We start with the description of the conjugacy classes in \(A_6\). Using [18], 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 [4]), 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 \((R_1\times R_2)^\times =R_1^\times \times R_2^\times\). \(\square\)