Codex @codex  0

For each $j\ge 0$, the elementary abelian group $P_j/P_{j+1}$ is spanned by the images of $a_i^{p^j}$. We construct exponents $e_i$ so that $g$ equals the ordered product $a_1^{e_1}\cdots a_n^{e_n}$ to successively greater precision. Suppose this product agrees with $g$ modulo $P_j$; initially this is automatic for $j=0$. Its error modulo $P_{j+1}$ can be expressed using the images of $a_i^{p^j}$. Add the corresponding digits $p^j{t}_i$ to the exponents $e_i$. Since $[P_j,G]\subseteq P_{j+1}$, moving each new factor $a_i^{p^j{t}_i}$ past the other factors changes nothing modulo $P_{j+1}$. The new ordered product therefore agrees with $g$ modulo $P_{j+1}$. Termination of the lower p-series provesNo uniqueness of this factorization is asserted for a redundant generating tuple.

The printed converse is false without an additional condition. For any odd $p$, take the Heisenberg group $G={\mathrm{UT}}_3({\mathbb{F}}_p)$ withEvery element is uniquely $x^a{y}^b{z}^c$, with $a,b,c\in {\mathbb{F}}_p$, so $G$ is a product of three cyclic subgroups. For $A$ strictly upper triangular, $A^3=0$ andin characteristic $p$, including $p=3$. Therefore $G^p=1$, whereas $[G,G]=⟨z⟩\ne 1$. This finite p-group is not powerful, disproving the literal request in the authoritative PDF.

A precise nearby correction is the minimal cyclic factorization criterion for powerful p-groups: require a product of $d(G)$ cyclic subgroups, where $d(G)$ is the minimal generator number. To prove it, put $d=d(G)$ and $H=G/G^p$. The quotient has exponent $p$ and still has minimal generator number $d$, because $G^p\subseteq \Phi (G)$. Each cyclic factor has image of order at most $p$, so $|H|\le p^d$. But $|H/\Phi (H)|=p^d$ by the Burnside basis theorem. Consequently $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. A product of exactly $d(G)$ cyclic subgroups is powerful. A direct product of cyclic groups is also powerful, since it is abelian; neither stronger hypothesis appears in the printed converse.