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Feb 3, 2012 at 19:38 vote accept grok
Jan 29, 2012 at 20:21 answer added Marty Isaacs timeline score: 8
Nov 12, 2011 at 18:14 comment added Paul Broussous It seems that Mackey's criterion for irreducibility of $\pi$ is not $I_H (\rho ) =H$, but $J_H (\rho )=H$, where $J$ denotes the intertwining ! You should have the following equality : $$ J_G (\pi ) =H. J_G (\rho ). H $$
Nov 10, 2011 at 17:36 comment added grok Sorry if I caused confusion. I don't want explicitly to assume that $N$ is normal in $G$; rather, by $I_G(\rho)$ I mean those elements of the normalizer of $N$ in $G$ that leave $\rho$ invariant; i.e. $I_G(\rho)$ is by necessity contained in $Norm_G(N)$.
Nov 10, 2011 at 9:33 comment added Derek Holt I think (assuming $N$ normal in $G$) your equation would be true for finite $N$, since if $I_H(\rho) = N$ then by Frobenius Reciprocity the irreducible representations of $N$ that induce up to $\rho^H$ are precisely the $|H:N|$ conjugates of $\rho$ under $H$. But you don't appear to be assuming that $N$ is finite, and I have no idea how much of standard representation theory of finite groups applies to infinite groups.
Nov 10, 2011 at 1:36 comment added José Figueroa-O'Farrill I took the liberty to edit the LaTeX to get the curly brackets to render. In MO it's probably best to use \lbrace and \rbrace instead of \{ and \}, respectively.
Nov 10, 2011 at 1:35 history edited José Figueroa-O'Farrill CC BY-SA 3.0
changed \{ \} to \lbrace \rbrace
Nov 9, 2011 at 21:42 comment added Alain Valette How do you define $I_G(\rho)$, as $N$ is not assumed to be normal in $G$? Did you mean that both $N$ and $H$ are normal in $G$?
Nov 9, 2011 at 21:32 history edited grok CC BY-SA 3.0
Made more compactthe condition; tried to fix the \{\}
Nov 9, 2011 at 20:49 history asked grok CC BY-SA 3.0