The unit group of group algebra F q SL ( 2 , Z 3 )

Let Fq be a finite field of characteristic p having q elements, where q = p and p ≥ 5. Let SL(2,Z3) be the special linear group of 2 × 2 matrices with determinant 1 over Z3. In this note we establish the structure of the unit group of FqSL(2,Z3). 2010 MSC: 16U60, 20C05


Introduction
Let F G be a group algebra of a finite group G over a field F and U(F G) be the group of units in F G. It is a classical problem to study units and their properties in group ring theory.The case, when G is a finite abelian group, the structure of F G is studied by Perlis and Walker in [14].In 2006, T. Hurley introduced a correspondence between group ring and certain ring of matrices (see [6]).As an application of units of a group ring, T. Hurley gave a method to construct convolutional codes from units in group ring (see [7]).
A lot of work has been done for finding the algebraic structure of the unit group U(F G) of a group algebra F G, when G is a finite non-abelian group.Here we are providing some literature survey for the same.For dihedral groups, the structure of the unit group U(F G) over a finite field F is discussed in [1,4,10,12].J. Gildea et.al.(see [3]) and R. K. Sharma et.al.(see [15]) have given the structure of the unit group U(F G), where G is alternating group A 4 .Unit group of algebra of circulant matrix has been discussed in [11,17].The unit group of group algebras of some non-abelian groups with small orders are established in [16,18,19]).
In this article, we are interested in studying the structure of the unit group of F q SL(2, Z 3 ) over a finite field of characteristic greater than 3.
This work was supported by IIT Delhi, India through GATE Senior Research Fellowship.Swati Maheshwari (Corresponding Author), R. K. Sharma; Department of Mathematics, Indian Institute of Technology Delhi, India (email: swatimahesh88@gmail.com, rksharmaiitd@gmail.com).

Preliminaries
The following results provide useful information about the decomposition of A/J(A), where A = F G, J(A) be its Jacobson radical and F being a field of characteristic p.For basic definitions and results, we refer to [13].We briefly introduce some definitions and notations those will be needed subsequently.Definition 2.1.An element g ∈ G is said to be p-regular if p o(g).Let s be the l.c.m. of the orders of the p-regular elements of G, ζ be a primitive s-th root of unity over F .Then T G,F be the multiplicative group consisting of those integers t, taken modulo s, for which ζ → ζ t defines an automorphism of Note that if u is a power of a prime such that (u, s) = 1 and c = ord s (u) is the multiplicative order of u modulo s, then Definition 2.2.If g ∈ G is a p-regular element, then the sum of all conjugates of g ∈ G is denoted by γ g and the cyclotomic F -class of g is defined to be the set Throughout this article, G = SL(2, Z 3 ).F q is a field of characteristic p, where q = p k and k is a positive integer.The conjugacy class of g ∈ G is denoted by [g].

Main result
We shall use the presentation of G given in [5], where a = 1 0 1 1 and b = 0 1 −1 0 .
We can see that G has 7 conjugacy classes as follows: representative elements in the class order of element [a] a, (ba) 4 , (ab) 4 , b −1 ab 3 [ab] ab, ba, a 2 ba 2 , ab 2  6 We have (p, |G|) = 1 and so J(F p = 0. Further, we discuss the decomposition of F p k G.
Theorem 3.1.Let F q be a finite field of characteristic p, where p ≥ 5. Then the Wedderburn decomposition of F q G is given by k is odd p ≡ 1 mod 3 and p ≡ ±1mod 4 Proof.Since F q G is semisimple, so it has the Wedderburn decomposition which is given by where for each i, n i ≥ 1and F i is a finite extension of F q .By using Lemma 2.5, we have Further, we find n i 's and F i 's.Since | G |= 24, hence any element g ∈ G is a p-regular element.For finding cyclotomic F q -classes of G, first we assume that k is even.We have Then by Chinese remainder theorem By using above observation, we have Therefore by using Equation (1), Proposition 2.3 and Theorem 2.4, we have for some n i ≥ 1.As dimension of F q G is 24, we get Using above equality, 1 ≤ n i ≤ 3. Clearly any n i = n j = 3 for 1 ≤ i = j ≤ 3 not possible.So the only possible choice for n i 's is Therefore the decomposition F q G is given by Now we consider the case when k is odd.We shall discuss this case into two parts 1. p ≡ 1 mod 3 and p ≡ ±1 mod 4 2. p ≡ −1 mod 3 and p ≡ ±1 mod 4 Case 1. Suppose k is odd with p ≡ 1 mod 3 and p ≡ ±1 mod 4.
Observe that p k ≡ p mod 4 and p k ≡ p mod 3.
Then by Chinese remainder theorem Since Hence n i 's and F i 's are same as above.So the decomposition of F q G is given by Suppose k is odd with p ≡ −1 mod 3 and p ≡ ±1 mod 4. Using the observation in case 1, we have Therefore by using Equation (1), Proposition 2.3 and Theorem 2.4, we have Clearly n 3 and n 4 can not be equal to 3.So the only possible choice for n i 's is n 1 = 2, n 2 = 3, n 3 = 1, n 4 = 2. Therefore the decomposition of F q G is given by F q G ∼ = F q ⊕ F q 2 ⊕ M (2, F q ) ⊕ M (2, F q 2 ) ⊕ M (3, F q ).Corollary 3.2.Let q = p k ,where p ≥ 5 is a prime.Then the structure of U(F q G) is given by condition on k U(F q G) k is even C 3 q−1 ⊕ GL(2, F q ) 3 ⊕ GL(3, F q ) k is odd p ≡ 1 mod 3 and p ≡ ±1mod 4 k is odd p ≡ −1 mod 3, ±1mod 4 C q−1 ⊕ C q 2 −1 ⊕ GL(2, F q ) ⊕ GL(2, F q 2 ) ⊕ GL(3, F q ) Proof.It follows by the fact that, if R and S are two rings then U(R ⊕ S) = U(R) ⊕ U(S).
S 1 , S 2 , . . ., S w are the cyclotomic F -classes of G, then with a suitable re-ordering of indices, | Si |= [K i : F ]. Lemma 2.5.[9, Observation 2.2.1, p.22] Let B 1 ,B 2 be two finite dimensional F -algebras such that B 2 is semisimple.If f : B 1 → B 2 is an onto homomorphism of F -algebras, then there exists a semisimple F -algebra such that