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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
show/hide this revision's text 2 added 82 characters in body

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 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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A question about the existence of a specific extension of a character.

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 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 $[\chi_H , \mu] \neq 0.$ 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