Minimal normal subgroup with order not dividing index is characteristic

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Statement

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

Related facts

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.