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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 $[\chi_H , \psi^{H}]=[\chi,\psi^{G}]=[\chi_{1+J^{2}},\psi] psi^{H}]=[\chi,\psi^{G}]=[\chi_{N},\psi] \neq 0$ Choose an irreducible constituent of $\psi^{H}$,called $\phi$ with $[\phi,\chi_H] \neq 0.$ On the one hand we have: $(\psi^{H})_{1+J^{2}} (\psi^{H})_{N} = |H:(1+J^{2})| H:(N)| \psi$ and on the other (since $[\phi,\psi^{H}]=[\phi|{1+J^{2}} [\phi,\psi^{H}]=[\phi|{N} , \psi]$):

$(\psi^{H})_{1+J^{2}} (\psi^{H})_{N} = a \phi +etc$ Comparing these 2 gives: $\phi|{1+J^{2}}=c \phi|{N}=c \psi$ for natural a and c and etc as a combination of other caracters. we have to show that c=1 to finish the proof. $c=[\phi|{1+J^{2}} c=[\phi|{N} , \psi]=[\phi,\psi^{H}]$,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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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 $[\chi_H , \psi^{H}]=[\chi,\psi^{G}]=[\chi_{1+J^{2}},\psi] \neq 0$ Choose an irreducible constituent of $\psi^{H}$,called $\phi$ with $[\phi,\chi_H] \neq 0.$ On the one hand we have: $(\psi^{H})_{1+J^{2}} = |H:(1+J^{2})| \psi$ and on the other (since $[\phi,\psi^{H}]=[\phi|{1+J^{2}} , \psi]$):

$(\psi^{H})_{1+J^{2}} = a \phi +etc$ Comparing these 2 gives: $\phi|{1+J^{2}}=c \psi$ for natural a and c and etc as a combination of other caracters. we have to show that c=1 to finish the proof. $c=[\phi|{1+J^{2}} , \psi]=[\phi,\psi^{H}]$,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.