Poisson suspensions without roots

The paper’s Theorem 1 asserts: if T is ergodic, satisfies property (⊥) (mutual singularity with its convolution powers), and T^n has exactly n invariant ergodic sets for every n>0, then both T and its Poisson suspension P(T) have no roots. The given proof argues: assuming a root R with R^n=T (taking n equal to the root order), commutation forces R to permute the n ergodic components of T^n so that RY=T^kY for some k; then TY=R^nY=T^{kn}Y=Y (since Y is T^n-invariant), contradicting the decomposition into n disjoint T-iterates. For P(T), the paper invokes Roy’s Proposition 5.2 (as quoted there) that (⊥) implies C(P(T))=P(C(T)), ensuring any root of P(T) must be Poissonian and hence correspond to a root of T, which does not exist . The candidate solution is correct and slightly more detailed: it supplements the paper’s short argument with a group-theoretic lemma on permutations (forcing gcd(k,n)=1 if π^k is an n-cycle) and explicitly uses the faithfulness (injectivity) and functoriality of the Poisson construction to pass roots between base and suspension. The two arguments agree on conclusions; the candidate gives a different, more explicit permutation-based proof of the “no roots for T” part and spells out the step P(R)^k=P(R^k), together with injectivity, to conclude R^k=T for the Poisson case. The construction section of the paper further exhibits rank-one T with the required properties and concludes P(T) has no roots (Theorem 2) .