# Minimal normal subgroup with order not dividing index is characteristic

## Statement

A minimal normal subgroup of a finite group whose order does not divide its index is characteristic.

## Facts used

1. Automorphisms preserve the property of being normal and hence of being minimal normal.
2. Normality is strongly intersection-closed
3. Product formula

## Proof

Given: A finite group $G$, a minimal normal subgroup $H$ of $G$. The order of $H$ does not divide the index $[G:H]$ of $H$ in $G$. An automorphism $\sigma$ of $G$.

To prove: $\sigma(H) = H$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $\sigma(H)$ is a minimal normal subgroup of $G$. Fact (1) $\sigma$ is an automorphism and $H$ is a minimal normal subgroup of $G$. Given-cum-fact direct.
2 $\sigma(H) \cap H$ is a normal subgroup of $G$. Fact (2) Step (1) $H$ is normal and $\sigma(H)$ is also normal by Step (1). Thus, by Fact (2), $\sigma(H) \cap H$ is normal.
3 Either $\sigma(H) \cap H$ is trivial, or $H = \sigma(H)$. Steps (1), (2) By Step (2), $\sigma(H) \cap H$ is normal. Since $H$ is minimal normal, the normal subgroup $\sigma(H) \cap H$ is either trivial or equals $H$. Similarly, since $\sigma(H)$ is minimal normal, the normal subgroup $\sigma(H) \cap H$ is either trivial or equals $\sigma(H)$. Combining these, we get the conclusion.
4 If $\sigma(H) \cap H$ is trivial, then $\langle H, \sigma(H) \rangle = H\sigma(H)$ is a normal subgroup of $G$ of order $|H|^2$. Fact (3) Step (1) By Step (1), $\sigma(H)$ is normal. So is $H$. Thus, their product is their join. If $H \cap \sigma(H)$ is trivial, the product has order $|H||\sigma(H)|$ by Fact (3), which simplifies to $|H|^2$ because $|H| = |\sigma(H)|$.
5 $|H|^2$ does not divide $|G|$. Fact (4) $|H|$ does not divide $[G:H]$. By Lagrange's theorem, $[G:H] = |G|/|H|$. $|H|$ dividing $|G|/|H|$ is equivalent to $|H|^2$ dividing $|G|$.
6 $\sigma(H) \cap H$ is nontrivial. Fact (4) Steps (4), (5) If $\sigma(H) \cap H$ is trivial, then Step (4) gives us a subgroup of $G$ of order $|H|^2$. By Fact (4), this would give that $|H|^2$ divides $|G|$. However, by Step (5), $|H|^2$ does not divide $|G|$.
7 $H = \sigma(H)$. Steps (3) and (6) Step-combination direct.