A question about the existence of a specific extension of a character. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:32:51Z http://mathoverflow.net/feeds/question/69653 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69653/a-question-about-the-existence-of-a-specific-extension-of-a-character A question about the existence of a specific extension of a character. trew 2011-07-06T18:21:41Z 2012-01-31T12:45:18Z <p>general situation:<br> 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?<br> 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:<br> Lemma 3.1: "Let G=1+J be a finite Algebra group and $\chi \in$ Irr(G).Then the following properties are equivalent:<br> 1.There exists a proper algebra group H and $\varphi \in Irr(H)$ such that $\varphi^{G}=\chi$.<br> 2.$\chi_{1+J^{2}}$ is not irreducible.<br> Proof:...<br> 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 $\chi_H$...."<br> 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</p> http://mathoverflow.net/questions/69653/a-question-about-the-existence-of-a-specific-extension-of-a-character/69681#69681 Answer by trew for A question about the existence of a specific extension of a character. trew 2011-07-06T23:24:24Z 2011-07-10T15:22:18Z <p>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: </p> <p>"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 $\lambda \in Lin(H/N)$ " </p> <p>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 $[\psi^{H} , \lambda \vartheta]=[\psi, (\lambda \vartheta)_N]=[\psi,\psi]=1$. </p> <p>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 $[\phi,\psi^{H}]=[\phi|{N} , \psi]$): </p> <p>$(\psi^{H})_{N} = a \phi +etc$ 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.</p> http://mathoverflow.net/questions/69653/a-question-about-the-existence-of-a-specific-extension-of-a-character/69720#69720 Answer by A Stasinski for A question about the existence of a specific extension of a character. A Stasinski 2011-07-07T13:55:20Z 2012-01-31T12:45:18Z <p>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 $[\chi_{H},\mu]\neq0$.</p> <p><strong>Edit:</strong> My previous statement was wrong as it left out the necessary hypothesis that $H/N$ be abelian. See Marty Isaacs's answer.</p> <p>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.</p> <p>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 $\mu$.</p> http://mathoverflow.net/questions/69653/a-question-about-the-existence-of-a-specific-extension-of-a-character/87078#87078 Answer by Marty Isaacs for A question about the existence of a specific extension of a character. Marty Isaacs 2012-01-30T22:04:28Z 2012-01-30T22:04:28Z <p>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 $H$.</p> <p>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 $\chi_H$.</p>