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 $V$.


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.

  • $\begingroup$ Thanks. One more quick question: on p73, when we're talking about $SL(2,p)$ and its Sylow $p$-subgroup $L$, why does $kL$ have $p-1$ simple modules of dimension 1, and why must all indecomposable $kL$-modules be uniserial? $\endgroup$ – Clinton Boys Aug 26 '11 at 0:43
  • $\begingroup$ @user The Sylow $p$-subgroup of $SL_2(\mathbb{F}_p)$ is cyclic of order $p$. There are many ways of seeing that the cyclic group of order $p$ has only one simple module, namely the trivial one. From there, it is also easy to see that every indecomposable has a unique trivial submodule, and uniseriality follows. I suggest that we don't misuse the comments for discussing a totally different question, since they are not at all suited for this purpose. If you can't fill in the details in the above argument, feel free to open another questions. $\endgroup$ – Alex B. Aug 26 '11 at 3:45

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