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