No common composition factor with quotient group is transitive

Statement
Suppose $$G$$ is a group of finite composition length, and $$H \le K \le G$$ are subgroups such that $$K$$ is a normal subgroup of $$G$$ having no common composition factor with $$G/K$$ and $$H$$ is a normal subgroup of $$K$$ having no common composition factor with $$K/H$$.

Then, $$H$$ is a normal subgroup of $$G$$ having no common composition factor with $$G/H$$.

Related facts

 * Normal Hall is transitive
 * No common composition factor with quotient group is quotient-transitive

Facts used

 * 1) uses::No common composition factor with quotient group implies characteristic
 * 2) uses::Characteristic of normal implies normal
 * 3) uses::Third isomorphism theorem

Proof
Given: $$G$$ of finite composition length, $$H \le K \le G$$ with $$H$$ normal in $$K$$ and $$K$$ normal in $$G$$, $$H$$ has no common composition factor with $$K/H$$ and $$K$$ has no common composition factor with $$G/K$$.

To prove: $$H$$ is normal in $$G$$ and $$H$$ has no common composition factors with $$G/H$$.

Proof:


 * 1) $$H$$ is normal in $$G$$: This follows from facts (1) and (2).
 * 2) $$H$$ has no common composition factor with $$G/H$$: Let $$C(L)$$ be the set of composition factors of a group $$L$$. Then, $$C(K) = C(H) \cup C(K/H)$$ and $$C(G/H) = C(G/K) \cup C(K/H)$$. By assumption, $$C(H)$$ and $$C(K/H)$$ are disjoint. Also, since $$C(H) \subseteq C(K)$$, and $$C(K)$$ and $$C(G/K)$$ are disjoint, $$C(H)$$ and $$C(G/K)$$ are disjoint. Thus, $$C(H)$$ is disjoint from the union $$C(G/H) = C(G/K) \cup C(K/H)$$.