# Do the irreducible modules of this finite group preserve a tensor product structure?

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!

• Are you assuming that $\mathbb{K}$ contains enough roots of unity (when finite)? Otherwise, for example when $\mathbb{K}$ does not contain $3$-rd roots of unity (say $\mathbb{K}=\mathbb{F}_5$), the irreducibe $\mathbb{K}$-modules of $A_4 \times C_3$ are not necessarily tensor products of the above form. Jun 15, 2016 at 10:04
• @FriederLadisch, Thank you for this remark, good point. Jun 20, 2016 at 12:27

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.

• Hi Geoff, Thanks for your answer which is very helpful. I need to read and think before I say too much but at the very least it seems that there is no obvious way to answer the question in the title above -- one needs to consider each representation in turn to test whether the tensor product structure is preserved. Do you think this is a fair comment? Jun 20, 2016 at 12:29
• Have been offline for a while. Yes, that is more or less true. Jun 22, 2016 at 21:12

Your group is a semidirect product of $S_4$ with the abelian group $C_\ell$. The action is $\pi x \pi^{-1}=\text{sign}(\pi) x$ (where $C_\ell$ is written additively). So over an algebraically closed field the Mackey machine gives you all irreducible representations: they are classified by pairs $(\chi,\zeta_\ell^i)$ where $0\le i\le \ell/2$ and $\chi$ is an irrep of $S_4$ if $i\in\{0,\ell/2\}$ ($5$ items) or an irrep of $A_4$ otherwise ($4$ items).

If the field is not algebraically closed then the irreps correspond to Galois orbits of irreps over the algebraic closure. Over finite fields all division rings are trivial. That means that an irrep over $\mathbb F_q$ lifted to $\overline{\mathbb F}_q$ is the direct sum of a Galois orbit of irreps (i.e., no multiplicities).

• Hi Friedrich,Thanks very much for your answer, I appreciate it. As I commented above on Geoff's answer, it seems that there is no obvious way to answer the question in the title above -- one needs to consider each representation in turn to test whether the tensor product structure is preserved... Jun 20, 2016 at 12:30