22 I. MEASURE PRESERVING AUTOMORPHISMS

[G, G] ⊆ N. Indeed if π : G → G/N is the canonical projection, every two

elements of π(H) commute, so since π(H) generates π(G),π(G) is abelian,

so [G, G] ⊆ N. Also notice that for each 0, the elements T ∈ [E] with

δu(T, 1) generate [E]. This follows immediately from 4.5 and the fact

that any involution in [E] is a product of involutions in [E] with arbitrarily

small support (or alternatively one can use the path connectedness of [E],

see 3.12). Finally fix T0 = 1 in N and then a set A of positive measure with

T0(A) ∩ A = ∅. Let = µ(A)/2. We will show that the commutator of any

two elements of [E] with δu(T, 1) is in N.

Fix S, T ∈ [E] with δu(T, 1),δu(S, 1) . Conjugating, if necessary, we

can assume that their supports are contained in A. Thus

S−1,T0T −1T0

−1

have disjoint supports, so commute. Consider

ˆ

T = (TT0T

−1)T0 −1

∈ N.

Then

[S, T] =

STS−1T −1

= ST(T0T

−1T0 −1)S−1(T0TT0 −1T −1)

= S(TT0T

−1T0 −1)S−1(T0TT0 −1T −1)

= S

ˆ

TS

−1(

ˆ)−1

T ∈ N.

✷

Note that conversely simplicity of [E] implies that E is ergodic. Indeed,

if A is an E-invariant Borel set which is neither null or conull, then the

elements of [E] supported by A form a non-trivial closed normal subgroup of

[E]. In fact, Bezuglyi-Golodets [BG1] have shown that in case E is aperiodic,

non-ergodic, all the nontrivial closed normal subgroups of [E] arise this way

by letting A vary over the invariant Borel sets.

We need one more lemma before embarking in the proof of 4.1.

Lemma 4.7. Let E be ergodic. Let 1 = T ∈ [E] be an involution, CT its

centralizer in [E]. Then [T] is the largest abelian normal subgroup of CT .

Proof. Note that for S, T ∈ Aut(X, µ),S[T

]S−1

= [ST

S−1].

So if

S ∈ CT ,

S[T]S−1

= [T], i.e., [T] ✁ CT . Clearly, as T is an involution, [T] is

abelian.

Let now N ✁ CT , N abelian. Consider the standard Borel space X/T =

X/ET , the projection π : X → X/T and the measure ν = π∗µ. Then E/T is

a countable, measure preserving, ergodic equivalence relation on (X/T, ν).

Clearly every S ∈ CT preserves the support of T and induces a map

S∗

in the

full group of E/T restricted to supp(T)/T; ρ(S) =

S∗

is clearly a surjective

homomorphism. So ρ(N) is a normal subgroup of [(E/T)|(supp(T)/T)],

which is simple, by 4.6. Since ρ(N) is abelian, ρ(N) = [(E/T)|(supp(T)/T)],

thus ρ(N) = {1}, i.e., N ⊆ ker(ρ) = CT ∩ [T] ⊆ [T], which completes the

proof. ✷

(C) We are now ready to give the proof of 4.1. It is clearly enough to

show (iv) ⇒ (i) and the “moreover” assertion.