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Background:

Let$G$ be a finite group. Fix a prime$p$. We say that$H\subseteq G$ isstrongly$p$-embedded if:

  1. $p$ divides$|H|$;
  2. For every$x\in G-H$,$p$ does not divide$|H\cap H^x|$.

Some facts about / properties of strongly$p$-embedded subgroups:

  1. If$x,y\in H$ are$p$-singular elements which are conjugate in$G$, then they are already conjugate in$H$.
  2. For every non-trivial$p$-subgroup$1<P\subseteq H$, we have$\mathbf{N}_G(P)\subseteq H$. (This may be taken as an alternate definition of strongly$p$-embedded subgroups.)
  3. A strongly$p$-embedded subgroup must contain a Sylow$p$-subgroup.
  4. If$G$ is a finite group with a cyclic Sylow$p$-subgroup$P$, let$C\subseteq P$ be the unique subgroup of order$p$. Then$\mathbf{N}_G(C)$ is strongly$p$-embedded.

Motivation:

Let$H\subseteq G$ be a Frobenius complement. Let$\text{cf}(H)$ denote the space of class functions on$H$, equipped with the usual inner product. Let:$$\text{cf}(H)^0=\{\theta\in\text{cf}(H)\mid \theta(1)=0\}$$For every$\theta\in\text{cf}(H)^0$, we have$(\theta)^G_H=\theta$.

The crux of the proof of Frobenius' theorem (that$H$ has a normal complement in$G$) is constructing a map$\alpha\mapsto\widehat{\alpha}$ from$\text{cf}(H)\rightarrow\text{cf}(G)$ such that:

  1. For$\alpha,\beta\in\text{cf}(H)$,$[\widehat{\alpha},\widehat{\beta}]=[\alpha,\beta]$. (We say that the map$\alpha\mapsto\widehat{\alpha}$ is anisometry.)
  2. For$\alpha\in\text{cf}(H)$,$(\widehat{\alpha})_H=\alpha$.
  3. For$\theta\in\text{cf}(H)^0$,$\widehat{\theta}=\theta^G$.
  4. If$\alpha$ is a character of$H$, then$\widehat{\alpha}$ is a character of$G$.

Now, let$H$ be a strongly$p$-embedded subgroup, and define:$$\text{cf}(H)^0=\{\theta\in\text{cf}(H)\mid \theta\text{ vanishes on $p$-regular elements}\}$$Then, again, for every$\theta\in\text{cf}(H)^0$, we have:$$(\theta^G)_H=\theta$$But now there is no guarantee that induction on$\text{cf}(H)^0$ can be extended to an isometry on the whole of$\text{cf}(H)$ that maps characters to characters.

The case of$G=A_5$:

Let$G=A_5$. Let$H\subseteq G$, with$H=\langle (1 2 3),(12)(45)\rangle\cong S_3$.

Observe that$H$ is the normalizer of$\langle (123)\rangle$, and$H$ is strongly$3$-embedded in$G$.

Below is the character table of$G=A_5$:

$$\begin{array}{|c|c|c|c|c|c|}\hline\text{class} & 1 & 2 & 3 & 5A & 5B \\ \hline\text{size} & 1 & 15 & 20 & 12 & 12 \\ \hline\chi_1 & 1 & 1 & 1 & 1 & 1 \\ \hline\chi_2 & 3 & -1 & 0 & \frac{1+\sqrt{5}}{2} & \frac{1-\sqrt{5}}{2} \\ \hline\chi_3 & 3 & -1 & 0 & \frac{1-\sqrt{5}}{2} & \frac{1+\sqrt{5}}{2} \\ \hline\chi_5 & 4 & 0 & 1 & -1 & -1 \\ \hline\chi_6 & 5 & 1 & -1 & 0 & 0 \\ \hline\end{array}$$

Below is the character table of$H\cong S_3$:

\begin{array}{|c|c|c|c|}\hline\text{class} & 1 & 2 & 3 \\ \hline\text{size} & 1 & 3 & 2 \\ \hline\rho_1 & 1 & 1 & 1 \\ \hline\rho_2 & 1 & -1 & 1 \\ \hline\rho_3 & 2 & 0 & -1 \\ \hline\end{array}

(I took the above character tables from the wonderful siteGroupNames.)

It turns out that the map$\mathfrak{g}:\text{cf}(H)\rightarrow \text{cf}(G)$ defined by:$$\mathfrak{g}(\rho_1)=\chi_1$$$$\mathfrak{g}(\rho_2)=\chi_5$$$$\mathfrak{g}(\rho_3)=\chi_6$$is in fact an isometry that extends induction on$\text{cf}(H)^0$.

Moreover, for every$\rho$ a character of$H$,$$\rho^G=\mathfrak{g}(\rho)+\Phi$$$$\mathfrak{g}(\rho)_H=\rho+\Gamma$$where$\Phi$ and$\Gamma$ are projective characters (characters of projective$RG$-modules which are free and finitely-generated as$R$-modules, where$R$ may be taken to be the ring of algebraic integers$\mathbb{A}\subseteq\mathbb{C}$ localized at a maximal ideal containing$p$.)

In particular,$\mathfrak{g}(\rho)$ agrees with$\rho^G$, and extends$\rho$, on$p$-singular elements.

This appears to be related to the Green correspondence, but there is no Green correspondence for$\mathbb{C}G$-modules. We could work instead with$R$-forms ($R$ defined above), but the same$\mathbb{C}G$-module can have different$R$-forms, and different$R$-forms will have different Green correspondents, which may have different characters (although they will all agree on$p$-singular elements.)

Questions:

  1. Is this a special case of a more general phenomenon, or just an anomaly?
  2. Can this be explained by the Green correspondence?
askedyesterday
semisimpleton's user avatar
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  • $\begingroup$You need to read about perfect isometries. When they exist, they preserve irreducible characters UP TO SIGN. These always exits for ${\rm SL}(2,2^{n})$ (between the whole group and the Sylow $2$-normalizer) by work of Rouquier.$\endgroup$Commentedyesterday
  • $\begingroup$@GeoffRobinson Is the isometry described in the original post (between the Sylow $3$-normalizer of $A_5$ and the whole group) an example of a perfect isometry?$\endgroup$Commentedyesterday
  • $\begingroup$I just did the calculations, and it is really a perfect isometry! Is it generally true that if there is a perfect isometry $I$ between a block $B$ of a subgroup $H$ and $B'$ of $G$, then for $\chi\in\text{Irr}(B)$, $\chi^G=\pm I(\chi)+\Phi$ where $\Phi$ is a projective character?$\endgroup$Commentedyesterday
  • $\begingroup$I think that perfect isometries are often more complicated than that.$\endgroup$Commentedyesterday

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