Maximal permutable implies normal

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This article gives the proof of a maximality equivalence. In other words, there are two subgroup properties: a stronger one (Normal subgroup (?)) and a weaker one (Permutable subgroup (?)). However, any subgroup maximal among the proper subgroups with the weaker property also has the stronger property, and is thus also maximal among proper subgroups with the stronger property.
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Suppose G is a group and H is maximal among the proper Permutable subgroup (?)s of G. Then, H is a normal subgroup of G.

Related facts


Facts used

  1. Permutability is strongly join-closed
  2. Product of conjugates is proper


Given: A group G, a subgroup H that is maximal among the proper permutable subgroups of G.

To prove: H is normal in G: Any conjugate K of H in G is contained in H.


  1. K is also permutable: This is because conjugation is an automorphism, so conjugate subgroups share the same properties.
  2. HK = \langle H, K \rangle is also permutable: This follows from fact (1).
  3. HK \ne G: This follows from fact (2).
  4. HK = H: Since H \le HK \le G and HK is permutable, maximality of H forces HK = H or HK = G. The latter case is ruled out by the previous step, so HK = H.
  5. K \le H: This follows from the previous step.


Textbook references

  • Subnormal subgroups of groups by John C. Lennox and Stewart E. Stonehewer, Oxford Mathematical Monographs, ISBN 019853552X, Page 213, Theorem 7.1.1, More info