Minimal normal subgroup with order not dividing index is characteristic
- Automorphisms preserve the property of being normal and hence of being minimal normal.
- Normality is strongly intersection-closed
- Product formula
Given: A finite group , a minimal normal subgroup of . The order of does not divide the index of in . An automorphism of .
To prove: .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||is a minimal normal subgroup of .||Fact (1)||is an automorphism and is a minimal normal subgroup of .||Given-cum-fact direct.|
|2||is a normal subgroup of .||Fact (2)||Step (1)||is normal and is also normal by Step (1). Thus, by Fact (2), is normal.|
|3||Either is trivial, or .||Steps (1), (2)||By Step (2), is normal. Since is minimal normal, the normal subgroup is either trivial or equals . Similarly, since is minimal normal, the normal subgroup is either trivial or equals . Combining these, we get the conclusion.|
|4||If is trivial, then is a normal subgroup of of order .||Fact (3)||Step (1)||By Step (1), is normal. So is . Thus, their product is their join. If is trivial, the product has order by Fact (3), which simplifies to because .|
|5||does not divide .||Fact (4)||does not divide .||By Lagrange's theorem, . dividing is equivalent to dividing .|
|6||is nontrivial.||Fact (4)||Steps (4), (5)||If is trivial, then Step (4) gives us a subgroup of of order . By Fact (4), this would give that divides . However, by Step (5), does not divide .|
|7||.||Steps (3) and (6)||Step-combination direct.|