Exercise 4.4 (b) for Isaacs' Character Theory of Finite Groups

Question

The exercise is as follows:

Let $G = HK$ be a finite group, with $H \subset C_G(K)$. Let $\rho \in \operatorname{Irr}(H)$ and $\psi \in \operatorname{Irr}(K)$ be such that $\rho \mid_{H\cap K}$ and $\psi\mid_{H\cap K}$ share an irreducible constituent. Prove that there exists a unique $\chi \in \operatorname{Irr}(G)$ such that $\chi \mid_H = \psi(1)\rho$ and $\chi \mid_K = \rho(1) \psi$.

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My attempt

Let $\phi: H \times K \to G$ be such that $\phi(h, k) = hk$. It's clear that $\phi$ is a surjective homomorphism. In particular, if $N = \ker \phi = \{(h, h^{-1}) \mid h \in H \cap K\}$, we can identify $G$ with a quotient of the direct product, meaning we can use lifting to obtain the character $\chi$; that is to say, the desired character will be induced by an irreducible character of $H \times K$ with $N$ contained in its kernel.

Any irreducible character of $H \times K$ is of the form $\alpha \times \beta$, for $\alpha \in \operatorname{Irr}(H)$ and $\beta \in \operatorname{Irr}(K)$. And $N \subset \ker (\alpha \times \beta)$ iff $\alpha(h)\beta(h^{-1}) = \alpha(1)\beta(1)$ for all $h \in H \cap K$. This is, in turn, equivalent to saying $\alpha(h) = m \omega_h$ and $\beta(h) = n \omega_h$, for some $o(h)$-root of unity $\omega_h \in \mathbb{C}$, for each $h \in H \cap K$.

The existence would be established, then, if we could prove $N \subset \ker (\rho \times \psi)$. Since they have a common constituent, we know: $$ \langle \rho \mid_{H \cap K}, \psi \mid_{H \cap K} \rangle = \frac{1}{|H \cap K|} \sum_{h \in H \cap K} \rho(h)\psi(h^{-1}) \neq 0$$

If we knew $N \subset \ker(\rho \times \psi)$, then the above sum would reduce to $\rho(1)\psi(1)$. At the same time, since $H \cap K$ is abelian, $\rho(1)$ is the number of irreducible constituents of $\rho \mid_{H \cap K}$. So $\rho$ and $\psi$ have $\rho(1)\psi(1)$ constituents in common, while $\rho$ has $\rho(1)$ constituents. This means $\psi(1) = 1$. The same logic applied to $\psi$ yields $\rho(1) = 1$, meaning $\rho \mid_{H \cap K} = \psi \mid_{H \cap K}$.

If we knew $\rho \mid_{H \cap K} = \psi \mid_{H \cap K}$ and $\rho(1) = 1$, in turn, we'd get $\rho(h)\psi(h^{-1}) = \rho(hh^{-1}) = \rho(1) = 1 = \rho(1)^2$, for all $h \in H \cap K$, implying $N \subset \ker(\rho \times \psi)$. So the two statements are actually equivalent, and it suffices to show the latter.

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This is where I get completely stuck. I'm not quite sure if there's a mistake in the above reasoning, but it seems farfetched to obtain that the two characters are both equal in $H \cap K$ and linear just from the fact that they share a constituent...

Any help is greatly appreciated!

Thanks in advance!

Answer 1

The key observation that has yet been made is that $H \cap K \subseteq Z(G)$, and more importantly, $H \cap K \subseteq Z(H)$ and $H \cap K \subseteq Z(K)$. Any irreducible representation maps the center of the group to scalar matrices with the scalars being roots of unity (see Lemma 2.25 in Isaacs', for instance). This implies that $\rho |_{H \cap K} = \rho(1)\lambda_1$ and $\psi |_{H \cap K} = \psi(1)\lambda_2$ for some linear characters $\lambda_1, \lambda_2 \in \mbox{Irr}(H\cap K)$. Now, we know that $\rho |_{H \cap K}$ and $\psi |_{H \cap K}$ share a constituent, but the only way this is possible is if $\lambda_1 = \lambda_2$.

We can now show that $\rho(g)\psi(g^{-1}) = \rho(1)\psi(1)$ for $g \in H \cap K$ using orthogonality.

$$ \dfrac{1}{|H \cap K|} \sum_{g \in H \cap K} \rho(g)\psi(g^{-1}) = \dfrac{\rho(1)\psi(1)}{|H \cap K|} \sum_{g \in H \cap K} \lambda(g)\lambda(g^{-1}) = \rho(1)\psi(1) $$

We now have that $\sum_{g \in H \cap K} \rho(g)\psi(g^{-1}) = | H \cap K|\rho(1)\psi(1)$, but the triangle inequality implies that $\rho(g)\psi(g^{-1}) = \rho(1)\psi(1)$ for all $g \in H \cap K$.

This implies that $\ker(\rho \times \psi)$ contains the kernel of the map from $H \times K$ into $G$, and so you can associate $\rho \times \psi$ with an irreducible character of $G$ as you've desired.

I think I'll leave uniqueness to you.

Answer 2

To prove that $N \subset \ker(\rho \times \psi)$ note that $0\not=\langle \rho \mid_{H \cap K}, \psi \mid_{H \cap K} \rangle$ implies \begin{equation} 0\not=\sum_{h \in H \cap K} \rho(h)\psi(h^{-1})=\sum_{x\in N}(\rho\times\psi)(x)=|N|\langle(\rho\times\psi)\mid_N,1_N \rangle. \end{equation} Since $1_N$ is an irreducible constituent of $(\rho\times\psi)\mid_N$ and $N\vartriangleleft H\times K$, Clifford's theorem (6.2 of Isaacs CTFG) tells us that $1_N$ is the only irreducible constituent of $(\rho\times\psi)\mid_N$ (since the $H\times K$-conjugates of $1_N$ are just $1_N$). As a consequence, \begin{equation} \ker(\rho\times\psi)\cap N=\ker((\rho\times\psi)\mid_N)=\ker1_N=N. \end{equation} Hence $N\subset\ker(\rho\times\psi)$, as desired.
Now, write $\rho\mid_{H\cap K}=\sum_i n_i\chi_i$, $\psi\mid_{H\cap K}=\sum_jm_j\lambda_j$ for some irreducible characters $\chi_i$, $\lambda_j$ of $H\cap K$ and some nonnegative integers $n_i$,$m_j$. We have that $H\cap K$ is abelian, thus \begin{equation} \sum_{i,j}n_im_j\langle\chi_i,\lambda_j\rangle=\langle \rho \mid_{H \cap K}, \psi \mid_{H \cap K} \rangle=\rho(1)\psi(1)=\sum_{i,j}n_i\chi_i(1)m_j\lambda_j(1)=\sum_{i,j}n_im_j. \end{equation} Since $\langle\chi_i,\lambda_j\rangle=1$ for each $i,j$ it follows that $\rho\mid_{H\cap K}$ and $\psi\mid_{H\cap K}$ have only one irreducible consituent, and it is the same one, but this does not imply that both characters are equal since we don't know if $\rho(1)=\psi(1)$.