The set agreement power of a shared object O describes O’s ability to solve set agreement problems: it is the sequence $$(n_1, n_2, {\ldots }, n_k, {\ldots })$$ such that, for every $$k\ge 1$$, using O and registers one can solve the k-set agreement problem among at most $$n_k$$ processes. It has been shown that the ability of an object O to implement other objects is not fully characterized by its consensus number (the first component of its set agreement power). This raises the following natural question: is the ability of an object O to implement other objects fully characterized by its set agreement power? We prove that the answer is no: every level $$n \ge 2$$ of Herlihy’s consensus hierarchy has two linearizable objects that have the same set agreement power but are not equivalent, i.e., at least one cannot implement the other. We also show that every level $$n \ge 2$$ of the consensus hierarchy contains a deterministic linearizable object $$O_n$$ with some set agreement power $$(n_1,n_2,\ldots ,n_k,\ldots )$$ such that being able to solve the k-set agreement problems among $$n_k$$ processes, for all $$k\ge 1$$, is not enough to implement $$O_n$$.
