4
$\begingroup$

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

$\endgroup$

3 Answers 3

6
$\begingroup$

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 $\chi_H$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\mu$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.