Alperin "Local Representation Theory" Lemma 9.7 Have a question about the proof of Lemma 9.7 from Alperin "Local Representation Theory", p. 69:
Lemma 7
If $U$ is an indecomposable $kG$-module with vertex $Q$ and trivial source and $H$ is any subgroup of $G$ then $U_H$ has an indecomposable summand with a vertex containing $Q\cap H$.


*

*Why is the source of $U$ equal to $k_Q$ and not $k$?

*Why must $k_Q\mid U_Q$ and why does this imply $k_{Q\cap H}\mid U_{Q\cap H}$?


Thanks in advance. For those who don't own the book the page is available on Google Books here. The notation $U\mid V$ for modules $U,V$ simply means $U$ is (isomorphic to) a direct summand of $V$.
 A: Maybe I misunderstand your first question. The source is by definition a module of the vertex, namely the module $M$ for which $U | M^G$. Trivial source in this context means trivial module of the vertex $Q$.
Now for the second question. The easy part is why $k_Q|U_Q$ implies $k_{Q\cap H}|U_{Q_H}$. If you write out the definitions of these symbols, the implication becomes trivial, using transitivity of restriction:
$$
\begin{align}
k_Q|U_Q & \Rightarrow U_Q = k_Q\oplus\ldots\\
& \Rightarrow U_{Q\cap H} = (U_Q)_{H\cap Q} = (k_Q)_{H\cap Q}\oplus \ldots= k_{H\cap Q}\oplus \ldots
\end{align}
$$
It remains to show that $k_Q|U_Q$. Again, let us write out the definitions. We have $(k_Q)^G = U\oplus \ldots$, since $k_Q$ is a source of $U$. So, restricting back to $Q$, $((k_Q)^G)_Q = U_Q\oplus\ldots$ Now use Mackey:
$$
((k_Q)^G)_Q = \bigoplus_{g\in Q\backslash G/Q}(k_{Q\cap Q^g})^Q = U_Q\oplus\ldots
$$
By Krull-Schmidt, $U_Q$ is a direct sum of trivial modules induced from various subgroups, so let's simply write $U_Q = \bigoplus_i (k_{H_i})^Q$, $H_i\leq Q$, and we want to show that one of these $H_i$ is $Q$ itself. Now, $Q$ being a vertex implies that $U|(U_Q)^G$ (indeed, one of the possible definitions of vertex is that $Q$ is the smallest such subgroup of a Sylow). So, $U\;|\;\bigoplus_i (k_{H_i})^G$, and since $U$ is indecomposable, it divides one of the summands. But hang on a minute, $Q$ was the smallest subgroup from which $U$ could divide an induction, so one of these $H_i$ must indeed be $Q$, as claimed.
