No common composition factor with quotient group is transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup having no common composition factor with its quotient group) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
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Statement

Suppose $G$ is a group of finite composition length, and $H\leq K\leq G$ 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. No common composition factor with quotient group implies characteristic
2. Characteristic of normal implies normal
3. Third isomorphism theorem

Proof=

Given: $G$ of finite composition length, $H\leq K\leq G$ 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 $C(G/H)=C(G/K)\cup C(K/H)$.