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explained dihedral case in more detail.
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Geoff Robinson
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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.

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.

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.

Source Link
Geoff Robinson
  • 44.4k
  • 5
  • 123
  • 169

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.