  Let $T$ be a quasinilpotent operator on a Banach space. Under assumptions of
a certain nonsymmetry in the growth of the resolvent of $T$, it is proved that
every operator in the commutant of $T$ is not unicellular. In particular, $T$
has nontrivial hyperinvariant subspaces. The proof is based on a modification
of the reasoning of [S].
