2 The p-Divisible kG-Module
Definition 2.1 For a kG-module V, if the dimension of any indecomposable direct summand of V is divisible by P, where $p={\rm char} k$, we say V is a p-divisible kG-module.
Notation The terminology of the p-divisible kG-module here is based on "absolutely" (see [2]), while for the algebraic closed field k, any indecomposable kG-module is already absolutely indecomposable therein, so that, in this paper, a p-divisible kG-module is essentially controlled by the prime p. We often denote a p-divisible kG-module with (p-divisible) for short. Obviously, the trivial kG-module k is not p-divisible.
Lemma 2.2 Let U, V be two p-divisible kG-modules,
W be a kG-module, and P be a proper p-subgroup of G, then
(1) any P-projective kG-module is p-divisible, particularly, any projective kG-module is p-divisible;
(2) $V^{*}$ is p-divisible;
(3) $U\oplus_{k} V$ is p-divisible, and vice versa;
(4) $U\otimes_{k} W$ is p-divisible;
(5) ${\rm Hom}_{k}(U, V)$ is p-divisible, in particular, ${\rm
End}_{k}(U)$ is p-divisible;
(6) any direct summand of U is p-divisible, especially, k is not the direct summand of U and not the direct summand of ${\rm End}_{k}(U)$.
Proof (1) It comes from the fact that any direct summand of the P-projective kG-module (respectively, the projective kG-module) remains to be P-projective (respectively, projective), and from the fact that the dimension of any indecomposable P-projective kG-module (respectively, any projective kG-module) can be divided by $|G: P|_{p}$
(respectively, by $|G|_{p}$) (see [5, Exercise 23.1, Exercise 21.2(a)]).
(2) First, we see that ${\rm dim}_{k}(V^{*})={\rm dim}_{k}(V)$ from the dual basis, secondly, we have $(V^{*})^{*}\cong V$ as kG-module, then if V is indecomposable, so is $V^{*}$, hence we have the result from the fact that all the dimensions of the indecomposable direct summands of $V^{*}$ are just that of V.
(3) We confirm this result because of Krull-Schmidt theorem.
(4) In the case that U is an indecomposable kG-module, so is $U\otimes_{k} W$ by [1, Corollary 4.3, Corollary 4.7] since k here is algebraically closed. For the general case it is also true because of $(3)$.
(5) It is true because of $(4)$ and the kG-module isomorphism ${\rm Hom}_{k}(U, V)\cong U^{*}\otimes_{k} V $.
(6) It follows from $(3)$.
Proposition 2.3 Let $G\geq H$ and V be a kG-module, if ${\rm Res}^{G}_{H}(V)$ is a p-divisible kH-module, then V is a p-divisible kG-module.
Proof It follows from Krull-Schmidt theorem.
Proposition 2.4 Let $G\geq H$ and V be a p-divisible kG-module, if H contains a Sylow p-subgroup of G, then ${\rm
Res}^{G}_{H}(V)$ is a p-divisible kH-module.
Proof Proof by contradiction. If ${\rm Res}^{G}_{H}(V)$ is not p-divisible, then p does not divide ${\rm
dim}_{k}(V_{1})$, and then $k| {\rm End}_{k}(V_{1})$ for some direct summand $V_{1}$ of ${\rm Res}^{G}_{H}(V)$ by [1, Corollary 4.7], we see that $k| {\rm End}_{k}({\rm Res}^{G}_{H}(V))$, so that
| ${\rm Ind}^{G}_{H}(k)|{\rm Ind}^{G}_{H}({\rm Res}^{G}_{H}({\rm End}_{k}(V))).$ |
On the other hand,
| ${\rm Ind}^{G}_{H}({\rm Res}^{G}_{H}({\rm End}_{k}(V)))\cong {\rm Ind}^{G}_{H}(k)\otimes_{k} {\rm End}_{k}(V)$ |
by Frobenius Reciprocity, so that ${\rm Ind}^{G}_{H}(k)\otimes_{k}
{\rm End}_{k}(V)$ is also p-divisible by Lemma 2.2(5)--(4), that is, ${\rm Ind}^{G}_{H}({\rm Res}^{G}_{H}({\rm End}_{k}(V)))$ is p-divisible, it means that ${\rm Ind}^{G}_{H}(k)$ is p-divisible by Lemma 2.2(3), it contradicts with the order of ${\rm Ind}^{G}_{H}(k)$ if H contains some Sylow p-subgroup of G.
For any $k(G/N)$-module V, where $G\unrhd N$, the inflation kG-module $\inf(V)$ has the same k-vector space as V, while its G-action is from the composition of the canonical group homomorphisms: $G\rightarrow G/N\rightarrow {\rm End}_{k}(V)$.
Proposition 2.5 Let N be a normal subgroup of G with $p||G:N|$, then V is a p-divisible $k(G/N)$-module if and only if $\inf(V)$ is a p-divisible kG-module.
Proof We see that V is an indecomposable $k(G/N)$-module if and only if $\inf(V)$ is an indecomposable kG-module, moreover, the dimensions of V and $\inf(V)$ are the same since as k-module they are the same, then V is p-divisible if and only if $\inf(V)$ is p-divisible.
Let $G\geq H$, a kG-module M is said to be H-projective, if there exists a kH-module N such that $M| {\rm
Ind}^{G}_{H}(N)$; moreover, if M is indecomposable, then the minimal subgroup H such that M is H-projective is unique under the G-conjugation, we call H the vertex of M, it must be a p-group, in the case, the kH-module N is called the corresponding source module (see [5, 6]).
Corollary 2.6 Let U be not a p-divisible kG-module, if P is a p-subgroup of G, then $U\otimes_{k} V$ is P-projective if and only if the kG-module V is P-projective.
Proof The proof of sufficiency is obvious; for the necessity, we see $k |(U^{*}\otimes_{k} U)$ as the proof of Proposition 2.4, then
| $k\otimes_{k} V|((U^{*}\otimes_{k} U)\otimes_{k} V), $ |
that is,
| $V| U^{*}\otimes_{k}(P\mbox{-projective}), $ |
hence, V is a direct summand of the P-projective kG-module, so that V is P-projective.
By Green indecomposability theorem (see [5, Corollary 23.6]),
${\rm Ind}^{P}_{Q}(k)$ is p-divisible if Q is the proper subgroup of the p-group P. In general, we have the following result.
Proposition 2.7 Let $G\geq H$, V be a kG-module, and U be a kH-module such that $V={\rm Ind}^{G}_{H}(U)$;
(1) if the Sylow p-subgroup of H is a proper p-subgroup of G, then V is p-divisible;
(2) if U is p-divisible and H contains some Sylow p-subgroup of G such that $p||G: H\cap\ ^{g}H|$ for any $g\in
G-H$, then V is p-divisible;
(3) if U is not p-divisible while V is p-divisible, then the Sylow p-subgroup of H is a proper p-subgroup of G.
Proof (1) By Lemma 2.2(1), ${\rm Ind}^{G}_{H}(U)$ is p-divisible since ${\rm Ind}^{G}_{H}(U)$ is P-projective, where P is a Sylow p-subgroup of H.
(2) We see that
| $\begin{eqnarray*} {\rm Res}^{G}_{H}(V)&&{\rm Res}^{G}_{H}({\rm Ind}^{G}_{H}(U))\\
&\cong& \bigoplus_{g\in[H\backslash G/H]}{\rm Ind}^{H}_{H\cap\
^{g}H}({\rm Res}^{H}_{H\cap\ ^{g}H}\ (^{g}U))\\
&&\bigoplus_{g\in [H\backslash G/H]}{\rm Ind}^{H}_{H\cap\ ^{g}H}\
(^{g}U)\\
&&U\oplus(\bigoplus_{1\neq g\in [H\backslash G/H]}{\rm Ind}^{H}_{H\cap\ ^{g}H}\ (^{g}U)), \end{eqnarray*}$ |
where each ${\rm
Ind}^{H}_{H\cap\ ^{g}H}\ (^{g}U)$ must be p-divisible by (1). So that ${\rm Res}^{G}_{H}(V)$ is p-divisible by Lemma 2.2 (3), and then V is also p-divisible by Proposition 2.3.
(3) On the contrary, if H contains a Sylow p-subgroup of G, then $U|{\rm Res}^{G}_{H}({\rm Ind}^{G}_{H}(U))$, it means that ${\rm Ind}^{G}_{H}(U)$ cannot be p-divisible. Indeed, if ${\rm Ind}^{G}_{H}(U)$ is p-divisible, so is
| ${\rm Res}^{G}_{H}({\rm Ind}^{G}_{H}(U))$ |
by Proposition 2.4, and so is U by Lemma 2.2 (3). Contradiction!
Recall that a subgroup H of G is strongly p-embedded if p divides the order of H but does not divide $|H\cap\ ^{x}H|$, for all $x\in G-H$. Note here that the strongly p-embedded subgroup H of G contains the normalizer in G of any p-subgroup of G, and such H exists whenever the Sylow p-subgroup of G is trivial intersection (that is, T.I set) (see [9]).
Corollary 2.8 (1) Let $G\geq H$, then ${\rm
Ind}^{G}_{H}(k)$ is p-divisible if and only if the Sylow p-subgroup of H is a proper p-subgroup of G;
(2) let $G\geq H$ and U be a p-divisible kH-module such that $V={\rm Ind}^{G}_{H}(U)$, if H is strongly p-embedded, then V is p-divisible.
Proof It follows from Proposition 2.7.
A kG-module M is called a p-permutation kG-module if ${\rm
Res}^{G}_{Q}(M)$ is a permutation module for every p-subgroup Q of G (see [3]); an endo-p-permutation kG-module N is a kG-module N with ${\rm End}_{k}(N)$ being a p-permutation kG-module under the conjugation action of G, in the case, if G is a p-group, we call it an endo-permutation module (see [5]); any p-permutation kG-module is an endo-p-permutation kG-module, and the p-permutation kG-module plays a crucial role for the equivalence of the categories of blocks of kG (see [7, 8]).
Theorem 2.9 Any indecomposable endo-p-permutation kG-module V with the vertex P is p-divisible if and only if P is the proper p-subgroup of G, moreover, in the case of P being the Sylow p-subgroup of G, P does not divide ${\rm
dim}_{k}(V), $ and ${\rm Res}^{G}_{P}(V)$ is a capped endo-permutation kP-module.
Proof If P is the Sylow p-subgroup of G, ${\rm
Res}^{G}_{P}(V)$ is an endo-permutation kP-module, and the source module of V has the vertex P and is the direct summand of ${\rm Res}^{G}_{P}(V)$, then ${\rm Res}^{G}_{P}(V)$ is a capped endo-permutation kP-module, so that, ${\rm dim}_{k}(V)\equiv \pm
1 ~(\mod\ p)$ by [5, Corollary 28.11], that is, p does not divide ${\rm dim}_{k}(V)$, V is not p-divisible.
On the contrary, if P is the proper p-subgroup of G, V is p-divisible by Lemma 2.2 (1).
Given a kG-module M, the Heller translate $\Omega(M)$ is the kernel of the projective cover $P_{M}\rightarrow M$, so that there is a short exact sequence of kG-modules $0\rightarrow
\Omega(M)\rightarrow P_{M}\rightarrow M\rightarrow 0$; since projective covers of M are unique up to isomorphism, $\Omega(M)$ is well-defined up to isomorphism; similarly, one may define $\Omega^{-1}(M)$ to be the cokernel of the injective hull $M\rightarrow I_{M}$. The Heller operator $\Omega$ provides a way to construct new indecomposable kG-modules from the old ones. Let $\Omega^{0}(M)$ be the non-projective kG-module such that $M=\Omega^{0}(M)\oplus_{k} (\rm projective)$,
$\Omega^{1}(M)=\Omega(M)$; one can obtain the nth-Heller translates
| $\Omega^{n}(M):=\Omega(\Omega^{n-1}(M))~~ (n\geq 2)$ |
and
| $\Omega^{n}(M):=\Omega^{-1}(\Omega^{n+1}(M))~~ (n\leq -2)$ |
(see [6]).
Lemma 2.10 Let V and W be kG-modules, then for any $m, n\in Z$, $\Omega^{m}(V)\otimes_{k} \Omega^{n}(W)\cong
\Omega^{m+n}(V\otimes_{k} W)\oplus_{k}$ (projective), particularly, there exists a projective kG-module U such that $V\otimes_{k} \Omega^{n}(k)\cong \Omega^{n}(V)\oplus_{k} U$.
Proof Set m=0, $W=k$ in the former, we can obtain the latter. While the latter is also Proposition 11.7.2 of [6], now we prove the former by the latter.
Set $V\otimes_{k} \Omega^{m}(k)\cong \Omega^{m}(V)\oplus_{k}X$ and $W\otimes_{k} \Omega^{n}(k)\cong \Omega^{n}(W)\oplus_{k}Y$, where X and Y are projective kG-modules. Then
| $\begin{eqnarray*}
&&(\Omega^{m}(V)\oplus_{k}X)\otimes_{k}(\Omega^{n}(W)\oplus_{k}Y)
\cong\Omega^{m}(V)\otimes_{k}\Omega^{n}(W)\oplus_{k}({\rm projective})\\
&\cong&(V\otimes_{k} \Omega^{m}(k))\otimes_{k}(W\otimes_{k}
\Omega^{n}(k)) \cong V\otimes_{k} W\otimes_{k}(\Omega^{m}(k)\otimes_{k} \Omega^{n}(k))\\
&\cong&
(V\otimes_{k} W)\otimes_{k}(\Omega^{m+n}(k)\oplus_{k}({\rm projective}))\\
&\cong& \Omega^{m+n}(V\otimes_{k} W)\oplus_{k}({\rm projective}), \end{eqnarray*}$ |
it means that $\Omega^{m}(V)\otimes_{k} \Omega^{n}(W)\cong
\Omega^{m+n}(V\otimes_{k} W)\oplus_{k} ({\rm projective})$ since $\Omega^{m+n}(V\otimes_{k} W)$ has no projective direct summands.
The following result shows that, $\Omega^{n}(-)$ permutates the isomorphism classes of indecomposable non-projective p-divisible kG-modules.
Theorem 2.11 Let V be a kG-module, then for any $n\in
Z$, V is p-divisible if and only if $\Omega^{n}(V)$ is p-divisible.
Proof If V is p-divisible, so is $V\otimes_{k}
\Omega^{n}(k)$, and then $\Omega^{n}(V)\oplus_{k} U$ is also p-divisible by Lemma 2.10, that is, $\Omega^{n}(V)$ is p-divisible for each $n\in Z$.
On the contrary, if $\Omega^{n}(V)$ is p-divisible, so is $\Omega^{n}(V)\oplus_{k} U$ in Lemma 2.10, and then $V\otimes_{k} \Omega^{n}(k)$ is p-divisible, so that $(V\otimes_{k} \Omega^{n}(k))\otimes_{k} \Omega^{-n}(k)$ is also p-divisible. While
| $\begin{eqnarray*} &&(V\otimes_{k}
\Omega^{n}(k))\otimes_{k} \Omega^{-n}(k)\cong V\otimes_{k}
(\Omega^{n}(k)\otimes_{k} \Omega^{-n}(k))\\ &\cong& V\otimes_{k}(\Omega^{0}(k\otimes_{k} k)\oplus_{k} U_{1})\cong V\otimes_{k}(k\oplus_{k} U_{1})\\ &\cong& V\oplus_{k}(V\otimes_{k} U_{1}), \end{eqnarray*} $ |
where $U_{1}$ is some projective kG-module. We have $V|(V\otimes_{k} \Omega^{n}(k))\otimes_{k}
\Omega^{-n}(k)$, then V is also p-divisible.
Corollary 2.12 Let $0\rightarrow U\rightarrow W\rightarrow
V\rightarrow 0$ be a short exact sequence of the kG-modules, if W is projective, then U is a p-divisible kG-module if and only if V is p-divisible.
Proof One can check the following kG-module isomorphisms by Schanuel$^{'}$s lemma
| $\begin{eqnarray*}
&&U\cong \Omega(V)\oplus_{k} ({\rm projective}), \\
&&V\cong \Omega^{-1}(U)\oplus_{k} ({\rm projective}), \end{eqnarray*}$ |
then the result follows from Theorem 2.11 and Lemma 2.2(1)(3).
A V-projective kG-module W is the one such that $W|(V\otimes_{k} X)$ for some kG-module X; the V-projective kG-module generalizes the notion of the (relative) projective kG-modules (see [4]).
Proposition 2.13 Let V be a p-divisible kG-module and W be a V-projective kG-module, then W is p-divisible; if moreover, V is Q-projective, where Q is a p-subgroup of G, then W is also Q-projective.
Proof If W is indecomposable and p does not divide ${\rm dim}_{k}(W)$, then $k|(W^{*}\otimes_{k} W)$, and then $k|(W^{*}\otimes_{k} V\otimes_{k} X)$ since W is V-projective, where X is a kG-module; while $W^{*}\otimes_{k} V\otimes_{k}
X$ is p-divisible by Lemma 2.2 (4), so that, k is p-divisible by Lemma 2.2 (3), too; it contradicts with Lemma 2.2 (6). In general, W is always p-divisible.
If V is Q-projective, so is $V\otimes_{k} X$ (see [5], Lemma 14.3), and then the direct summand W is also Q-projective.
Let $G\geq H\geq N_{G}(P)$, where P is a Sylow p-subgroup of G. The following result shows that the class of the indecomposable p-divisible modules is closed under the bijection of Green correspondence between the indecomposable kG-modules and the indecomposable kH-modules whenever H is strongly p-embedded in G.
Theorem 2.14 Let $G\geq H\geq N_{G}(P)$, where P is a Sylow p-subgroup of G, if H is strongly p-embedded, then Green correspondence between the indecomposable kG-modules and the indecomposable kH-modules induces a bijection between the isomorphism classes of indecomposable p-divisible kG-modules and that of indecomposable p-divisible kH-modules.
Proof In the case of H being strongly p-embedded, for the indecomposable p-divisible kG-module V with the vertex P, ${\rm Res}^{G}_{H}(V)$ is a p-divisible kH-module by Proposition 2.4, it means that the Green correspondent of V remains to be p-divisible by Lemma 2.2 (3); similarly, if U is a p-divisible kH-module with the vertex P, then ${\rm
Ind}^{G}_{H}(U)$ is p-divisible by Corollary 2.8, and then the Green correspondent of U remains to be p-divisible by Lemma 2.2 (3) again.
Since H contains the normalizer of any proper p-subgroup Q of G (see [9]), Green correspondence sets up a bijection between the isomorphism classes of the indecomposable kG-modules with the vertex Q and that of the indecomposable kH-modules with the same vertex Q, moreover, these indecomposable modules with the vertex Q are p-divisible (Lemma 2.2 (1)).
Sum up the above, for any p-subgroup Q of G, whether or not it is a proper p-subgroup, the indecomposable p-divisible modules with the vertex Q are closed under the bijection of Green correspondence between the indecomposable kG-modules and the indecomposable kH-modules, so that, Green correspondence between the indecomposable kG-modules and the indecomposable kH-modules induces a bijection between the isomorphism classes of indecomposable p-divisible kG-modules and that of indecomposable p-divisible kH-modules.
2017, Vol. 37 
PDF