ISSN 2148-838Xhttps://doi.org/10.13069/jacodesmath.935938 J. Algebra Comb. Discrete Appl. 8(2) • 59–71 Received: 30 July 2020 Accepted: 20 October 2020 Journal of Algebra Combinatorics Discrete Structures and Applications On unit group of finite semisimple group algebras of non-metabelian groups of order 108 Research Article Gaurav Mittal, Rajendra. K. Sharma Abstract: In this paper, we characterize the unit groups of semisimple group algebras FqG of non-metabelian groups of order 108, where Fq is a field with q = pk elements for some prime p > 3 and positive integer k. Upto isomorphism, there are 45 groups of order 108 but only 4 of them are non-metabelian. We consider all the non-metabelian groups of order 108 and find the Wedderburn decomposition of their semisimple group algebras. And as a by-product obtain the unit groups. 2010 MSC: 16U60, 20C05 Keywords: Unit group, Finite field, Wedderburn decomposition 1. Introduction Let Fq denote a finite field with q = pk elements for odd prime p > 3, G be a finite group and FqG be the group algebra. The study of the unit groups of group algebras is a classical problem and has applications in cryptography [4] as well as in coding theory [5] etc. For the exploration of Lie properties of group algebras and isomorphism problems, units are very useful see, e.g. [1]. We refer to [11] for elementary definitions and results about the group algebras and [2, 14] for the abelian group algebras and their units. Recall that a group G is metabelian if there is a normal subgroup N of G such that both N and G/N are abelian. The unit groups of the finite semisimple group algebras of metabelian groups have been well studied. In this paper, we are concerned about the unit groups of the group algebras of non-metabelian groups. Let us first mention the available literature in this direction. From [13], we know that all the groups up to order 23 are metabelian. The only non-metabelian groups of order 24 are S4 and SL(2, 3) and the unit group of their group algebras have been discussed in [7, 9]. Further, from [13], we also Gaurav Mittal (Corresponding Author); Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee, India (email: gmittal@ma.iitr.ac.in). R. K. Sharma; Department of Mathematics, Indian Institute of Technology Delhi, New Delhi, India (email: rksharma@maths.iitd.ac.in). 59 https://orcid.org/0000-0001-8292-9646 https://orcid.org/0000-0001-5666-4103 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 deduce that there are non-metabelian groups of order of 48, 54, 60, 72, 108 etc. The unit groups of the group algebras of non-metabelian groups up to order 72 have been discussed in [10, 12]. The unit group of the semisimple group algebra of the non-metabelian group SL(2, 5) has been discussed in [15]. The main motive of this paper is to characterize the unit groups of FqG, where G represents a non-metabelian group of order 108. It can be verified that, upto isomorphism, there are 45 groups of order 108 and only 4 of them are non-metabelian. We deduce the Wedderburn decomposition of group algebras of all the 4 non-metabelian groups and then characterize the respective unit groups. The rest of the paper is organized in following manner: We recall all the basic definitions and results to be needed in our work in Section 2. Our main results on the characterization of the unit groups are presented in third section and the last section includes some discussion. 2. Preliminaries Let e denote the exponent of G, ζ a primitive eth root of unity. On the lines of [3], we define IF = {n | ζ 7→ ζn is an automorphism of F(ζ) over F}, where F is an arbitrary finite field. Since, the Galois group Gal ( F(ζ),F ) is a cyclic group, for any τ ∈ Gal ( F(ζ),F ) , there exists a positive integer s which is invertible modulo e such that τ(ζ) = ζs. In other words, IF is a subgroup of the multiplicative group Z∗e (group of integers which are invertible with respect to multiplication modulo e). For any p-regular element g ∈ G, i.e. an element whose order is not divisible by p, let the sum of all conjugates of g be denoted by γg, and the cyclotomic F-class of γg be denoted by S(γg) = {γgn | n ∈ IF}. The cardinality of S(γg) and the number of cyclotomic F-classes will be incorporated later on for the characterization of the unit groups. Next, we recall two important results from [3]. First one relates the number of cyclotomic F-classes with the number of simple components of FG/J(FG). Here J(FG) denotes the Jacobson radical of FG. Second one is about the cardinality of any cyclotomic F-class in G. Theorem 2.1. The number of simple components of FG/J(FG) and the number of cyclotomic F-classes in G are equal. Theorem 2.2. Let ζ be defined as above and j be the number of cyclotomic F-classes in G. If Ki, 1 ≤ i ≤ j, are the simple components of center of FG/J(FG) and Si, 1 ≤ i ≤ j, are the cyclotomic F-classes in G, then |Si| = [Ki : F] for each i after suitable ordering of the indices if required. To determine the structure of unit group U(FG), we need to determine the Wedderburn decomposi- tion of the group algebra FG. In other words, we want to determine the simple components of FG. Based on the existing literature, we can always claim that F is one of the simple component of decomposition of FG/J(FG). The simple proof is given here for the completeness. Lemma 2.3. Let A1 and A2 denote the finite dimensional algebras over F. Further, let A2 be semisimple and g be an onto homomorphism between A1 and A2, then we must have A1/J(A1) ∼= A3 + A2, where A3 is some semisimple F-algebra. Proof. From [6], we have J(A1) ⊆ Ker(g). This means there exists F-algebra homomorphism g1 from A1/J(A1) to A2 which is also onto. In other words, we have g1 : A1/J(A1) 7−→ A2 defined by g1(a + J(A1)) = g(a), a ∈ A1. As A1/J(A1) is semisimple, there exists an ideal I of A1/J(A1) such that A1/J(A1) = ker(g1) ⊕ I. 60 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 Our claim is that I ∼= A2. For this to prove, note that any element a ∈ A1/J(A1) can be uniquely written as a = a1 + a2 where a1 ∈ ker(g1),a2 ∈ I. So, define g2 : A1/J(A1) 7→ ker(g1) ⊕A2 by g2(a) = (a1,g1(a2)). Since, ker(g1) is a semisimple algebra over F and A2 is an isomorphic F-algebra, claim and the result holds. Above lemma concludes that F is one of the simple components of FG provided J(FG) = 0. Now, we recall a result which characterize the set IF defined in the beginning of this section. For proof, see Theorem 2.21 in [8]. Theorem 2.4. Let F be a finite field with prime power order q. If e is such that gcd(e,q) = 1, ζ is the primitive eth root of unity and |q| is the order of q modulo e, then modulo e, we have IF = {1,q,q2, . . . ,q|q|−1}. Next result is Proposition 3.6.11 from [11] and is useful for the determination of commutative simple components of the group algebra FqG. Theorem 2.5. If RG is a semisimple group algebra, then RG ∼= R ( G/G′ ) ⊕ ∆(G,G′), where G′ is the commutator subgroup of G, R ( G/G′ ) is the sum of all commutative simple components of RG, and ∆(G,G′) is the sum of all others. We end this section by recalling a Proposition 3.6.7 from [11] which is a generalized version of the last result. Theorem 2.6. Let RG be a semisimple group algebra and H be a normal subgroup of G. Then RG ∼= R ( G/H ) ⊕ ∆(G,H), where ∆(G,H) is a left ideal of RG generated by the set {h− 1 : h ∈ H}. 3. Unit group of FqG where G is a non-metabelian group of order 108 The main objective of this section is to characterize the unit group of FqG where G is a non- metabelian group of order 108. Upto isomorphism, there are 4 non-metabelian groups of order 108 namely: (1) G1 = ((C3 ×C3) o C3) o C4. (2) G2 = ((C3 ×C3) o C3) o C4. (3) G3 = ((C3 ×C3) o C3) o (C2 ×C2). (4) G4 = C2 × (((C3 ×C3) o C3) o C2). Here G1 and G2 are two non-isomorphic groups formed by the semi-direct product of (C3×C3) oC3 and C4 which will be clear once we discuss the presentation of these groups. We consider each of these 4 groups one by one and discuss the unit groups of their respective group algebras along with the Wedderburn decompositions in subsequent subsections. Throughout this paper, we use the notation [x,y] = x−1y−1xy. 3.1. The group G1 = ((C3 ×C3) o C3) o C4 Group G1 has the following presentation: G1 = 〈x,y,z,w,t | x2y−1, [y,x], [z,x]z−1, [w,x]w−1, [t,x], y2, [z,y], [w,y], [t,y], z3, [w,z]t−1, [t,z], w3, [t,w], t3〉. Also G1 has 20 conjugacy classes as shown in the table below. 61 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 r e x y z w t xy xt yz yw yt zw t2 xyt xt2 yzw yt2 z2w xyt2 yz2w s 1 9 1 6 6 1 9 9 6 6 1 6 1 9 9 6 1 6 9 6 o 1 4 2 3 3 3 4 12 6 6 6 3 3 12 12 6 6 3 12 6 where r, s and o denote the representative of conjugacy class, size and order of the representative of the conjugacy class, respectively. From the above discussion, it is clear that exponent of G1 is 12. Also G′1 ∼= (C3×C3) oC3 with G1/G′1 ∼= C4. Next, we discuss the unit group of the group algebra FqG1 when p > 3. Theorem 3.1. The unit group of FqG1, for q = pk, p > 3 where Fq is a finite field having q = pk elements is as follows: 1. for any p and k even or pk ≡ 1 mod 12 with k odd, we have U(FqG1) ∼= (F∗q) 4 ⊕GL2(Fq)8 ⊕GL3(Fq)8. 2. for pk ≡ 5 mod 12 with k odd, we have U(FqG1) ∼= (F∗q) 4 ⊕GL2(Fq)8 ⊕GL3(Fq2 )4. 3. for pk ≡ 7 mod 12 with k odd, we have U(FqG1) ∼= (F∗q) 2 ⊕F∗q2 ⊕GL2(Fq) 8 ⊕GL3(Fq)4 ⊕GL3(Fq2 )2. 4. for pk ≡ 11 mod 12 with k odd, we have U(FqG1) ∼= (F∗q) 2 ⊕F∗q2 ⊕GL2(Fq) 8 ⊕GL3(Fq2 )4. Proof. Since FqG1 is semisimple, using Lemma 2.3 we get FqG1 ∼= Fq ⊕t−1r=1 Mnr (Fr), for some t ∈ Z. (1) First assume that k is even which means for any prime p > 3, we have pk ≡ 1 mod 3 and pk ≡ 1 mod 4. Using Chinese remainder theorem, we get pk ≡ 1 mod 12. This means |S(γg)| = 1 for each g ∈ G1 as IF = {1}. Hence, (1), Theorems 2.1 and 2.2 imply that FqG1 ∼= Fq ⊕19r=1 Mnr (Fq). (2) Incorporating Theorem 2.5 with G′1 ∼= (C3 ×C3) o C3 in (2) to obtain FqG1 ∼= F4q ⊕ 16 r=1 Mnr (Fq), where nr ≥ 2 with 104 = 16∑ r=1 n2r. (3) Above equation gives the only possibility (28, 38) for the values of n′rs where a b means (a,a, · · · ,b times) and therefore, (3) implies that FqG1 ∼= F4q ⊕M2(Fq) 8 ⊕M3(Fq)8. (4) Now we consider that k is odd. We shall discuss this possibility in the following four cases: Case 1: pk ≡ 1 mod 3 and pk ≡ 1 mod 4. In this case, Wedderburn decomposition is given by (4). Case 2. pk ≡ 5 mod 12. In this case, we have S(γt) = {γt,γt2}, S(γxt) = {γxt,γxt2}, S(γyt) = {γyt,γyt2}, S(γxyt) = {γxyt,γxyt2}, 62 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 and S(γg) = {γg} for the remaining representatives g of conjugacy classes. Therefore, (1) and Theorems 2.1, 2.2 imply that FqG1 ∼= Fq ⊕11r=1 Mnr (Fq) ⊕ 15 r=12 Mnr (Fq2 ). Since G1/G′1 ∼= C4, we have FqC4 ∼= F4q. This with above and Theorem 2.5 yields FqG1 ∼= F4q ⊕ 8 r=1 Mnr (Fq) ⊕ 12 r=9 Mnr (Fq2 ), nr ≥ 2 with 104 = 8∑ r=1 n2r + 2 12∑ r=9 n2r. (5) Further, consider the normal subgroup H1 = 〈t〉 of G1 having order 3 with K1 = G1/H1 ∼= (C3 × C3) o C4. The quotient group K1 has 12 conjugacy classes as shown in the table below. Here elements of K1 are cosets, for instance, x ∈ K1 is xH1 but we keep the same notation. r e x y z w xy yz yw zw yzw z2w yz2w s 1 9 1 2 2 9 2 2 2 2 2 2 o 1 4 2 3 3 4 6 6 3 6 3 6 It can be verified that for all the representatives g of K1, |S(γg)| = 1. Therefore, from Theorems 2.1 and 2.2, we have FqK1 ∼= Fq ⊕11r=1 Mtr (Fq), tr ∈ Z. Observe that K1/K′1 ∼= C4. So, Theorem 2.5 implies that FqK1 ∼= F4q ⊕ 8 r=1 Mtr (Fq), with 32 = 8∑ r=1 t2r, tr ≥ 2. This gives us the only choice (28) for values of t′rs and therefore, Theorem 2.5 and (5) yields FqG1 ∼= F4q ⊕M2(Fq) 8 ⊕4r=1 Mnr (Fq2 ), nr ≥ 2 with 36 = 4∑ r=1 n2r. Above leaves us with the only choice (34) for values of n′rs which means the required Wedderburn decomposition is FqG1 ∼= F4q ⊕M2(Fq) 8 ⊕M3(Fq2 )4. Case 3. pk ≡ 7 mod 12. In this case, we have S(γx) = {γx,γxy}, S(γxt) = {γxt,γxyt}, S(γxt2 ) = {γxt2,γxyt2}, and S(γg) = {γg} for the remaining representatives g of conjugacy classes. Therefore (1) and Theorems 2.1, 2.2 imply that FqG1 ∼= Fq ⊕13r=1 Mnr (Fq) ⊕ 16 r=14 Mnr (Fq2 ). Since G1/G′1 ∼= C4, we have FqC4 ∼= F2q ⊕Fq2. This with above and Theorem 2.5 yields FqG1 ∼= F2q ⊕Fq2 ⊕ 12 r=1 Mnr (Fq) ⊕ 14 r=13 Mnr (Fq2 ), nr ≥ 2 with 104 = 12∑ r=1 n2r + 2 14∑ r=13 n2r. (6) 63 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 Again consider the normal subgroup H1 of G1. In this case, it can be verified that |S(γg)| = 1 for all the representatives g of K1 except x and xy for which S(γx) = {γx,γxy}. Therefore, employing Theorems 2.1 and 2.2 to obtain FqK1 ∼= Fq ⊕9r=1 Mtr (Fq) ⊕Mt10 (Fq2 ), tr ∈ Z. Since K1/K′1 ∼= C4, above and Theorem 2.5 imply that FqK1 ∼= F2q ⊕Fq2 ⊕ 8 r=1 Mtr (Fq), with 32 = 8∑ r=1 t2r, tr ≥ 2. This gives us the only choice (28) for values of t′rs. Hence, Theorem 2.6 and (6) imply that FqG1 ∼= F2q ⊕Fq2 ⊕M2(Fq) 8 ⊕4r=1 Mnr (Fq) ⊕ 6 r=5 Mnr (Fq2 ), with 72 = 4∑ r=1 n2r + 2 6∑ r=5 n2r. Above leaves us with the only choice (36) for values of n′rs which means the required Wedderburn decomposition is FqG1 ∼= F2q ⊕Fq2 ⊕M2(Fq) 8 ⊕M3(Fq)4 ⊕M3(Fq2 )2. Case 4. pk ≡ 11 mod 12. In this case, we have S(γt) = {γt,γt2}, S(γxt) = {γxt,γxyt2}, S(γyt) = {γyt,γyt2}, S(γxyt) = {γxyt,γxt2}, S(γx) = {γx,γxy}, and S(γg) = {γg} for the remaining representatives g of conjugacy classes. Therefore, (1) and Theorems 2.1, 2.2 imply that FqG1 ∼= Fq ⊕9r=1 Mnr (Fq) ⊕ 14 r=10 Mnr (Fq2 ). Since G1/G′1 ∼= C4, we have FqC4 ∼= F2q ⊕Fq2. This with above and Theorem 2.5 yields FqG1 ∼= F2q ⊕Fq2 ⊕ 8 r=1 Mnr (Fq) ⊕ 12 r=9 Mnr (Fq2 ), nr ≥ 2 with 104 = 8∑ r=1 n2r + 2 12∑ r=9 n2r. (7) In this case, again we have |S(γg)| = 1 for all representatives g of K1 except x and xy which means the Wedderburn decomposition of FqK1 is same as obtained in case 3, i.e. FqK1 ∼= F2q ⊕Fq2 ⊕M2(Fq) 8. Now employ Theorem 2.6 and (7) to obtain FqG1 ∼= F2q ⊕Fq2 ⊕M2(Fq) 8 ⊕4r=1 Mnr (Fq2 ), with 36 = 4∑ r=1 n2r. This leaves us with the only choice (34) for values of n′rs which means the required Wedderburn decom- position is FqG1 ∼= F2q ⊕Fq2 ⊕M2(Fq) 8 ⊕M3(Fq2 )4. 64 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 3.2. The group G2 = ((C3 ×C3) o C3) o C4 Group G2 has the following presentation: G2 = 〈x,y,z,w,t | x2y−1, [y,x], [z,x]w−2, [w,x]t−1w−1z−2, [t,x], y2, [z,y]z−1, [w,y]t−1w−1, [t,y], z3, [w,z]t−1, [t,z], w3, [t,w], t3〉. Further, G2 has 14 conjugacy classes as shown in the table below. r e x y z w t xy xz xt yw yt t2 xyz xyzw s 1 9 9 12 12 1 9 9 9 9 9 1 9 9 o 1 4 2 3 3 3 4 12 12 6 6 3 12 12 From above discussion, clearly the exponent of G2 is 12. Also G′2 ∼= (C3 ×C3) o C3 and G2/G′2 ∼= C4. Next, we discuss the unit group of FqG2 when p > 3. Theorem 3.2. The unit group of FqG2, for q = pk, p > 3 where Fq is a finite field having q = pk elements is as follows: 1. for any p and k even or pk ≡ 1 mod 12 with k odd, we have U(FqG2) ∼= (F∗q) 4 ⊕GL3(Fq)8 ⊕GL4(Fq)2. 2. for pk ≡ 5 mod 12 with k odd, we have U(FqG2) ∼= (F∗q) 4 ⊕GL4(Fq)2 ⊕GL3(Fq2 )4. 3. for pk ≡ 7 mod 12 with k odd, we have U(FqG2) ∼= (F∗q) 2 ⊕F∗q2 ⊕GL3(Fq) 4 ⊕GL4(Fq)2 ⊕GL3(Fq2 )2. 4. for pk ≡ 11 mod 12 with k odd, we have U(FqG2) ∼= (F∗q) 2 ⊕F∗q2 ⊕GL4(Fq) 2 ⊕GL3(Fq2 )4. Proof. Since FqG2 is semisimple, we have FqG2 ∼= Fq ⊕t−1r=1 Mnr (Fr), for some t ∈ Z. (8) First assume that k is even which means for any prime p > 3, pk ≡ 1 mod 12. This means |S(γg)| = 1 for each g ∈ G2. Hence, (8), Theorems 2.1 and 2.2 imply that FqG2 ∼= Fq ⊕13r=1 Mnr (Fq). Using Theorem 2.5 with G′2 ∼= (C3 ×C3) o C3 to obtain FqG2 ∼= F4q ⊕ 10 r=1 Mnr (Fq), where nr ≥ 2 with 104 = 10∑ r=1 n2r. (9) Above equation gives us four possibilities (28, 62), (25, 32, 4, 52), (24, 34, 4, 6) and (38, 42) for the values of n′rs. Further, consider the normal subgroup H2 = 〈t〉 of G2 having order 3 with K2 = G2/H2 ∼= (C3 ×C3) o C4. It can be verified that K2 has 6 conjugacy classes as shown in the table below. r e x y z w xy s 1 9 9 4 4 9 o 1 4 2 3 3 4 65 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 Also, for all the representatives g of K2, |S(γg)| = 1 which means by Theorems 2.1 and 2.2, we have FqK2 ∼= Fq ⊕5r=1 Mtr (Fq), tr ∈ Z. Observe that K2/K′2 ∼= C4. This with above and Theorem 2.4 imply that FqK2 ∼= F4q ⊕ 2 r=1 Mtr (Fq), with 32 = 2∑ r=1 t2r, tr ≥ 2. This gives us the only choice (42) for values of t′rs. Therefore, Theorem 2.6 and (9) imply that (3 8, 42) is the correct choice for values of n′rs and therefore, we have FqG2 ∼= F4q ⊕M3(Fq) 8 ⊕M4(Fq)2. (10) Now we consider that k is odd. We shall discuss this possibility in the following four cases: Case 1: pk ≡ 1 mod 12. In this case, Wedderburn decomposition is given by (10). Case 2. pk ≡ 5 mod 12. In this case, we have S(γt) = {γt,γt2}, S(γxz) = {γxz,γxt}, S(γyw) = {γyw,γyt}, S(γxyz) = {γxyz,γxyzw}, and S(γg) = {γg} for the remaining representatives g of conjugacy classes. Therefore, (8) and Theorems 2.1, 2.2 imply that FqG2 ∼= Fq ⊕5r=1 Mnr (Fq) ⊕ 9 r=6 Mnr (Fq2 ). Since G2/G′2 ∼= C4, we have FqC4 ∼= F4q. This with above and Theorem 2.5 yields FqG2 ∼= F4q ⊕ 2 r=1 Mnr (Fq) ⊕ 6 r=3 Mnr (Fq2 ), nr ≥ 2 with 104 = 2∑ r=1 n2r + 2 6∑ r=3 n2r. (11) Further, again consider the normal subgroup H2 = 〈t〉 of G2. It can be verified that for all the represen- tatives g of K2, |S(γg)| = 1 which means (as earlier) FqK2 ∼= F4q ⊕M4(Fq) 2. This with Theorem 2.6 and (11) imply that FqG2 ∼= F4q ⊕M4(Fq) 2 ⊕4r=1 Mnr (Fq2 ), nr ≥ 2 with 36 = 4∑ r=1 n2r. Above leaves us with the only choice (34) for values of n′rs which means the required Wedderburn de- composition is FqG2 ∼= F4q ⊕M4(Fq) 2 ⊕M3(Fq2 )4. Case 3. pk ≡ 7 mod 12. In this case, we have S(γx) = {γx,γxy}, S(γxz) = {γxz,γxyzw}, S(γxt) = {γxt,γxyz}, and S(γg) = {γg} for the remaining representatives g of conjugacy classes. Therefore, (8) and Theorems 2.1, 2.2 imply that FqG2 ∼= Fq ⊕7r=1 Mnr (Fq) ⊕ 10 r=8 Mnr (Fq2 ). Since G2/G′2 ∼= C4, we have FqC4 ∼= F2q ⊕Fq2. This with Theorem 2.5 yields FqG2 ∼= F2q ⊕Fq2 ⊕ 6 r=1 Mnr (Fq) ⊕ 8 r=7 Mnr (Fq2 ), nr ≥ 2 with 104 = 6∑ r=1 n2r + 2 8∑ r=7 n2r. (12) 66 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 Further, it can be verified that |S(γg)| = 1 for all the representatives g of K2 except x and xy. For these, we have S(γx) = {γx,γxy} which means by Theorems 2.1 and 2.2, we have FqK2 ∼= Fq ⊕3r=1 Mtr (Fq) ⊕Mt4 (Fq2 ), tr ∈ Z. Incorporating K2/K′2 ∼= C4 with Theorem 2.5 to obtain FqK2 ∼= F2q ⊕Fq2 ⊕ 2 r=1 Mtr (Fq), with 32 = 2∑ r=1 t2r, tr ≥ 2. This gives us only choice (42) for values of t′rs. Therefore, Theorem 2.6 and (12) yields FqG2 ∼= F2q ⊕Fq2 ⊕M4(Fq) 2 ⊕4r=1 Mnr (Fq) ⊕ 6 r=5 Mnr (Fq2 ), with 72 = 4∑ r=1 n2r + 2 6∑ r=5 n2r. Above leaves us with the only choice (36) for values of n′r which means the required Wedderburn decom- position is FqG2 ∼= F2q ⊕Fq2 ⊕M4(Fq) 2 ⊕M3(Fq)4 ⊕M3(Fq2 )2. Case 4. pk ≡ 11 mod 12. In this case, we have S(γt) = {γt,γt2}, S(γx) = {γx,γxy}, S(γxz) = {γxz,γxyz}, S(γxt) = {γxt,γxyzw}, S(γyw) = {γyw,γyt}, and S(γg) = {γg} for the remaining representatives g of conjugacy classes. Therefore, (8) and Theorems 2.1, 2.2 imply that FqG2 ∼= Fq ⊕3r=1 Mnr (Fq) ⊕ 8 r=4 Mnr (Fq2 ). Since G2/G′2 ∼= C4, we have FqC4 ∼= F2q ⊕Fq2. This with above and Theorem 2.5 yields FqG2 ∼= F2q ⊕Fq2 ⊕ 2 r=1 Mnr (Fq) ⊕ 6 r=3 Mnr (Fq2 ), nr ≥ 2 with 104 = 2∑ r=1 n2r + 2 6∑ r=3 n2r. (13) Further, it can be verified that |S(γg)| = 1 for all the representatives g of K2 except x and xy which means (as in case 3), FqK2 ∼= F2q ⊕Fq2 ⊕M4(Fq) 2. Now employ Theorem 2.6 and (13) to obtain FqG2 ∼= F2q ⊕Fq2 ⊕M4(Fq) 2 ⊕4r=1 Mnr (Fq2 ), with 36 = 4∑ r=1 n2r. Above leaves us with the only choice (34) for values of n′r which means the required Wedderburn decom- position is FqG2 ∼= F2q ⊕Fq2 ⊕M4(Fq) 2 ⊕M3(Fq2 )4. 67 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 3.3. The group G3 = ((C3 ×C3) o C3) o (C2 ×C2) Group G3 has the following presentation: G3 = 〈x,y,z,w,t | x2, [y,x], [z,x], [w,x]w−1, [t,x]t−1, y2, [z,y]z−1, [w,y], [t,y]t−1, z3, [w,z]t−1, [t,z], w3, [t,w], t3〉. Further, G3 has 11 conjugacy classes as shown in the table below. rep 1 x y z w t xy xz yw zw xyt size of class 1 9 9 6 6 2 9 18 18 12 18 order of rep 1 2 2 3 3 3 2 6 6 3 6 From above discussion, clearly the exponent of G3 is 6. Also G′3 ∼= (C3×C3)oC3 with G3/G′3 ∼= C2×C2. Next, we discuss the unit group of FqG3 when p > 3. Theorem 3.3. The unit group of FqG3, for q = pk, p > 3 where Fq is a finite field having q = pk elements is as follows: U(FqG3) ∼= (F∗q) 4 ⊕GL2(Fq)4 ⊕GL4(Fq) ⊕GL6(Fq)2. Proof. Since FqG3 is semisimple, we have FqG3 ∼= Fq ⊕t−1r=1 Mnr (Fr), for some t ∈ Z. (14) First assume that k is even which means for any prime p > 3, pk ≡ 1 mod 6. This means |S(γg)| = 1 for each g ∈ G3. Hence, (14), Theorems 2.1 and 2.2 imply that FqG3 ∼= Fq ⊕10r=1 Mnr (Fq). Incorporating Theorem 2.5 with G′3 ∼= (C3 ×C3) o C3 in above to obtain FqG3 ∼= F4q ⊕ 7 r=1 Mnr (Fq), where nr ≥ 2 with 104 = 7∑ r=1 n2r. (15) Above equation gives us four possibilities (24, 4, 62), (23, 32, 5, 7), (2, 32, 42, 52) and (34, 42, 6) for the values of n′rs. Further, consider the normal subgroup H3 = 〈t〉 of G3 having order 3 with K3 = G3/H3 ∼= S3 ×S3. It can be verified that K3 has 9 conjugacy classes as shown in the table below. r e x y z w xy xz yw zw s 1 3 3 2 2 9 6 6 4 o 1 2 2 3 3 2 6 6 3 Further, for all the representatives g of K3, |S(γg)| = 1 which means by Theorems 2.1 and 2.2, we have FqK3 ∼= Fq ⊕8r=1 Mtr (Fq), tr ∈ Z. Observe that K3/K′3 ∼= C2 ×C2. So, above and Theorem 2.5 imply that FqK3 ∼= F4q ⊕ 5 r=1 Mtr (Fq), with 32 = 5∑ r=1 t2r, tr ≥ 2. 68 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 This gives us the only choice (24, 4) for values of t′rs. Therefore, from Theorem 2.6 and (15), we conclude that (24, 4, 62) is the correct choice for n′rs which means FqG3 ∼= F4q ⊕M2(Fq) 4 ⊕M4(Fq) ⊕M6(Fq)2. (16) Now we consider that k is odd. We shall discuss this possibility in the following two cases: Case 1: pk ≡ 1 mod 6. In this case, Wedderburn decomposition is given by (16). Case 2. pk ≡ 5 mod 6. In this case, we have S(γg) = {γg} for all the representatives g of conjugacy classes. Therefore, Wedderburn decomposition is again given by (16). 3.4. The group G4 = C2 × (((C3 ×C3) o C3) o C2 Group G4 has the following presentation: G4 = 〈x,y,z,w,t | x2, [y,x], [z,x]z−1, [w,x]w−1, [t,x], y2, [z,y], [w,y], [t,y]t−1, z3, [w,z]t−1, [t,z], w3, [t,w], t3〉. Further, G4 has 20 conjugacy classes as shown in the table below. r e x y z w t xy xt yz yw yt zw t2 xyt xt2 yzw yt2 z2w xyt2 yz2w s 1 9 1 6 6 1 9 9 6 6 1 6 1 9 9 6 1 6 9 6 o 1 2 2 3 3 3 2 6 6 6 6 3 3 6 6 6 6 3 6 6 From above discussion, clearly the exponent of G4 is 6. Also G′4 ∼= (C3×C3)oC3 with G4/G′4 ∼= C2×C2. Next, we discuss the unit group of FqG4 when p > 3. Theorem 3.4. The unit group of FqG4, for q = pk, p > 3 where Fq is a finite field having q = pk elements is as follows: 1. for any p and k even or pk ≡ 1 mod 6 with k odd, we have U(FqG4) ∼= (F∗q) 4 ⊕GL2(Fq)8 ⊕GL3(Fq)8. 2. for pk ≡ 5 mod 6 with k odd, we have U(FqG4) ∼= (F∗q) 4 ⊕GL2(Fq)8 ⊕GL3(Fq2 )4. Proof. Since FqG4 is semisimple, we have FqG4 ∼= Fq ⊕t−1r=1 Mnr (Fr), for some t ∈ Z. (17) First assume that k is even which means for any prime p > 3, pk ≡ 1 mod 6. This means |S(γg)| = 1 for each g ∈ G4. Hence, (17), Theorems 2.1 and 2.2 imply that FqG4 ∼= Fq ⊕19r=1 Mnr (Fq). Using Theorem 2.5 with G′4 ∼= (C3 ×C3) o C3 in above to obtain FqG4 ∼= F4q ⊕ 16 r=1 Mnr (Fq), where nr ≥ 2 with 104 = 16∑ r=1 n2r. (18) Above equation gives us the only possibility (28, 38) for values of n′rs which means the required Wedderburn decomposition is FqG4 ∼= F4q ⊕M2(Fq) 8 ⊕M3(Fq)8. (19) 69 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 Now we consider that k is odd. We shall discuss this possibility in the following two cases: Case 1: pk ≡ 1 mod 6. In this case, Wedderburn decomposition is given by (19). Case 2. pk ≡ 5 mod 6. In this case, we have S(γt) = {γt,γt2}, S(γxt) = {γxt,γxt2}, S(γyt) = {γyt,γyt2}, S(γxyt) = {γxyt, γxyt2}, and S(γg) = {γg} for all the remaining representatives g of conjugacy classes. Hence, (17), Theorems 2.1 and 2.2 imply that FqG4 ∼= Fq ⊕11r=1 Mnr (Fq) ⊕ 15 r=12 Mnr (Fq2 ). Above with Theorem 2.5 yields FqG4 ∼= F4q ⊕ 8 r=1 Mnr (Fq) ⊕ 12 r=9 Mnr (Fq2 ), where nr ≥ 2 with 104 = 8∑ r=1 n2r + 2 12∑ r=9 n2r. (20) Further, consider the normal subgroup H4 = 〈t〉 of G4 having order 3 with K4 = G4/H4 ∼= C2 × ((C3 × C3) o C2). It can be verified that K4 has 12 conjugacy classes as shown in the table below. r e x y z w xy yz yw zw yzw z2w yz2w s 1 9 1 2 2 9 2 2 2 2 2 2 o 1 2 2 3 3 2 6 6 3 6 3 6 It can be seen that for all the representatives g of K4, |S(γg)| = 1 which means by Theorems 2.1 and 2.2, we have FqK4 ∼= Fq ⊕11r=1 Mtr (Fq), tr ∈ Z. Observe that K4/K′4 ∼= C2 ×C2. This with above and Theorem 2.5 imply that FqK4 ∼= F4q ⊕ 8 r=1 Mtr (Fq), with 32 = 8∑ r=1 t2r, tr ≥ 2. This gives us the only choice (28) for values of t′rs. Therefore, Theorem 2.6 and (20) yields FqG4 ∼= F4q ⊕M2(Fq) 8 ⊕4r=1 Mnr (Fq2 ), nr ≥ 2 with 36 = 4∑ r=1 n2r. Above leaves us with the only choice (34) for values of n′rs which means the required Wedderburn decomposition is FqG4 ∼= F4q ⊕M2(Fq) 8 ⊕M3(Fq2 )4. 4. Discussion We have characterized the unit groups of semisimple group algebras of 4 non-metabelian groups having order 108 and the results are verified using GAP. Clearly, the complexity in the calculation of Wedderburn decomposition upsurges with the increase in order of the group and we need to look into the Wedderburn decompositions of the quotient groups. The technique used for obtaining the Wedderburn decomposition works well provided the group has non-trivial normal subgroups of small order. 70 G. Mittal, R. K. Sharma / J. Algebra Comb. Discrete Appl. 8(2) (2021) 59–71 References [1] A. Bovdi, J. 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Mittal, On the unit group of a semisimple group algebra FqSL(2,Z5), Math Bohemica (2021). 71 https://doi.org/10.14232/actasm-013-510-1 https://doi.org/10.14232/actasm-013-510-1 https://doi.org/10.1080/00927870802103503 https://doi.org/10.1504/IJICOT.2009.024047 https://doi.org/10.1504/IJICOT.2009.024047 https://doi.org/10.1007/s10474-007-6169-4 https://doi.org/10.1007/s10474-007-6169-4 https://doi.org/10.1017/CBO9781139172769 https://doi.org/10.1017/CBO9781139172769 https://doi.org/10.13069/jacodesmath.83854 https://doi.org/10.13069/jacodesmath.83854 https://doi.org/10.14232/actasm-014-311-2 https://doi.org/10.14232/actasm-014-311-2 https://doi.org/10.21136/MB.2021.0116-19 https://doi.org/10.21136/MB.2021.0116-19 https://doi.org/10.1002/mana.19800950102 https://doi.org/10.21136/MB.2021.0104-20 https://doi.org/10.21136/MB.2021.0104-20 Introduction Preliminaries Unit group of FqG where G is a non-metabelian group of order 108 Discussion References