Supersolvable implies every nontrivial normal subgroup contains a cyclic normal subgroup

Property-theoretic statement
The group property of being a supersolvable group is stronger than, or implies the property of being a group in which every nontrivial normal subgroup contains a cyclic normal subgroup.

Statement with symbols
Let $$G$$ be a supersolvable group. Then, if $$N$$ is a nontrivial normal subgroup of $$G$$, there exists a (nontrivial) cyclic normal subgroup of $$G$$ contained in $$N$$.

Proof
Given: A supersolvable group $$G$$, with a nontrivial normal subgroup $$N$$

To prove: There exists a nontrivial cyclic normal subgroup of $$G$$ contained in $$N$$

Proof: By assumption, there exists a normal series for $$G$$:

$$\{ e \} = K_0 \le K_1 \le K_2 \le \dots \le K_n = G$$

where each $$K_i/K_{i-1}$$ is a cyclic group and each $$K_i$$ is normal in $$G$$.

Now consider the series:

$$\{ e \} = K_0 \cap H \le K_1 \cap H \le \dots \le K_n \cap H = H$$

It's clear that each of the $$K_i \cap H$$ is normal in $$G$$ (since normality is intersection-closed), and moreover, each $$(K_i \cap H)/(K_{i-1} \cap H)$$ is cyclic (since it can be identified with a subgroup of the cyclic group $$K_i/K_{i-1}$$). Pick the first $$i$$ such that $$K_i \cap H$$ is nontrivial. Such an $$i$$ exists, because $$K_n \cap H = H$$ is nontrivial. Then, this is a cyclic normal subgroup of $$G$$ contained in $$H$$.