To prove (1), we only need to check that
\(B\) is an almost finite projective
\(A\)-module, as
(24.3) already implies that
\(B\) is an almost finite projective
\(B \otimes_A B\)-module. By
(24.3), the idempotent element of
\(\prod_{g \in G} B\) that picks out the identity component is an almost element of
\(B \otimes_A B\text{.}\) Consequently, for each
\(\eta \in \frakm\text{,}\) we may multiply by
\(\eta\) to get a genuine element
\(e_\eta \in B \otimes_A B\) satisfying
\(e_\eta^2 = e_\eta\) that kills the kernel of
\(\mu\) and projects to
\(\eta \in B\text{.}\) Write
\(e_\eta = \sum_{i=1}^n b_i \otimes b'_i\) for some
\(b_i, b_i' \in B\text{;}\) we then have
\(\sum_{i=1}^n \gamma(b_i) b_i' = 0\) for
\(\gamma \in G \setminus \{e\}\) and
\(\sum_{i=1}^n b_i b_i' = \eta\text{.}\)
Define the trace map \(t_{B/A}\colon B \to A\) as the sum over \(G\)-conjugates. Then
\begin{equation*}
\sum_i t_{B/A}(bb_i) b'_i = \eta b \qquad (b \in B).
\end{equation*}
In other words, the composition
\begin{equation*}
B \stackrel{b \mapsto (t_{B/A}(b b_i))_i}{\longrightarrow} A^n \stackrel{(a_i) \mapsto \sum a_i b'_i}{\longrightarrow} B
\end{equation*}
is multiplication by \(\eta\text{;}\) since \(\eta \in \frakm\) was arbitrary, this proves that \(B\) is an almost finite projective \(A\)-module.
To prove (2), we first apply (1) to deduce that \(C \to B\) is almost finite étale. We then check that the canonical map \(C \otimes_A B \to \prod_{G/H} B\) is an almost isomorphism: we can check this after tensoring over \(C\) with \(B\text{,}\) in which case we have almost isomorphisms
\begin{equation*}
B \otimes_C (C \otimes_A B) = B \otimes_A B \to \prod_G B \to \prod_{G/H} (B \otimes_C B) = B \otimes_C \prod_{G/H} B.
\end{equation*}
Thus the map
\(A \to C\) becomes almost finite étale after tensoring over
\(A\) with
\(B\text{,}\) and so by
Remark 24.2.10 is itself almost finite étale. (Compare
[4], Proposition 9.1.)