is even;F2 ⊕ F, if p ≡ 2(mod 3) and n is odd.

Then, by Lemma 2.2, Lemma 2.4 and Lemma 2.9, we have

FG1∼=

{(F ⊕ F ⊕ F )K4, if p ≡ 1 (mod 3) or n is even;(F2 ⊕ F )K4, if p ≡ 2 (mod 3) and n is odd.

∼=

{FK4 ⊕ FK4 ⊕ FK4, if p ≡ 1 (mod 3) or n is even;F2K4 ⊕ FK4, if p ≡ 2 (mod 3) and n is odd.

Hence, if p ≡ 1 (mod 3) or n is even,FG2

∼= F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ⊕ F ;

if p ≡ 2 (mod 3) and n is odd,FG2

∼= F2 ⊕ F2 ⊕ F2 ⊕ F2 ⊕ F ⊕ F ⊕ F ⊕ F.


4. The unit groups of FQ12 and FD12

In [16], the structure of U (FA4) was studied by R.K. Sharma, J.B. Srivastavaand M. Khan in 2007. In this section, we give the structures of U (FQ12) andU (FD12).

Lemma 4.1. Let p ≥ 5 be a prime and F a finite field of characteristic pwith |F | = pn. Then FS3

∼= F ⊕ F ⊕M(2, F ).

Proof. Assume p ≥ 5. Since p ∤ |S3| , by Lemma 2.5 and Lemma 2.6, wehave

FS3∼= M(n1,D1)⊕M(n2,D2)⊕ · · · ⊕M(nr,Dr),

where D′is are finite dimentional division algebras over F . Thus D′

is arefinite fields, as F is finite. Since FS3 is non-commutative, there exists a k suchthat nk > 1. So


FS3∼= F ⊕ F ⊕M(2, F ) or F2 ⊕M(2, F ).

We know that C1 = {1}, C2 = {a, a2}, and C3 = {b, ab, a2b} are all conju-

gacy classes of S3, and then C1 = 1, C2 = a+ a2, C3 = b+ ab+ a2b, and by [8],

Z(FS3) = F C1 + F C2 + F C3. It is easy to verify that

C1pn

= C1, C2pn

= C2.

Since 1 + a+ a2 ∈ Z(FS3) and p is odd, we can conclude that

C3pn

= (b+ ab+ a2b)pn

= (1 + a+ a2)pn

bpn

= b+ ab+ a2b = C3.

Thus, we have xpn

= x, ∀ x ∈ Z(FS3) and therefore FS3∼= F ⊕ F ⊕

M(2, F ).

Theorem 4.2. Let U (FQ12) be the unit group of FQ12 of the generalizedquaternion group of order 12, Q12, over a finite field F of positive characteristicp and |F | = pn. Let V1 = 1 + J(FQ12), where J(FQ12) is the Jacobson radicalof the group ring FQ12.

(1) If p = 2 and V2 = 1 + ω(H), where H = {1, a3}, then

(a) U (FQ12)/V1∼= GL(2, F ) × F ∗

(b) The structure of V1 is determined as following:

(i) V1/V2 is an elementary Abelian 2−group of order 2n;

(ii) V1 is a nilpotent group of class 2.

(2) If p = 3, then

(a) U (FQ12)/V1∼=

{F ∗ × F ∗ × F ∗ × F ∗ if n is even;F ∗2 × F ∗ × F ∗ if n is odd.

(b) V1 is an elementary 3-group of order 38n and a nilpotent group of class2.

(c) U (FQ12) is a nilpotent group of class 3.

(3) If p > 3, then

(a) U (FQ12) ∼= GL(2, F ) × GL(2, F ) × F ∗ × F ∗ × F ∗ × F ∗, if p ≡1, 5(mod 12) or n is even;

(b) U (FQ12) ∼= GL(2, F )×GL(2, F )×F ∗2 ×F ∗×F ∗, if p ≡ 7, 11(mod 12)

and n is odd.

Proof. (1) Let p = 2 and |F | = 2n.(a) Since a3 ∈ Z(FQ12), we know that H = {1, a3} is a normal subgroup of Q12.As Q12/H ∼= S3, we have FS3

∼= F (Q12/H) ∼= FQ12/ω(H) so that dimF (ω(H))= 6. Further, we get ω2(H) = 0. By Lemma 2.3, we obtain ω(H) ⊆ J(FQ12).Hence, J(FS3) ∼= J(FQ12/ω(H)) ∼= J(FQ12)/ω(H). By calculation, we derivethat J(FS3) = Fα, where α = 1 + a + a2 + b + ab + a2b. So dimFJ(FS3) =1 and J(FS3)

2 = 0. Thus we have dimFJ(FQ12) = 7 and (J(FQ12)/ω(H))2

= 0. This implies (J(FQ12))2 ⊆ ω(H). Hence, dimF (FQ12/J(FQ12)) = 5.


Since FQ12/J(FQ12) ∼= (FQ12/ω(H))/(J(FQ12)/ ω(H)) ∼= FS3/J(FS3), andFS3/J(FS3) is semisimple, dimF (FS3/J(FS3)) = 5, then FS3/J(FS3) ∼= F ⊕F ⊕ F ⊕ F ⊕ F or M(2, F ) ⊕ F . Since γ = (1 − a)ab(1 + a) ∈ FS3/J(FS3),then γ 6= 0 but γ2 = 0. Thus, we have FS3/J(FS3) ∼= M(2, F ) ⊕ F . Hence,FQ12/J(FQ12) is non-commutative, FQ12/J(FQ12) ∼= M(2, F ) ⊕ F . Thus

U (FQ12)/V1∼= U (FQ12/J(FQ12)) ∼= GL(2, F ) × F ∗.

(b) Since (J(FQ12))2 ⊆ ω(H), for any y ∈ V1/V2, let y = v1V2, where

v1 ∈ V1, then y2 = v21V2. Let v1 = 1 + x, x ∈ J(FQ12), we get v21 = (1 + x)2

= 1 + x2 ∈ V2 . So o(y) = 2. Hence V1/V2 is an elementary Abelian 2-group.Further, |V1| = |J(FQ12)| = 27n and |V2| = |ω(H)| = 26n. This proves (i) asV1/V2 = 2n.

Now, observe that (J(FQ12))2 ⊆ ω(H) ⊆ Z(FQ12) and ω2(H) = 0, so that

(J(FQ12)4 ⊆ ω2(H) = 0. For ζ, η ∈ J(FQ12), since (1 + ζ)(1−ζ + ζ2 − ζ3) =

1−ζ4 = 1, then (1 + ζ)−1 = 1 −ζ + ζ2 − ζ3. Hence (1 + ζ)−1 ≡ 1 − ζ (modZ(FQ12)). Similarly (1 + η)−1 ≡ 1− η (mod Z(FQ12)). Then we have

(1 + ζ, 1 + η) ≡ (1 + ζ)−1 (1 + ζ)−1 (1 + ζ) (1 + η) mod Z(FQ12)≡ (1− ζ)(1− η) (1 + ζ) (1 + η) mod Z(FQ12)≡ (1− ζ − η) (1 + ζ + η) mod Z(FQ12)≡ 1 mod Z(FQ12)

Thus γ2(V1) ⊆ Z(FQ12) and hence γ3(V1) = (1). Consequently, V1 is anilpotent group of class 2.

(2) Let p = 3 and |F | = 3n.(a) It is easy to see that the commutator subgroup of Q12 is Q′

12 = {1, a2, a4}.Since Q12/Q

′12

∼= C4, we have FC4∼= F (Q12/Q

′12)

∼= FQ12/ω(Q′12) so that

dimFω(Q′12) = 8. Further, ω(Q′

12)2 = (1 + a2 + a4) FQ12 and ω3(Q′

12) = 0, byLemma 2.3, we have ω(Q′

12) ⊆ J(FQ12). Hence, J(FC4) ∼= J(FQ12/ω(Q′12))

∼=J(FQ12)/ω(Q

′12).

Since char(F ) = 3, by Lemma 3.3, we have

FC4∼=

{F ⊕ F ⊕ F ⊕ F if n is even;F2 ⊕ F ⊕ F if n is odd.

Thus we have dimFJ(FC4) = 0 and J(FC4) = 0. Hence, J(FQ12) ⊆ ω(Q′12).

So J(FQ12) = ω(Q′12). Hence

FQ12/J(FQ12) ∼=

{F ⊕ F ⊕ F ⊕ F if n is even;F2 ⊕ F ⊕ F if n is odd.

Thus,

U (FQ12)/V1∼= U (FQ12/J(FQ12)) ∼=

{F ∗ × F ∗ × F ∗ × F ∗ if n is even;F ∗2 × F ∗ ×F∗ if n is odd.


(b) Observe that ω2(Q′12) = (1 + a2 + a4) FQ12 ⊆ Z(FQ12) and ω3(Q′

12)= 0. ∀ ζ, η ∈ ω(Q′

12), we have

(1 + ζ, 1 + η) ≡ (1 + ζ)−1 (1 + η)−1 (1 + ζ) (1 + η) mod Z(FQ12)

≡ (1 −ζ) (1 −η) (1 + ζ) (1 + η) mod Z(FQ12)

≡ (1 −ζ − η) (1 + ζ + η) mod Z(FQ12)

≡ (1 + 2ζ + 2η) (1 + ζ + η) mod Z(FQ12)

≡ 1 mod Z(FQ12)

Thus γ2(V1) ⊆ Z(FQ12) and γ3(V1) = (1). Hence, V1 is a nilpotent groupof class 2.

Also, dimFJ(FQ12) = 8, then |V1| = 38n. Since J(FQ12)3 = ω3(Q′

12) =0, then the order of any nontrivial element of V1 is 3. So V1 is an elementary3−group of order 38n.

(c) Since U (FQ12)/V1∼= U (FC4) is an Abelian group, we have U (FQ12)

′ ⊆V1, therefore U (FQ12)

′′ ⊆ V ′1 ⊆ Z(FQ12). Hence U (FQ12) is a nilpotent

group of class 3.

(3) Assume p > 3. Since p ∤ |Q12|, by Lemma 2.5 and Lemma 2.6, we haveFQ12

∼= M(n1,D1)⊕M(n2,D2)⊕ · · · ⊕M(nr,Dr)

where D′is are finite dimentional division algebras over F . Thus D′

is arefinite fields, as F is finite. Since FQ12 is non-commutative, there exists a ksuch that nk > 1, so that nk will be either 2 or 3. Further, dimFZ(FQ12) = 6,all of the possible structures of the group ring FQ12 are given by

FQ12∼= M(2, F ) ⊕M(2, F ) ⊕ F ⊕ F ⊕ F ⊕ F or

M(2, F2)⊕ F ⊕ F ⊕ F ⊕ F or

M(2, F2)⊕ F2 ⊕ F ⊕ F or

M(2, F2)⊕ F2 ⊕ F2 or

M(2, F2)⊕ F3 ⊕ F or

M(2, F2)⊕ F4 or

M(2, F ) ⊕M(2, F ) ⊕ F2 ⊕ F ⊕ F or

M(2, F ) ⊕M(2, F ) ⊕ F2 ⊕ F2 or

M(2, F ) ⊕M(2, F ) ⊕ F3 ⊕ F or

M(2, F ) ⊕M(2, F ) ⊕ F4

Obviously, every element of Z(FQ12) is an F -linear combination of conju-gacy class sums of Q12. Observe the following regarding these conjugacy classsums equations:

C1pn

= C1, C3pn

= C3 and C4pn

= C4

Next, we need to show other equations.

(i) If p ≡ 1 (mod 12), then pn ≡ 1 (mod 12), for all n. Then we computethat


C2pn

= (a+ a5)pn

= a · apn−1 + a5 · (a5)p

n−1 = a+ a5 = C2

C5pn

= [b(1 + a2 + a4)]pn

= bpn−1 · [b(1 + a2 + a4)] = b(1 + a2 + a4) = C5

C6pn

= [ab(1 + a2 + a4)]pn

= (ab)pn−1 · [ab(1 + a2 + a4)] = ab(1 + a2 + a4) = C6

Thus xpn

= x, for all x ∈ Z(FQ12). In particular, if x ∈ U (Z(FQ12)),then o(x) | (pn− 1). Hence

FQ12∼= M(2, F ) ⊕M(2, F ) ⊕ F ⊕ F ⊕ F ⊕ F .

(ii) If p ≡ 5, 7, 11 (mod 12) and n is even, then clearly pn ≡ 1 (mod 12).Hence, in this case , we also have

FQ12∼= M(2, F ) ⊕M(2, F ) ⊕ F ⊕ F ⊕ F ⊕ F .

(iii) If p ≡ 5 (mod 12) and n is odd, note that o(a) = 6, o(b) = 4, andn = 2k + 1, where k = (n − 1)/2, then we can verify that:

Ci

pn

= Ci, for 1 ≤ i ≤ 6.

Hence, in this case, we also haveFQ12

∼= M(2, F ) ⊕M(2, F ) ⊕ F ⊕ F ⊕ F ⊕ F .

(iv) If p ≡ 7, 11 (mod 12) and n is odd, then we can verify that:

Ci

p2n

= Ci, for 1 ≤ i ≤ 6, but C5pn

6= C5.

So, in this case, xp2n

= x, for any x ∈ Z(FQ12), i.e., o(x) | (p2n− 1), for any

x ∈ U (Z(FQ12)). Thus the occurrences of F3 and F4 in the presentation ofFQ12 are impossible. Hence the possible presentations of the group ring FQ12

are the followings:FQ12

∼= M(2, F2)⊕ F ⊕ F ⊕ F ⊕ F orM(2, F2)⊕ F2 ⊕ F ⊕ F or

M(2, F2)⊕ F2 ⊕ F2 or

M(2, F ) ⊕M(2, F ) ⊕ F2 ⊕ F ⊕ F or

M(2, F ) ⊕M(2, F ) ⊕ F2 ⊕ F2

Since FQ12/ω(H) ∼= FS3 and FQ12 is semisimple, by Lemma 4.1, we havethat FQ12

∼= M(2, F )⊕M(2, F ) ⊕ F2 ⊕ F ⊕ F .Hence, we have

(a) U (FQ12) ∼= GL(2, F ) × GL(2, F ) × F ∗ × F ∗ × F ∗ × F ∗, if p ≡1, 5(mod 12) or n is even;

(b) U (FQ12) ∼= GL(2, F )×GL(2, F )×F ∗2 ×F ∗×F ∗, if p ≡ 7, 11(mod 12)

and n is odd.


Theorem 4.3. Let U (FD12) be the unit group of FD12 of the dihedralgroup of order 12, D12, over a finite field F of positive characteristic p and


|F | = pn . Let V1 = 1 + J(FD12), where J(FD6) is the Jacobson radical of thegroup ring FD12.

(1) If p = 2, then

(a) U (FD12)/V1∼= GL(2, F ) × F ∗ ;

(b) The structure of V1 is determined as:

(i) V1/V2 is an elementary Abelian 2−group of order 2n;

(ii) V1 is a nilpotent group of class 2.

(2) If p = 3, then

(a) U (FD12)/V1∼= F ∗ × F ∗ × F ∗ × F ∗;

(b) V1 is an elementary 3-group of order 3 8n and a nilpotent group of class2;

(c) U (FD12) is a nilpotent group of class 3.

(3) If p > 3, then

U (FD12) ∼= GL(2, F ) ×GL(2, F ) × F ∗ × F ∗ × F ∗ × F ∗.

Proof. (1)(a) Let char(F ) = 2 with |F | = 2n. We know that a3 ∈ Z(FD12)and H = {1, a3} is a normal subgroup of D12. It is easy to verify that D12/H ∼=S3. Then we get that U (FD12)/V1

∼= GL(2, F ) × F ∗, which is similar to (1)(a) of Theorem 4.2.

The proof of (b) is Similar to the proof of (1) (b) of Theorem 4.2.

(2)(a) Let char(F ) = 3 and |F | = 3n. It is easy to see that the commutatorsubgroup of D12 is D′

12 = {1, a2, a4} and D12/D′12

∼= K4. By Lemma 2.8and Lemma 2.1, we have that J(FD12) = J(FD′

12)FD12 = ω(FD′12)FD12 =

ω(D′12), so FD12/J(FD12) ∼= FD12/ω(D

′12)

∼= FK4∼= F⊕F⊕F⊕F by Lemma

2.9.

ThusU (FD12)/V1

∼= U (FD12/J(FD12)) ∼= F ∗ × F ∗ × F ∗ × F ∗.

In addition, a similar argument of (2)(b) and (c) of Theorem 4.2 can beapplied to here.

(3) Assume p > 3 and |F | = pn. Since p ∤ |D12|, by Lemma 2.5 and Lemma2.6, we have

FD12∼= M(n1, C1)⊕M(n2,H)⊕ · · · ⊕M(nr, Cr)

where C ′is are finite dimentional division algebras over F . Thus C ′

is arefinite fields, as F is finite. Since FD12 is non-commutative, there exists a ksuch that nk > 1, so that nk will be either 2 or 3. Further, since dimFZ(FD12)= 6, we will get the following possibilities:

FD12∼= M(2, F ) ⊕M(2, F ) ⊕ F ⊕ F ⊕ F ⊕ F or

M(2, F2)⊕ F ⊕ F ⊕ F ⊕ F or

M(2, F2)⊕ F2 ⊕ F ⊕ F or


M(2, F2)⊕ F2 ⊕ F2 or

M(2, F2)⊕ F3 ⊕ F or

M(2, F2)⊕ F4 or

M(2, F ) ⊕M(2, F ) ⊕ F ⊕ F ⊕ F ⊕ F or

M(2, F ) ⊕M(2, F ) ⊕ F2 ⊕ F ⊕ F or

M(2, F ) ⊕M(2, F ) ⊕ F2 ⊕ F2 or

M(2, F ) ⊕M(2, F ) ⊕ F3 ⊕ F or

M(2, F ) ⊕M(2, F ) ⊕ F4

We know that every element of Z(FD12) is an F -linear combination ofconjugacy class sums of D12. Since p > 3, we have p ≡ 1 or 5 (mod 6) and thenpn ≡ 1 or 5 (mod 6) for any positive integer n. By straightforward computing,we get that:

Ci

Pn

= Ci, for any 1 ≤ i ≤ 6.

Thus xpn

= x, for any x ∈ Z(FD12). In particular, if x ∈ U (Z(FD12)),then o(x) | (pn− 1). Hence

FD12∼= M(2, F )⊕M(2, F ) ⊕ F ⊕ F ⊕ F ⊕ F .

Hence, in this case, we have

U (FD12) ∼= GL(2, F ) ×GL(2, F ) × F ∗ × F ∗ × F ∗ × F ∗.


Acknowledgments

This research was supported by the National Natural Science Foundation ofChina (11161006, 11171142), the Guangxi Natural Science Foundation (2011GXNSFA018139, 2010GXNSFB013048, 0991102) and Guangxi “New Century1000 Talent Project”.

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