On unit group of ﬁnite semisimple group algebras of non-metabelian groups of order 108

: In this paper, we characterize the unit groups of semisimple group algebras F q G of non-metabelian groups of order 108 , where F q is a ﬁeld with q = p k 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.


Introduction
Let F q denote a finite field with q = p k elements for odd prime p > 3, G be a finite group and F q G 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 S 4 and SL (2,3) and the unit group of their group algebras have been discussed in [7,9]. Further, from [13], we also 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 F q G, 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.

Preliminaries
Let e denote the exponent of G, ζ a primitive e th root of unity. On the lines of [3], we define 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) = ζ s . In other words, I F 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 γ g , and the cyclotomic F-class of γ g be denoted by S(γ g ) = {γ g n | n ∈ I F }. The cardinality of S(γ g ) 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 2.2. Let ζ be defined as above and j be the number of cyclotomic F-classes in G. If K i , 1 ≤ i ≤ j, are the simple components of center of FG/J(FG) and S i , 1 ≤ i ≤ j, are the cyclotomic F-classes in G, then |S i | = [K i : 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 decomposition 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 A 1 and A 2 denote the finite dimensional algebras over F. Further, let A 2 be semisimple and g be an onto homomorphism between A 1 and A 2 , then we must have Proof. From [6], we have J(A 1 ) ⊆ Ker(g). This means there exists F-algebra homomorphism g 1 from A 1 /J(A 1 ) to A 2 which is also onto. In other words, we have As A 1 /J(A 1 ) is semisimple, there exists an ideal I of A 1 /J(A 1 ) such that Our claim is that I ∼ = A 2 . For this to prove, note that any element a ∈ A 1 /J(A 1 ) can be uniquely written as a = a 1 + a 2 where a 1 ∈ ker(g 1 ), a 2 ∈ I. So, define g 2 : A 1 /J(A 1 ) → ker(g 1 ) ⊕ A 2 by g 2 (a) = (a 1 , g 1 (a 2 )).
Since, ker(g 1 ) is a semisimple algebra over F and A 2 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 I F 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 e th root of unity and |q| is the order of q modulo e, then modulo e, we have I F = {1, q, q 2 , . . . , 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 F q G.
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. 3. Unit group of F q G where G is a non-metabelian group of order 108 The main objective of this section is to characterize the unit group of F q G where G is a nonmetabelian group of order 108. Upto isomorphism, there are 4 non-metabelian groups of order 108 namely: Here G 1 and G 2 are two non-isomorphic groups formed by the semi-direct product of (C 3 × C 3 ) C 3 and C 4 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 Group G 1 has the following presentation: Also G 1 has 20 conjugacy classes as shown in the table below.
r e x y z w t xy xt yz yw yt zw t 2 xyt xt 2 yzw yt 2 z 2 w xyt 2 yz 2 w 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 G 1 is 12. Also Next, we discuss the unit group of the group algebra F q G 1 when p > 3.
Theorem 3.1. The unit group of F q G 1 , for q = p k , p > 3 where F q is a finite field having q = p k elements is as follows: 1. for any p and k even or p k ≡ 1 mod 12 with k odd, we have 3. for p k ≡ 7 mod 12 with k odd, we have Proof. Since F q G 1 is semisimple, using Lemma 2.3 we get First assume that k is even which means for any prime p > 3, we have p k ≡ 1 mod 3 and p k ≡ 1 mod 4.
Using Chinese remainder theorem, we get p k ≡ 1 mod 12. This means |S(γ g )| = 1 for each g ∈ G 1 as I F = {1}. Hence, (1), Theorems 2.1 and 2.2 imply that Above equation gives the only possibility (2 8 , 3 8 ) for the values of n r s where a b means (a, a, · · · , b times) and therefore, (3) implies that Now we consider that k is odd. We shall discuss this possibility in the following four cases: Case 1: p k ≡ 1 mod 3 and p k ≡ 1 mod 4. In this case, Wedderburn decomposition is given by (4). Case 2. p k ≡ 5 mod 12. In this case, we have and S(γ g ) = {γ g } for the remaining representatives g of conjugacy classes. Therefore, (1) and Theorems 2.1, 2.2 imply that This with above and Theorem 2.5 yields Further, consider the normal subgroup H 1 = t of G 1 having order 3 with The quotient group K 1 has 12 conjugacy classes as shown in the table below. Here elements of K 1 are cosets, for instance, x ∈ K 1 is xH 1 but we keep the same notation.
r e x y z w xy yz yw zw yzw z 2 w yz 2 w 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 K 1 , |S(γ g )| = 1. Therefore, from Theorems 2.1 and 2.2, we have Observe that This gives us the only choice (2 8 ) for values of t r s and therefore, Theorem 2.5 and (5) yields Above leaves us with the only choice (3 4 ) for values of n r s which means the required Wedderburn decomposition is Case 3. p k ≡ 7 mod 12. In this case, we have for the remaining representatives g of conjugacy classes. Therefore (1) and Theorems 2.1, 2.2 imply that This with above and Theorem 2.5 yields Again consider the normal subgroup H 1 of G 1 . In this case, it can be verified that |S(γ g )| = 1 for all the representatives g of K 1 except x and xy for which S(γ x ) = {γ x , γ xy }. Therefore, employing Theorems 2.1 and 2.2 to obtain Since K 1 /K 1 ∼ = C 4 , above and Theorem 2.5 imply that This gives us the only choice (2 8 ) for values of t r s. Hence, Theorem 2.6 and (6) imply that Above leaves us with the only choice (3 6 ) for values of n r s which means the required Wedderburn decomposition is Case 4. p k ≡ 11 mod 12. In this case, we have S(γ x ) = {γ x , γ xy }, and S(γ g ) = {γ g } for the remaining representatives g of conjugacy classes. Therefore, (1) and Theorems 2.1, 2.2 imply that This with above and Theorem 2.5 yields In this case, again we have |S(γ g )| = 1 for all representatives g of K 1 except x and xy which means the Wedderburn decomposition of F q K 1 is same as obtained in case 3, i.e.
Now employ Theorem 2.6 and (7) to obtain This leaves us with the only choice (3 4 ) for values of n r s which means the required Wedderburn decomposition is

The group
Group G 2 has the following presentation: Further, G 2 has 14 conjugacy classes as shown in the table below.
Theorem 3.2. The unit group of F q G 2 , for q = p k , p > 3 where F q is a finite field having q = p k elements is as follows: 1. for any p and k even or p k ≡ 1 mod 12 with k odd, we have 2. for p k ≡ 5 mod 12 with k odd, we have 3. for p k ≡ 7 mod 12 with k odd, we have 4. for p k ≡ 11 mod 12 with k odd, we have Proof. Since F q G 2 is semisimple, we have First assume that k is even which means for any prime p > 3, p k ≡ 1 mod 12. This means |S(γ g )| = 1 for each g ∈ G 2 . Hence, (8), Theorems 2.1 and 2.2 imply that Above equation gives us four possibilities (2 8 , 6 2 ), (2 5 , 3 2 , 4, 5 2 ), (2 4 , 3 4 , 4, 6) and (3 8 , 4 2 ) for the values of n r s. Further, consider the normal subgroup H 2 = t of G 2 having order 3 with It can be verified that K 2 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 Also, for all the representatives g of K 2 , |S(γ g )| = 1 which means by Theorems 2.1 and 2.2, we have Observe that K 2 /K 2 ∼ = C 4 . This with above and Theorem 2.4 imply that This gives us the only choice (4 2 ) for values of t r s. Therefore, Theorem 2.6 and (9) imply that (3 8 , 4 2 ) is the correct choice for values of n r s and therefore, we have Now we consider that k is odd. We shall discuss this possibility in the following four cases: Case 1: p k ≡ 1 mod 12. In this case, Wedderburn decomposition is given by (10). Case 2. p k ≡ 5 mod 12. In this case, we have and S(γ g ) = {γ g } for the remaining representatives g of conjugacy classes. Therefore, (8) and Theorems 2.1, 2.2 imply that This with above and Theorem 2.5 yields Further, again consider the normal subgroup H 2 = t of G 2 . It can be verified that for all the representatives g of K 2 , |S(γ g )| = 1 which means (as earlier) This with Theorem 2.6 and (11) imply that Above leaves us with the only choice (3 4 ) for values of n r s which means the required Wedderburn decomposition is Case 3. p k ≡ 7 mod 12. In this case, we have for the remaining representatives g of conjugacy classes. Therefore, (8) and Theorems 2.1, 2.2 imply that This with Theorem 2.5 yields Further, it can be verified that |S(γ g )| = 1 for all the representatives g of K 2 except x and xy. For these, we have S(γ x ) = {γ x , γ xy } which means by Theorems 2.1 and 2.2, we have Incorporating K 2 /K 2 ∼ = C 4 with Theorem 2.5 to obtain This gives us only choice (4 2 ) for values of t r s. Therefore, Theorem 2.6 and (12) yields Above leaves us with the only choice (3 6 ) for values of n r which means the required Wedderburn decomposition is Case 4. p k ≡ 11 mod 12. In this case, we have for the remaining representatives g of conjugacy classes. Therefore, (8) and Theorems 2.1, 2.2 imply that This with above and Theorem 2.5 yields Further, it can be verified that |S(γ g )| = 1 for all the representatives g of K 2 except x and xy which means (as in case 3), Now employ Theorem 2.6 and (13) to obtain Above leaves us with the only choice (3 4 ) for values of n r which means the required Wedderburn decomposition is Group G 3 has the following presentation: Further, G 3 has 11 conjugacy classes as shown in the table below.
Next, we discuss the unit group of F q G 3 when p > 3.
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 K 3 , |S(γ g )| = 1 which means by Theorems 2.1 and 2.2, we have Observe that K 3 /K 3 ∼ = C 2 × C 2 . So, above and Theorem 2.5 imply that This gives us the only choice (2 4 , 4) for values of t r s. Therefore, from Theorem 2.6 and (15), we conclude that (2 4 , 4, 6 2 ) is the correct choice for n r s which means Now we consider that k is odd. We shall discuss this possibility in the following two cases: Case 1: p k ≡ 1 mod 6. In this case, Wedderburn decomposition is given by (16). Case 2. p k ≡ 5 mod 6. In this case, we have S(γ g ) = {γ g } for all the representatives g of conjugacy classes. Therefore, Wedderburn decomposition is again given by (16).

3.4.
The group Group G 4 has the following presentation: Further, G 4 has 20 conjugacy classes as shown in the table below.
Next, we discuss the unit group of F q G 4 when p > 3.
Theorem 3.4. The unit group of F q G 4 , for q = p k , p > 3 where F q is a finite field having q = p k elements is as follows: 1. for any p and k even or p k ≡ 1 mod 6 with k odd, we have 2. for p k ≡ 5 mod 6 with k odd, we have Proof. Since F q G 4 is semisimple, we have First assume that k is even which means for any prime p > 3, p k ≡ 1 mod 6. This means |S(γ g )| = 1 for each g ∈ G 4 . Hence, (17), Theorems 2.1 and 2.2 imply that Above equation gives us the only possibility (2 8 , 3 8 ) for values of n r s which means the required Wedderburn decomposition is Now we consider that k is odd. We shall discuss this possibility in the following two cases: Case 1: p k ≡ 1 mod 6. In this case, Wedderburn decomposition is given by (19). Case 2. p k ≡ 5 mod 6. In this case, we have S(γ t ) = {γ t , γ t 2 }, S(γ xt ) = {γ xt , γ xt 2 }, S(γ yt ) = {γ yt , γ yt 2 }, S(γ xyt ) = {γ xyt , γ xyt 2 }, and S(γ g ) = {γ g } for all the remaining representatives g of conjugacy classes. Hence, (17), Theorems 2.1 and 2.2 imply that Above with Theorem 2.5 yields Further, consider the normal subgroup H 4 = t of G 4 having order 3 with K 4 = G 4 /H 4 ∼ = C 2 × ((C 3 × C 3 ) C 2 ). It can be verified that K 4 has 12 conjugacy classes as shown in the table below.
r e x y z w xy yz yw zw yzw z 2 w yz 2 w 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 K 4 , |S(γ g )| = 1 which means by Theorems 2.1 and 2.2, we have Observe that K 4 /K 4 ∼ = C 2 × C 2 . This with above and Theorem 2.5 imply that Above leaves us with the only choice (3 4 ) for values of n r s which means the required Wedderburn decomposition is

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.