Write
Under the improved bounds allowed in the question, and
Moreover , whose size is at most by the higher product bound for an approximate group. Thus
The Ruzsa covering lemma supplies of size at most such that
Taking the to be the factors in this last product gives
where and every is a -approximate group in generating a subgroup of step less than .
We induct on the nilpotency class . For , the group is Abelian, so take and no . For , part (i) writes
with small and every of class at most . Apply the induction hypothesis to each . Since and its approximation parameter is , all resulting small sets lie in , all their sizes are at most
and all resulting approximate groups lie in and have approximation parameter . Their generated subgroups are abelian groups.
There are factors at each of at most induction levels. Absorbing the resulting products of the bounds into the notation gives
Keeping the factors in the order supplied by the induction yields the required product of the and containing .
Use the product from part (ii). Move each selected element of each small set to the left, conjugating every approximate-group factor that it crosses. For each tuple , the corresponding part of the product is therefore contained in
where every is a conjugate subset of some . Conjugation preserves cardinality, the approximation parameter, and the property that the generated subgroup is abelian. The conjugating elements belong to , so after enlarging the implicit constant.
The number of tuples is at most
These translated products cover , so one has size at least the reciprocal fraction of . For that tuple, put . Then
where , and each is a -approximate group generating an abelian subgroup.

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