Maximal implies normal or abnormal

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This article gives the statement, and possibly proof, of a result according to which any Maximal subgroup (?) of a group satisfies exactly one of the following two subgroup properties: Normal subgroup (?) and Abnormal subgroup (?)
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Statement

Any maximal subgroup of a group is either normal or abnormal.

Related facts

Weaker facts

Definitions used

Maximal subgroup

Further information: Maximal subgroup

A proper subgroup H of a group G is termed a maximal subgroup if there is no proper subgroup of G properly containing H. In other words, if K is a subgroup of G such that H \le K \le G, then H = K or K = G.

Normal subgroup

Further information: Normal subgroup

A subgroup H of a group G is termed normal in G if it satisfies the following equivalent conditions:

  • For any g \in G, the subgroup H^g := g^{-1}Hg is equal to H.
  • The normalizer of H in G equals G.

Abnormal subgroup

Further information: Abnormal subgroup

A subgroup H of a group G is termed abnormal in G if, for any g \in G, g \in \langle H, H^g \rangle where H^g := g^{-1}Hg.

Proof

Given: A group G, a maximal subgroup H of G.

To prove: H is either normal or abnormal in G.

Proof: Let N_G(H) be the normalizer of H in G. Then, H \le N_G(H) \le G. Thus, either N_G(H) = G, or H = N_G(H). In the former case, H is normal in G.

In the latter case, H = N_G(H). Now, pick any g \in G. Consider the subgroup H^g. There are three cases:

  • H^g is not contained in H: By maximality of H, \langle H, H^g \rangle = G, so g \in \langle H, H^g \rangle.
  • H^g = H: Thus, g \in N_G(H), and since H = N_G(H), we get g \in H. Thus, g \in \langle H, H^g \rangle.
  • H^g is a proper subgroup of H: H is a proper subgroup of H^{g^{-1}}, forcing H^{g^{-1}} = G, which would imply that H = G^g = G, a contradiction.