Decomposition of an induced representation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:23:43Z http://mathoverflow.net/feeds/question/118742 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118742/decomposition-of-an-induced-representation Decomposition of an induced representation edward-poe 2013-01-12T16:51:28Z 2013-01-13T18:56:03Z <p>If there is a finite group $G$ with a cyclic normal subgroup $C_n$, one can describe the indecomposable representations of $G$ through induction. How does $Ind_{C_n}^G$ decompose? For representations over fields, I know that Clifford's theory whould help. But what happens if the representations should be over a ring? I am interested expecially in the p-adic ring $Z_p$. </p> <p>Best regards</p> http://mathoverflow.net/questions/118742/decomposition-of-an-induced-representation/118839#118839 Answer by Geoff Robinson for Decomposition of an induced representation Geoff Robinson 2013-01-13T18:56:03Z 2013-01-13T18:56:03Z <p>Not really an answer, but this is already difficult in the complex case, when $C$ is a central subgroup. For example, if $C = Z(G),$ and we induce a faithful irreducible $C$-module to $G,$ the number of distinct irreducible constituents is bounded above by the number of conjugacy classes of $G/C,$ but I don't know many other general statements- the number can be as low as $1$ (when we have a character of so-called ``central type"). To understand the general case of inducing an indecomposable $RC$-module to $G$ for a general commutative ring $R$ when $C$ ne not be central, Mackey decompoosition is certainly a helpful tool ( the induced module restricts freely to any subgroup meeting $C$ only in the identity, and more generally, the restriction to a subgroup $H$ of the induced module can be reasonably described by knowing how the original indecomposable restricts to $C \cap H,$ though precise details may require care. When working over $\mathbb{Z}_{p}$, Green's indecomposability theorem may also be useful.</p>