 Solution

Assume condition (iii), and factor $m={\prod }_j{f}_j^{e_j}$. If some $e_j\ge 2$, the primary decomposition theorem gives the nonzero summand $V_j=\ker f_j(T{)}^{e_j}$ on which the minimal polynomial is exactly $f_j^{e_j}$. The operator $f_j(T)$ cannot vanish on this whole summand, or its restriction would have minimal polynomial dividing $f_j$. Choose $v\in V_j$ with $f_j(T)v\ne 0$. Thencontradicting (iii) with $p=f_j$ and $n=e_j$. Hence every $e_j=1$, proving (iii) implies (ii).