# A finite nilpotent group has supersolvable series

Let $G$ be a finite nilpotent group. If G has a normal series $$<e>=N_0\leq N_1\leq ... \leq N_k=G$$ such that $G_i\lhd G$ and $G_{i+1}/G_i$ is cyclic for all $i$, then G is called supersolvable.

I want to prove that a finite nilpotent group $G$ has a normal series above whose factors are cyclic. How should I construct desired series?

• The key is that the series can be taken to be central, so all intermediate subgroups between $N_i$ and $N_{i+1}$ are normal in $G$. – egreg Dec 21 '16 at 22:18
• Every finitely generated nilpotent group is supersolvable. It doesn't need to be finite. – Derek Holt Dec 21 '16 at 22:34

## 1 Answer

A finite nilpotent group is the direct product of its Sylow subgroups, so it suffices to show in the case when $G$ is a $p$-group, say of order $p^n$.

To this end, we argue by induction. It is trivial when $n=1$, so let $n>1$. Then $G$ has non-trivial center (a common fact about $p$-groups), so we can choose an element $x\in Z(G)$ of order $p$. Let $N=\langle x\rangle$, and note $N$ is normal in $G$ since $N\leqslant Z(G)$. Then $G/N$ has order $p^{n-1}$, so by induction has a super-solvable series. Lifting back gives a supersolvable series for $G$, and we are done.