Maximal implies normal or abnormal

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

Weaker facts

 * Maximal implies normal or self-normalizing
 * Maximal implies normal or contranormal

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
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
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.