## Introduction

Let $$q=p^k$$ for some prime p and $$k\in \mathbb {N}$$. Let $$\mathbb {F}_q$$ denote the finite field of cardinality q. For any group G, let $$\mathbb {F}_qG$$ denote the group algebra of G over $$\mathbb {F}_q$$. We will follow  for basic notations. The group of units of $$\mathbb {F}_qG$$ has applications in different areas, including the construction of convolutional codes (see [5,6,7]) and solving problems in combinatorial number theory (see ) et cetera. This necessitates finding the explicit structure of the group of units of $$\mathbb {F}_qG$$.

In , the author has described units of $$\mathbb {F}_qG$$, where G is a p-group. The authors of  complete the study of unit groups of semisimple group algebras of all groups up to order 120, except that of the symmetric group $$S_5$$ and groups of order 96. In Remark 3.7 of this article, we complete the characterization for the group $$S_5$$. However, the complexity of the problem increases if the group has a larger size, as that requires solving an equation of type $$\displaystyle \sum \limits _{i=1}^nk_in_i^2=|G|$$. The most commonly used tricks include identifying a normal subgroup H of G and considering the algebra $$\mathbb {F}_q(G/H)$$ inside $$\mathbb {F}_qG$$. This is not possible if the group G does not have any normal subgroup. This is why we will be using representations of the group $$A_6$$ to solve the problem in case of the same. See [16, 17] et cetera for more exposition about other groups.

The objective of this article is twofold. We start by investigation of $$\mathbb {F}_qS_n$$ where $$p>n$$. This is mainly a consequence of the representation theory of $$S_n$$ over $${\mathbb {C}}$$ and the connection between the Brauer characters of the group when $$p>n$$ and the ordinary characters over $${\mathbb {C}}$$. We state the characterization in this case in Theorem 3.6. The group of units of the semisimple algebras $$\mathbb {F}_qA_5$$ and $$\mathbb {F}_q\text {SL}(3,2)$$ have been characterized in [1, 12], respectively. In this article, we look at the next non-Abelian simple group $$A_6$$, the alternating group on six letters. We give a complete characterization of $$\mathbb {F}_qA_6$$ for the case $$p\ge 7$$ in Theorem 4.8.

The rest of the article is organized as follows: in “Preliminaries” section, we give some basic definitions and results. “Units of $$\mathbb {F}_{p^k}S_n$$ for $$p \not \mid n$$3” section is about the general description of representations of $$S_n$$ over an arbitrary field of characteristic $$p>n$$ and deducing the structure of $${\mathcal {U}}(\mathbb {F}_{p^k}S_n)$$ for $$p>n$$. In section 4, we present the result about $$\mathbb {F}_{p^k}A_6$$ where $$p\ge 7$$.

## Preliminaries

We start by fixing some notations. Already mentioned notations from section 1 are adopted. For a field extension $$E/\mathbb {F}_q$$, $$\text {Gal}(E/\mathbb {F}_q)$$ will denote the Galois group of the extension. For $$m\in \mathbb {N}$$, the notation $$\text {M}(m,R)$$ denotes the ring of $$m\times m$$ matrices over R and $$\text {GL}(m,R)$$ will denote the set of all invertible matrices in $$\text {M}(m,R)$$. For a ring R, the set of units of R will be denoted by $$R^\times$$. Let Z(R) and J(R) denote the center and the Jacobson radical, respectively. If G is a group and $$x\in G$$, then [x] will denote the conjugacy class of x in G. For the group ring $$\mathbb {F}_q G$$, the group of units will be denoted as $$\mathcal {U}(\mathbb {F}_q G)$$. For the notations on projective spaces, we follow .

We say an element g $$\in$$ G is a $$p'$$-element if the order of g is not divisible by p. Let e be the exponent of the group G and $$\eta$$ be a primitive rth root of unity, where $$e=p^fr$$ and $$p\not \mid r$$. Let

\begin{aligned} I_{\mathbb {F}_q} = \left\{ l~(\text {mod }e):\text { there exists } \sigma \in \text {Gal}(\mathbb {F}_q(\eta )/\mathbb {F}_q)\text { satisfying } \sigma (\eta ) = \eta ^l\right\} . \end{aligned}

### Definition 2.1

For a $$p'$$-element $$g \in G$$, the cyclotomic $$\mathbb {F}_q$$-class of g, denoted by $$S_{\mathbb {F}_q}(\gamma _g)$$, is defined as $$\left\{ \gamma _{g^l} : l \in I_{\mathbb {F}_q} \right\}$$, where $$\gamma _{g^l}\in \mathbb {F}_q G$$ is the sum of all conjugates of $$g^l$$ in G.

Then, we have the following results, which are crucial in determining the Artin–Wedderburn decomposition of $$\mathbb {F}_qG$$.

### Lemma 2.2

[2, Proposition 1.2] The number of simple components of $$\mathbb {F}_qG/J(\mathbb {F}_qG)$$ is equal to the number of cyclotomic $$\mathbb {F}_q$$-classes in G.

### Definition 2.3

Let $$\pi$$ be a representation of a group G over a field F. $$\pi$$ is said to be absolutely irreducible if $$\pi ^E$$ is irreducible for every field $$F\subseteq E$$, where $$\pi ^E$$ is the representation $$\pi \otimes E$$ over E.

### Definition 2.4

A field F is a splitting field for G if every irreducible representation of G over F is absolutely irreducible.

### Lemma 2.5

[2, Theorem 1.3] Let n be the number of cyclotomic $$\mathbb {F}_q$$-classes in G. If $$L_1,L_2,\cdots ,L_n$$ are the simple components of $$Z(\mathbb {F}_qG/J(\mathbb {F}_qG))$$ and $$S_1,S_2,\cdots ,S_n$$ are the cyclotomic $$\mathbb {F}_q$$-classes of G, then with a suitable reordering of the indices,

\begin{aligned} |S_i| = [L_i:\mathbb {F}_q]. \end{aligned}

### Lemma 2.6

[11, Lemma 2.5] Let K be a field of characteristic p and let $$A_1$$, $$A_2$$ be two finite dimensional K-algebras. Assume $$A_1$$ to be semisimple. If g : $$A_2$$ $$\longrightarrow$$ $$A_1$$ is a surjective homomorphism of K-algebras, then there exists a semisimple K-algebra l such that $$A_2/J(A_2)$$ $$=$$ $$l\oplus A_1$$.

We need the following lemmas from our previous work to compute some components of the Artin–Wedderburn decomposition of $$\mathbb {F}_qG$$, for a finite group G under consideration.

### Lemma 2.7

[1, Lemma 3.1] Let G be a group of order n and $$\mathbb {F}$$ be a field of characteristic $$p>0$$. Let G acts on $$\{1,2,\cdots ,k\}$$ doubly transitively. Set $$G_{i}=\{g\in G:g\cdot i=i\}$$ and $$G_{i,j}=\{g\in G:g\cdot i=i,g\cdot j=j\}$$. Then, the $$\mathbb {F}G$$ module

\begin{aligned} W=\left\{ x\in \mathbb {F}^k:\displaystyle \sum \limits _{i=1}^kx_i=0\right\} \end{aligned}

is an irreducible $$\mathbb {F}G$$ module if $$p\not \mid k,p\not \mid |G_{1,2}|$$.

### Lemma 2.8

[1, Corollary 3.8] Let G be a finite group, K be a finite field of characteristic $$p>0$$, $$p\not \mid |G|$$. Suppose there exists an n dimensional irreducible representations of G over k. Then, M(nk) appears as one of the components of the Artin-Wedderburn decomposition of the semisimple algebra $$\mathbb {F}_qG$$.

## Units of $$\mathbb {F}_{p^k}S_n$$ for $$p \not \mid n$$

We start the section by talking about representations of $$S_n$$ over a finite field. We define the Brauer character and state some important results about representations over an arbitrary field. See  for further details.

Let E be a field of characteristic p. We choose a ring of algebraic integers A in $${\mathbb {C}}$$ such that $$E = A/M$$, where M is a maximal ideal of A containing pA. Take f to be the natural map $$A \longrightarrow E$$. Take $$W=\{z\in {\mathbb {C}} | z^m=1\text { for some } m \in {\mathbb {Z}}\text { with } p\not \mid m \}$$ (note that $$W \subseteq A$$). Now let $$\pi$$ be a representation of a finite group G over E. Let S be the set of $$p'$$ elements of G. For $$\alpha \in S$$, let $$\epsilon _1,\epsilon _2,\ldots ,\epsilon _l\in E^\times$$ be the eigenvalues of $$\pi (\alpha )$$ with multiplicities. Then, for every i, there exists a unique $$u_i \in W$$ such that $$f(u_i)= \epsilon _i$$. Define $$\phi$$ : $$S \longrightarrow {\mathbb {C}}$$ as $$\phi (\alpha ) = \Sigma u_i$$. Then, $$\phi$$ is called the Brauer character of G afforded by $$\pi$$.

### Remark 3.1

The description of Brauer character comes along with a choice of a maximal ideal M of A.

Suppose $$\pi _1,\pi _2,\ldots ,\pi _k$$ are all the non-isomorphic irreducible representations of G over E up to isomorphism. Let $$\phi _i$$ be the Brauer character afforded by $$\pi _i$$. Then, $$\phi _i's$$ are called irreducible Brauer characters and we denote by IBr(G) the set {$$\phi _i$$}. We denote by Irr(G) the set of irreducible characters of G over $$\mathbb {C}$$. We have the following results.

### Lemma 3.2

[8, Theorem 15.13] We have $$IBr(G) = Irr(G)$$ whenever $$p\not | |G|$$.

For the rest of this section, take $$G= S_n$$, the symmetric group on n letters. We say a partition $$\lambda = (\lambda _1,\lambda _2,\cdots ,\lambda _l)$$ of n is p-singular if for some j we have $$\lambda _{j+1}=\lambda _{j+2}= \ldots =\lambda _{j+p}$$. If a partition is not p-singular, it is called p-regular. Then we have the following.

### Lemma 3.3

[9, Theorem 11.5] If F is a field of characteristic p, then as $$\lambda$$ varies over the p-regular partitions, $$D^{\lambda }$$ varies over the complete set of inequivalent irreducible $$FS_n$$-modules, where $$D^\lambda =\dfrac{S^\lambda }{S^\lambda \cap (S^{\lambda })^{\perp }}$$ and $$S^\lambda$$ denotes the Specht module corresponding to the partition $$\lambda$$. Moreover, every field is a splitting field for $$S_n$$.

### Proof

The proof follows immediately from the fact that every partition of n is a p-regular partition. $$\square$$

### Lemma 3.4

The dimensions of non-isomorphic irreducible representations of $$S_n$$ over E coincide with the dimensions of non-isomorphic irreducible representations of $$S_n$$ over $$\mathbb {C}$$ when characteristic of the field E is greater than n.

### Proof

Since the dimension of a representation is as same as the value of the corresponding character $$\chi$$ at the identity element of the group, the result follows from Lemma 3.2. $$\square$$

### Proposition 3.5

Let $$S_n$$ denote the symmetric group on n letters and $$\mathbb {F}_{p^k}$$ be a finite field where $$p>n$$. Then,

\begin{aligned} \mathbb {F}_{p^k}S_n\cong \displaystyle \bigoplus _{\chi \in Irr(G)}\text {M}(\chi (1),\mathbb {F}_{p^k}). \end{aligned}

### Proof

Since being a semisimple algebra $$\mathbb {C}S_n\cong \displaystyle \bigoplus _{\chi \in Irr(G)}\text {M}(\chi (1),\mathbb {C})$$, the result follows from Lemmas 2.8, 3.2 and 3.4. $$\square$$

### Theorem 3.6

Let $$S_n$$ denote the symmetric group on n letters and $$\mathbb {F}_{p^k}$$ be a finite field where $$p>n$$. Then,

\begin{aligned} \mathcal {U}(\mathbb {F}_{p^k}S_n)\cong \displaystyle \bigoplus _{\chi \in Irr(G)}\text {GL}(\chi (1),\mathbb {F}_{p^k}). \end{aligned}

### Proof

This follows immediately from Proposition 3.5 and the fact that given two rings $$R_1,R_2$$, we have $$(R_1\times R_2)^\times =R_1^\times \times R_2^\times$$. $$\square$$

### Remark 3.7

Theorem 3.6 improves the result of  and proves that when $$p>5$$, unit group of $$\mathbb {F}_{p^k}S_5$$ is $$\mathcal {U}(\mathbb {F}_{p^k}S_5)$$ given by

\begin{aligned} \mathbb {F}_{p^k}^\times \oplus \mathbb {F}_{p^k}^\times \oplus \text {GL}(4,\mathbb {F}_{p^k})\oplus \text {GL}(4,\mathbb {F}_{p^k})\oplus \text {GL}(5,\mathbb {F}_{p^k}) \oplus \text {GL}(5,\mathbb {F}_{p^k})\oplus \text {GL}(6,\mathbb {F}_{p^k}) \end{aligned}

### Remark 3.8

For an irreducible representation $$\chi$$ of $$S_n$$ over a field of characteristic $$p>n$$, this is characterized by a partition $$\lambda$$ of n. The value $$\chi (1)$$ can be calculated as the number of standard Young tableaux of shape $$\lambda$$.

## Units of $$\mathbb {F}_{p^k}A_6$$ for $$p \ge 7$$

We start with the description of the conjugacy classes in $$A_6$$. Using , the group has 7 conjugacy classes, of which the representatives are given by $$(1),a=(1,2)(3,4),b=(1,2,3),c=(1,2,3)(4,5,6),d=(1,2,3,4)(5,6),e=(1,2,3,4,5)$$ and $$f=(1,2,3,4,6).$$ We have the following relations:

\begin{aligned} \text {for all }&g\not \in [e]\cup [f],~[g]=[g^{-1}],\end{aligned}
(4.1)
\begin{aligned} \text {and }&[e]=[e^4],[e^2]=[e^3]=[f]. \end{aligned}
(4.2)

### Proposition 4.1

Let $$\mathbb {F}_{q}$$ be a field of characteristic $$p\ge 7$$ and $$G=A_6$$. Then, the Artin–Wedderburn decomposition of $$\mathbb {F}_qG$$ is one of the following:

$$\mathbb {F}_q\oplus \bigoplus \limits _{i=1}^6M(n_i,\mathbb {F}_q)$$,

$$\mathbb {F}_q\oplus \bigoplus \limits _{i=1}^4M(n_i,\mathbb {F}_q)\oplus M(n_5,\mathbb {F}_{q^2})$$

### Proof

Since $$p\ge 7$$, we have $$p\not \mid |A_6|$$; by Maschke’s theorem we have $$J(\mathbb {F}_qG)=0$$ . Hence, Wedderburn decomposition of $$\mathbb {F}_qG$$ is isomorphic to $$\bigoplus \limits _{i=1}^nM({n_i},K_i)$$, where for all $$1\le i\le n$$, we have $$n_i > 0$$ and $$K_i$$ is a finite extension of $$\mathbb {F}_q$$.

Firstly, from Lemma 2.6, we have

\begin{aligned} \mathbb {F}_qG \cong \mathbb {F}_q \bigoplus \limits _{i=1}^{n-1}M({n_i},K_i), \end{aligned}
(4.3)

taking g to be the map $$g(\sum \limits _{x\in A_6}\alpha _xx)=\sum \limits _{x\in A_6}\alpha _x$$. Now to compute these $$n_i$$’s and $$K_i$$’s we calculate the cyclotomic $$\mathbb {F}_q$$ classes of G. Note that $$p^k\equiv \pm 1\mod 4,p^k\equiv \pm 1\mod 3$$ for any prime p. Hence, $$S_{\mathbb {F}_q}(\gamma _g)=\{\gamma _g\}$$ whenever $$g\not \in [e]\cup [f]$$ (by Equation 4.1). Hence, we have to consider $$S_{\mathbb {F}_q}(\gamma _g)$$ in the other cases.

When $$p\equiv \pm 1\mod 5$$, $$S_{\mathbb {F}_q}(\gamma _e)=\{\gamma _e\}$$ and $$S_{\mathbb {F}_q}(\gamma _f)=\{\gamma _f\}$$, by Eq. 4.2 and the fact that $$p^k\equiv \pm 1\mod 5$$. Thus, by Lemmas 2.2 and 2.5, there are seven cyclotomic $$\mathbb {F}_q$$-classes and $$[K_i:\mathbb {F}_q]=1$$ for all $$1\le i\le 6$$. This gives that in this case the Artin–Wedderburn decomposition is

\begin{aligned} \mathbb {F}_q\oplus \bigoplus \limits _{i=1}^6M(n_i,\mathbb {F}_q). \end{aligned}

When $$p\equiv \pm 2\mod 5$$ and k is even, then $$p^k\equiv - 1\mod 5$$. Similarly, in this case the Artin–Wedderburn decomposition is

\begin{aligned} \mathbb {F}_q\oplus \bigoplus \limits _{i=1}^6M(n_i,\mathbb {F}_q). \end{aligned}

Lastly, when $$p\equiv \pm 2\mod 5$$ and k is odd, then $$p^k\equiv \pm 2\mod 5$$ and $$S_{\mathbb {F}_q}(\gamma _e)=\{\gamma _e,\gamma _f\}$$ by Eq. 4.2. Thus, by Lemmas 2.2 and 2.5, there are six cyclotomic $$\mathbb {F}_q$$-classes and $$[K_i:\mathbb {F}_q]=1$$ for all $$1\le i\le 4$$, $$[K_5:\mathbb {F}_q]=2$$ . In this case, the Artin–Wedderburn decomposition is

\begin{aligned} \mathbb {F}_q\oplus \bigoplus \limits _{i=1}^4M(n_i,\mathbb {F}_q)\oplus M(n_5,\mathbb {F}_{q^2}) . \end{aligned}

$$\square$$

Since $$\dim \mathbb {F}_{q}A_6=|A_6|=360$$, Proposition 4.1 gives that the $$n_i$$’s should satisfy $$n_1^2+n_2^2+n_3^2+n_4^2+n_5^2+n_6^2=359$$ or $$n_1^2+n_2^2+n_3^2+n_4^2+2n_5^2=359$$. Since these equations do not have a unique solution, we find some of the $$n_i$$’s using representations of $$A_6$$ over $$\mathbb {F}_q$$ and invoke Lemma 2.7 to reach a unique solution for the mentioned equations. We have the following results.

### Lemma 4.2

The group $$S_6$$ has four inequivalent irreducible representations of degree 5, which on restriction on $$A_6$$ give two inequivalent irreducible representations of $$A_6$$ over $$\mathbb {F}_{p^k}$$ for $$p\ge 7$$. Moreover, these irreducible representations are obtained from two non-isomorphic doubly transitive actions on a set of 6 points.

### Proof

Note that $$S_6$$ acts on $$T=\{1,2,3,4,5,6\}$$ doubly transitively. Hence, by Lemma 2.7, we get an irreducible representation of degree 5. Since tensoring with sign representation gives irreducible representations, we get two inequivalent irreducible representations of degree 5 of $$S_6$$, say $$\pi _1$$ and $$\pi _2$$.

For the other two irreducible representations of dimension 5, we consider the outer automorphism of $$S_6$$, say $$\varphi$$, given on generators as follows:

\begin{aligned} \varphi ((1,2))&=(1,2)(3,4)(5,6)\\ \varphi ((2,3))&=(1,3)(2,5)(4,6)\\ \varphi ((3,4))&=(1,5)(2,6)(3,4)\\ \varphi ((4,5))&=(1,3)(2,4)(5,6)\\ \varphi ((5,6))&=(1,6)(2,5)(3,4). \end{aligned}

This gives another doubly transitive action on T, which is not isomorphic to the previous action. Thus, we get another 5 dimensional irreducible representation, say $$\pi _3$$. Tensoring $$\pi _3$$ with the sign representations, we get $$\pi _4$$ which is a 5-dimensional irreducible representation of $$S_6$$ different from $$\pi _3$$. By considering the characters of the corresponding representations, we see that $$\pi _1,\pi _2,\pi _3$$ and $$\pi _4$$ are all distinct.

Since $$A_6$$ acts doubly transitively on T via the restrictions of these two actions, we obtain two non-isomorphic 5-dimensional irreducible representations of $$A_6$$. $$\square$$

### Corollary 4.3

The algebra $$\mathbb {F}_qA_6$$ has two components to be $$\text {M}(5,\mathbb {F}_q)$$ for $$p\ge 7$$.

### Proof

Immediately follows from Lemmas 4.2 and 2.7. $$\square$$

### Corollary 4.4

There does not exist any 4 dimensional irreducible representations of $$A_6$$ over $$\mathbb {F}_{p^k}$$ for $$p\ge 7$$.

### Proof

From Lemma 3.3, we know that any field $$\mathbb {F}_{p^k},p\ge 7$$ is a splitting field of $$S_6$$. Hence, by Proposition 3.5, we have degrees of irreducible representations of $$S_6$$ are $$\{1,5,9,10,16\}$$.

Recall that by Frobenius reciprocity we have the following bijection

\begin{aligned} \text {Hom}_{\mathbb {F}_qS_6}(\text {Ind}V,W)\cong \text {Hom}_{\mathbb {F}_qA_6}(V,\text {Res}W), \end{aligned}

where Ind, Res denote the induction functor, restriction functor, respectively. Here V is an irreducible representation of $$A_6$$ and W is an irreducible representation of $$S_6$$. Suppose $$A_6$$ has an irreducible representation V with $$\dim V=4$$. Since $$[S_6:A_6]=2$$, we have that $$\dim \text {Ind}V=8$$. Since $$S_6$$ does not have any irreducible representation of dimension 8, the induced representation splits. Being $$\dim \text {Ind}V=8$$, $$\text {Ind}(V)$$ does not have any component of dimensions 9, 10 and 16. Now, let us assume that $$\dim W=5$$, then by Lemma 4.2, $$\text {Res}W$$ is an irreducible representation. Hence $$\text {Hom}_{\mathbb {F}_qA_6}(V,\text {Res}W)=0$$, which implies that $$\text {Ind}V$$ does not have any irreducible component of dimension 5. Similarly, $$\text {Ind}V$$ does not have any irreducible component of dimension 1. This completes the proof. $$\square$$

### Corollary 4.5

The algebra $$\mathbb {F}_qA_6$$ has one component to be $$\text {M}(9,\mathbb {F}_q)$$ for $$p\ge 7$$.

### Proof

The group $$A_6$$ being isomorphic to $$\text {PSL}(2,\mathbb {F}_9)$$ acts doubly transitively on a set with 10 points (see ), hence the conclusion. $$\square$$

### Corollary 4.6

We have $$(n_1,n_2,n_3,n_4,n_5,n_6)=(5,5,9,8,8,10)$$ or $$(n_1,n_2,n_3,n_4,n_5)=(5,5,9,10,8)$$ up to permutation.

### Proof

Since $$A_6$$ has one 1-dimensional, two 5-dimensional and one 9-dimensional irreducible representations, we can assume that $$n_1=5,n_2=5,n_3=9$$. Hence, we are left with the equation

\begin{aligned} n_4^2+n_5^2+n_6^2=228\text { or }n_4^2+2n_5^2=228. \end{aligned}

Then, $$(n_4,n_5,n_6)\in \{(4,4,14),(8,8,10)\},(n_4,n_5)\in \{(14,4),(10,8)\}$$. Hence, the result is obvious from Corollary 4.4. $$\square$$

### Proposition 4.7

Let $$\mathbb {F}_{p^k}$$ be a field of characteristic $$p\ge 7$$ and $$A_6$$ denote the alternating group on six letters. Then, the Artin–Wedderburn decomposition of $$\mathbb {F}_{p^k}A_6$$ is

\begin{aligned} \mathbb {F}_q\oplus \text {M}(5,\mathbb {F}_q)\oplus \text {M}(5,\mathbb {F}_q)\oplus \text {M}(9,\mathbb {F}_q)\oplus \text {M}(10,\mathbb {F}_q)\oplus \text {M}(8,\mathbb {F}_{q^2}), \end{aligned}

when $$p\equiv \pm 2\mod 5,k\equiv 1\mod 2$$ and

\begin{aligned} \mathbb {F}_q\oplus \text {M}(5,\mathbb {F}_q)\oplus \text {M}(5,\mathbb {F}_q)\oplus \text {M}(8,\mathbb {F}_q)\oplus \text {M}(8,\mathbb {F}_q)\oplus \text {M}(9,\mathbb {F}_q)\oplus \text {M}(10,\mathbb {F}_q), \end{aligned}

otherwise.

### Proof

Follows from Proposition 4.1 and Corollary 4.6. $$\square$$

### Theorem 4.8

Let $$\mathbb {F}_{p^k}$$ be a field of characteristic $$p\ge 7$$ and $$A_6$$ denote the alternating group on six letters. Then, the unit group of the algebra, $$\mathcal {U}(\mathbb {F}_{p^k}A_6)$$ is

\begin{aligned} \mathbb {F}_q^\times \oplus \text {GL}(5,\mathbb {F}_q)\oplus \text {GL}(5,\mathbb {F}_q)\oplus \text {GL}(9,\mathbb {F}_q)\oplus \text {GL}(10,\mathbb {F}_q)\oplus \text {GL}(8,\mathbb {F}_{q^2}), \end{aligned}
(4.4)

when $$p\equiv \pm 2\mod 5,k\equiv 1\mod 2$$ and

\begin{aligned} \mathbb {F}_q^\times \oplus \text {GL}(5,\mathbb {F}_q)\oplus \text {GL}(5,\mathbb {F}_q)\oplus \text {GL}(8,\mathbb {F}_q)\oplus \text {GL}(8,\mathbb {F}_q)\oplus \text {GL}(9,\mathbb {F}_q)\oplus \text {GL}(10,\mathbb {F}_q), \end{aligned}
(4.5)

otherwise.

### Proof

This follows immediately from Proposition 4.7 and the fact that given two rings $$R_1,R_2$$, we have $$(R_1\times R_2)^\times =R_1^\times \times R_2^\times$$. $$\square$$