I am interested in a particular group $G$, where
$$ (A\_4\times C\_\ell) \lhd G \lhd S\_4 \times D\_\ell$$
Here, $C\_\ell$ is cyclic, $D\_\ell$ is dihedral of order $2\ell$, and the two inclusions both have index $2$. Since
$$ (S\_4\times D\_\ell)/(A\_4\times C\_\ell) \cong C\_2\times C\_2,$$
my description of $G$ could be one of three groups. To be clear, I am interested in the "diagonal one", i.e. the one that is not a direct product of two proper subgroups.


Now I am interested in the irreducible modules of $G$ over a field $\mathbb{K}$. I want to know what happens when $\mathbb{K}$ is either $\mathbb{C}$, or else is a finite field of order coprime to $|G|$.


If I understand correctly, the irreducible $\mathbb{K}$-modules of $A\_4\times C\_\ell$ can be expressed as tensor products $N\_1\otimes N\_2$ where $N\_1$ is an irreducible for $A\_4$, and $N\_2$ is an irreducible for $C\_\ell$. Likewise for $S\_4\times D\_\ell$ (this is just a property of groups that are direct products).


My group $G$ lies in between these two group but is itself not **quite** a direct product, so I want to know what can be said. For instance, I can decompose an irreducible for $G$ into $1$ or $2$ irreducibles for $A\_4\times C\_\ell$, both of which will themselves be tensor products -- will the action of $G$ preserve this structure?


More generally, I would like recommendations for texts that discuss the representation theory of direct products of finite groups over finite fields.


Thanks!



Do the irreducible modules of this finite group preserve a tensor product structure?
This is all about Clifford's theorem in the complex case, which I treat in detail. The Klein $4$ subgroup that is acting on $A\_{4}\times C\_{\ell}$ contains three involutions: one inverts the $C\_{\ell}$ but centralizes the $A\_{4}$, one induces the outer automorphism of $A\_{4}$, but centralizes the $C\_{\ell}$, and the third inverts $C\_{\ell}$ and induces the outer automorphism of $A\_{4}$. You have indicated that it is the third case you are concerned with. You have not indicated whether $\ell$ is even or odd, and that makes a difference.


Now $A\_{4}$ has $4$ complex irreducible characters. The trivial character and the irreducible character of degree $3$ are stable under the outer outer automorphism of $A\_{4}$, so each extends in two ways to $S\_{4}$. The other two irreducible characters are complex conjugates, and are interchanged by the 
outer automorphism of $S\_{4}$. Hence each one of them induces irreducibly to an irreducible character of $S\_{4}$, and these two induced characters are equal.
 ( This explains the irreducible character degrees $1,1,3,3,2$ for $S\_{4}$ by the way).


If $\ell$ is odd, only the trivial character of $C\_{\ell}$ is stable under inversion, and this extends in two ways to a linear character of $D\_{\ell}$. 
For the remaining $\frac{\ell-1}{2}$ pairs of complex conjugate linear characters, we obtain one irreducible character of $D\_{\ell}$ for each pair by inducing either one to $D\_{\ell}$.


If $\ell$ is even, there are two real-valued linear characters of $C\_{\ell}$,which are both stable under inversion, and each extends in two ways to a linear character of $D\_{\ell}$. 
For the remaining $\frac{\ell-2}{2}$ pairs of complex conjugate linear characters, we obtain one irreducible character of $D\_{\ell}$ for each pair by inducing either one to $D\_{\ell}$.


Clifford's theorem allows you to extend this analysis to $G$: Each irreducible character of $A\_{4} \times C\_{\ell}$ has the form $\alpha \otimes \beta,$ where $\alpha$ is an irreducible character of $A\_{4}$ and $\beta$ is an irreducible character of $C\_{\ell}$. The dichotomy is that $\alpha \otimes \beta$ induces irreducibly to $G$, except when $\alpha \otimes \beta$ is $G$-stable, in which case it extends in two ways to $G$ which differ only multiplication by the non-trivial linear character of $G/(A\_{4} \times C\_{\ell}).$


By the earlier analysis, the character $\alpha \otimes \beta$ is $G$-stable 
exactly when $\alpha$ is either trivial or of degree $3$ AND $\beta$ is a real valued irreducible character of $C\_{\ell}$.


Hence there are either $2$ or $4$ stable irreducible characters when $\ell$ is respectively odd or even. Each of these (half of which have degree $1$ and the other half have degree $3$) extends in two ways to $G$. There are 
$\ell-1$ or $\ell-2$ non-$G$-stable irreducible characters of $A\_{4} \times C\_{\ell}$ when $\ell$ is respectively odd or even and these give rise to 
$\frac{\ell-1}{2}$ or $\frac{\ell-2}{2}$ induced irreducible characters of degree $6$ for $G$ and the same number of irreducible characters of degree
$2$.


Working over a finite field which is not algebraically closed field $\mathbb{K}$ is slightly more complicated because Schur's Lemma takes a more complicated form. Probably the easiest way in practice is to extend $\mathbb{K}$ to a finite splitting field $\mathbb{F}$ , describe all the (absolutely) irreducible modules
for $G$ over $\mathbb{F}$, and then note that there is one irreducible $\mathbb{K}G$-module for each orbit of absolutely irreducible modules under ${\rm Gal}(\mathbb{F}/\mathbb{K})$.


The character of direct products and Clifford's Theorem are covered in most representation theory texts: Curtis and Reiner (1962) or Isaacs ( ~1970) both
treat these, though Curtis and Reiner probably has more about the case of general fields.


