# Mackey Maschine for semidirect product

Hello, in this post

Representations of semidirect product over $C_p$

there was the question how to get irreducible representations of a semicirect product over finite fields. I am working on the same topic and confused about the answer to use Mackey's theorem. As far as I know this works just for algebraically closed fields, i.e. not for representations over $\mathbb{Z} / p \mathbb{Z}$. Are there better theorems? I tried to decompose the groupring $(\mathbb{Z} / p \mathbb{Z}) [G]$ into its simple components for $G$ dihedral with order $2p'$, $p'$ another prime, but it worked just in some special cases. Is there a general way to write $$(\mathbb{Z} / p \mathbb{Z})[G] \cong M_1 \oplus \cdots \oplus M_r$$ for the above group, $M_i$ simple $(\mathbb{Z} / p \mathbb{Z})$-modules? What would $s$ be?

Regards, Matt Brown

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Some part of Clifford theory (=Mackey Machine) works over any field, at least according to Wikipedia: en.wikipedia.org/wiki/Clifford_theory But you are right that my reference for the other question is wrong, since Weintraub explains the situation only over $\mathbb{C}$. Clifford seems to work out part of theory for any ground field, and later restrict its attention to algebraic closed fields: jstor.org/stable/1968599. – Marc Palm Aug 6 '12 at 9:19

It depends what that person meant by "Mackey Machine". Mackey decomposition, which tells you how to decompose a module induced from one subgroup when restricted to another subgroup works over pretty much any ring. Clifford's theorem, which tells how an irreducible representation of a group $G$ restricts to a normal subgroup $N$ (as a fixed multiple of a sum of $G$-conjugate irreducible representations of $N$) does not really need algebraic closure of the ground ring. If the $G$-conjugate irreducible representations of $N$ form an orbit of size greater than one, then one can reduce the situation to an intermediate subgroup, strictly containing $N$, but smaller than $G$. However, when we have an irreducible $G$-module whose restriction to $N \lhd G$ is a multiple of a single irreducible module $U$ for $N,$ then the analysis can become delicate when working over a field $F$ which is not algebraically closed. The reason is that the $N$-endomorphism ring of $U$ might not just consist of scalars, as would be the case if we were working over an algebraically closed field. We do have an action of $G$ on ${\rm End}_{F}(U),$ which is induced by inner automorphims of the matrix ring. However, this asscoiates an inner automorphism to $g \in G$ which is only determined up to multiiplication by an element of ${\rm End}_{FN}(U).$ Hence we do not necessarily get a homomorphism from $G/N$ to ${\rm PG}L(m,F)$ where $m = {\rm dim}_{F}(U)$ from the original representation.
Later edit: I don't know if the question has changed, but I just read it again. The statement about he group algebra doesn't mention $s$, only $r$, but then asks what $s$ would be. That may just be a typo. As long as $p^{\prime}$ is odd and different from $p,$ the group algebra $FG$ is still semisimple for $G$ dihedral with $2p^{\prime}$ elements and $F = \mathbb{Z}/p\mathbb{Z}.$ The number of isomorphism types of simple $FG$-module can be calculated in that case. Let $C$ be the cyclic subgroup of index $2$ of $G.$ A faithful simple $FG$-module $V$ can restrict in several ways to and $FC$-module. It can remain simple or it can restrict as two copies of a single simple. If $C = \langle c \rangle,$ and $c^{-1}$ can not be written in the form $c^{p^{t}}$ for some integer $t,$ then the split will be as two different simples; otherwise it will remain simple, or it might split as a direct sum of two copies of a single simple for $FN.$ But the latter cas can't occur, otherwise the endomorphism ring ${\rm End}_{FG}(V)$ is a finite division ring, using the more precise version of Schur's Lemma, henceis a field. On the other hand ${\rm End}_{FN}(V)$ is isomorphic to a $2 \times 2$-matrix ring over a field, and is acted on by $G/C$, a group of order $2,$ as an inner automorphism. The inner automorphism can't be scalar, as its fixed points are a field, wheeas they are the whole matrix ring, a contradiction. Nor can it be non-scalar of order $2$, otherwise its fixed-points contain non-trivial idempotents, again contrary to its fixed points being a field. Once the faithful simple modules are dealt with, the non-faithful one follow using (mathematical) induction. For a more general semidirect product, say $G = NT$, then it is necessary to understand how $T$ acts on various matrix ring over fields, ( if working with simple $FG$-modules for finite fields $F$), the fields in question being fields of the form ${\rm End}_{FN}(U)$ for simple $FN$-modules $U$. That's assuming characteristic of $F$ coprime to $|G|$. If the characteristic of $F$ is not coprime to $|G|$, the simple $FG$ modules can be described in similar terms, but the group algebra is of course no longer semisimple.