# A5 is the unique simple non-abelian group of smallest order

## Contents

## Statement

The following are true:

- The alternating group of degree five, denoted , is a simple non-Abelian group of order .
- It is, up to isomorphism, the only simple non-Abelian group of order .
- There is no simple non-Abelian group of smaller order.

## Facts used

- Prime power order implies not centerless
- The basic condition on Sylow numbers:
- Prime divisor greater than Sylow index is Sylow-unique
- Order is product of Mersenne prime and one more implies normal Sylow subgroup
- Facts specific to simple non-Abelian groups (they're all closely related):
- simple non-Abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup of finite index
- Sylow number equals index of Sylow normalizer
- Order of simple non-Abelian group divides half the factorial of index of proper subgroup
- Order of simple non-Abelian group divides half the factorial of every Sylow number

- Order is product of three distinct primes implies normal Sylow subgroup
- A5 is simple

## Proof that there is no simple non-Abelian group of smaller order

### Proof using Sylow's theorem

We eliminate all possible orders less than using the information from Sylow's theorems. First, some preliminary observations.

- If the order is a prime power, then fact (1) tells us that the group has a nontrivial center. Hence, it cannot be a simple non-Abelian group. This eliminates the orders .
- If the order is of the form where is a prime and , then fact (3) tells us that the group is not simple, since it has a nontrivial normal Sylow subgroup. This eliminates the orders . In toto, we have eliminated: .
- If the order is or , then fact (4) tells us that the group has a nontrivial normal subgroup. The total list of eliminated numbers is now:

. The list of numbers not eliminated is: .

- Of these remaining numbers, the following can be eliminated because a direct application of the congruence and divisibility conditions (fact (2)) yields a Sylow-unique prime divisor:
- : Here, .
- : Here again, .

Thus, the list is shortened to .

- We use fact (5.4) to eliminate the numbers and a slight modification of it is :
- : Here, we have or . Fact (5.4) shows that since does not divide in either case, a group of order is not simple.
- : Here, we have or . Fact (5.4) shows that since does not divide in either case, a group of order is not simple.
- : Here, we have or . Fact (5.4) shows that sinc e does not divide in either case, a group of order is not simple.

This leaves only one number: .

- : For this, we use fact (6).

### Proof using Burnside's theorem

Using Burnside's theorem, the order of a simple non-Abelian group must have at least three distinct prime factors. The only numbers less than satisfying this are and . Both of them can be eliminated using the methods discussed above.

## Proof that the alternating group of degree five is simple

Refer fact (7).

## Proof that there is only one simple group of order sixty, isomorphic to the alternating group of degree five

**Given**: A simple group of order sixty.

**To prove**: is isomorphic to the alternating group of degree five.

**Proof**: The key idea is to prove that has a subgroup of index five. After that, we use the fact that is simple to complete the proof.

- The number of -Sylow subgroups is either or : By fact (2) (the congruence and divisibility conditions on Sylow numbers), we have . By fact (4), cannot be or . Thus, or .
- If , there is a subgroup of index five: The number of -Sylow subgroups equals the index of the normalizer of any -Sylow subgroup (fact (5.2)). Thus, there is a subgroup of index five.
- If , there is a subgroup of index five:
- We first consider the case that any two -Sylow subgroups intersect trivially: In this case, there are non-identity elements in -Sylow subgroups. This leaves other non-identity elements. We also know that by the congruence and divisibility conditions, and fact (4) again forces . Thus, there are non-identity elements in -Sylow subgroups. But , a contradiction.
- Thus, there exist at least two -Sylow subgroups that intersect nontrivially. Suppose and are two -Sylow subgroups whose intersection, is nontrivial. The -Sylow subgroups are of order , hence Abelian, so are Abelian. Thus, and . This yields that . If , then , so the center is nontrivial. We thus get a proper nontrivial normal subgroup, a contradiction. Thus, is a proper subgroup of . Lagrange's theorem forces that has index either three or five in . cannot have index three, by fact (5.3). Thus, must have index five.

- has a subgroup of index five: Note that steps (2) and (3) show that for both possible values of , has a subgroup of index five.
- is isomorphic to a subgroup of : This follows fron fact (5.1).
- is isomorphic to : By order considerations, the order of equals that of , so . Thus, is isomorphic to .