We have already known that the inverse of Lagrange's Theorem is a right fact about for example abelian or nilpotent finite groups. How can I show that:
If $G$ be finite and supersolvable$^*$ and $n\mid|G|$, then $G$ has a subgroup of order $n$?
$^*$ A group is supersolvable if there exists normal subgroups $N_i$ with $1=N_0\subseteq N_1\subseteq ...\subseteq N_r=G$ where each factor $\frac{N_i}{N_{i-1}}$ is cyclic for $1\leq i\leq r$.
Thanks for any hint to start.