How to find out if a semilinear representation is irreducible (possibly with gap)

Question

Let be $\Gamma=\Gamma(2^6)$ the semilinear group on $GF(2^6)$, namely the group of semilinear mappings $\tau_{a,\sigma}\colon x\to ax^\sigma$ for $x \in GF(2^6)$ as a vector space over $GF(2)$, $a \in GF(2^6)^\times$ and $\sigma \in \mathcal{G}=Gal(GF(2^6)|GF(2))$. We denote $\Gamma_0$ the set of mapping $\tau_{a,1}$. Then $\Gamma_0$ is cyclic and $\Gamma/\Gamma_0\simeq\mathcal{G}$ is cyclic. If $V$ is the additive group of $GF(2^6)$, then $V$ is an irreducible faithful $\Gamma$-module. The group $\Gamma$ has order $(2^6-1)\cdot 6$. Let be $L \le \Gamma$ of order $18$ such that $L_0=L\cap \Gamma_0$ has order $9$. Then $L$ is a dihedral group of order $18$ and $V$ is an faithful $L$-module. I interested to know if $V$ is irreducible as $L$-module. Maybe it is easy to find it out theoretically, even though I am not able to do that right now. Nevetheless, since I want to improve my gap-system knowledge, I tried to do with it.

Firstly, the representations of $L$ over $GF(2)$ are three, of degree $1$, $2$ and $6$.

gap> IrreducibleRepresentations(DihedralGroup(18));
[ Pcgs([ f1, f2, f3 ]) -> [ [ [ 1 ] ], [ [ 1 ] ], [ [ 1 ] ] ], 
  Pcgs([ f1, f2, f3 ]) -> [ [ [ -1 ] ], [ [ 1 ] ], [ [ 1 ] ] ], 
  Pcgs([ f1, f2, f3 ]) -> 
    [ [ [ 0, 1 ], [ 1, 0 ] ], [ [ E(3), 0 ], [ 0, E(3)^2 ] ], 
      [ [ 1, 0 ], [ 0, 1 ] ] ], 
  Pcgs([ f1, f2, f3 ]) -> 
    [ [ [ 0, 1 ], [ 1, 0 ] ], [ [ -E(9)^4-E(9)^7, 0 ], [ 0, -E(9)^2-E(9)^5 ] ]
        , [ [ E(3), 0 ], [ 0, E(3)^2 ] ] ], 
  Pcgs([ f1, f2, f3 ]) -> 
    [ [ [ 0, 1 ], [ 1, 0 ] ], [ [ E(9)^4, 0 ], [ 0, E(9)^5 ] ], 
      [ [ E(3), 0 ], [ 0, E(3)^2 ] ] ], 
  Pcgs([ f1, f2, f3 ]) -> 
    [ [ [ 0, 1 ], [ 1, 0 ] ], [ [ E(9)^7, 0 ], [ 0, E(9)^2 ] ], 
      [ [ E(3), 0 ], [ 0, E(3)^2 ] ] ] ]

The one of degree $2$ is absolutely irreducible and the one of degree $6$, if considered over $GF(2^3)$ should split into three absolutely irreducible represetation. So, a priori, there is no way to exclude that my semilinear representation of $L$ is irreducible. On the other hand, if I can identify the natural representation of the semilinear group $\Gamma$, there is hope that I can find $L$ in terms of matrices and then use a command suggested by Hulpke in one other Stack Exchange question to test irreducibility.

I have a function (not done by myself) to calculate semilinear groups.

semilinear := function(p, n)
local vectors, permgl, frob, permfrob, gamma; 
vectors := AsList(GF(p^n)^1); 
permgl := Action(GL(1, p^n), vectors, OnRight);
frob := FrobeniusAutomorphism(GF(p^n));
permfrob := Permutation(frob, vectors, OnTuples); 
gamma := ClosureGroup(permgl, permfrob);
return(gamma); 
end;

Unfortunately, there are many (nine) representations of $\Gamma$ with dimention $6$ over $GF(2)$, as tells me the command

IrreducibleRepresentations(semilinear(2,6),GF(2));

Moreover, the group in the output of the function semilinear() does not act on $GF(2^6)$. Indeed, the group returned seems to be a permutation on a different set.

    OrbitsDomain(semilinear(2,6));
[ [ 2, 11, 12, 13, 14, 16, 15, 18, 17, 21, 19, 5, 25, 23, 22, 31, 20, 28, 6, 
      26, 38, 24, 34, 33, 32, 51, 29, 3, 45, 27, 8, 40, 39, 36, 35, 58, 7, 
      55, 53, 52, 30, 48, 4, 46, 43, 42, 41, 37, 62, 60, 59, 10, 56, 54, 49, 
      9, 47, 44, 64, 63, 61, 57, 50 ] ]

As semilinear groups appear quite often, this is a problem that I already encountered. How can I build in gap a semilinear group along with their natural modules?

You may guess I am not very much confident with GAP, especially in group actions. So, to whom that wants to aswer this question, please inlude a little of explaination.

Answer 1

I will use the notation $\mathbb{F}_q$ for the finite field with $q$ elements.

Even ignoring $\mathrm{Gal}(\mathbb{F}_{q^f}\mid\mathbb{F}_q)$, many subgroups of $\mathbb{F}_{q^f}^\times$ act irreducibly on $\mathbb{F}_{q^f}$ as a $\mathbb{F}_q$-vector space. Indeed, $\mathbb{F}_{q^f}$ is an irrep of $\langle\omega\rangle$ iff it is a simple $\mathbb{F}_q(\omega)$-module, which is true iff $\mathbb{F}_q(\omega)=\mathbb{F}_{q^f}$, which is in turn true iff $\omega$ is not in any proper subfield. In the case of $q^f=2^6$, we can verify a $9$th root of unity is not contained in the subfields of sizes $2^2$ or $2^3$ because its order does not divide $2^2-1$ or $2^3-1$.

I'd be curious which subgroups of the semilinear group act irreducibly in general.

Dunno how to help with GAP unfortunately, sorry.