Real-valued character in Block with cyclic defect has at most two constituents modulo $p$

Question

Let $G$ be a finite group and let $(K,R,k)$ be a $p$-modular system (large enough for $G$ etc.) and consider a block algebra $B \subseteq RG$ with cyclic defect group.

My question is about the following statement found for example in Feit's 'Representation Theory of Finite Groups' Chapter 7 Theorem 2.24

If $\chi \in \text{Irr}(B)$ is real-valued, then $\widehat{\chi} = \chi\mid_{G_{p'}}$ is the sum of at most two irreducible Brauer characters.

My goal is to prove a version of this statement where we replace the invariance under complex conjugation by other more general operations. I believe the proof by Feit can easily be adapted to show what I have in mind but there are some doubts which I would like to have clarified.

The proof given in Feit's book is roughly as follows (if I understand it correctly):

1. Let $\varphi_1, \dots, \varphi_s$ denote the distinct irreducible Brauer characters which are constituents of $\widehat{\chi}$.
2. For every $1 \leq i \leq s$ there exists an $RG$-lattice $X_i$ such that $K \otimes X_i$ affords $\chi$, $k \otimes X_i$ is indecomposable and serial and its socle affords $\varphi_i$.
3. There exists an ordering $\varphi^{(1)}, \dots, \varphi^{(s)}$ of the $\varphi_1, \dots, \varphi_s$ such that if $\varphi^{(t)} = \varphi_i$, then the composition factors of $k \otimes X_i$ are in ascending order: $\varphi^{(t)}, \varphi^{(t + 1)}, \dots $ with superscripts read modulo $s$. This is Theorem (2.23).

This is the preparation needed. Now starts the real proof:

4. If $\chi$ is real valued, then $(k \otimes X_i^*) \cong k \otimes X_j$ for some $j$.

Feit claims to use Theorem 2.20 here, I am not sure if he wants to say that every $RG$-lattice $X$ for which $K \otimes X$ affords $\chi$ and $k \otimes X$ is indecomposable satisfies $k \otimes X \cong k \otimes X_i$ for some $i$. But this is not the main problem I have.

5. Without loss of generality we may assume $\varphi^{(1)} = \varphi_j$. Let $n$ be such that ${\varphi^{(n)}}^* = \varphi^{(1)}$. It follows that ${\varphi^{(n)}}^*,{\varphi^{(n-1)}}^* \dots$ and ${\varphi^{(1)}}, {\varphi^{(2)}} \dots $ both describe the composition factors of $k \otimes X_j$ in ascending order. We thus have ${\varphi^{(n - k)}}^* = \varphi^{(k +1)}$

For the next part I will cite from Feit's book:

6. Hence $\varphi^{(n - t)} = {\varphi^{(n-t)}}^*$ if and only if $n - t = t + 1$ (mod $s$) or $2t = n - 1$ (mod $s$).There are at most two solutions to this congruence which implies the result.

According to Feit, this finishes the proof.

To make use of what he proved in 6. we would have to have $\varphi^{(n -t)} = {\varphi^{(n-t)}}^*$ for all $t$ and it is not clear to me where this identity comes from and I do not see any reason why it should be true. Wouldn't this imply that every module affording $\varphi_i$ would be isomorphic to its dual? Does this follow from something I have missed? Any clarification is appreciated.

Answer 1

The problem here does not lie in the mathematics but in my reading capapabilities. In fact, the correct Theorem 2.24 says

If $\chi \in \text{Irr}(B)$ is real-valued, then there are at most two real-valued irreducible Brauer characters which are constituents of $\widehat{\chi} = \chi\mid_{G_{p'}}$.

For such a character $\varphi$ we have ${\varphi}^* = \varphi$ and this is exactly what is used in 6.