A nonabelian supersolvable group has a noncentral abelian normal subgroup, also in its nonabelian quotients

Let $K⊴G$ and put $\bar{G}:=G\mathrm{/}K$, ${\bar{G}}_j:=G_{j}K\mathrm{/}K$ for $j=0,\dots ,r$. Each ${\bar{G}}_j$ is a subgroup of $\bar{G}$ containing ${\bar{G}}_{j-1}$ and is normal in $\bar{G}$, because $G_j⊴G$ and $K⊴G$ ([F2], [F3]); moreover ${\bar{G}}_0=K\mathrm{/}K=1$ and ${\bar{G}}_r=G\mathrm{/}K=\bar{G}$. For each $j$ the map $\phi :{\bar{G}}_j\to G_{j}K\mathrm{/}{G}_{j-1}K$, $gK\mapsto g{G}_{j-1}K$, is well defined and a surjective homomorphism with kernel ${\bar{G}}_{j-1}=G_{j-1}K\mathrm{/}K$, so by [F5] the quotient ${\bar{G}}_j\mathrm{/}{\bar{G}}_{j-1}$ is isomorphic to $G_{j}K\mathrm{/}{G}_{j-1}K$; that group is in turn a quotient of the prime-order group $G_j\mathrm{/}{G}_{j-1}$, since $G_j\to G_{j}K\mathrm{/}{G}_{j-1}K$ is surjective with $G_{j-1}$ in its kernel. Hence every factor ${\bar{G}}_j\mathrm{/}{\bar{G}}_{j-1}$ has order $1$ or prime.

Now assume that $G$ is nonabelian, so $G=G_r⊈Z(G)$ by [F4], while $G_0=1\subseteq Z(G)$. Let $i\in \{1,\dots ,r\}$ be the least index with $G_i⊈Z(G)$; such an index exists because $G_r⊈Z(G)$, and $i\ge 1$ because $G_0=1$ is central. By minimality $G_{i-1}\subseteq Z(G)$.

By step 1.2 the subgroup $G_{i-1}$ is contained in $Z(G)$, hence in $Z(G_i)$; by [F1] the factor $G_i\mathrm{/}{G}_{i-1}$ has prime order, so $G_i=G_{i-1}⟨x⟩$ for any $x\in G_i\setminus G_{i-1}$ by [A1], and every element of $G_i$ is of the form $g{x}^k$ with $g\in G_{i-1}$. Two such elements $g{x}^k$ and $g^{\mathrm{\prime }}{x}^{k^{\mathrm{\prime }}}$ commute, because $g,g^{\mathrm{\prime }}\in G_{i-1}\subseteq Z(G_i)$ and powers of $x$ commute with each other. Hence $G_i$ is abelian.

Thus $A:=G_i$ is abelian by step 2.1, it is normal in $G$ by hypothesis, and $A=G_i⊈Z(G)$ by the choice of $i$ in step 1.2. This is the asserted abelian normal subgroup of $G$ and proves the first assertion.

Finally assume that $\bar{G}=G\mathrm{/}K$ is nonabelian, so $\bar{G}={\bar{G}}_r⊈Z(\bar{G})$ by [F4] while ${\bar{G}}_0=1\subseteq Z(\bar{G})$. Let $j$ be the least index with ${\bar{G}}_j⊈Z(\bar{G})$; then $j\ge 1$, ${\bar{G}}_{j-1}\subseteq Z(\bar{G})$, and ${\bar{G}}_j\mathrm{\ne }{\bar{G}}_{j-1}$, so by step 1.1 the factor ${\bar{G}}_j\mathrm{/}{\bar{G}}_{j-1}$ has prime order. The computation of step 2.1, applied with the pair $(G_i,G_{i-1})$ replaced by the pair $({\bar{G}}_j,{\bar{G}}_{j-1})$ inside $\bar{G}$, shows in the same way that ${\bar{G}}_j$ is abelian, while ${\bar{G}}_j⊈Z(\bar{G})$ by the choice of $j$. Hence the nonabelian quotient $\bar{G}$ also contains an abelian normal subgroup that is not contained in its center, which is the second assertion.