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general situation:
Let $ N \leq G $ be a subgroup,and let $ \chi \in Irr(G) $ be an irreducible character of G such that $\chi_N $ is not irreducible( i dont think that this is really needed) and let $ \psi \in Irr(N) $ be an irreducible constituent of $\chi_N$. Assume there is an subgroup H of G with $ N \leq H \leq G $ to which $ \psi $ is extendible. Then there is a character $ \mu \in Irr(H) $ with $ \mu_N =\psi$ and $[\chi_H , \mu] \neq 0.$ NOTE: the problem here is to find such an extension with $[\chi_H , \mu] \neq 0.$ Is this true in general?
I have the following specific situation: Let J be a nilpotent finitedimensional algebra over a finite field. Then G=1+J(called finite algebra group) is a p-group,$N=1+J^{2}$ is a normal subgroup. In the paper "On characters and commutators of finite algebra groups" written by Halasi,he writes:
Lemma 3.1: "Let G=1+J be a finite Algebra group and $\chi \in $ Irr(G).Then the following properties are equivalent:
1.There exists a proper algebra group H and $\varphi \in Irr(H)$ such that $\varphi^{G}=\chi$.
2.$\chi_{1+J^{2}}$ is not irreducible.
Proof:...
Assume now that $\chi_{1+J^{2}}$ is not irreducible and let $\psi \in Irr(1+J^{2})$ be a constituent of $\chi_{1+J^{2}}$.Let H be a maximal algebra subgroup such that $\psi $ is extendible to H.Then H $ \neq $G.We choose a $\varphi \in Irr(H) $such that $\varphi$ is an extension of $\psi$ and $\varphi$ is a constituent of $\chi_H$...."
I wonder why there is such a $\varphi$ .Is it true in the general situation or what properties of 1+J and $1+J^{2}$ or H(being maximal) are used here? Thanks for helping

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3 Answers 3

Given $N \subseteq H \subseteq G$ and an irreducible character $\psi$ of $N$ that has an extension to $H$, it is NOT in general true that every irreducible character $\chi$ of $G$ that lies over $\psi$ must lie over some extension of $\psi$ to $H$. Probably the easiest counterexample is to take $N = 1$ and $H = G$, where $G$ is any nonabelian group. Let $\psi$ be the principal character of $N$. Then of course, $\psi$ extends to $H$, but if $\chi$ is any nonlinear irreducible character of $G$ then $\chi$ does not lie over any extension of $\psi$ to $H$.

A case where it is true that $\chi$ must lie over an extension of $\psi$ to $H$ is where $N$ is normal in $H$ and $H/N$ is abelian. In that case, every character if $H$ lying over $\psi$ is an extension of $\psi$, so $\mu$ can be taken to be an arbitrary irreducible constituent of $\chi_H$.

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Hi, I just saw a theorem in a paper of Isaacs which could be useful. Assume additional to the general case(with same notation)that N is normal in H and H/N is abelian. Then the lemma:

"Let N be a normal subgroup of H and H/N abelian.Let $\vartheta \in Irr(H) $and $\psi \in Irr(N)$ and $[\psi,\vartheta_N] \neq 0$.Then every Z $\in Irr(H)$ with $[Z_N,\psi] \neq 0$ has the form $Z=\lambda \vartheta$ for a linear $\lambda \in Lin(H/N) $ "

tells us that every irreducible constituent of $ \psi^{H} $ has the form $\lambda \vartheta$,where $\lambda \in Lin(H/N)$ because if Z is an irreducible constituent of $ \psi^{H} $ then $[Z,\psi^{H}]=[Z_N,\psi]$.We can then write $ \psi^{H} $ as a sum of all different $\lambda \vartheta$ with multiplicity one,because $[\psi^{H} , \lambda \vartheta]=[\psi, (\lambda \vartheta)_N]=[\psi,\psi]=1$.

We have $[\chi_H , \psi^{H}]=[\chi,\psi^{G}]=[\chi_{N},\psi] \neq 0$ Choose an irreducible constituent of $ \psi^{H} $,called $\phi$ with $[\phi,\chi_H] \neq 0.$ On the one hand we have: $(\psi^{H})_{N} = |H:(N)| \psi $ and on the other (since $[\phi,\psi^{H}]=[\phi|{N} , \psi]$):

$(\psi^{H})_{N} = a \phi +etc$ Comparing these 2 gives: $\phi|{N}=c \psi$ for natural a and c and etc as a combination of other caracters. we have to show that c=1 to finish the proof. $c=[\phi|{N} , \psi]=[\phi,\psi^{H}]$,but we showed above that $\psi^{H}$ has only irreducible constituents with multiplicity one. I am thankful for proofreading this or giving hint to avoid the additional assumptations.

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Let $N$ be a normal subgroup of a finite group $G$, $\chi\in\mathrm{Irr}(G)$, and $\psi$ be an irred constituent of $\chi_{N}$. Let $H$ be a subgroup containing $N$ such that $H/N$ is abelian and $\psi$ is extendible to $H$. Then there exists a $\mu\in\mathrm{Irr}(H)$ such that $\mu_{N}=\psi$ and $[\chi_{H},\mu]\neq0$.

Edit: My previous statement was wrong as it left out the necessary hypothesis that $H/N$ be abelian. See Marty Isaacs's answer.

Below is my old argument. When $H/N$ is not abelian it does not go through because it is not necessarily true that $\mu_{N}=\psi$; in general $\mu_{N}$ will be a number of copies of $\psi$. When $H/N$ is abelian there is a much simpler argument, as given in Marty Isaacs's answer.

The natural context for this statement is Clifford theory for finite groups (see Section 6 in Isaacs's book). To see why this is true, first note that $H$ must be a subgroup of the stabiliser $S$ of $\psi$ in $G$. Moreover, there exists a $\rho\in\text{Irr}(S)$ such that $\rho_{N}$ contains $\psi$ and $\rho^{G}=\chi$. Since $\rho_{N}$ contains $\psi$, there exists an irred constituent $\mu$ of $\rho_{H}$ such that $\mu_{N}$ contains $\psi$. Since $\psi$ is extendible to $H$, a theorem of Gallagher's (Isaacs's book (6.17)) implies that $\mu_{N}=\psi$. Since $\mu^{S}$ contains $\rho$, $\mu^{G}$ contains $\chi$, and so $\chi_{H}$ contains $\mu$.

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