Permutable implies subnormal in finite

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., Permutable subgroup (?)) must also satisfy the second subgroup property (i.e., Subnormal subgroup (?)). In other words, every permutable subgroup of finite group is a subnormal subgroup of finite group.
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Statement

Any permutable subgroup of a finite group is a subnormal subgroup. In particular, it is a permutable subnormal subgroup.

Related facts

Similar facts

• Conjugate-permutable implies subnormal in finite
• Permutable implies ascendant
• Permutable implies locally subnormal
• Permutable implies subnormal in finitely generated

Generalizations of proof technique

• Max-equivalent to normal and intermediate subgroup condition implies subnormal in finite: Other examples of this proof technique are:
  • Conjugate-permutable implies subnormal in finite

Facts used

1. Maximal permutable implies normal
2. Permutability satisfies intermediate subgroup condition

Proof

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

To prove: $H$ is subnormal in $G$.

Proof: We prove this by induction on the order of $G$.

If $H=G$, we are done. Otherwise:

1. Let $K$ be a maximal element among the proper permutable subgroups of $G$ containing $H$. By fact (1), $K$ is normal in $G$.
2. By fact (2), $H$ is permutable in $K$, so induction on the order yields that $H$ is subnormal in $K$.
3. Thus, $H$ is subnormal in $K$ which is normal in $G$, so $H$ is subnormal in $G$.