No common composition factor with quotient group is transitive

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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.

Related facts

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 \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).