Finite and any two maximal subgroups intersect trivially implies not simple non-abelian
- Finite non-abelian and every proper subgroup is abelian implies not simple
- Finite non-nilpotent and every proper subgroup is nilpotent implies not simple
- Classification of cyclicity-forcing numbers
- Classification of abelianness-forcing numbers
- Schmidt-Iwasawa theorem: This states that if every proper subgroup of a finite group is nilpotent, the group itself is a finite solvable group
- The trivial subgroup is maximal if and only if the group is a group of prime order, which is a simple abelian group.
- Lagrange's theorem
- Group acts as automorphisms by conjugation: Thus, conjugates of a maximal subgroup are maximal and all have the same order.
- Size of conjugacy class of subgroups equals index of normalizer
- In a finite non-cyclic group, every element is contained in a maximal subgroup. This basically follows from the fact that cyclic iff not a union of proper subgroups.
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We prove the statement by contradiction.
Given: A finite simple non-abelian group of order such that any two distinct maximal subgroups of intersect trivially.
To prove: A contradiction.
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||Every maximal subgroup of is self-normalizing||Fact (1)||is simple non-abelian||[SHOW MORE]|
|2||If is a maximal subgroup of with elements, and has elements, the union of all conjugates of in has elements||Facts (2), (3), (4)|| is finite, has order
Any two maximal subgroups of intersect trivially
|Step (1)||[SHOW MORE]|
|3||The union of conjugates of any one maximal subgroup has at least of the elements, but does not have all the elements||Step (2)||[SHOW MORE]|
|4||We have the desired contradiction, because cardinality considerations force to have more than one conjugacy class of maximal subgroups, but there is not enough room for two conjugacy classes of subgroups.||Fact (5)||Any two maximal subgroups of intersect trivially.||Step (3)||[SHOW MORE]|