Theorem 1

Let G be a p-group and $p^r$ its order. There exists a sequence of subgroups of G

such that $(G, G^k) \subset G^{k+1}, 1 \leq k \leq r,$ and $G^k/G^{k+1}, 1 \leq k \leq r,$ is cyclic of order $p$.

The theorem is true for $G = \{e\}$. We prove it by induction on $\operatorname{Card}(G)$. Let Z be the centre of G, $x \neq e$ an element of Z (Corollary to Proposition 11) and $p^s, s \neq 0,$ the order of x. Then $x^{p^s - 1}$ is an element of order $p$ and therefore Z contains a subgroup $G'$ which is cyclic of order $p$. By the induction hypothesis, the group $G' = G/G'$ has a series of subgroups $({G'}^k)_{1 \leq k \leq r}$ with the required properties. Let $\pi : G \to G'$ be the canonical homomorphism. The sequence of subgroups of G defined by $G^k = \pi^{-1}({G'}^k), 1 \leq k \leq r, G^{r+1} = \{e\}$ is a solution for $G^k/G^{k+1}$ is isomorphic to ${G'}^k/{G'}^{k+1}$ for $1 \leq k \leq r$ (§ 4, no. 7, Theorem 4).