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.
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