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Let $\rho$ be an irreducible representation of a group $N$, and let $G,H$ be groups with $N$ of finite index in $H$ and $H$ normal in $G$. Let $\pi=\rho^H$ be the induced representation of $\rho$ to $H$; I'd like to understand the isotropy of $\pi$ in $G$, that is, $I_G(\pi)=\lbrace g\in G:\pi^g\sim \pi\rbrace$, those $g$ such that $\pi$ and $h\mapsto\pi(g^{-1}hg)$ are equivalent.

Guess: if $I_H(\rho)=N$, that is, if $\pi$ is irreducible, then $$I_G(\pi)=H\cdot I_G(\rho).$$

It's easy to see that "$\supseteq$" holds.

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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 $G$? –  Alain Valette Nov 9 '11 at 21:42
    
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. –  José Figueroa-O'Farrill Nov 10 '11 at 1:36
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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. –  Derek Holt Nov 10 '11 at 9:33
    
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 $Norm_G(N)$. –  grok Nov 10 '11 at 17:36
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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 : $$ J_G (\pi ) =H. J_G (\rho ). H $$ –  Paul Broussous Nov 12 '11 at 18:14
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1 Answer

up vote 7 down vote accepted

The "guess" is wrong; here is a counterexample. Take $G$ to be dihedral of order 16. Let $H$ be one of the two copies of the dihedral group of order 8 in $G$, and let $N$ be one of the two copies of the Klein fours group in $H$. Let $\pi$ be the unique irreducible character of degree $2$ of $H$, and let $\rho$ be one of the two linear constituents of the restriction $\pi_N$.

Now $\rho^H = \pi$ is irreducible, and since $\pi$ is unique, it is invariant in $G$. We argue, however, that $G \ne HT$, where $T = I_G(\rho)$. By definition, $T$ is contained in the normalizer in $G$ of $N$, and of course, $N$ is also normal in $H$. Thus if $G = HT$, it would follow that $N$ is normal in $G$. This is not the case, however. One way to see that $N \not\kern -2pt\triangleleft\ G$ is to observe that otherwise $G/C$ would be embedded in ${\rm Aut}(G)$, where $C$ is the centralizer of $N$. This would force $|C| \ge 8$, and that would imply that $N$ is contained in an abelian subgroup of $G$ of order $8$. The only abelian subgroup of oirder $8$ in $G$, however, is cyclic, so does not contain $N$.

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