# 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 $H \le K \le G$ 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 $G/H$.

## 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 $C(G/H) = C(G/K) \cup C(K/H)$.