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.

Related facts

Equivalence of definitions of finite characteristically simple group

Facts used

Automorphisms preserve the property of being normal and hence of being minimal normal.
Normality is strongly intersection-closed
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.