Problem 7.3 I. Martin Isaacs's Character Theory Of Finite Groups

Question

Let $H\subseteq G$ and $\xi\in Irr(H)$. Suppose $(\xi-\xi(1)1_H)^G=\theta$ and $[\theta,\theta]=1+\xi(1)^2$. Show that there exists $N\lhd G$ with $N\cap H=\ker\xi$ and every $x\in G-N$ conjugate to some element of $H$.

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My attempt:

I am trying to imitate the proof of the Frobenius theorem(Theorem 7.2).

Obviously $ \xi\ne 1_H $ since $ [\theta,\theta]>1 $.

Let $ \chi=\theta+\xi(1)1_G $, it is a generalized character of $ G $, we claim that $ \chi\in Irr(G) $ by showing that $ [\chi,\chi]_G=1 $:

\begin{align*} [\chi,\chi]_G&=[\theta,\theta]+2\xi(1)[\theta,1_G]+\xi(1)^2\\ &=1+\xi(1)^2+2\xi(1)[\xi^G-\xi(1)(1_H)^G, 1_G]+\xi(1)^2\\ &=1+2\xi(1)^2+2\xi(1)[\xi^G, 1_G]-2\xi(1)^2[(1_H)^G, 1_G]\\ &=1+2\xi(1)^2+0-2\xi(1)^2\\ &=1. \end{align*} Also, $ \chi(1)=\theta(1)+\xi(1)=\xi(1)>0 $, it is indeed an irreducible character of $ G $.

Suppose we have shown that $ \chi|_H=\xi $.

Let $N=\ker\chi$.

We have $ N\lhd G $ and we claim that $ N\cap H=\ker\xi $. Since if $ h\in N\cap H $, then $ \xi(h)=\chi(h)=\chi(1)=\xi(1) $, we know that $ h\in\ker\xi $. Conversely, if $ \xi(x)=\xi(1) $, then of course $ x\in H $ and $ x\in\ker\xi\subseteq N $. Hence $ N\cap H=\ker\xi $.

To show that every $ x\in G-N $ conjugate to some element of $ H $, it suffices to show that $ G\setminus N\subseteq\bigcup_{g\in G} H^g $ which is equivalent to show that $ G\setminus\bigcup_{g\in G} H^g\subseteq N $. For any $ y\in G\setminus\bigcup_{g\in G} H^g $, and $ \chi|_H=\xi\in Irr(H)\setminus\{1_H\} $, we have $ \chi(y)-\chi(1)=\theta(y)=(\xi-\xi(1)1_H)^G(y)=0 $ by the definition of induced characters.

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But the problem is, I do not know how to show that $ \chi|_H=\xi $. We do not have any TI set here and I have tried to compute $[\chi|_H,\xi]=[\xi^G,\xi^G]-\xi(1)[(1_H)^G,\xi^G]$. And it suffices to show that $[\chi|_H,1_H]=0$.

Answer 1

I figured it out.

Obviously $ \xi\ne 1_H $ since $ [\theta,\theta]>1 $.

Let $ \chi=\theta+\xi(1)1_G $, it is a generalized character of $ G $, we claim that $ \chi\in Irr(G) $ by showing that $ [\chi,\chi]_G=1 $:

\begin{align*} [\chi,\chi]_G&=[\theta,\theta]+2\xi(1)[\theta,1_G]+\xi(1)^2\\ &=1+\xi(1)^2+2\xi(1)[\xi^G-\xi(1)(1_H)^G, 1_G]+\xi(1)^2\\ &=1+2\xi(1)^2+2\xi(1)[\xi^G, 1_G]-2\xi(1)^2[(1_H)^G, 1_G]\\ &=1+2\xi(1)^2+0-2\xi(1)^2\\ &=1. \end{align*} Also, $ \chi(1)=\theta(1)+\xi(1)=\xi(1)>0 $, it is indeed an irreducible character of $ G $.

Note that \begin{align*} [\chi_H,\xi]_H&=[\chi,\xi^G]\\ &=[\xi^G-\xi(1)(1_H)^G+\xi(1)1_G,\xi^G]\\ &=[\xi^G,\xi^G]-\xi(1)[(1_H)^G,\xi^G]+\xi(1)[1_G,\xi^G]\\ &=[\xi^G,\xi^G]-\xi(1)[(1_H)^G,\xi^G] \end{align*} since $ [1_G,\xi^G]=[1_H,\xi]=0 $.

Note that $ [(1_H)^G,1_G]=[1_H,1_H]=1 $. Let $ (1_H)^G=1_G+\sum_{i=1}^l a_i\varphi_i+\sum_{i=1}^{m} b_i\psi_i $, $ \xi^G=\sum_{i=1}^{l} a_i'\varphi_i+\sum_{i=1}^{m'} c_i\lambda_i $ where $ b_i,c_i,l,m,m'\in\mathbb Z_{\ge 0} $, $ a_i,a_i',i\in\mathbb{Z}_{>0} $ and $ \varphi_i,\psi_i,\lambda_i\in Irr(G) $ are distinct irreducible characters of $ G $ for all $ i $. So $ \varphi_1,...,\varphi_l $ are the common irreducible constituents of $ (1_H)^G $ and $ \xi $. When $ l=0 $, there are no common irreducible constituents.

Therefore $$ [\chi_H,\xi]=\sum_{i=1}^{l}a_i'^2+\sum_{i=1}^{m'}c_i^2-\xi(1)\sum_{i=1}^l a_ia_i'. $$

On the other hand, \begin{align*} \chi&=\xi^G-\xi(1)(1_H)^G+\xi(1)1_G\\&=\sum_{i=1}^{l} a_i'\varphi_i+\sum_{i=1}^{m'} c_i\lambda_i-\xi(1)\left(\sum_{i=1}^l a_i\varphi_i+\sum_{i=1}^{m} b_i\psi_i\right) . \end{align*}

Since $ \chi\in Irr(G) $, only one of the irreducible character above will survive. But $ \xi(1)>0 $ forces all $ b_i $ are zero. Hence $$ \chi=\sum_{i=1}^{l} (a_i'-\xi(1)a_i)\varphi_i+\sum_{i=1}^{m'} c_i\lambda_i .$$

The first case is $ \chi=\lambda_i $ for some $ i\in [m'] $, we suppose W.L.G. that $ c_1=1 $ and $ c_i=0 $ for all $ i=2,...,m' $ and $ a_i'=\xi(1)a_i $ for all $ i\in [l] $. Then $$ [\chi_H,\xi]=\sum_{i=1}^l\xi(1)^2a_i^2+1-\xi(1)\sum_{i=1}^la_i(\xi(1)a_i)=1 $$ which implies $ \chi_H=\xi $ since $ \xi\in Irr(H) $ and $ \chi_H(1)=\chi(1)=\xi(1) $.

The other case is $ \chi=\varphi_i $ for some $ i\in l $. We suppose W.L.G. that $ \chi=\varphi_1 $. Thus, $ c_i=0 $ for all $ i\in [m'] $, $ a_1'=\xi(1)a_1+1 $ and $ a_i'=\xi(1)a_i $ for all $ i=2,...,l $. We compute that \begin{align*} [\chi_H,\xi]&=(\xi(1)a_1+1)^2+\sum_{i=2}^l (\xi(1)a_i)^2-\xi(1)a_1(\xi(1)a_1+1)-\xi(1)\sum_{i=2}^la_i(\xi(1)a_i)\\ &=\xi(1)^2a_1^2+2\xi(1)a_1+1-\xi(1)^2a_1^2-\xi(1)a_1\\ &=\xi(1)a_1+1. \end{align*}

Note that $ \xi(1)\ge 1 $ and $ a_1\ge 1 $ so $ [\chi_H,\xi]\ge 2 $. But this can never happen since $ \chi(1)=\xi(1) $ and $ \xi\in Irr(H) $, $ \chi\in Irr(G) $.

So we conclude that $ \chi_H=\xi $.

Let $ N=\ker\chi $.

We have $ N\lhd G $ and we claim that $ N\cap H=\ker\xi $. Since if $ h\in N\cap H $, then $ \xi(h)=\chi(h)=\chi(1)=\xi(1) $, we know that $ h\in\ker\xi $. Conversely, if $ \xi(x)=\xi(1) $, then of course $ x\in H $ and $ x\in\ker\xi\subseteq N $. Hence $ N\cap H=\ker\xi $.

To show that every $ x\in G-N $ conjugate to some element of $ H $, it suffices to show that $ G\setminus N\subseteq\bigcup_{g\in G} H^g $ which is equivalent to show that $ G\setminus\bigcup_{g\in G} H^g\subseteq N $. For any $ y\in G\setminus\bigcup_{g\in G} H^g $, and $ \chi|_H=\xi\in Irr(H)\setminus\{1_H\} $, we have $ \chi(y)-\chi(1)=\theta(y)=(\xi-\xi(1)1_H)^G(y)=0 $ by the definition of induced characters. We are done.

Answer 2

Here is a more streamlined solution with notation that sticks closer to the text (not the problem) in Isaacs' book. It also includes the solution to Problem (7.4).

Notation If $\varphi $ is a generalized character of some subgroup $H \leq G$, we write $\varphi_0=\varphi-\varphi(1)1_H$. Observe that $\varphi_0(1)=0$, and in the notation of M.I. Isaacs CTFG page 107, $\varphi \in \mathbb{Z}[Irr(H)]^\circ$. Since $\varphi$ is a generalized character, also $\varphi_0$ and $\varphi_0^G$ are generalized characters of $H$ and $G$ respectively. If $\varphi \neq 1_H$ then $[\varphi_0,\varphi_0]=1+\varphi(1)^2$.

Proposition Let $H \lt G$ and let $\varphi \in Irr(H)$, with $[\varphi_0^G,\varphi_0^G]=1+\varphi(1)^2$. Then the following hold.
$(a)$ There exists an $N \unlhd G$, with $H \cap N=ker(\varphi)$
$(b)$ $(G-\bigcup_{g \in G}H^g) \cup \{1\}\subseteq N$, in other words, every element outside $N$ is conjugate to some element of $H$.

Proof The start of the proof is similar to the proof of how a normal complement to a Frobenius complement can be constructed. Observe that $[\varphi_0^G,1_G]=[\varphi_0,1_H]=-\varphi(1)$. Hence $\varphi_0^G=\varphi^* - \varphi(1)1_G$, for some generalized character $\varphi^*$ of $G$ with $[\varphi^*,1_G]=0$. Observe that $\varphi_0^G(1)=|G:H|\varphi_0(1)=0=\varphi^*(1)-\varphi(1)$, so $\varphi^*(1)=\varphi(1) \gt 0$. Using this we get $$1+\varphi(1)^2=[\varphi_0^G,\varphi_0^G]=[\varphi^*-\varphi(1)1_G,\varphi^*-\varphi(1)1_G]=[\varphi^*,\varphi^*]+\varphi(1)^2$$ whence $[\varphi^*,\varphi^*]=1$ and since $\varphi(1) \gt 0$ we conclude that $\varphi^* \in Irr(G)$.

To summarize: $\varphi_0^G=\varphi^G-\varphi(1)1_H^G=\varphi^*-\varphi(1)1_G$, so $\varphi^G-\varphi^*=\varphi(1)(1_H^G-1_G)$. Now let us take a closer look at the character $1_H^G-1_G$. Since $[1_H^G,1_G]=[1_H,1_H]=1$, we can write $1_H^G=1_G+\Delta$, with $\Delta$ a (possibly reducible) non-zero character of $G$ with $[\Delta,1_G]=0$. In addition (see also lemmas (2.21) and (5.11) in CTFG), observe that $ker(1_H^G)=core_G(H)=ker(1_G) \cap ker(\Delta)=ker(\Delta)$. We can rewrite $$\varphi^G=\varphi^*+\varphi(1)\Delta$$ and from this it follows at this point that $core_G(ker(\varphi))=ker(\varphi^*) \cap ker(\Delta)=ker(\varphi^*) \cap core_G(H)$. But we are going to prove something stronger, namely that even $ker(\varphi)=ker(\varphi^*) \cap H$, and then $N=ker(\varphi^*)$ is the normal subgroup we are looking for.

Now $\Delta=\sum_{\chi \in Irr(G)}a_\chi\chi$ for some non-negative integers $a_{\chi}$, where $a_{1_G}=0$ but not all $a_{\chi}$'s can be zero. We are going to show that $\varphi^*$ extends $\varphi$, that is $\varphi^*_H=\varphi$. First of all $$[\varphi^*_H,\varphi]=[\varphi^*,\varphi^G]=[\varphi^*,\varphi^*+\varphi(1)\Delta]=1+\varphi(1)[\varphi^*,\Delta]=1+\varphi(1)a_{\varphi^*}$$ yielding $$\varphi^*_H=(1+\varphi(1)a_{\varphi^*})\varphi+\Gamma$$ where $\Gamma$ is a character of $H$ with $[\Gamma,\varphi]=0$ or $\Gamma=0$. It follows that $\varphi^*(1)=\varphi(1)=\varphi(1)+\varphi(1)^2a_{\varphi^*}+\Gamma(1)$, so $\varphi(1)^2a_{\varphi^*}+\Gamma(1)=0$. Since this last equation concerns non-negative integers and of course $\varphi(1) \gt 0$, this can only happen when $a_{\varphi^*}=0$ and $\Gamma=0$. Hence $\varphi^*_H=\varphi$, which implies $ker(\varphi^*_H)=H \cap ker(\varphi^*)=ker(\varphi)$. This proves (a).

To prove (b) pick an $x \in G-\bigcup_{g \in G}H^g$ (note that this set is non-empty since $H$ is proper). Then by the definition of induced characters we have $\varphi_0^G(x)=0=\varphi^*(x)-\varphi(1)$. Since $\varphi^*$ and $\varphi$ have the same degree it follows that $\varphi^*(x)=\varphi^*(1)$, that is $x \in ker(\varphi^*)$ and we are done.

Corollary (Isaacs Problem (7.4))
Let $H \lt G$ then the following are equivalent.
$(a)$ $H$ is a Frobenius complement.
$(b)$ Induction to $G$ is an isometry on $\mathbb{Z}[Irr(H)]^\circ$.

Proof (a) $\Rightarrow$ (b). This is clear from Lemma (7.7) in Isaacs' CTFG with $X=H$ (note that in this case $N_G(X)-X=\emptyset$, since $H=N_G(H)$).
(b) $\Rightarrow$ (a) Let $\varphi \in Irr(H)$, $\varphi \neq 1_H$. Then $\varphi_0 \in \mathbb{Z}[Irr(H)]^\circ$ and $[\varphi_0,\varphi_0]=1+\varphi(1)^2$. Since $[\varphi_0^G,\varphi_0^G]=[\varphi_0,\varphi_0]=1+\varphi(1)^2$, we can apply Problem (7.3). So each $\varphi$ has an extension to $G$ called $\varphi^*$, where it is understood that $1_H$ extends to $1_G$. Now put $M=\bigcap_{\varphi \in Irr(H)}ker(\varphi^*)$. Then $M \unlhd G$. Since $ker(\varphi)=H \cap ker(\varphi^*)$ for all $\varphi \in Irr(H)$ , we get $M \cap H=1$. We now use a counting argument to show that in fact $G=MH$ and then we can apply Problem (7.1) (d) $\Leftrightarrow$ (f) in combination with the Problem(7.3). Problem (7.3) yields the set $G - \bigcup_{g \in G}H^g \subseteq ker(\varphi^*)$ for all $\varphi \in Irr(H)$. It follows that $G-\bigcup_{g \in G}H^g \subseteq M$, that is $G-M \subsetneq \bigcup_{g \in G}H^g$ (note $1 \notin G-M$). Now $$\#( \bigcup_{g \in G}H^g) \leq |G:N_G(H)|(|H|-1)+1 \leq |G:H|(|H|-1)+1=|G|-|G:H|+1$$ We conclude that $\#(G-M)=|G|-|M| \lt |G|-|G:H|+1$, so $|G:H| \lt |M|+1$, whence $|G:H| \leq |M|$. Then $|G| \leq |M| \cdot |H|=|MH|$, since $M \cap H=1$. This implies $G=MH$ and we conclude that $H$ is indeed a Frobenius complement.