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

## Statement

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.

## 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.

## Proof

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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.

Proof:

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
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]