Conjugate-permutable implies subnormal in finite

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Conjugate-permutable subgroup (?)) must also satisfy the second subgroup property (i.e., Subnormal subgroup (?)). In other words, every conjugate-permutable subgroup of finite group is a subnormal subgroup of finite group.
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Statement

Any conjugate-permutable subgroup of a finite group is subnormal.

Related facts

Facts used

  1. Conjugate-permutability satisfies intermediate subgroup condition
  2. Maximal conjugate-permutable implies normal

Proof

Given: A finite group G, a conjugate-permutable subgroup H of G.

To prove: H is a subnormal subgroup of G.

Proof:

  1. (Fact used: fact (1)): Define a descending chain as follows: K_0 = G, and if K_i properly contains H, K_{i+1} is a maximal element among the proper conjugate-permutable subgroups of K_i that contain H.
    1. (Well-definedness): Note that by fact (1), H is conjugate-permutable in K_i, so the collection of proper conjugate-permutable subgroups of K_i containing H is nonempty. Since G is finite, it has a maximal element.
    2. (Terminates at H in finitely many steps): The chain K_i is a strictly descending chain of subgroups until it reaches H. Since G is finite, it terminates in finitely many steps at H. Thus, there exists n such that K_n = H.
    3. (Fact used: fact (2)): By definition, K_{i+1} is a maximal conjugate-permutable subgroup of K_i, so fact (2) tells us that K_{i+1} is normal in K_i.
    4. (Conclusion): The K_is thus form a subnormal series for H in G, making H a subnormal subgroup of G.