# 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.

## 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.