1.2. THE WEISFEILER RADICAL 3
(iii) If g is irreducible and transitive, and [g−i, g−1] = g−(i+1) for all
i 1, then for any ideal J of g either J M(g) or J g−.
(iv) If g is 1-transitive (1.3), then M(g) = 0.
Proof. (i) We know from the calculations above that M(g) is an ideal
of g. If x = x−q + ··· + x−1 M(g), then one can see by bracketing with
homogeneous elements of positive degree that the homogeneous components
x−i g−i of x belong to M(g) also, and the quotient algebra g := g/M(g)
is graded.
(ii) Suppose that g is irreducible and transitive. If J is any ideal of g
contained in g− and if J ∩g−1 = 0, then J ∩g−1 = g−1 by irreducibility, since
J g−1 is a g0-submodule of g−1. But if J g−1, then J [g−1, g1] = 0
by transitivity, to contradict J g−. Hence, when g is irreducible and
transitive, every ideal J of g contained in g− has trivial intersection with g−1.
In particular, since M(g) g− and M(g) is graded, we must have M(g)

i≥2
g−i. It follows that g/M(g) will inherit the properties of irreducibility
and transitivity.
Now suppose that J is an ideal of g contained in g− and let x−q + ··· +
x−1 J, where x−i g−i for all i. Then it must be that x−i
Mi(g)
for
i = 1, . . . , q. Consequently J M(g)

i≥2
g−i, and M(g) is the sum
of all the ideals of g contained in g−. Any ideal of g = g/M(g) contained
in g− has the form K where K is an ideal of g contained in g−. But then
K M(g) so K = 0. Hence, M(g) = 0.
(iii) Under the hypotheses in (iii), suppose that J is an ideal of g, and
define the following subspaces of g:
(1.9) Yj :=



y gj y + z J for some z
i≤j−1
gi



.
Then Yj is a g0-submodule of gj for each j. Now either J g− (and hence
J M(g) by (ii)) or there exists an element x = x−q + ··· + xk J
with xk = 0 and k 0. Since in the second case (ad
g−1)k+1(x)
= 0 by
the transitivity of g, we see that J g− = 0 and Y−1 = 0. Hence by
irreducibility, Y−1 must equal g−1. Assume we have shown that Y−i = g−i.
Then Y−(i+1) [Y−i, g−1] = [g−i, g−1] = g−(i+1) so that Y−(i+1) = g−(i+1).
In particular, J g−q. Suppose that J g−t for all t with i + 1 t q.
Since Y−i = g−i, it follows that J g−i. Consequently J g−, as asserted.
(iv) Suppose
Mk(g)
= 0 for 0 k i. Then
[Mi(g),
g1]
Mi−1(g)
= 0.
Thus, if g is 1-transitive, we must have
Mi(g)
= 0 also. Therefore, 1-
transitivity implies M(g) =
Mi(g)
= 0.
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