Maximal implies normal or abnormal

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 $H\leq K\leq G$ , 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, $H\leq N_{G}(H)\leq G$ . 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.