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A subset $X$ of a finite group $G$ is said to be prime-power-independent ifeach element in $X$ has prime power order and there is no proper subset $Y$ of$X$ with $\langle Y, \Phi(G)\rangle = \langle X, \Phi(G)\rangle$, where$\Phi(G)$ is the Frattini subgroup of $G$. A group $G$ is $\mathcal{B}_{pp}$ ifall prime-power-independent generating sets for $G$ have the same cardinality.We prove that, if $G$ is $\mathcal{B}_{pp}$, then $G$ is solvable. Pivoting onsome recent results of Krempa and Stocka, this yields a complete classificationof $\mathcal{B}_{pp}$-groups.

Independent sets of generators of prime power order