16 MARTIN W. LIEBECK AND GARY M. SEITZ
Since A! is the unique SL2 in CXv(T{)°, we deduce that A' X n X
0
. Thus X n X
0
contains (ATi, A'), which is equal to X , proving the result in this case.
Now let X = G2,V = 10 ® 10^). Again let X
0
be a closed complement to V in
X 7 . Pick a fundamental subgroup A = A\ in X . The nontrivial composition factors
of i on F are 1 ® l^q\ 1 and 1^). Since p 3, none of these extends the trivial
module by 1.8, so as above we may assume that A X f l X
0
. Also
Cxv(A)0
VbA',
where V0 - Cy{A) and A! - Cx(A) = A\. Moreover Vo is the irreducible A'-module
2 ® 2^), so again using 1.8 and 1.5, we may take AA! X 0 X0.
Let T2 be a maximal torus in AA!. Then Cxv(T
2
= ViT2, where Vi = Cy(T
2
) is
a 1-space; and there is an element w £ Nx(T2) of order 3 which centralizes V\. Thus
JVjfV^ ) contains a unique subgroup isomorphic to T2(w), from which it follows that
T2(w) X 0 Xo- Therefore X D X
0
contains (AA'^w), which is equal to X . This
completes the proof.
Proposition 1.16 Let X Y be simple algebraic groups in characteristic p, with
( X , y , p ) = (2?
n
,I}
n
+i,2) (ra 2), (^4,^6,3) or (C3,As,3), m eac/i case ^ e natural
embedding. Let A = Ai, Ai or X2 in the respective cases, and V Vy (A). Then there
is just one class of closed complements to V in XV.
Proof. In each case there is an X-composition series 0 V\ V2 V with V\ and
V/V2 both trivial 1-dimensional X-modules; moreover, V is indecomposable for X
(see [LS2,1.6]). Write W V/V\, and suppose that there is more than one class ol
closed complements in XW. The proof of 1.5 implies that either there is a rational
indecomposable extension of W by the trivial X-module, or X = Cn and p 2. The
latter is not the case, by hypothesis. Hence W extends the trivial module. Writing
V2/V\ VxifJ*), 1.6 now implies that
Homx(rad(W;r(AA))?ii')
has dimension at least
2, which is a contradiction.
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