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

## 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.

## 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.

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.

1. 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 .
2. 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.
3. If , there is a subgroup of index five:
1. 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.
2. 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.
4.  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.
5.  is isomorphic to a subgroup  of : This follows fron fact (5.1).
6.  is isomorphic to : By order considerations, the order of  equals that of , so . Thus,  is isomorphic to .