Finite and any two maximal subgroups intersect trivially implies not simple non-abelian

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Suppose G is a finite group with the property that any two distinct maximal subgroups of G intersect trivially. Then, G is not a simple non-abelian group.

Related facts

Direct applications

Indirect applications

Facts used

  1. The trivial subgroup is maximal if and only if the group is a group of prime order, which is a simple abelian group.
  2. Lagrange's theorem
  3. Group acts as automorphisms by conjugation: Thus, conjugates of a maximal subgroup are maximal and all have the same order.
  4. Size of conjugacy class of subgroups equals index of normalizer
  5. 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 G of order n such that any two distinct maximal subgroups of G intersect trivially.

To prove: A contradiction.


Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Every maximal subgroup of G is self-normalizing Fact (1) G is simple non-abelian [SHOW MORE]
2 If A is a maximal subgroup of G with a elements, and G has n elements, the union of all conjugates of A in G has n - (n/a) + 1 elements Facts (2), (3), (4) G is finite, has order n
Any two maximal subgroups of G intersect trivially
Step (1) [SHOW MORE]
3 The union of conjugates of any one maximal subgroup has at least 1 + (n/2) of the elements, but does not have all the elements Step (2) [SHOW MORE]
4 We have the desired contradiction, because cardinality considerations force G 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 G intersect trivially. Step (3) [SHOW MORE]