数理解析研究所講究録
第 1784 巻 2012 年 77-95

77

RELATIVE PROJECTIVE COVERS AND THE BRAUERCONSTRUCTION OVER FINITE GROUP ALGEBRAS
北海道教育大学旭川校

奥山

哲郎 (TETSURO OKUYAMA)

HOKKAIDO UNIVERSITY OF EDUCATION, ASAHIKAWA CAMPUS

Some properties of relative projective covers of modules in the modular representation
theory of finite groups will be discussed. Especially, we study effects of the Brauer constructions for relative projective covers of p-permutation modules. We also discuss someuse of our results to investigate derived equivalences in the principal block algebras offinite groups with Sylow p-subgroup isomorphic to $M_{n+1}(p),$ odd.
In my lecture, we only talked on the sections 1 and 2 below. We include the sections
3 and 4 which provide some results for proofs of theorems in sections 2. Section 5 is also
included to give another examples with finite groups with Sylow p-subgroups $M_{n+1}(p)$ .
Let be an algebraically closed field of characteristic $p>0$ .
$p$

$k$

1. $M_{n+1}(p)$
Let

$p$

be odd and$n$be an integer with $n\geqq 2$ . The p-group $M_{n+1}(p)=P$ of order$p^{n+1}$

is presented by$M_{n+1}(p)=P$ $;=\{x,$ $y|y^{p^{n}}=1=x^{p},$$P$$xyx^{-1}=y^{1+p^{n-1}}\rangle$

has a unique maximal elementary abelian p-subgroup $\{x,$$Q=\{y\}$ ,$y^{p^{n-1}}\rangle$. Set$R=\{x\}$

We fix an integer $s\in Z$ which has multiplicative order $p-1$ in the residue ring $Z/p^{n}$ Z. Notice then that has multiplicative order $p-1$ in $Z_{p}=Z/pZ$ also.
has an automorphismof order $p-1$ which sends
so that we have a group.
$P$

$s$

$t_{0}$$P\rangle\triangleleft\{t_{0}\}$$x\mapsto x,$$y\mapsto y^{s}$$;=\{x, y, t_{0}|y^{p^{n}}=1=x^{p}, xyx^{-1}=y^{1+p^{n-1}}, t_{0}^{p-1}=1, t_{0}^{-1}xt_{0}=x, t_{0}^{-1}yt_{0}=y^{s}\}$In the following discussion, fix a positive divisor $e\geqq 2$ of $p-1$ and set$p= \frac{p-1}{e}$$t=t_{0}^{\ell}$where. And set
$H=P\lambda\{t\}\cong P\lambda Z_{e}$1.1. Some Complexes of $kH$ -modules. $kH$ has simple modules $S(i),$ $i\in Z/eZ$ (allof dimension 1). We can name the simples so that the following facts hold.
$e$$Ext_{kH}^{1}(S(i), S(i+1))\neq 0$ ,Let denote a projective cover of $S(i)$ by $P(i)$ .

$S(O)=k_{H}$78
By a result of Okuyama and Sasaki [7], we have a (chain) complex$X^{\cdot}(1)$of $kH$ -modules(1.1)$xi(1)$$X_{k}^{\cdot}(1)$:$arrow P(1)$$S(1)$$–arrow$$arrow P(1)\oplus P(1)arrow\Omega^{-2e}(S(1))arrow 0arrow\cdots$: . . . $arrow\Omega^{-2(k-1)e}(S(1))arrow P(1)\oplus P(1)arrow P(1)\oplus P(1)arrow\Omega^{-2ke}(S(1))arrow 0arrow\cdots$$S(2)$

satisfying:$H_{1}(X_{\dot{k}}(1))\cong H_{2}(X_{\dot{k}}(1))=$$S(-1)$

and $H_{0}(X_{\dot{k}}(1))\cong H_{3}(X_{\dot{k}}(1))=0$ for each$S(0)$$1\leqq k\leqq\ell$

where the last nonzero terms are in degree .
$0$Tilting. Let be an arbitraly symmetric algebra over and $\{S(i)$ ;$I\}$ be the set of simple A-modules. $L\overline{etP(i)}$beaprojective cover of $S(i)$ .
Take a (nomempty proper) subset of . For each $i\in I$ , construct a complex $P(i)\in$$C^{b}(P(A))$ of projective A-modules as follows.

1.2. Richard

$A$

$s$

$k$

$I$

$I_{0}$:$arrow 0arrow P(j)\lambda_{j}arrow 0arrow 0arrowarrow..\cdot.\cdot$$j\not\in I_{0},$.

$arrow 0arrow 0arrow R(j)arrow P(i)$$P^{\cdot}(j)P^{\cdot}(i):$where for$i\in$$R(j)arrow P(j)\lambda_{j}$$j\not\in I_{0}i\in I_{0}$is a minimal one satisfying that(1). $R(j)$ is a direct sum of $P(i),$(2). Composition factors of Cok$i\in I_{0}$$\lambda_{j}$are $S(k)$ for some$k\not\in I_{0}$Set$P^{\cdot}(I_{0})= \oplus\sum_{i\in I}P^{\cdot}(i)$Then in the homotopy category $K^{b}(P(A))$ of complexes of projective A-modules,$P^{\cdot}(I_{0})$is a tilting complex for

$A$

Set
$B=End_{K^{b}(P(A))}(P^{\cdot}(I_{0}))$is a derived equivalent algebra to A. is also a symmetric algebra and simple
B-modules are also parametrized by the set . Let $Q(i)$ be a projective indecomposable B. Let $T(i)$ be the simple B-moduleofmodule corresponding to the summand$M(I_{0})$$(A,
B)$$T(i)$(with no bimodule projective. There is a-bimodule
corresponding tosatisfying the following.
summand) constructed from the complex(1). Both of AM
are projective and a functor
andso that

$B$

$B$

$I$

$P^{\cdot}(i)$$P^{\cdot}(I_{0})$$P^{\cdot}(I_{0})$$(I_{0})$$M(I_{0})_{B}$$F_{0}=F(I_{0})$: $mod-Aarrow mod-B$ ,$V\mapsto V\otimes_{A}M(I_{0})$gives a stable equivalence of Morita type between mod-A and $mod-B$ .
$F_{0}(S(j))=T(j)$ for(2.1). For.
$P(i)\subset
W(i)\subset
P(i)$(2.2). For, let Soc
be the largest submodule of $P(i)$ such that
. Then $F_{0}(P(i)/W(i))=$all the composition factors of $W(i)/S(i)$ are $S(k)$ for some$T(i)$ . for.
$j\not\in I_{0},$$j\not\in I_{0}$$i\in I_{0}$$k\not\in I_{0}$$i\in I_{0}$79
We call the procedure above the Richard Tilting with respect to the set. Thegiven above is called the associated functor of the tiltng. The dual argument
functorto the above discussion is also valid which we call the dual Richard Tilting.
$I_{0}\subset I$$s$$F_{0}$$s$1.2.1. Examples. For $kH$ , apply Richard‘s tiltings with repect to the set {1} twice.
First do the Richard tilting with respect to the set {1}. And then for the resulting
new algebra, do the Richard‘s tilting with respect to the set {1}.
Letbe the resulting algebra and let $S(i)_{2}$ (resp. $P(i)_{2}$ ) be a simple (resp. projective)-module corresponding to $S(i),$ $i\in I$ . Let $F^{2}:mod-kHarrow mod-A_{2}$ be the associatedfunctor. Then by the existence of the complex $xi(1)$ in (1.1), we have$s$$A_{2}$$A_{2}$

Lemma 1.1.
$F^{2}(S(i))=S(i)_{2},$$i\neq 1$,$F^{2}(S(1))=\Omega^{2e}(S(1)_{2})$The existence of comlexes $X_{\dot{k}}(1)(1\leqq k\leqq\ell)$ implies the following. For each with, do the Richard‘s tiltings with repect to the set {1}times.
Let
be the resulting algebra and let $S(i)_{2k}$ (resp. $P(i)_{2k}$ ) be a simple (resp. projective)-module corresponding to $S(i),$ $i\in I$ . Let $F^{2k}:mod-kHarrow mod-A_{2k}$ be theassociated functor. Then$k$

$1\leqq k\leqq\ell$$\underline{2k}$$A_{2k}$$A_{2k}$Lemma 1.2.
$F^{2k}(S(i))=S(i)_{2k},$$i\neq 1$,

The discussion above is valid for any fixed$F^{2k}(S(1))=\Omega^{2ke}(S(1)_{2k})$$i_{0}\in I$.

Lemma 1.3. Let
and be an integer with $1\leqq k\leqq P$ .
(1) There exists an algebra $B$ derived equivalent to $kH$ satisfying the following. Let$T(i),$ $i\in I$ be the set of simple B-modules and: mod $-kHarrow$ mod-B be theassociated stable equivalence. Then$i_{0}\in I$$k$$F^{*}$$F^{*}(S(i))=T(i),$$i\neq i_{0}$,$F^{*}(S(i_{0}))=\Omega^{2ke}(T(i_{0}))$(2) There exists an algebm $C$ dewived equivalent to $kH$ satisfying the following. Let$U(i),$ $i\in I$ be the setof simple C-modules and : mod $-kHarrow$ mod-C be theassociated stable equivalence. Then
$F_{*}$$F_{*}(S(i))=U(i),$$i\neq i_{0}$,$F_{*}(S(i_{0}))=\Omega^{-2ke}(U(i_{0}))$1.3. Relative Projective Covers. Set$K=R\cross\langle t\}=\langle x\}\cross\langle t\}\subset H$

and$P_{R}(i)=(S(i)\downarrow_{K})\uparrow^{H}=P_{R}(0)\otimes S(i)$Then we have a canonical surjection and a canonical injection$P_{R}(0)arrow\mu S(0)arrow 0$,

ノ$0arrow S(0)arrow\iota$ $P_{R}(0)$is so called an (relative) R-projective cover of $S(O)=k_{H}$ and l ノ is an (relative)
R-injective hull of $S(O)=k_{H}$ .

$\mu$

80
For any $kH$ -module , an R-projective cover (R-injective hull) of is obtained as asummand of the sequence obtained by tensoring with the above sequences. Let) be the kernel (resp. cokernel) of an R-projective cover (resp. R-injective(resp.
hull) of . In particular, we have the following short eaxact sequences,
$V$

$V$

$\Omega_{R}(V)$$\Omega_{R}^{-1}(V)$$V$$0arrow\Omega_{R}(S(0))arrow P_{R}(0)arrow\mu S(O)arrow 0$,$0arrow S(1)arrow\nu P_{R}(0)arrow\Omega_{R}^{-1}(S(O))arrow 0$

The heart $H_{R}(0)$ of $P_{R}(S(0))=P_{R}(k_{H})$ is defined by$H_{R}(0)=Ker\mu/{\rm Im} v$

1.3.1. Examples. By a result of Okuyama and Sasaki [7], we have$\Omega_{R}^{2}(S(0))\cong\Omega^{-2(p-1)}(S(1))$

Actually, we can show that an R-projective cover of$\Omega_{R}(S(0))$

has the form

$0arrow\Omega^{-2(p-1)}(S(1))arrow P(1)\oplus P_{R}(1)arrow\Omega_{R}(S(0))arrow 0$

so that we have the complex(1.2)$X_{0}^{\cdot}$$X_{\dot{0}}$of $kH$ -modules of the form

: . . . $arrow 0arrow\Omega^{-2(p-1)}(S(1))arrow P(1)\oplus P_{R}(1)arrow H_{R}(0)arrow 0arrow\cdots$

which satisfies that
$H_{1}(X_{0}^{\cdot})=S(0)$where $H_{R}(0)$ is in degree$0$,$H_{2}(X_{0}^{\cdot})=0=H_{0}(X_{0}^{\cdot})$term. Set
$F_{*}(1)=\Omega^{-1}\Omega_{R}(H_{R}(0))$

Then by the sequence (1.2), we have the complex$X^{\cdot}$of $kH$ -modules of the form

: . . . $arrow 0arrow\Omega^{-2(p-1)}(S(1))arrow P(1)\oplus P(1)arrow F_{*}(1)arrow 0arrow\cdots$
which satisfies that
$H_{2}(X^{\cdot})=0=H_{0}(X^{\cdot})$$H_{1}(X^{\cdot})=S(0)$ ,and$X^{\cdot}$(1.3)$F_{*}(1)\subset P(-1)\oplus P(1)$where$F_{*}(1)$is in degree

$0$

term.

Assume that $e=2$ . Do the Richard
times.

$s$tiltings with repect to the set {1} $p=2k+1$

Let
be the resulting algebra and let $S(i)_{0}$ (resp. $P(i)_{0}$ ) be a simple (resp. projective)-module corresponding to $S(i),$ $i\in I$ . Let $F^{0}:mod-kHarrow mod-A_{0}$ be the associatedin (1.1) and $X$ in (1.3), we have thefunctor. Then by the existence of complexes
following lemma.
$A_{0}$$A_{0}$$X_{\dot{k}}(1)$Lemma 1.4. Assume that $e=2$ . Then in the notatios above, we have$F^{0}(S(0))=S(0)_{0},$$i\neq 1$,$F^{0}(F_{*}(1))=S(1)_{0}$81
2. EXAMPLE $SL(2, q)$
The example here is one discussed by Holloway-Koshitani-Kunugi [4].
Letbe a prime power and be an odd prime such that divides $q_{1}+1$ . Write$q_{1}+1=p^{n-1}l’,$ $(p, l’)=1,$ $n\geqq 2$ . Set. Then $q+1=p^{n}\ell$ for some positive integer
with $(p, l)=1$ .
$p$$q_{1}$$p$$q=q_{1^{p}}$

$p$

Set
$G_{0}=SL(2, q),$ $C_{0}=SL(2, q_{1})$ .

$R=\mathcal{G}(GF(q)/GF(q_{1}))=\langle x\}$,

$G=R\ltimes G_{0}$Letbe a Borel subgroup ofwhere $|T_{0}|=(q-1)$ and $|U_{0}|=q$ . We havean R-invariant subgroupof order $q+1$ such thatis of order $q_{1}+1$
and $B_{0}\cap F_{0}=Z(G_{0})$ .
Letbe a Sylow p-subgroup ofand set $P=R\ltimes P_{0}$ . We have that $P\cong M_{n+1}(p)$ .
We use notations introduced in the beginning of the talk.
So $Q=P_{0}$ . Set $H=N_{G}(Q)=N_{G}(P_{0})$ . Then $H/O_{p’}(H)$ is our $H$ with $e=2$ . Set$H_{0}=N_{G_{0}}(P_{0})=H\cap G_{0}$ .
$B_{0}=T_{0}\ltimes U_{0}$$G_{0}$$F_{0}\supset Z(G_{0})$$P_{0}\subset F_{0}$$F_{0}\cap C_{0}$$G_{0}$

2.1.
. The principal block algebraof
has a cyclic defect group andis well understood. It is known thatand the principal block $B_{0}(kH_{0})$ are derived
equivalent. A two sided tilting complex for
and $B_{0}(kH_{0})$ due to Rouquier isgiven as follows. Set$B_{0}(kG_{0})$$B_{0}(kG_{0})$$kG_{0}$$B_{0}(kG_{0})$$B_{0}(kG_{0})$$A=B_{0}(kG_{0})$$B_{0}(kG_{0})$,$B=B_{0}(kH_{0})$and $B_{0}(kH_{0})$ have two simple modules$B_{0}(kG_{0})$:$\phi_{0}=k_{G_{O}}$,$\phi_{1}$,$\dim_{k}\phi_{1}=q-1$:,, $\dim_{k}T_{1}=1$$B=B_{0}(kH_{0})$ is a symmetric Nakayama algebra of length.
Let $P(\phi_{i})(i=0, 1)$ be a projective cover ofand $P(T_{i})(i=0, 1)$ be a projective
cover of .
is a $(A, B)$ -bimodule (a $(kG_{0},$-bimodule). As usual, we can regard as $k[G\cross H]-$
module. Let
be a Brou\’e-Puig indecomposable $(A, B)$ -summand of . As a$k[G\cross H]$ -module,
is a Scott module with vertex $\triangle P_{0}=\{(a, a) ; a\in P_{0}\}\subset G_{0}\cross H_{0}$ .
Actually, for the group $GL(2, q),$ $M_{0}=A$ . Notice also thatare both projective.
A functor$F:mod-Aarrow mod-B$ ,gives a stable equivalence of Morita type between $mod-A$ and $mod-B$ .
We can see that a-projective cover of $k=k_{G\cross H}$ has the form
$B_{0}(kH_{0})$$T_{0}=k_{H_{0}}$$T_{1}$$p^{n}$$\phi_{i}$$T_{i}$$A$$A$$kH_{0})$$A$$M_{0}$$M_{0}$$AM_{0},$$M_{B}$$V\mapsto V\otimes_{A}M_{0}$$\triangle P_{0}$$M_{0}arrow\pi karrow 0$

and Top $Ker\pi=\phi_{1}^{*}\otimes_{k}T(1)$ whereis a left-module. Let$P(T_{1})arrow\lambda Ker\piarrow 0$$Ker\pi$
be a projective cover ofand consider the following complexof $(A, B)$ -bimodules.
$\phi_{1}^{*}=Hom_{k}(\phi_{1}, k)$$kG_{0}$$M^{\cdot}$(2.1)$M^{\cdot}$: ...

$arrow 0arrow P(\phi_{1})^{*}\otimes_{k}P(T_{1})arrow\lambda M_{0}arrow 0arrow\cdots$$P(\phi_{1})^{*}\otimes_{k}$82
The complex$M^{\cdot}$

satisfies the following conditions.

$M^{\cdot}\otimes_{B}M^{*}\cong A[0]\oplus Z^{\cdot}$

in $C^{b}(mod-A^{\sigma p}\otimes A)$ andplex of projective $(A, A)$ -bimodules andbimodules.

,$C^{b}(mod-B^{\sigma\rho}\otimes B)$$W^{\cdot}$$M^{*}\otimes_{A}M^{\cdot}\cong B[0]\oplus W^{\cdot}$is a contractible comis a contractible complex of projective $(B, B)-$, respectively where$Z^{\cdot}$$T_{1}$$T_{0}$$F(\phi_{0})=T_{0}$,$F(\phi_{1})=\phi_{1}\otimes_{A}M_{0}=$of length $p^{n}-2$:$T_{0}$$T_{1}$

and$\phi_{0}\otimes_{A}M^{\cdot}$: . . . $arrow 0arrow 0arrow T_{0}arrow 0arrow\cdots$: . . . $arrow 0arrow P(T_{1})arrow\pi_{1}F(T_{1})arrow 0arrow\cdots$
where $P(T_{1})arrow\pi_{1}F(T_{1})arrow 0$ is a projective cover of $F(T_{1})$ .
is a complex obtained by the Richard
As a complex of projective B-modules,
tilting for the algebra with respect to the set $I_{0}=1\subset I=\{0,1\}$ . We have
: . . . $arrow 0arrow P(T_{1})arrow P(T_{0})arrow 0arrow\cdots$$\phi_{1}\otimes_{A}M^{\cdot}$$M_{\dot{B}}$

$s$

$B$

$P(\phi_{0})\otimes_{A}M^{\cdot}$$P(\phi_{1})\otimes_{A}M^{\cdot}$: . . . $arrow 0arrow P(T_{1})arrow 0arrow 0arrow\cdots$

and$M_{B}\cong P^{\cdot}(0)\oplus(q-1)P^{\cdot}(1)$is a -module and has a. $M_{0}^{x}=M_{0}$ so that
2.2. Let$(B_{0}(kG),
B_{0}(kH))$ -module.
indecomposable
Brou\’e-Puiga.
isvertexsuch thatwith vertex-moduleThere exists a p-permutationof. So it is natural to ask whether we can construct a complex
form
of
the
-modules: . . . $arrow 0arrow X_{0}arrow\mu M_{0}arrow 0arrow\cdots$
gives a twosided tilting. If such a complex exits, thensuch thatcomplex for $B_{0}(kG)$ and $B_{0}(kH)$ .
However, we can no have such a complex.
$\Gamma=(G_{0}\cross H_{0})\triangle R\subset G\cross H$$\triangle P$$k\Gamma$$M_{0}$$M=M_{0}\uparrow^{G\cross H}$$k\Gamma$$\triangle R$$X_{0}$$X_{0}\downarrow_{G\cross H}=$$X^{\cdot}$$P(\phi_{1})^{*}\otimes_{k}P(T_{1})$$k\Gamma$$X^{\cdot}$$X^{\cdot}\uparrow^{G\cross H}$$X^{\cdot}\downarrow_{G_{0}\cross H_{0}}\cong M$

2.3. Recall that $C_{0}=SL(2, q_{1})=C_{G_{0}}(R)$ and $N_{G}(R)=R\cross C_{0}$ . The principal block
algebra $B_{0}(kC_{0})$ has a cyclic defect group $Q_{0}=C_{Q}(R)$ and the stmcture of $B_{0}(kC_{0})$ is
described in the entiely same way as in $B_{0}(kG_{0})$ . $B_{0}(kC_{0})$ and $B_{0}(kN_{C_{0}}(Q_{0}))$ have two
simple modules$B_{0}(kC_{0}):\theta_{0}=k_{C_{0}}$ ,, $\dim_{k}\theta_{1}=q_{1}-1$$B_{0}(kN_{C_{0}}(Q_{0}))$,, $\dim_{k}T_{1}’=1$:$B_{0}(kC_{0})$ and $B_{0}(kN_{C_{0}}(Q_{0}))$ are derived equivalent. Let
be a Brou\’e-Puig indecomposcomplex
has
for them so thatbe the twosided tilting
able module for them and let
the form: . . . $arrow 0arrow P(\theta_{1})^{*}\otimes P(T_{1}’)arrow N_{0}arrow 0arrow\cdots$$\theta_{1}$$T_{0}’=k_{N_{C_{0}}(Q_{0})}$$T_{1}’$$N_{0}$$N_{\dot{0}}$$N_{0}^{\cdot}$$N_{\dot{0}}$83Using the isomorphism
twosided tilting complex$(C_{0}\cross N_{C_{0}}(Q_{0}))\triangle(R)/\triangle R=C_{0}\cross N_{C_{0}}(Q_{0})$$N^{\cdot}$(2.2)$N^{\cdot}$, we can lift

$N_{\dot{0}}$for $B_{0}(kN_{G}(R))$ and $B_{0}(N_{H}(R))$ .
: . . . $arrow 0arrow Yarrow Narrow 0arrow\cdots$

to a

is a Brou\’e-Puig indecomposable $(B_{0}(kN_{G}(R)), B_{0}(kN_{H}(R)))$-module. The Brauer
costructions for $M$ with respect tois $N$ . If we set, then $X(\triangle R)=Y$ .
And we can construct a complex of $(B_{0}(kG), B_{0}(kH))$ -bimodulesof the form
$N$

$X=X_{1}\uparrow^{G\cross H}$$\triangle R$$X^{\cdot}$(2.3): . . . $arrow 0arrow Xarrow\mu Marrow 0arrow\cdots$$X^{\cdot}$such that.

$X^{\cdot}(\triangle R)\cong N$$X^{\cdot}(\triangle R)$

satisfies the following conditions.

$X^{\cdot}\otimes_{B_{0}(kH)}X^{*}\cong B_{0}(kG)[0]\oplus Z^{\cdot}$,$x\cdot*\otimes_{B_{0}(kG)}X^{\cdot}\cong B_{0}(kH)[0]\oplus W^{\cdot}$

in $C^{b}(mod-B_{0}(kG)^{o\rho}\otimes B_{0}(kG))$ and $C^{b}(mod-B_{0}(kH)^{op}\otimes B_{0}(kH))$ , respectively whereis a complex of projective $(B_{0}(kG), B_{0}(kG))$ -bimodules andis a contractible complex of projective $(B_{0}(kH), B_{0}(kH))$ -bimodules. A way of construction of $X$ by
is a(verry special type of) gluing methods of Rouquier.
If we take a suitable projective $(B_{0}(kG), B_{0}(kH))$ -bimodule $X’$ and a map $X’arrow\nu M$
such that$X\oplus X’arrow M\mu\oplus\nuarrow 0$(exact)
Then the complex$Z^{\cdot}$$W^{\cdot}$$Y^{\cdot}$

(2.4)

$X’$

. ..
:

. $arrow 0arrow X\oplus X’arrow M\mu\oplus\nuarrow 0arrow\cdots$

has the same properties as forwhere the complexes
concentraited in degree . In particular, if we set$X^{\cdot}$$Z^{\cdot}$

and$W^{\cdot}$

have homologies$0$$M_{1}=\Omega^{-1}(Ker(\mu\oplus\nu))$, then A functor$F_{1}:mod-B_{0}(kG)arrow mod-B_{0}(kH)$ ,$V\mapsto V\otimes_{B_{0}(kG)}M_{1}$gives a stable equivalence of Morita type between $mod-B_{0}(kG)$ and $mod-B_{0}(kH)$ . We
have the following lemma.

Lemma 2.1.
$F_{1}(\varphi_{0})=S(0)$,$F_{1}(\varphi_{1})=\Omega^{-1}\Omega_{R}(S(0))$

Thus by Lemma 1.4, the following result follows.
Corollary 2.2 (Holloway-Koshitani-Kunugi [4]).
$B_{0}(kG)$ and $B_{0}(kH)$ are derived equivalent.
The procedure of Richard tilting in the previous section implies that the resulting
twosided tilting complex has the following form$s$$arrow 0arrow X_{p}arrow X_{p-1}arrow\cdotsarrow X_{2}arrow X_{1}\oplus X_{1}’arrow Marrow 0arrow\cdots$

The results in this section are obtained through the discussions with Koshitani andKunugi.

84
3. RELATIVE PROJECTIVE COVERS AND BRAUER CONSTRUCTION
be a finite group and ec be a nonempty
3.1. Relative Projective Coveres. Let$M$$kG$-module , a short exact sequence $M$ ; $0arrow Narrow$family of subgroups of . For a$Xarrow Marrow 0$ of $kG$ -module is called-projective cover of $M$ if it satisfies(1) $X$ is X-projective,(2) ‘the sequence $M$ is X-split.
For a $kG$-module $M$ , a minimal X- projective cover of $M$ exists and is uniquely determined
up to isomorphism of exact sequences. An arbitraly -projective cover contains a minimal
one as a summand of exact sequences. If the above sequence $M$ is minimal, then we denote$N$ by. $M$ is X-projective if and only if $\Omega_{X}(M)=0$ . For $kG$ -modules $M$ and $M’$ ,$\Omega_{X}(M\oplus M’)=\Omega_{X}(M)\oplus\Omega_{X}(M)$ .
Let $H$ be a subgroup of and set $\mathfrak{Y}=X^{G}\cap H=\{A^{g}\cap H ; g\in G, A\in ac\}$ . Then
; $0arrow N\iota_{H}arrow X\iota_{H}arrow M\iota_{H}arrow 0$ is athe short exact sequence of $kH$ -module$kH$, not necessarily minimal even if $M$ is-module-projective presentation of aminimal.
$G$

$G$

$X$

$X$

$\Omega_{X}(M)$$G$$M\downarrow H$$M\downarrow H$$\mathfrak{Y}$3.2. Brauer Construction.
be a p-subgroup of . Then a functor called the
be a finite group and3.3. Let
Brauer construction with respect to ;$-(Q)$ : mod $(kG)arrow mod(kN_{G}(Q)/Q)$
$G$

$Q$

$G$

$Q$

is defined by$M(Q)=M^{Q}/( \sum_{R\subsetneq Q}Tr_{R,Q}(M^{R}))$and is called the Brauer
The canonical epimorphism from $M^{Q}arrow M(Q)$ is denoted byhomomorphism with respect to $Q$ .
: $Marrow N$ is a kG-homomorphism, $f$ induces a
If $M$ and $N$ are $kG$ -modules and$kN_{G}(Q)/Q$-homomorphism $f(Q)$ : $M(Q)arrow N(Q)$ . We denote $f(Q)$ by $Br_{Q}(f)$ . The
Green correspondence with respect to $(G, N_{G}(Q), Q)$ gives a bijection between the setof isomorphism classes of indecomposable p-permutation $kG$-modules with vertex $Q$ andthe set of isomorphism classes of indecomposable projective $kN_{G}(Q)/Q$-modules. If $X$
is an indecomposable p-permutation $kG$-modules with vertex , then the corresponding
indecomposable projective $kN_{G}(Q)/Q$ -module is the Brauer construction $X(Q)$ .
$Br_{Q}$$f$$Q$

is a permutation $kQ$ -module. If $M(Q)$ has a projective
Lemma 3.1. Assume that$kN_{G}(Q)/Q$ -summandwith vertex such that, then $M$ has a Q-projective summand$V(Q)=U$ .
$M\downarrow Q$

$V$

$U$

Proof. $kN_{G}(Q)$ -module$N_{G}(Q)$$M\downarrow N_{G}(Q)$ofand a p-subgroupis normal in . Let

assume that$Q$

satisfies the assumption in the lemma for the group. Thus by a theorem of Burry-Carlson, we may$N_{G}(Q)$

$Q$

$G$

$Q$

$Xarrow fMarrow 0$$0arrow Marrow gY$

is a perbe a Q-projective cover and a Q-injective hull of $M$ , respectively. As
mutation module, $X$ and are Q-projective, $l\succ permutationkG$ -modules. In particular,$M\downarrow Q$

$Y$

85
and $Y(Q)$ are projective $kG/Q$ -modules. As the sequence above are Q-split, wehave exact sequences,$X(Q)$$X(Q)arrow M(Q)f(Q)arrow 0$ ,$0arrow M(Q)arrow g(Q)Y(Q)$

There exists a primitive idempotent $e\in kG$ such that $e[Q]kG\cong U$ . Thus there exists anelement $m\in M^{Q}$ such that $me=m$ and $\overline{m}kG=U$ whereis the image of$m\in M^{Q}$ in $M(Q)$ . We can take an element $x\in X^{Q}$ such that $f(x)=m$
and $xe=x$ .
$X=X_{0}\oplus
X_{1}$$kG/Q$
Writewhere
is a projective-module and each indecomposablesummand of
has a vertex properly contained in . And write $x=x_{0}+x_{1}$ with.
Then $x_{0}e=x,$ $x_{1}e=x_{1}$ and. Thusand.
As
is a $kG/Q$ -module and $x_{0}e=e,$ $x_{0}kG$ is a homomorphic image of $[Q]ekG$ and we
can conclude that $x_{0}kG\cong[Q]ekG\cong U$ . Set $V=x_{0}kG$ . Then is a direct summandof(and of $X$ ). Thus we have proved that we have a direct sum decomposition of$kG$ -modules$\overline{m}\in M(Q)$$X_{0}$$Q$$X_{1}$$x_{i}\in X_{i}$$\overline{f(x_{0})}kG\cong U$$\overline{m}=\overline{f(x_{0})}$$x_{1} \in\sum_{R\subseteq Q}Tr_{R,Q}(X_{1}^{R})$$X_{0}$$V$$X_{0}$$X=V\oplus V’$such that $V\cong U,$ $f(Q)(V(Q))=U\subset M(Q)$ and $f(Q)\downarrow_{V(Q)}:V(Q)arrow M(Q)$ induces
isomorphisms$f(Q)\downarrow_{V(Q)}:V(Q)arrow U$

Write $Y=Y_{0}\oplus Y_{1}$ whereis a projective $kG/Q$ -module and each indecomposablesummand of
has a vertex properly contained in . And write $g(m)=y_{0}+y_{1}$ with. Then $y_{0}e=y_{0}$ and. By the similar argument as above, it$Y_{0}$$Q$$Y_{1}$$y_{i}\in Y_{i}$$g(Q)(\overline{m})=\overline{y_{0}}\in Y_{0}(Q)$follows that $y_{0}kG\cong[Q]ekG\cong U$ and we have a direct sum decomposition of $kG$ -modules$Y=W\oplus W’$

such that $W\cong U,$ $g(Q)(U)=W(Q)\subset Y(Q)$ .
Let : $Varrow X,$ : $Yarrow W$ be the injection and projection with respect to the above
decompositions and consider the maps $f’=fo\lambda$ : $Varrow M$ and $g’=\mu og:Marrow W$ . Then$f’(Q)=f(Q)0\lambda(Q)$ and $g’(Q)=\mu(Q)og(Q)$ . By the discussions above, the composite$g’(Q)of’(Q)$ : $V(Q)arrow M(Q)arrow W(Q)$ is an isomorphism. As $(g’of’)(Q)=g’(Q)of’(Q)$ ,it follows that the map $g’of$ : $Varrow Marrow W$ is an isomorphism and that is isomorphic
to a summand of M.
$\lambda$

$\mu$

$V$

$\square$Lemma 3.2. Assume thatis a permutation $kQ$ -module and let $0arrow Narrow Xarrow$$Marrow 0$ be a Q-projective coverof M. Then $0arrow N(Q)arrow X(Q)arrow M(Q)arrow 0$ is aminimal projective cover of a $kN_{G}(Q)/Q$ -module $M(Q)$ .
$M\downarrow Q$

Proof. As the sequence $0arrow Narrow Xarrow Marrow 0$ is Q-split, the resulting sequence$0arrow N(Q)arrow X(Q)arrow M(Q)arrow 0$ is exact and a projective presentation of a $kN_{G}(Q)/Q-$

module $M(Q)$ . We also have thatis a permutation module. By Lemma 3.1, $N(Q)$$[]$has no projective $kN_{G}(Q)/Q$ -summand and the lemma follows.
$N\downarrow Q$

The results in this section are obtained by joint works with Kunugi.

86
4. FINITE GROUPS WITH SYLOW p–SUBGROUP $M_{n+1}(p)$
Let
order

$p$

be an odd prime and $n\geqq 2$ be an integer. Consider the $r$ group $M_{n+1}(p)=P$ ofgiven in Section 1. We use notations in Section 1. And set$p^{n+1}$$P_{0}=\langle y\}$$R=\langle x\}$,,$z;=y^{p^{n-1}}=[x, y]$$Z(P)=\langle y^{p}\}$,,$Z;=\langle z\rangle$$Q;=C_{P}(R)=R\cross Z(P)$For an integer ,$i$$(y^{i}x)^{p}=y^{ip}$Thus$\Omega_{p}(P)=\langle x,$$z\rangle$,$0\leqq i\leqq p-1$

$x\sim Pxz^{i},$and it follows that a nontrivial subgroup

$S$

of

$P$

contains,$Z$$Z(P)=\{y^{p})$or is conjugate to

$R$

in

$P$

.

Let
be a finite group with Sylow subgroup $P=M_{n+1}(p)$ such that there exists asatisfying that
normal subgroup$G$$G_{0}$$G=R\ltimes G_{0}$,$G_{0}\cap P=P_{0}$Set$H=N_{G}(P_{0})=R\ltimes N_{G_{O}}(P_{0})$,$N_{G_{0}}(P_{0})=H_{0}$Then $N_{G}(P)\subset N_{G}(P_{0})=H$ and $H/O_{p’}(C_{G}(P))$ isomorphic to a subgroup of$\{t_{0}\}\ltimes P$.

Set$e=|H/PC_{G}(P)|$ ,$H=\{t, PC_{G}(P)\}$so that $t^{e}\in O_{p’}(C_{G}(P))$ and $H/O_{p’}(H)$ is the group $H$ in Section 1.
In this section, we shall be concerned with the principal block algebras $B_{0}(kG)$ and$B_{0}(kH)$ of $kG$ and $kH$ .
Notice that $G$ and $H$ have the same p-local structure.

4.1. p-Locals. Let $M$ be a Brou\’e-Puig indecomposable $k[G\cross H]$ -direct summand of. As we are working on the principal block case, $M$ is aof $B_{0}(kG)$ with vertex
for. We investigate Brauer constructions
Scott $k[G\cross H]$ -module with vertex$P$
nontrivial subgroups of .
$\triangle P$$M(\triangle S)$$\triangle P$$S$

4.1.1. . By a theorem of Burnside,
particular,$Z$$C_{G_{0}}(Z)$is p-nilpotent and therefore so is $C_{G}(Z)$ . In

$N_{G}(Z)=O_{p’}(C_{G}(Z))N_{H}(Z)$and $M(\triangle Z)=B_{0}(kC_{G}(Z))=B_{0}(kC_{H}(Z))$ .

4.1.2.
$S\cap G_{0}$. Let $S\subset P$ withis cyclic.

$S\supset Z$$S\supset Z$. Then $N_{G}(S)\subset N_{G}(Z)$ because $Z\subset S\cap G_{0}$ andThus $M(\triangle S)=B_{0}(kC_{G}(S))=B_{0}(kC_{H}(S))$ .

87
4.1.3.

$R$. We can see that
$N_{G}(R)=R\ltimes N_{G_{0}}(R)=R\cross C_{G_{0}}(R)=C_{G}(R)$Set $C=C_{G}(R)$ and $C_{0}=C_{G_{0}}(R)$ . Then $Q=R\cross Z(P)$ is a Sylow p-subgroup of$Z(P)$ is a Sylow p-subgroup of. We also have that

$C$

and

$C_{0}$$N_{H}(R)=C_{H}(R)=R\cross C_{H_{0}}(R)$Set $K_{0}=C_{H_{0}}(R)$ . Then $N_{C_{0}}(Z)=K_{0}O_{p’}(C_{C_{0}}(Z))$ by the following facts.
$C_{H_{0}}(R)\subset N_{C_{0}}(Z(P))$,$N_{C_{0}}(Z(P))=C_{H_{0}}(R)O_{p’}(N_{C_{0}}(Z(P)))$$N_{C_{0}}(Z(P))\subset N_{C_{0}}(Z)=\{t, C_{C_{0}}(Z)\}$,$N_{C_{0}}(Z)=N_{C_{0}}(Z(P))O_{p’}(C_{C_{0}}(Z))$

As $(kC_{0}, kN_{C_{0}}(Z))$ -module, $B_{0}(kC_{0})=N’\oplus$ proj. where $N’$ is a Brou\’e-Puig module for$B_{0}(kC_{0})$ and $B_{0}(kN_{C_{0}}(Z))$ . Thus by the result above, as $(kC_{0}, kK_{0})$ -module,$B_{0}(kC_{0})=N_{0}\oplus proj$.

(Actually, in the situation here,
is indecomposable and has a vertex$N_{0}=N’)$ .
gives a stable equivalence between $B_{0}(kC_{0})$ and $B_{0}(kK_{0})$ . By a result ofRouquier, there exists a two terms Rickard complex
for $B_{0}(kC_{0})$ and $B_{0}(kK_{0})$ of thefollowing form,

where$\triangle Z(P)$$N_{0}$$N_{0}$$Y_{0}$$Y_{O};\cdotsarrow 0arrow Y_{0}arrow N_{0}\nu_{0}arrow 0arrow\cdots$

where
then

is a projective-module. If
can be taken from a direct summand of$Q_{0}arrow N_{0}\nu_{0}’arrow 0$$k[C_{0}\cross K_{0}]$$Y_{0}$$Y_{0}$$B_{0}(kC)=B_{0}(kC_{G}(R))=kR\otimes_{k}B_{0}(kC_{0})$ ,$Q_{0}$

andis a projective cover of. We know that

$N_{0}$,$n_{0}=t\text{ノ_{}0}’\downarrow Y_{0}$$B_{0}(kC_{H}(R))=kR\otimes_{k}B_{0}(kK_{0})$Thus as $k[C\cross C_{H}(R)]$ -module,$B_{0}(kC)=N\oplus\triangle R$-proj.

where $N$ is a Brou\’e-Puig module for $B_{0}(kC)$ and $B_{0}(kC_{H}(R))$ . As $N_{G\cross H}(\triangle R)=C_{G}(R)\cross$$C_{H}(R)$ , we have $M(\triangle R)=N$ .
The complex
can be lifted to a Rickard complex for $B_{0}(kC_{G}(R))$ and $B_{0}(kC_{H}(R))$, the inflated
as follows. By the canonical epimorphism
complexof-modules ofcan be constructed.
$Y_{0}$$\triangle R(C_{0}\cross K_{0})/\triangle R\cong C_{0}\cross K_{0}$$\overline{Y_{0}}$$k[\triangle R(C_{0}\cross K_{0})]$$\overline{Y_{0}}$; ...

$Y_{O}$$arrow 0arrow\overline{Y_{0}}arrow\overline{N_{0}}\overline{\nu_{0}}arrow 0arrow\cdots$

Then the induced complex$B_{0}(C_{G}(R))$ and $B_{0}(C_{H}(R))$ . The degree
to $N=M(\triangle R)$ . Thus $Y$ has the form

is the desired Rickard complex forand is isomorphic
term of $Y$ is$Y=\overline{Y_{0}}\uparrow^{C_{G}(R)\cross C_{H}(R)}$

$Y$

where
Let
projective$Y=\tilde{Y_{0}}\uparrow^{C_{G}(R)\cross C_{H}(R)}$$Q_{0}arrow\nu_{0}’N_{0}arrow 0$$0$$N_{0}\uparrow^{C_{G}(R)\cross C_{H}(R)}$; . . . $arrow 0arrow Yarrow\nu M(\triangle R)arrow 0arrow\cdots$.

be a projective cover ofas before so that $Q_{0}=Y_{0}\oplus Z_{0}$ for some
. Set
-module and$k[C_{0}\cross K_{0}]$$Q=\overline{Q_{0}}\uparrow^{C_{G}(R)\cross C_{H}(R)},$$N_{0}$$\nu_{0}=\nu_{0}’\downarrow Y_{0}$$Y=\overline{Y_{0}}\uparrow^{C_{G}(R)\cross C_{H}(R)},$$Z=\overline{Z_{0}}\uparrow^{C_{G}(R)\cross C_{H}(R)}$88
Then the resulting sequence $Qarrow N\nu’arrow 0$ is a$Y\oplus Z$

and$v=v’\downarrow Y$-projective cover of $N=M(\triangle R),$ $Q=$$\triangle R$

.

. Let
By our construction, each indecomposable summand of has a vertex$M$$X’(\Delta R)arrow$$Marrow 0$ be a.
Then
Brauerconstmctionofits-projective cover$M(\triangle R)arrow 0$ is a
. Thus we have a-module-projective cover ofdecomposition $X’=X\oplus W$ of $k[G\cross H]$ -modules such that each indecomposable summandand setof $X$ has a vertex
and $X(\Delta R)=Y$ . Now set$Y$$\triangle R$$X’arrow\mu’$$\triangle R$$M(\triangle R)$$kN_{G\cross H}(\triangle R)$$\triangle R$$\triangle R$$\mu=\mu’\downarrow x$

X $;\cdotsarrow 0arrow Xarrow\mu Marrow 0arrow\cdots$
Then by our construction we have $X(\triangle R)=$ Y. And for $S\subset P$ with $S\supset Z$ , we have$X(\triangle S)=M(\triangle S)$ .
Now a result of Rouquier says the following fact.

Lemma 4.1. The complex X induces a stable equivalence of Rickard type between $B_{0}(kG)$and $B_{0}(kH)$ .

4.1.4. Stable Equivalence. Let $W$ be the $(B_{0}(kG), B_{0}(kH))$ -bimodule given in the previous subsections. And let $Parrow\lambda’Warrow 0$ be a projective cover of $W$ so that we have anexact sequence of $(B_{0}(kG), B_{0}(kH))$ -bimodule$X\oplus Parrow\lambda Marrow 0$(exact). Set $M_{0}=\Omega^{-1}(Ker\lambda)$ so that we have an exact
sequence of $(B_{0}(kG), B_{0}(kH))$ -modules of the form
where$\lambda=$$(\mu$, vo$\lambda’)$with$\nu=\mu’\downarrow W$$0arrow Xarrow M\oplus P_{0}arrow M_{0}arrow 0$

where$P_{0}$

is a projective $(B_{0}(kG), B_{0}(kH))$ -bimodule.
$0arrow Ker\lambdaarrow X\oplus Parrow^{\lambda}Marrow 0$$0arrow Ker\lambda$$fo\downarrow$$f_{1}\downarrow$$\Vert$—$P_{0}\oplus Parrow M_{0}arrow 0$

Lemma 4.2. The $functor-\otimes_{B_{0}(kG)}M_{0}$ : mod $-B_{0}(kG)arrow$ mod $-B_{0}(kH)$ gives a stableequivalence of Morita type between $B_{0}(kG)$ and $B_{0}(kH)$ .
Set $A=B_{0}(kG)$ and $B=B_{0}(kH)$ .
For a nonprojective indecomposable A-module , Let $F(V)$ be a nonprojective Bsummand of
so that $F(V)$ is indecomposable and $V\otimes_{A}M_{0}=F(V)\oplus$ proj.
Assume thatis a permutation $kR$-module and is not R-projective. Notice that a-module $V(R)$ has no projective summand by Lemma 3.1. Assume, furthermore that
$V(R)$ is simple. Then a B-module $F(V)$ is obtained by the following way.
$V$

$V\otimes_{A}M_{0}$$V\downarrow R$$kC_{0}$

Set $A_{0}=B_{0}(kC_{G}(R)/R)=B_{0}(kC_{0})$ and $B_{0}=B_{0}(kK_{0})$ . Then by a result of PuigRickard [11],$(V\otimes_{A}M)(R)\cong V(R)\otimes_{A_{0}}M(\triangle R)=V(R)\otimes_{A_{0}}N_{0}$and$(V\otimes_{A}X)(R)\cong V(R)\otimes_{A_{0}}X(\triangle R)=V(R)\otimes_{A_{0}}Y_{0}$89as$B_{0}(kK_{0})$-modules. Thus$(V\otimes_{A}X)(R)\cong V(R)\otimes_{A_{0}}Y_{0}$By the discussion on p-locals, we can write $V\otimes_{A}M=F’(V)\oplus U’$ where $F’(V)$ isindecomposable and $U’$ is R-projective. Then$F’(V)(R)\oplus U’(R)=(V\otimes_{A}M)(R)\cong V(R)\otimes_{A_{0}}N_{0}$As we are assuming that $V(R)$ is simple,
is indecomposabe. In particular,$U’(R)=0$ and $U’$ is projective. If we set $U=V(R)\otimes_{A_{0}}N_{0}$ , then by a result of Rouquier[13, 14], one of the following occurs.
$Uarrow 0arrow\cdots$$(*.1)$: ...
$V(R)\otimes_{A_{0}}N_{0}$$arrow 0arrow$$V(R)\otimes_{A_{0}}Y_{O}$$V(R)\otimes_{A_{0}}Y_{0}$:. . . $arrow 0arrow$$0arrow$$Q(U)arrow\rho$$(*.2)$$Uarrow 0arrow\cdots$

where $Q(U)arrow\rho Uarrow 0$ is a projective cover of a $B_{0}(kK_{0})$ -module
We have proved the following lemma.

$U$.

Lemma 4.3. Let be an indecomposable $B_{0}(kG)$ -module such thatis a permutation$kR$ -module and is not R-projective.
Assume, furthermore that $V(R)$ is simple. Then
$F(V)=V\otimes_{B_{0}(kG)}M$ or $F(V)=\Omega^{-1}\Omega_{R}(V\otimes_{B_{0}(kG)}M)$ according to the case $(*.1)$ occurs
or the case $(*.2)$ occurs.
$V$$V\downarrow R$

Corollary 4.4. Let
be a $B_{0}(kG)$ -module satisfying the conditions in the Lemma and$Hom_{k}(V,k)\otimes_{k}V=k_{G}\oplus
V’$
assume thatfor some R-projective $kG$ -module $V’$ .
$V$

Then

$Hom_{k}(F(V), k)\otimes_{k}F(V)=k_{H}\oplus V_{0}$is an endo-trivial $kP$ -module.
for some projective $kH$ -module . In particular,
Proof. By our construction of the functor , we can write$F(V)\downarrow P$$V_{0}$$F$$Hom_{k}(F(V), k)\otimes_{k}F(V)=k_{H}\oplus V_{0}$is an R-projective p-permutation $kH$-module. Thus it suffices to show that
$V_{0}(R)=0$ . We use the notations in the discussion before the lemma.
By a result of Puig-Richard,where$V_{0}$$k_{C_{0}}\oplus V_{0}(R)=(Hom_{k}(V, k)\otimes_{k}V)(R)\cong Hom_{k}(V(R), k)\otimes_{k}V(R)$as$kC_{0}$$(*)$-modules. Then for $U=V(R)\otimes_{A_{0}}N_{0}$ ,$k_{C_{0}}\oplus U_{0}\cong Hom_{k}(U, k)\otimes_{k}U$as$kC_{0}$-modules where$U_{0}$

is a projective$kK_{0}$-module. As $kC_{H}(R)$ -modules, we also have$k_{C_{H}(R)}\oplus U_{1}\cong Hom_{k}(\Omega_{R}(U), k)\otimes_{k}\Omega_{R}(U)$whereis an R-projective $kC_{H}(R)$ -module. By $(*)$ , we can see that a source of $B_{0}(kC_{0})-$$V(R)$
moduleis. By properties of Roquier‘s complexor, the case $(*.1)$
occurs if a source of $V(R)$ is
and the case $(*.2)$ occurs if a source of $V(R)$ is.
$U_{1}$$k_{Z(P)}$$\Omega(k_{Z(P)})$$k_{Z(P)}$$Y_{O}$$\Omega(k_{Z(P)})$

90
If the case $(*.1)$ occurs, then is a simple $B_{0}(kC_{H}(R))$ -module. If the case $(*.2)$ occurs,
as $kC_{H}(R)$ -module is a simple $B_{0}(kC_{H}(R))-$ofthen an R-projective cover$B_{0}(kC_{H}(R))$-modules are one dimensional. Thus $U_{0}=0$ in
module. Notice that simple$[]$
the case $(*.1)$ and $U_{1}=0$ in the case $(*.2)$ .
$U$

$U$

$\Omega_{R}(U)$5. EXAMPLES
with Sylow p-subgroup $M_{n+1}(p)$ where we
We shall give some examples of groups$B_{0}(kG)$-modules satisfy the assumption in Lemma 4.3.
could check that simple
isomorphic to $SL(2, q),$ $SU(3, q^{2})$ and $Sp(4, q)$
Our groups are constructed fromfor suitably chosen prime power such that $p|q+1,$ $p|q^{2}-q+1$ and $p|q^{2}+1$ ,have cyclic Sylow p-subgroups and the Brauer trees ofrespectively. These groups$B_{0}(kG_{0})$ are the following shapes. In the figures,is an ordinary irreducible charactersof degree . See the paper by Fong and Srinivasan [3].
$G$

$V$

$G$

$G_{0}$$q$$G_{0}$$\chi_{k}$$k$$SL(2, q),$ $p|q+1$$SU(3, q^{2}),$$p|q^{2}-q+1$$(s=q+1)$$Sp(4, q),$ $p|q^{2}+1$$(s= \frac{1}{2}(q+1)^{2}, t=\frac{1}{2}(q-1)^{2}, u=(q^{2}-1))$

5.1. $SL(2, q)$ . Let be a prime power and be an odd prime such that divides $r+1$ .
Write $r+1=p^{n-1}l’,$ $(p, l’)=1,$ $n\geqq 2$ . Set $q=r^{p}$ . Then $q+1=p^{n}l$ for some positive
integer with $(p, P)=1$ .
Set
$r$

$p$

$p$

$p$

$G_{0}=SL(2, q),$ $C_{0}=SL(2, r)$ .

$R=\mathcal{G}(GF(q)/GF(r))=\langle x\rangle$,$G=R\ltimes G_{0}$Let
where $|T_{0}|=q-1$ and $|U_{0}|=q$ . We have an
be a Borel subgroup of$q+1$
of orderR-invariant subgroupsuch thatis of order $r+1$ and$B_{0}=T_{0}\ltimes U_{0}$$G_{0}$$F_{0}\supset Z(G_{0})$$B_{0}\cap F_{0}=Z(G_{0})$$F_{0}\cap C_{0}$.

and set $P=R\ltimes P_{0}$ . We have that $P\cong M_{n+1}(p)$ .
be a Sylow p-subgroup of$B_{0}(kG_{0})$ and $B_{0}(kC_{0})$ have two simple modules

Let$P_{0}\subset F_{0}$$B_{0}(kG_{0})$:$\phi_{0}=k_{G_{0}}$$G_{0}$,$\phi_{1}$,$\dim_{k}\phi_{1}=q-1$,$B_{0}(kC_{0})$:$\theta_{0}=k_{C_{0}}$,$\theta_{1}$,$\dim_{k}\theta_{1}=r-1$

91
A simple moduleis the heart of a projective cover $P(\phi_{0})=P(k_{G_{0}})$ .

$\phi_{1}$$\phi_{0}$$P(\phi_{0})=k_{B_{0}}\uparrow^{G_{0}}$and is uniserial of the form

$P(\phi_{0})=\phi_{1}$$\phi_{0}$The entirely same thing occurs for a projective coverof .
.
Set $B=R\ltimes B_{0}$ andis an extension of $P(k_{G})$ and therefore
is uniserial of length 3 with form
$Q(\theta_{0})$$P_{R}(k_{G})=k_{B}\uparrow^{G}$$\theta_{0}$$P_{R}(k_{G})$$\varphi_{0}$$P_{R}(k_{G})=\varphi_{1}$$\varphi_{0}$

where
and
$kR$-module and$\varphi_{0}=k_{G}$$\varphi_{1}1_{G_{0}}=\phi_{1}$. It is not hard to see that

$P_{R}(k_{G})(R)=Q(k_{C_{0}})$,$\varphi_{1}\downarrow R$

is a permutation$\varphi_{1}(R)=\theta_{1}$5.2. $SU(3, q^{2})$ . Let be a prime power and be a prime with $p\geqq 5$ such that divides
Write $r^{2}-r+1=p^{n-1}\ell’,$ $(p, \ell’)=1,$ $n\geqq 2$ . Set $q=r^{p}$ . Then $q^{2}-q+1=p^{n}\ell$
for some positive integer with $(p, l)=1$ .
$r$

$p$

$p$

$r^{2}-r+1$ .

$p$

Set

$G_{0}=SU(3, q^{2}),$ $C_{0}=SU(3, r^{2})$.

$R=\mathcal{G}(GF(q)/GF(r))=\langle x\}$,

$G=R\ltimes G_{0}$Let $B_{0}=T_{0}\ltimes U_{0}$ be a Borel subgroup ofwhere $|T_{0}|=(q+1)(q-1)$ and $|U_{0}|=q^{3}$ .
We have an R-invariant subgroupof order $q^{2}-q+1$ such thatis oforder $r^{2}-r+1$ and $B_{0}\cap F_{0}=Z(G_{0})$ .
Letbe a Sylow p-subgroup ofand set $P=R\ltimes P_{0}$ . We have that $P\cong M_{n+1}(p)$ .
$B_{0}(kG_{0})$ and $B_{0}(kC_{0})$ have three simple modules$G_{0}$$F_{0}\supset Z(G_{0})$$P_{0}\subset F_{0}$$G_{0}$:$B_{0}(kG_{0})$$B_{0}(kC_{0})$Simple modules

5.2.1.

$\phi_{1}$$F_{0}\cap C_{0}$:$\phi_{1}$$\phi_{0}=k_{G_{0}}$,$\theta_{0}=k_{C_{0}}$,and$\phi_{2}$$\phi_{1}$,$\dim_{k}\phi_{1}=q^{3}-1$$\theta_{1}$,$\dim_{k}\theta_{1}=r^{3}-1$,$\phi_{2}$,$\theta_{1}$,,$\dim_{k}\phi_{2}=q(q-1)$$\dim_{k}\theta_{1}=r(r-1)$

are described as follows.

. A simple module$\phi_{1}$is the heart of a projective cover $P(\phi_{0})=P(k_{G_{0}})$ .
$\phi_{0}$$P(\phi_{0})=k_{B_{0}}\uparrow^{G_{0}}$and is uniserial of the form

$P(\phi_{0})=\phi_{1}$$\phi_{0}$The same thing occurs for a projective cover
Set $B=R\ltimes B_{0}$ and.
is uniserial of length 3 with form
$P_{R}(k_{G})=k_{B}\uparrow^{G}$.
is an extension of $P(k_{G})$ and therefore$Q(\theta_{0})$$P_{R}(k_{G})$of$\theta_{0}$$\varphi_{0}$$P_{R}(k_{G})=\varphi_{1}$$\varphi_{0}$

where
and
$kR$-module and$\varphi_{0}=k_{G}$$\varphi_{1}\iota_{G_{0}}=\phi_{1}$. It is not hard to see that

$P_{R}(k_{G})(R)=Q(k_{C_{0}})$,$\varphi_{1}(R)=\theta_{1}$$\varphi_{1}\downarrow R$

is a permutation

92
and ,
5.2.2. . Set $B=R\ltimes B_{0}$ . By the knowledge of the character tables ofof dimension $q(q-1)$ and the-modulewe can see that there exists a simple$G_{0}$$\phi_{2}$$B_{0}(kG_{0})$$B_{0}$$\phi_{2}$does not-module which is R-invariant.
is a simple
restriction
$kB$
group
has a cyclic defectandis. The block ofwhich covers
containa p’-group. Thus Alperin-Brauer-Dade-Glauberman theory can be applied. Notice thatis a Borel subgroup of $C_{0}=SU(3, r^{2})$ .
is a p-group. The Brauer-Glauberman
toas $|B$ :has a unique extension$Z(C_{U_{0}}(R))$
in its kernel. Thus $\dim_{k}\theta’=r(r-1)$ .
correspondent of does not contain
We see that $\dim_{k}\phi’-\dim_{k}\theta’=q(q-1)-r(r-1)\equiv 0(mod p)$ . Thus the extensionof has a trivial source module and$kB_{0}$$\phi’=\phi_{2}\downarrow B_{0}$$\phi’\downarrow Z(U_{0})$$R$$\phi’$$k_{Z(U_{0})}$$B_{0}$$C_{B_{0}}(R)$$\phi’$$\varphi’$$B$$\varphi’$$\theta’$$B_{0}|$$\phi’$$\phi’$$\varphi’(R)=\theta’$also has a unique extension tois a-module andand$\phi_{2}$$\varphi_{2}$$\varphi_{2}(R)$$B_{0}(kC_{0})$$G$. Then$\varphi_{2}\downarrow B$$\varphi_{2}(R)\iota_{B\cap C_{0}}=\theta’$. Thus.
-module must be simple

is an extension of. Such a$B_{0}(kC_{0})$$\phi’$$\varphi_{2}1_{B}=\varphi’$$\varphi_{2}(R)=\theta_{2}$The simple $B_{0}(kG_{0})$ -moduleThus$\phi_{2}$is self-dual and we can see that$\varphi_{2}^{*}\otimes\varphi_{2}=k_{G}\oplus$$\phi_{2}\otimes\phi_{2}=k_{G_{0}}\oplus$

defect blocks.
$0$

R-projective

5.3. $Sp(4, q)$ . Let be a prime power and be a prime with $p\geqq 5$ such that divides. Write $r^{2}+1=p^{n-1}l’,$ $(p, l’)=1,$ $n\geqq 2$ . Set $q=r^{p}$ . Then $q^{2}+1=p^{n}l$ for somepositive integer with $(p, l)=1$ .
Set
$r$

$p$

$p$

$r^{2}+1$$\ell$$G_{0}=Sp(4, q),$ $C_{0}=Sp(4, r)$.

$R=\mathcal{G}(GF(q)/GF(r))=\langle x\rangle$,$G=R\ltimes G_{0}$Let $B_{0}=T_{0}\ltimes U_{0}$ be a Borel subgroup ofwhere $|T_{0}|=(q-1)^{2}$ and $|U_{0}|=q^{4}$ . Let$W=N_{G}(T)/T=\{w_{a},$be the Weyl group of whereis a reflection corresponding
to a long root. We have an R-invariant subgroupof order $q^{2}+1$ such thatisof order $r^{2}+1$ and $B_{0}\cap F_{0}=Z(G_{0})$ .
Letbe a Sylow p-subgroup ofand set $P=R\ltimes P_{0}$ . We have that $P\cong M_{n+1}(p)$ .
$B_{0}(kG_{0})$ and $B_{0}(kC_{0})$ have four simple modules$G_{0}$$G$$w_{b}\rangle$$w_{a}$$F_{0}\cap C_{0}$$F_{0}$$P_{0}\subset F_{0}$$B_{0}(kG_{0})$$G_{0}$:$\phi_{0}=k_{G_{0}}$$\phi_{2},$$B_{0}(kC_{0})$:Simple modules$\phi_{1},$$\phi_{1},$$\dim_{k}\phi_{1}=\frac{1}{2}q(q+1)^{2}-1$$\dim_{k}\phi_{2}=q^{4}-\frac{1}{2}q(q+1)^{2}+1$$\theta_{0}=k_{C_{0}}$$\theta_{2}$,,$\theta_{1}$,,$\dim_{k}\theta_{2}=r^{4}-\frac{1}{2}r(r+1)^{2}+1$$\phi_{2}$and$\phi_{3}$$\dim_{k}\phi_{3}=\frac{1}{2}q(q-1)^{2}$$\phi_{3},$$\dim_{k}\theta_{1}=\frac{1}{2}r(r+1)^{2}-1$,,

are described as follows.

$\theta_{3}$,,,$\dim_{k}\theta_{3}=\frac{1}{2}r(r-1)^{2}$

93
5.3.1.
. Let. A simple module$\phi_{1},$be a maximal parabolic subgroup ofis the heart of a projective cover $P(\phi_{0})=P(k_{G_{0}})$ . We have$B_{0}\subset K_{0}=\langle w_{a},$$\phi_{2}$$G_{0}$$\phi_{1}$$B_{0}\rangle=L_{0}\ltimes V_{0}$$k_{K_{0}}\uparrow^{G_{0}}=P(k_{G_{0}})\oplus P_{0}’$where $P’$ is a simple projectiveof the form

$kG_{0}$-module of dimension$\frac{1}{2}q(q^{2}+1)$.

$P(\phi_{0})$is uniserial$P(\phi_{0})=\phi_{1}\phi_{0}$$\phi_{0}$The same thing occurs for a projective cover
Set $K=R\ltimes K_{0}$ . Then

$Q(\theta_{0})$of$\theta_{0}$.

$k_{K}\uparrow^{G}=P_{R}(k_{G})\oplus P’$

where
form

$P_{R}(k_{G})$is an extension of $P(k_{G})$ . In particular,

is uniserial of length 3 with

$P_{R}(k_{G})$$\varphi_{0}$$P_{R}(k_{G})=\varphi_{1}$$\varphi_{0}$

where
and
$kR$ -module and$\varphi_{0}=k_{G}$$\varphi_{1}\downarrow c_{0}=\phi_{1}$. It is not hard to see that

$P_{R}(k_{G})(R)=Q(k_{C_{0}})$Write$k_{B_{0}}\uparrow^{K_{0}}=k_{K_{0}}\oplus\rho_{0}$.

$\rho_{0}$,$\varphi_{1}\downarrow R$

is a permutation$\varphi_{1}(R)=\theta_{1}$is the Steinberg module of $K_{0}/V_{0}=L_{0}=GL(2, q)$ . We have$\rho_{0}\uparrow^{G_{0}}=P(\phi_{1})\oplus P_{0}’’$

where$P_{0}’’$

is a simple projective$kG_{0}$-module of dimension$\frac{1}{2}q(q^{2}+1)$.

$P(\phi_{1})$has the form

$P(\phi_{1})=\phi_{0}\oplus\phi_{2}\phi_{1}$$\phi_{1}$for some simple-module . The same thing occurs for a projective coverof
and we have a simple-module .
It is not hard to see thathas a unique extension to $K$ and is a p-permutaion
module. And we have$kG_{0}$$Q(\theta_{1})$$\phi_{2}$$kC_{0}$$\theta_{2}$$\rho_{0}$$\rho$$\rho$$\rho\uparrow^{G}=P_{R}(\phi_{1})\oplus P’’$

where

$P_{R}(\phi_{1})$is an extension of$P(\phi_{1})$. In particular,$P_{R}(\phi_{1})$has the form$\varphi_{1}$$P_{R}(\phi_{1})=\varphi_{0}\oplus\varphi_{2}$$\varphi_{0}$where$\varphi_{2}=\phi_{2}$. It is not hard to see that$\varphi_{2}\downarrow R$$P_{R}(\phi_{1})(R)=Q(\theta_{1})$,is a permutation $kR$-module and$\varphi_{2}(R)=\theta_{2}$$\theta_{1}$94
5.3.2.

$\phi_{3}$. By the knowledge of the character tables of$G_{0}$

and$K_{0}$, we can see thatand the restrictionof dimension-modulethere exists a simple
does not contain-module which is R-invariant.
is a simple
is a p’-group.
. The block of $kK$ which covers has a cyclic defect group andis aThus Alperin-Brauer-Dade-Glauberman theory can be applied. Notice that
$C_{0}=Sp(4,r))$.
maximal parabolic subgroup of
is a p-group. The Brauer-Glauberman
to $K$ as $|K$ :has a unique.
correspondent of does not contain $Z(C_{U_{0}}(R))$ in its kernel. Thus
$B_{0}(kG_{0})$$\frac{1}{2}q(q-1)^{2}$$\phi_{3}$$kK_{0}$$\phi’=\phi_{3}\downarrow K_{0}$$\phi’\downarrow Z(U_{0})$$R$$\phi’$$k_{Z(U_{0})}$$K_{0}$$C_{K_{0}}(R)$$extension\backslash \varphi’$$\phi’$$\theta’$$K_{0}|$$\dim_{k}\theta’=\frac{1}{2}r(r-1)^{2}$$\phi’$We see that $\dim_{k}\phi’-\dim_{k}\theta’=\frac{1}{2}(q(q-1)^{2}-r(r-1)^{2})\equiv 0(mod p)$ . Thus theextension of has a trivial source module and$\varphi’$$\phi’$$\varphi^{l}(R)=\theta’$toalso has a unique extension
is a $B_{0}(kC_{0})$ -module andand$\phi_{3}$$\varphi_{3}$$G$. Then.
is an extension of . Thus
. Such a $B_{0}(kC_{0})$ -module must be simple$\phi’$$\varphi_{3}\downarrow K$$\varphi_{3}(R)\downarrow_{K\cap C_{0}}=\theta’$$\varphi_{3}(R)$$\varphi_{3}1_{K}=\varphi’$$\varphi_{3}(R)=\theta_{3}$The simple $B_{0}(kG_{0})$ -module$\phi_{3}$is self-dual and we can see that$\phi_{3}\otimes\phi_{3}=k_{G_{O}}\oplus$

defect$0$

blocksThus$\varphi_{3}^{*}\otimes\varphi_{3}=k_{G}\oplus R$-projective

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