Proof. See Lemma 1.3 in unique?, factorisation?. ◻

Proof. See Theorem 3 in [2]. ◻

Proof. Clearly \(\chi|_S \in R(S)\), all that remains is to show that \(\chi|_S\) is \(\mathop{\mathrm{\mathcal{F}}}\)-stable. Because \(\mathop{\mathrm{\mathcal{F}}}\) is realised by \(G\), \(\text{Hom}_{\mathop{\mathrm{\mathcal{F}}}}(P,Q) = \text{Hom}_{G}(P,Q)\) so \(s' \in s^{\mathop{\mathrm{\mathcal{F}}}} \iff s' \in s^G\) for all \(s, s' \in S\). Since \(\chi \in \text{cl}(G)\), we have that for any two \(s, s' \in S\) with \(s' \in s^G\) then \(\chi|_S(s) = \chi|_S(s')\). ◻

Proof. Lemma 2.1 in completion?, Theorem?. ◻

Proof. See Corollary 2.2 of completion?, Theorem?. ◻

Proof. See Corollary 2.9 in unique?, factorisation?. ◻

Proof. By definition, if there is an \(\mathop{\mathrm{\mathcal{F}}}_1\times\mathop{\mathrm{\mathcal{F}}}_2\)-isomorphism sending \((p_1, p_2) \mapsto (q_1, q_2)\) it is of the form \((\phi_1, \phi_2) \in \text{Iso}_{\mathop{\mathrm{\mathcal{F}}}_1}(P_1, Q_1)\times\text{Iso}_{\mathop{\mathrm{\mathcal{F}}}_2}(P_2, Q_2)\) where \(P_i \leq S_i\) containing \(p_i\) and likewise for \(Q_i\) and \(q_i\). So \(p_i \in {q_i}^{\mathop{\mathrm{\mathcal{F}}}_1}\) with \(\phi_i\) being a map in \(\mathop{\mathrm{\mathcal{F}}}_1\) such that \(\phi_i(p_i) = q_i\).

If we now assume \(p_i \in {q_i}^{\mathop{\mathrm{\mathcal{F}}}_i}\) with \(\phi_i\) being the isomorphism mapping \(p_i\) to \(q_i\). Define \(P_i, Q_i\) as before, then \((\phi_1, \phi_2) \in \text{Iso}_{\mathop{\mathrm{\mathcal{F}}}_1 \times \mathop{\mathrm{\mathcal{F}}}_2}((P_1, P_2), (Q_1, Q_2))\), so \((p_1, p_2) \in (q_1, q_2)^{\mathop{\mathrm{\mathcal{F}}}_1 \times \mathop{\mathrm{\mathcal{F}}}_2}\). ◻

Proof. It is clear that \(|C_{S_1 \times S_2}((x_1, x_2))| = |C_{S_1}(x_1) \times C_{S_2}(x_2)|\). So if \(x_i\) is fully \(\mathop{\mathrm{\mathcal{F}}}_i\)-centralised, then \(|C_{S_1 \times S_2}((x_1, x_2))| =|C_{S_1}(x_1)||C_{S_2}(x_2)| \geq |C_{S_1}(x_1')||C_{S_2}(x_2')| = |C_{S_1 \times S_2}((x_1', x_2'))|\) for all \(x_i' \in x_i^{\mathop{\mathrm{\mathcal{F}}}}\). Because \((x_1, x_2)^{\mathop{\mathrm{\mathcal{F}}}} = x_1^{\mathop{\mathrm{\mathcal{F}}}_1} \times x_2^{\mathop{\mathrm{\mathcal{F}}}_2}\) by Lemma 16, we have our result. ◻

Proof. We first assume \(\chi_i\) are \(\mathop{\mathrm{\mathcal{F}}}_i\)-stable. If \((p_1, p_2) \in (q_1, q_2)^{\mathop{\mathrm{\mathcal{F}}}_1\times\mathop{\mathrm{\mathcal{F}}}_2}\) then \(p_1 \in q_1^{\mathop{\mathrm{\mathcal{F}}}_1}\) and \(p_2 \in q_2^{\mathop{\mathrm{\mathcal{F}}}_2}\). Then since \(\chi_1, \chi_2\) are \(\mathop{\mathrm{\mathcal{F}}}_1\)-stable and \(\mathop{\mathrm{\mathcal{F}}}_2\)-stable respectively, we have \(\chi(p_1, p_2) = \chi_1(p_1)\chi_2(p_2) = \chi_1(q_1)\chi_2(q_2) = \chi(q_1, q_2)\). Thus \(\chi\) is \(\mathop{\mathrm{\mathcal{F}}}_1\times\mathop{\mathrm{\mathcal{F}}}_2\)-stable.

Now, take an \(\mathop{\mathrm{\mathcal{F}}}_1 \times \mathop{\mathrm{\mathcal{F}}}_2\)-stable character \(\chi\). Since \(\chi \in R(S_1 \times S_2)\) we can decompose it as \(\chi = \chi_1\chi_2\) with \(\chi_i \in R(S_i)\).

Let \((p_1, p_2) \in (q_1, q_1)^{\mathop{\mathrm{\mathcal{F}}}_1 \times \mathop{\mathrm{\mathcal{F}}}_2}\), so \(\chi(p_1, p_2) = \chi(q_1, q_2)\) and \(\chi_1(p_1)\chi_2(p_2) = \chi_1(q_1)\chi_2(q_2)\). So if we set \(p_2 = 1\) we have \(\chi_1(p_1)\chi_2(1) = \chi_1(q_1)\chi_2(1) \Rightarrow \chi_1(p_1) = \chi_1(q_1)\). Since \((p_1, 1) \in (q_1,1)^{\mathop{\mathrm{\mathcal{F}}}_1 \times \mathop{\mathrm{\mathcal{F}}}_2} \iff p_1 \in q_1^{\mathop{\mathrm{\mathcal{F}}}_1}\) we can conclude that \(\chi_1\) is \(\mathop{\mathrm{\mathcal{F}}}_1\)-stable. An identical argument holds for \(\chi_2\) by setting \(p_1 = 1\) and the result is shown. ◻

Proof. By Theorem 18 we have for any \(\chi \in R(\mathop{\mathrm{\mathcal{F}}})\) that \(\chi = \chi_1\chi_2\) for some \(\chi_i \in R(\mathop{\mathrm{\mathcal{F}}}_i)\), hence \[\chi = \chi_1\chi_2 = \left(\sum_{\psi\in B_1} \alpha_{\psi}\psi\right)\left(\sum_{\mu \in B_2} \alpha_{\mu}\mu\right) = \sum_{\psi \in B_1}\sum_{\mu \in B_2} \alpha'_{\psi\mu}\psi\mu\] So \(B_1B_2\) is a generating set for \(R(\mathop{\mathrm{\mathcal{F}}})\). We have the equality of matrices: \[\begin{align} (X_{B_1}(\mathop{\mathrm{\mathcal{F}}}_1) \otimes X_{B_2}(\mathop{\mathrm{\mathcal{F}}}_2))_{(\psi, \mu) \in B_1B_2, (s_1, s_2) \in S_1 \times S_2} &= \psi(s_1)\mu(s_2) \\ &= \psi\mu(s_1, s_2) \\ &= (X_{B_1B_2}(\mathop{\mathrm{\mathcal{F}}}))_{(\psi,\mu) \in B_1 \times B_2, (s_1, s_2) \in S_1 \times S_2} \end{align}\] And then since the Kronecker product of two full rank matrices is of full rank, we conclude that \(B_1B_2\) is a basis for \(R(\mathop{\mathrm{\mathcal{F}}})\). ◻

Proof. We begin by computing \(\text{det}(X_{B_1\otimes B_2}(\mathop{\mathrm{\mathcal{F}}})) = \text{det}(X_{B_1}(\mathop{\mathrm{\mathcal{F}}}_1)) \otimes \text{det}(X_{B_2}(\mathop{\mathrm{\mathcal{F}}}_2))\). The eigenvalues of \(A \otimes B\) is the set of products \(a_ib_j\) where \(a_i\) and \(b_j\) are eigenvalues of \(A, B\) with multiplicity. Hence: \[\text{det}(A \otimes B) = \prod_{i=1}^{\text{dim}(A)}\prod_{j=1}^{\text{dim}(B)}(a_ib_j) = \left(\prod_{i=1}^{\text{dim}(A)} a_i\right)\left(\prod_{i=1}^{\text{dim}(B)} b_i\right) = \text{det}(A)^{\text{dim}(B)}\text{det}(B)^{\text{dim}(A)}\] Thus \[\begin{align} \text{det}(X_{B_1 \otimes B_2}(\mathop{\mathrm{\mathcal{F}}})) = \text{det}(X_{B_1}(\mathop{\mathrm{\mathcal{F}}}_1))^{|\text{cl}(\mathop{\mathrm{\mathcal{F}}}_2)|}\text{det}(X_{B_2}(\mathop{\mathrm{\mathcal{F}}}_2))^{|\text{cl}(\mathop{\mathrm{\mathcal{F}}}_1)|} \end{align}\] Where \(\text{dim}(X_{B_i}(\mathop{\mathrm{\mathcal{F}}})) = |\text{cl}(\mathop{\mathrm{\mathcal{F}}}_i)|\) from Theorem 12. Since we’re assuming Conjecture A holds for \(\mathop{\mathrm{\mathcal{F}}}_i\), we have: \[\begin{align} \text{det}(X_{B_1 \otimes B_2}(\mathop{\mathrm{\mathcal{F}}})) &= \left(\prod_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}_1)} |C_{S_1}(x_K)|\right)^{|\text{cl}(\mathop{\mathrm{\mathcal{F}}}_1)|}\left(\prod_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}_2)}(|C_{S_2}(x_K)|\right)^{|\text{cl}(\mathop{\mathrm{\mathcal{F}}}_2)|} \\ &= \prod_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}_1)}\prod_{J \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}_2)}|C_{S_1}(x_K)||C_{S_2}(x_J)| \\ &= \prod_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}_1)}\prod_{J \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}_2)}|C_{S_1 \times S_2}((x_K, x_J))| \end{align}\] Since \(x_K, x_J\) are fully \(\mathop{\mathrm{\mathcal{F}}}_1, \mathop{\mathrm{\mathcal{F}}}_2\)-centralised respectively, \(x_L := (x_K, x_J)\) is fully \(\mathop{\mathrm{\mathcal{F}}}_1 \times \mathop{\mathrm{\mathcal{F}}}_2\)-centralised by Lemma 17, combined with \[\prod_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}_1)}\prod_{J \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}_2)}|C_{S_1 \times S_2}((x_K, x_J))| = \prod_{L \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})}|C_{S_1 \times S_2}(x_L)|\] the result is shown. ◻

Proof. Note that \(\text{cl}_{q'}(G) = \text{cl}_{p}(G)\) and \(\text{cl}(\mathop{\mathrm{\mathcal{F}}}) = \text{cl}_{p}(G) \cap S\). Now take \(\psi \in \text{IBr}_q(G)\). Because \(S\) is a \(q'\)-group, we have a bijection \(\psi \mapsto \psi'\) between \(\text{IBr}_q(S)\) and \(\text{Irr}(S)\) given by Brauer lifting (see Theorem 2.12 in [5], Theorem 43.ii in [6]). It is clear that \(\psi|_S' = \psi'|_S\).

Because restrictions of Brauer characters are Brauer characters and \(S\) is a \(q'\)-group we have \(\psi'|_S \in R(S)\) for \(\psi \in \text{IBr}_q(G)\), furthermore \(\psi'\) is invariant on \(\text{cl}_{q'}(G) = \text{cl}_p(G)\) so we have \(\psi'|_S \in R(\mathop{\mathrm{\mathcal{F}}})\).

As all Sylow \(p\)-subgroups are conjugate, \(x \in \bigcup_{K \in \text{cl}_{p}(G)} K\) is conjugate to some \(\tilde{x} \in S\). We define an additive map \(\phi\) extending \(\mathop{\mathrm{\mathcal{F}}}\)-stable characters to class functions on \(\text{cl}_{q'}(G)\) by \(\phi(\chi)(x) := \chi(\tilde{x})\), Since \(\chi\) is \(\mathop{\mathrm{\mathcal{F}}}\)-stable this map is well defined and does not depend on the choice of \(\tilde{x}\) and it is clear that \(\phi\) is a ring homomorphism.

We show that the \(\chi \in \text{Ind}(\mathop{\mathrm{\mathcal{F}}})\) is the restriction to \(S\) of the lift of an irreducible \(q\)-Brauer character of \(G\). Firstly we note that \(\chi\) is the restriction of a non-virtual \(q\)-Brauer character \(\psi\) of \(G\) because \(\psi|_S\) is still non-virtual, and then because Brauer lifting is a ring homomorphism that maps \(\text{IBr}_q(S) \rightarrow \text{Irr}(S)\) bijectively, lifts of non-virtual \(q\)-Brauer characters of \(S\) are non-virtual characters.

If \(\chi \in \text{Ind}(\mathop{\mathrm{\mathcal{F}}})\) with \(\chi = \psi'_1|_S+\psi'_2|_S\) for two \(\psi_1, \psi_2 \in \text{IBr}_q(G)\) then this immediately contradicts the \(\mathop{\mathrm{\mathcal{F}}}\)-indecomposablity of \(\chi\), hence \(\chi = \psi'|_S\) for a single \(\psi \in \text{IBr}_q(G)\). ◻

Proof. Let \(g \in G\), \(s \in S\), then: \[\begin{align} \sum_{\psi \in B} \overline{\Phi_{\psi}(g)}\psi(s) &= \sum_{\psi \in B}\sum_{\chi \in \text{Irr}(G)} d_{\chi\psi}\overline{\chi(g)}\psi(s) = \sum_{\chi \in \text{Irr}(G)} \overline{\chi(g)}\sum_{\psi \in B}d_{\chi\psi}\overline{\psi(s)} = \sum_{\chi \in \text{Irr}(G)} \overline{\chi(g)}\chi(s) \\ &= \delta_{g^{G}s^{G}}|C_G(x)| \end{align}\] By column orthogonality. Writing out \(\overline{P^T}Q\) gives us: \[\begin{align} (\overline{P^T}X(\mathop{\mathrm{\mathcal{F}}}))_{K, K' \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})} &= \sum_{\psi \in B}\overline{\Phi_{\psi}(x_K)}\psi(x_{K'}) = \delta_{K,K'}|C_G(x_K)| \\ &\Rightarrow \overline{P^T}X(\mathop{\mathrm{\mathcal{F}}}) = \text{Diag}_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})}(|C_G(x_K)|) = \Delta \end{align}\] Since \(|C_G(x_K)| \neq 0\) for any \(K\), \(\Delta\) is of full rank. We know that \(P\) is square by Theorem 12, therefore \[|\text{cl}(\mathop{\mathrm{\mathcal{F}}})| = \text{rk}(\Delta) \leq \text{min}(\text{rk}(P), \text{rk}(X(\mathop{\mathrm{\mathcal{F}}}))) \leq |\text{cl}(\mathop{\mathrm{\mathcal{F}}})| \Rightarrow \text{rk}(P) \text{ and } \text{rk}(X(\mathop{\mathrm{\mathcal{F}}})) = |\text{cl}(\mathop{\mathrm{\mathcal{F}}})|\] and both results are proven. ◻

Proof. Let \(g \in G\) such that \(g\) is not a \(p\)-element. Then we have that \(g^G \cap S = \varnothing\), hence \(\sum_{\psi \in B} \overline{\Phi_{\psi}(g)}\psi(s) = 0\) for all \(s \in S\) by Lemma 25. By the linear independence of \(B\), we conclude that \(\Phi_{\psi}(g) = 0\) for all \(\psi\). ◻

Proof. Let \(q \neq p\) be a prime dividing \(|G|\) and \(Q \in \text{Syl}_q(G)\), then \(\langle \Phi_{\psi}|_Q, 1_Q \rangle_Q = \frac{1}{|Q|}\sum_{q \in Q} \Phi_{\psi}|_Q(q)\) which is just \(\frac{\Phi_{\psi}(1)}{|Q|}\) because \(Q \cap S = 1\) and \(\Phi_{\psi}\) is zero everywhere else in \(Q\). Since \(\Phi_{\psi}\) is a virtual character of \(G\), \(\langle \Phi_{\psi}|_Q, 1_Q\rangle \in \mathop{\mathrm{\mathbb{Z}}}\), and so \(|Q|\) divides \(\Phi_{\psi}(1)\).

Because the above argument holds for every \(q \neq p\) dividing \(|G|\), \(|G|_{p'}\) must divide \(\Phi_{\psi}(1)\). ◻

Proof. By Lemma 26 we know \(\Phi_{\psi} \in \text{zcf}_p(G)\) for all \(\psi \in B\), and by Lemma 25 \(P\) is full rank hence \(\{\Phi_{\psi}\}_{\psi \in B}\) is linearly independent. Combining this with Theorem 12 we have \(\text{rk}(\text{zcf}_p(G)) = |\text{cl}(\mathop{\mathrm{\mathcal{F}}})| = |B| = \text{rk}(\langle \{\Phi_{\psi}\}_{\psi \in B} \rangle)\) and we’re done. ◻

Proof. \[\rho_G|_S = \sum_{\chi \in \text{Irr}(G)} \chi(1)\chi|_S = \sum_{\chi \in \text{Irr}(G)}\left(\chi(1)\sum_{\psi \in B}d_{\chi\psi}\psi\right) = \sum_{\psi \in B}\left(\psi\sum_{\chi \in \text{Irr}(G)}d_{\chi\psi}\chi(1)\right) = \sum_{\psi \in B} \psi\Phi_{\psi}(1)\] Now \(\rho_G|_S = [G : S]\rho_S\) thus \[\rho_S = \sum_{\psi \in B}\frac{\Phi_{\psi}(1)}{[G:S]}\psi\] If these coefficients were not integers, then because \(\rho_S \in R(\mathop{\mathrm{\mathcal{F}}})\) there exists unique integers \(a_{\psi}\) giving a decomposition \(\rho_S = \sum_{\psi \in B} a_{\psi}\psi\). Furthermore, \(\mathop{\mathrm{\mathbb{Q}}}\) is flat as a \(\mathop{\mathrm{\mathbb{Z}}}\)-module, so \(B\) is also a basis of \(\mathop{\mathrm{\mathbb{Q}}}\otimes R(\mathop{\mathrm{\mathcal{F}}})\). But then we’d have the linear relation \(0 = \rho_S-\rho_S = \sum_{\psi \in B} (a_{\psi}-\frac{\Phi_{\psi}(1)}{[G:S]})\psi\) in \(\mathop{\mathrm{\mathbb{Q}}}\otimes R(\mathop{\mathrm{\mathcal{F}}})\), which is a contradiction. So \(\frac{\Phi_{\psi}(1)}{[G:S]} = a_{\psi} \in \mathop{\mathrm{\mathbb{Z}}}\). ◻

Proof. \[\begin{align} \langle \Phi_{\psi}, \Phi_{\mu} \rangle_G &= \sum_{\chi \in \text{Irr}(G)}\sum_{\chi' \in \text{Irr}(G)}d_{\psi\chi}d_{\mu\chi'}\langle \chi, \chi' \rangle = \sum_{\chi \in \text{Irr}(G)} d_{\psi\chi}d_{\mu\chi} \\ &= (D^TD)_{\psi\mu} = c_{\psi\mu} \end{align}\] ◻

Proof. By definition of the Cartan numbers, \(c_{\psi\mu} = \displaystyle\sum_{\chi \in \text{Irr}(G)} d_{\chi\psi}d_{\chi\mu}\), thus: \[\Phi_{\psi}|_S = \sum_{\chi \in \text{Irr}(G)} d_{\chi\psi}\chi|_S = \sum_{\chi \in \text{Irr}(G)}\sum_{\mu \in B} d_{\chi\psi}d_{\chi\mu}\mu = \sum_{\mu \in B} c_{\psi\mu}\mu\] ◻

Proof. Write \(C'\) for the matrix \(\displaystyle(C')_{\psi, \mu \in B} := \sum_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})} \frac{\psi(x_K)\overline{\mu(x_K)}}{|C_G(x_K)|}\). Then \[\begin{align} (C'C(\mathop{\mathrm{\mathcal{F}}}))_{\mu\psi} &= \sum_{\theta \in B} C'_{\mu\theta}C(\mathop{\mathrm{\mathcal{F}}})_{\theta\psi} \\ &= \sum_{\theta \in B}\sum_{K \in\text{cl}(\mathop{\mathrm{\mathcal{F}}})} \frac{\mu(x_K)\overline{\theta(x_K)}}{|C_G(x_K)|}c_{\theta\psi} \\ &= \sum_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})}\frac{\mu(x_K)\sum_{\theta \in B} c_{\theta\psi}\overline{\theta(x_K)}}{|C_G(x_K)|} \\ &= \sum_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})} \frac{\mu(x_K)\overline{\Phi_{\psi}|_S(x_K)}}{|C_G(x_K)|} \quad \text{ (applying Corollary \ref{proj32restriction})} \end{align}\] Recall that we write \(\Delta := \text{Diag}_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})}(|C_G(x_K)|)\), then we know from Lemma 25 that \[X(\mathop{\mathrm{\mathcal{F}}})\overline{P^T} = \Delta \Rightarrow (X(\mathop{\mathrm{\mathcal{F}}})\Delta^{-1}\overline{P^T})_{\mu, \psi} = \sum_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})} \frac{\mu(x_K)\overline{\Phi_{\psi}(x_K)}}{|C_G(x_K)|} = \delta_{\mu,\psi}\] Since \(x_K \in S\) we may replace \(\Phi_{\psi}\) in the above expression with \(\Phi_{\psi}|_S\), hence \(C'C = I_{|\text{cl}(\mathop{\mathrm{\mathcal{F}}})|}\) and the result is proven. ◻

Proof. For \(\chi \in R(S)\) we define \(\tilde{\chi}\) given by \[\tilde{\chi}(g) := \begin{cases} |G|_{p'}\chi(g) & g \text{ is a p-element} \\ 0 & \text{ otherwise} \end{cases}\] \(\tilde{\chi} \in R(G)\) by Brauer’s characterisation of characters. Hence for any \(\mu, \psi \in B\), we have \(\langle \tilde{\mu}, \tilde{\psi} \rangle_{G} \in \mathop{\mathrm{\mathbb{Z}}}\). Now since \(\tilde{\mu}, \tilde{\psi}\) are \(0\) on anything that isn’t a \(p\)-element: \[\langle \tilde{\mu}, \tilde{\psi} \rangle_G = \sum_{K \in \text{cl}(G)} \frac{\tilde{\mu}(x_K)\overline{\tilde{\psi}(x_K)}}{|C_G(x_K)|} =\sum_{K \in \text{cl}(G), K \cap S \neq \varnothing}\frac{\tilde{\mu}(x_K)\overline{\tilde{\psi}(x_K)}}{|C_G(x_K)|}\] All Sylow \(p\)-subgroups are conjugate in \(G\), so we are free to choose our \(G\)-conjugacy class representatives \(x_K\) to lie in \(S\): \[\begin{align} \sum_{K \in \text{cl}(G), K \cap S \neq \varnothing}\frac{\tilde{\mu}(x_K)\overline{\tilde{\psi}(x_K)}}{|C_G(x_K)|} &= \sum_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})}\frac{\tilde{\mu}(x_K)\overline{\tilde{\psi}(x_K)}}{|C_G(x_K)|} \\ &= |G|_{p'}^2 \sum_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})}\frac{\mu(x_K)\overline{\psi(x_K)}}{|C_G(x_K)|} \\ &= |G|_{p'}^2(C(\mathop{\mathrm{\mathcal{F}}})^{-1})_{\mu\psi} \in \mathop{\mathrm{\mathbb{Z}}} \end{align}\] And the result is shown. ◻

Proof. Using the notation from the proof of Lemma 35, define \(M\) with \((M)_{\mu, \psi \in B} = \langle \tilde{\mu}, \tilde{\psi} \rangle_G\), which is \(|G|^2_{p'}C(\mathop{\mathrm{\mathcal{F}}})^{-1}\) by Lemma 35. So \(C(\mathop{\mathrm{\mathcal{F}}})M = |G|^2_{p'}I_{|\text{cl}(\mathop{\mathrm{\mathcal{F}}})|}\). Since \(M\) is in integer matrix, it has an integer determinant. Furthermore, \(C(\mathop{\mathrm{\mathcal{F}}})\) is an integer matrix by Remark 31 and thus also has an integer determinant. Therefore \[\text{det}(C(\mathop{\mathrm{\mathcal{F}}}))= \frac{|G|_{p'}^{2|\text{cl}(\mathop{\mathrm{\mathcal{F}}})|}}{\text{det}(M)} \in \mathop{\mathrm{\mathbb{Z}}}\Longrightarrow \text{det}(C(\mathop{\mathrm{\mathcal{F}}})) \text{ is coprime to } p\] ◻

Proof. We surpress our notation and write \(X, D, C\) for \(X(\mathop{\mathrm{\mathcal{F}}}), D(\mathop{\mathrm{\mathcal{F}}}), C(\mathop{\mathrm{\mathcal{F}}})\) in this proof.

Let \(\chi \in \text{Irr}(G)\) and \(K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})\), then: \[(DX)_{\chi K} = \sum_{\mu \in B} D_{\chi\mu}X_{\mu K} = \sum_{\mu \in B} d_{\chi \mu}\mu(x_K) = \chi|_S(x_K)\] So by Lemma 25, we have that \((\overline{(DX)^T}DX)_{ij} = \delta_{ij}|C_G(x_{K_j})|\). Therefore \[\text{det}(\overline{(DX)^T}DX) = \prod_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})}|C_G(x_K)|\] Finally \[\begin{align} \text{det}(\overline{(DX)^T}DX) &= \text{det}(\overline{D^T}D)\text{det}(\overline{X^T}X) = \text{det}(D^TD)|\text{det}(X)|^2 =\text{det}(C)|\text{det}(X)|^2 \\ &\Rightarrow |\text{det}(X)|^2 = \frac{\prod_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})} |C_G(x_K)|}{\text{det}(C)} \end{align}\] ◻

Proof. Since \(X(\mathop{\mathrm{\mathcal{F}}})\) is a matrix of character values, \(X(\mathop{\mathrm{\mathcal{F}}}) \in M_{|\text{cl}(\mathop{\mathrm{\mathcal{F}}})|}(\mathcal{R})\) where \(\mathcal{R} \subset \mathop{\mathrm{\mathbb{C}}}\) is the ring of algebraic integers. So by Lemma 37 and Remark 31, \(|\text{det}(X(\mathop{\mathrm{\mathcal{F}}}))|^2 \in \mathop{\mathrm{\mathbb{Q}}}\cap \mathcal{R} = \mathop{\mathrm{\mathbb{Z}}}\). ◻

Proof. Given two bases \(B, B'\) of \(R(\mathop{\mathrm{\mathcal{F}}})\), there is a change of basis matrix \(M \in GL(|B|, \mathop{\mathrm{\mathbb{Z}}})\) such that \(MX_B(\mathop{\mathrm{\mathcal{F}}}) = X_{B'}(\mathop{\mathrm{\mathcal{F}}})\), hence \(\text{det}(M)\text{det}(X_B(\mathop{\mathrm{\mathcal{F}}})) = \pm\text{det}(X_{B}(\mathop{\mathrm{\mathcal{F}}})) = \text{det}(X_{B'}(\mathop{\mathrm{\mathcal{F}}}))\) and therefore \(|\text{det}(X_B(\mathop{\mathrm{\mathcal{F}}}))|^2 = |\text{det}(X_{B'}(\mathop{\mathrm{\mathcal{F}}}))|^2\). ◻

Proof. Set \(B = \text{Ind}(\mathop{\mathrm{\mathcal{F}}})\), by Lemma 41 it is enough to show Conjecture A for this basis.

By Lemma 2.16 in unique?, factorisation?, \(R(\mathop{\mathrm{\mathcal{F}}})\) is freely generated by \(\text{Ind}(\mathop{\mathrm{\mathcal{F}}})\) when \(\mathop{\mathrm{\mathcal{F}}}\) is transitive, so by Corollary 13 we have \(|\text{Ind}(\mathop{\mathrm{\mathcal{F}}})| = |\text{cl}(\mathop{\mathrm{\mathcal{F}}})| = 2\). It is clear to see that \(\text{Ind}(\mathop{\mathrm{\mathcal{F}}}) = \{1_S, \rho_S-1_S\}\) where \(\rho_S\) is the regular character of \(S\) and \(1_S\) is the trivial character, hence \[X := X_{\text{Ind}(\mathop{\mathrm{\mathcal{F}}})}(\mathop{\mathrm{\mathcal{F}}}) = \begin{pmatrix} 1 & 1 \\ |S|-1 & -1 \end{pmatrix}\] So \(|\text{det}(X)|^2 = |S|^2\). Now we label the two \(\mathop{\mathrm{\mathcal{F}}}\)-conjugacy classes \(K_1 = \{1\}, K_2 = S-K_1\). Because \(S\) is a \(p\)-group, \(\mathcal{Z}(S)\) is non trivial. Therefore \(K_2 \cap \mathcal{Z}(S) \neq \varnothing\), so a fully \(\mathop{\mathrm{\mathcal{F}}}\)-centralised representative of \(K_2\) is central. Hence \(\prod_{K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})} |C_S(x_K)| = |S|^2\) and we are done. ◻

Proof. By Corollary 2.10 in [5], we have that \(|\text{IBr}_p(G)| = |\text{cl}_{p'}(G)|\), combined with the fact that \(S\) is a \(p\)-group, we have that \(|\text{IBr}_p(S)| = |\text{cl}_{p'}(S)| = 1\), and so we have \(\text{IBr}_q(S) = \{1_S\}\). This character is obviously \(\mathop{\mathrm{\mathcal{F}}}\)-stable, and so we have our result. ◻

Proof. By Lemma 2.4, Theorem 1.19 in [5] we have that \(\text{IBr}_q(G)\) is linearly independent over \(\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\). Thus \(\text{rk}\langle \text{IBr}_{\ell}(G) \rangle_{\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}} = |\text{cl}_{\ell'}(G)| = \text{rk}(\text{cf}_{\ell'}(G)) \Rightarrow \text{rk}\langle \text{IBr}_{\ell}(G) \rangle_{\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}} = \text{cl}_{\ell'}(G)\), the result then follows upon restricting to \(S\). ◻

Proof. See Lemma 5.2.2 in [7]. ◻

Proof. For any \(s \in S\) we define the evaluation map \(e_s \colon R(\mathop{\mathrm{\mathcal{F}}}, \ell) \otimes \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}} \rightarrow \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\) with \(e_s(\psi) = \psi(s)\). It is clear that these maps are ring homomorphisms.

Take a set of \(\mathop{\mathrm{\mathcal{F}}}\)-conjugacy class representatives \(x_K\) and consider the set \(\mathscr{E} := \{e_{x_K} \colon K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})\}\). We aim to show that each \(e_{x_K}\) is distinct, so \(\mathscr{E}\) is linearly independent over \(\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\) by Lemma 49.

By Lemma 8, there exists a finite group \(G\) such that \(\mathop{\mathrm{\mathcal{F}}}\) is realised by \(G\). Consider the set \(\text{IBr}_{\ell}(G)|_S\), these are \(\ell\)-Brauer characters of \(S\) by Lemma 2.2 in [5]. They are also \(\mathop{\mathrm{\mathcal{F}}}\)-stable because \(\mathop{\mathrm{\mathcal{F}}}\) is realised by \(G\) and so are contained in \(R(\mathop{\mathrm{\mathcal{F}}}, \ell)\).

Again, because \(\mathop{\mathrm{\mathcal{F}}}\) is realised by \(G\) and \(\ell \neq p\), we have for all \(G\)-classes \(K\) of elements of \(S\) that \(K \cap S \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})\). Hence by Lemma 48 the indicator functions \(\pi_{\ell}(i_K)\) for \(K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})\) are in the \(\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\)-span of \(\text{IBr}_{\ell}(G)|_S\), and are therefore in \(R_v(\mathop{\mathrm{\mathcal{F}}}, \ell) \otimes \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\).

Because \(e_{x_K}\) are indicator functions, we have \(e_{x_K}(\pi_{\ell}(i_{K'})) = \delta_{K,K'}\) for any \(K, K' \in \text{cl}(\mathop{\mathrm{\mathcal{F}}})\) hence each \(e_{x_K}\) is distinct and \(\mathscr{E}\) is linearly independent over \(\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\). Since \(R(S, \ell)\) is a finite dimensional \(\mathop{\mathrm{\mathbb{F}}}_{\ell}\)-algebra, \(R(\mathop{\mathrm{\mathcal{F}}}, \ell) \otimes \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\) is a finite dimensional \(\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\)-algebra. Now we have: \[\text{rk}_{\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}}(\langle\mathscr{E}\rangle) = |\text{cl}(\mathop{\mathrm{\mathcal{F}}})| \leq \text{rk}_{\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}}(\text{Hom}(R(\mathop{\mathrm{\mathcal{F}}}, {\ell}) \otimes \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}, \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}})) = \text{rk}_{\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}}(R(\mathop{\mathrm{\mathcal{F}}}, {\ell}) \otimes \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}) = \text{rk}_{\mathop{\mathrm{\mathbb{F}}}_{\ell}}(R(\mathop{\mathrm{\mathcal{F}}}, {\ell}))\] So \(\text{rk}_{\mathop{\mathrm{\mathbb{F}}}_{\ell}}(R(\mathop{\mathrm{\mathcal{F}}}, \ell)) \geq |\text{cl}(\mathop{\mathrm{\mathcal{F}}})|\). Recalling Definition 47, we also have \(R(\mathop{\mathrm{\mathcal{F}}}, \ell) \leq \langle \pi_{\ell}(i_k) \colon K \in \text{cl}(\mathop{\mathrm{\mathcal{F}}}) \rangle\), which has a rank of \(|\text{cl}(\mathop{\mathrm{\mathcal{F}}})|\). Therefore, \(\text{rk}_{\mathop{\mathrm{\mathbb{F}}}_{\ell}}(\mathop{\mathrm{\mathcal{F}}}, \ell) = |\text{cl}(\mathop{\mathrm{\mathcal{F}}})|\) as desired. ◻

Proof. Take \(\chi \in \text{IBr}_{\ell}(G)\), we may Brauer lift (see Theorem 43.ii in [6]) \(\chi\) to a \(\chi' \in R(G)\). Then because \(\chi'|_S\) is \(\mathop{\mathrm{\mathcal{F}}}\)-stable we may apply Proposition 11 to write \(\chi'|_S = \sum_{\psi \in B} c_{\psi}\psi\) for some \(c_{\psi} \in \mathop{\mathrm{\mathbb{Z}}}\).

Because each \(c_{\psi} \in \mathop{\mathrm{\mathbb{Z}}}\subset \mathcal{R}\), \(\pi_{\ell}(c_{\psi})\) is well defined. By Theorem 43 in [6] we know that \(\pi_{\ell}(\chi') = \chi\). So \(\pi_{\ell}(\chi'|_S) = \chi|_S = \sum_{\psi \in B} \pi_{\ell}(c_{\psi})\pi_{\ell}(\psi)\).

By Lemma 48 we have that \(\langle \text{IBr}_{\ell}(G)|_S \rangle_{\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}} = R(\mathop{\mathrm{\mathcal{F}}}, {\ell}) \otimes \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}\), in particular: \[R(\mathop{\mathrm{\mathcal{F}}}, {\ell}) \otimes \overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}} = \langle \text{IBr}_{\ell}(G)|_S \rangle_{\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}} = \langle \pi_{\ell}(B) \rangle_{\overline{\mathop{\mathrm{\mathbb{F}}}_{\ell}}}\] Thus \(\pi_{\ell}(B)\) spans \(R(\mathop{\mathrm{\mathcal{F}}}, {\ell}) \otimes \overline{\mathop{\mathrm{\mathbb{F}}}}_{\ell}\) as a \(\overline{\mathop{\mathrm{\mathbb{F}}}}_{\ell}\)-vector space, which has rank \(|\text{cl}(\mathop{\mathrm{\mathcal{F}}})| = |B|\), so \(|B| = |\pi_{\ell}(B)|\) and \(\pi_{\ell}(B)\) is linearly independent. ◻

Proof. This follows as in Theorem 39: \(|\text{det}(X(\mathop{\mathrm{\mathcal{F}}}))|^2 \in \mathop{\mathrm{\mathbb{Z}}}\) and \(\pi_{\ell}(\text{det}(X(\mathop{\mathrm{\mathcal{F}}}))) \neq 0\) for all primes \(\ell \neq p\) by Corollary 38 and Corollary 51 respectively. Then we again have that \(|\text{det}(X(\mathop{\mathrm{\mathcal{F}}}))|^2 = p^{\alpha}\) for some \(\alpha \in \mathbb{N}\) as before. ◻