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

From Groupprops

## Contents

## Statement

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

## Related facts

### Direct applications

- Finite non-abelian and every proper subgroup is abelian implies not simple
- Finite non-nilpotent and every proper subgroup is nilpotent implies not simple

### Indirect applications

- Classification of cyclicity-forcing numbers
- Classification of abelianness-forcing numbers
- Schmidt-Iwasawa theorem

## Facts used

- 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.

## Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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.

**Proof**:

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] |