4. THE RECONSTRUCTION THEOREM 21
induced transformation. So T [E] and T(T
)−1
is periodic, thus, by 4.3,
S = T(T
)−1
= [U, V ] = U1V1, for some U, V, U1,V1 [E],U1,V1 involutions.
Also supp(T ) = A and µ(A)
1
2
µ(supp(T)) =
1
2
, as A T(A) = ∅, so
δu(T , 1)
1
2
.
Fix now T [E]. All transformations below are in [E]. Then, by 4.4,
T = S0T1, where S0 is a commutator and δu(T1, 1) = µ(supp(T1))
1
2
.
We concentrate on T1. Split X = X1 X2 . . . , with µ(Xi) =
2−i,
supp(T1) X1. Then, by 4.4 again, T1 = S1T2, where S1 is a commu-
tator of elements with support contained in X1,T2 has support also con-
tained in X1, and δu(T2, 1)
1
4
. Then find an involution U1,2 supported
by X1 X2, with U1,2(supp(T2)) X2, thus, as in general W (supp(V )) =
supp(WV W
−1),
we have supp(U1,2T2U1,2) X2. Let T2 = U1,2T2U1,2, so
that T1 = S1U1,2T2U1,2 = S1(U1,2T2U1,2T2
−1)T2
= S1S1,2T2, where S1,2 is
a commutator of elements supported by X1 X2. Continuing this way, we
write T2 = S2S2,3T3, where S2 is a commutator of elements supported by X2
and S2,3 a commutator of elements supported in X2 ∪X3 and T3 is supported
by X3, etc. Then if
¯n
S = SnSn,n+1,
¯n
S is the product of two commutators
supported by Xn Xn+1,T1 =
¯1T2,T2
S =
¯2T3,...
S , with Tn supported by
Xn. Let P1 be equal to
¯2n+1,n
S 0, on X2n+1 X2n+2 and P2 be equal to
¯2n+2,n
S 0, on X2n+2 X2n+3 and the identity otherwise. Then
P1 =
¯1
S
¯3
S
¯5
S . . . = (T1T2
−1)(T3T4 −1)(T5T6 −1)
. . .
= (T1T3T5 . . . )(T2
−1T4 −1T6 −1
. . . )
and
P2 =
¯2
S
¯4
S
¯6
S . . . = (T2T3
−1)(T4T5 −1)(T6T7 −1)
. . .
= (T2T4T6 . . . )(T3
−1T5 −1T7 −1
. . . ).
Thus T1 = P1P2 and P1,P2 are each a product of 2 commutators.
Lemma 4.5. Let E be ergodic. Then every element of [E] is a product of
10 involutions in [E].
Proof. Let T [E]. All transformations below are in [E]. As in the
preceding proof, we write T = S0T1, where δu(T1, 1)
1
2
and, by 4.4, S0 a
product of 2 involutions. Then, proceeding as before, we have T1 = S1T2 =
S1(T2U1,2(T2)−1)U1,2(U1,2T2U1,2), with S1 a product of 2 involutions, and
note that
U1,2,T2U1,2(T2)−1
are involutions. So T1 = S1S1,2T2, where S1,2 =
(T2U1,2(T2)−1)U1,2
is the product of two involutions. Thus, continuing as in
the rest of the proof, T1 is the product of two transformations each of which
is the product of 4 involutions and we are done.
We now have as a consequence of 4.2 the following result.
Theorem 4.6. If E is ergodic, then [E] is a simple group.
Proof. Let N [E],N = {1}. We will show that N = [E]. Notice that
for any group G and N G, if H generates G and [H, H] N, then also
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