Abelian normal subgroup of core-free maximal subgroup is contranormal implies derived subgroup of whole group is monolith

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Suppose G is a Primitive group (?), M is a Core-free maximal subgroup (?) of G (or, the stabilizer of a point for a faithful primitive group action of G), and A is an Abelian normal subgroup (?) of M that is a Contranormal subgroup (?) of G: G is generated by the conjugates of A in it. Then, the Commutator subgroup (?) [G,G] is the monolith of G, i.e., every nontrivial normal subgroup of G contains the commutator subgroup of G.

Facts used

  1. Second isomorphism theorem
  2. Abelianness is quotient-closed


Given: A group G, a core-free maximal subgroup M of G. An abelian normal subgroup A of M that is contranormal in G. A nontrivial normal subgroup N of G.

To prove: G/N is abelian.


  1. MN = G: Since M is core-free and N is nontrivial normal, M does not contain N. Since M is maximal, MN = G.
  2. AN is normal in G: Clearly, N \le AN \le N_G(AN). Also, A is normal in M and N is normal in <mah>G</math>, so M \le N_G(AN). Thus, MN \le N_G(AN), so N_G(AN) = G by step (1).
  3. AN = G: By assumption, A is contranormal. Thus, the only normal subgroup of G containing A is G. So, by step (2), AN is normal in G.
  4. G/N \cong A/(A \cap N) is abelian: This follows from facts (1) and (2).


Textbook references