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

Statement

The following are true:

• The alternating group of degree five, denoted $A_5$, is a simple non-Abelian group of order $60$.
• It is, up to isomorphism, the only simple non-Abelian group of order $60$.
• There is no simple non-Abelian group of smaller order.

Facts used

1. Prime power order implies not centerless
2. The basic condition on Sylow numbers:
   1. Congruence condition on Sylow numbers
   2. Divisibility condition on Sylow numbers
3. Prime divisor greater than Sylow index is Sylow-unique
4. Order is product of Mersenne prime and one more implies normal Sylow subgroup
5. Facts specific to simple non-Abelian groups (they're all closely related):
   1. simple non-Abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup of finite index
   2. Sylow number equals index of Sylow normalizer
   3. Order of simple non-Abelian group divides half the factorial of index of proper subgroup
   4. Order of simple non-Abelian group divides half the factorial of every Sylow number
6. Order is product of three distinct primes implies normal Sylow subgroup
7. 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 $60$ 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 $2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41,43,47,49,53,59$.
• If the order is of the form $p^km$ where $p$ is a prime and $m < p$, then fact (3) tells us that the group is not simple, since it has a nontrivial normal Sylow subgroup. This eliminates the orders $6,10,14,15,18,20,21,22,26,28,33,34,35,38,39,42,44,46,50,51,52,54,55,57,58$. In toto, we have eliminated: $2,3,4,5,6,7,8,9,10,11,13,14,15,16,17,18,19,20,21,22,23,25,26,27,28,29,31,32,33,34,35,37,38,39,41,42,43,44,46,47,49,50,51,52,53,54,55,57,58,59$.
• If the order is $12$ or $56$, then fact (4) tells us that the group has a nontrivial normal subgroup. The total list of eliminated numbers is now:

$2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,25,26,27,28,29,31,32,33,34,35,37,38,39,41,42,43,44,46,47,49,50,51,52,53,54,55,56,57,58,59$. The list of numbers not eliminated is: $24,30,36,40,45,48$.

• 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:
  • $40 = 2^3 \cdot 5$: Here, $n_5 = 1$.
  • $45 = 3^2 \cdot 5$: Here again, $n_5 = 1$.

Thus, the list is shortened to $24,30,36,48$.

• We use fact (5.4) to eliminate the numbers $36, 48$ and a slight modification of it is $24$:
  • $36 = 2^2 \cdot 3^2$: Here, we have $n_3 = 1$ or $n_3 = 4$. Fact (5.4) shows that since $36$ does not divide $n_3!/2$ in either case, a group of order $36$ is not simple.
  • $48 = 2^4 \cdot 3$: Here, we have $n_2 = 1$ or $n_2 = 3$. Fact (5.4) shows that since $48$ does not divide $n_2!$ in either case, a group of order $48$ is not simple.
  • $24 = 2^3 \cdot 3$: Here, we have $n_2 = 1$ or $n_2 = 3$. Fact (5.4) shows that sinc e$24$ does not divide $n_2!/2$ in either case, a group of order $24$ is not simple.

This leaves only one number: $30$.

• $30 = 2 \cdot 3 \cdot 5$: 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 $60$ satisfying this are $30$ and $42$. 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 $G$ of order sixty.

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

Proof: The key idea is to prove that $G$ has a subgroup of index five. After that, we use the fact that $A_5$ is simple to complete the proof.

1. The number of $2$-Sylow subgroups is either $5$ or $15$: By fact (2) (the congruence and divisibility conditions on Sylow numbers), we have $n_2 = 1,3,5,15$. By fact (4), $n_2$ cannot be $1$ or $3$. Thus, $n_2 = 5$ or $n_2 = 15$.
2. If $n_2 = 5$, there is a subgroup of index five: The number of $2$-Sylow subgroups equals the index of the normalizer of any $2$-Sylow subgroup (fact (5.2)). Thus, there is a subgroup of index five.
3. If $n_2 = 15$, there is a subgroup of index five:
   1. We first consider the case that any two $2$-Sylow subgroups intersect trivially: In this case, there are $(4-1) \cdot 15 = 45$ non-identity elements in $2$-Sylow subgroups. This leaves $15$ other non-identity elements. We also know that $n_3 = 1, 4, 10$ by the congruence and divisibility conditions, and fact (4) again forces $n_3 = 10$. Thus, there are $(3-1) \cdot 10 = 20$ non-identity elements in $3$-Sylow subgroups. But $45 + 20 = 65 > 60$, a contradiction.
   2. Thus, there exist at least two $2$-Sylow subgroups that intersect nontrivially. Suppose $P$ and $Q$ are two $2$-Sylow subgroups whose intersection, $R = P \cap Q$ is nontrivial. The $2$-Sylow subgroups are of order $4$, hence Abelian, so $P, Q$ are Abelian. Thus, $P \le C_G(R)$ and $Q \le C_G(R)$. This yields that $S = \langle P, Q \rangle \le C_G(R)$. If $S = \langle P, Q \rangle = G$, then $R \le Z(G)$, so the center is nontrivial. We thus get a proper nontrivial normal subgroup, a contradiction. Thus, $S = \langle P, Q \rangle$ is a proper subgroup of $G$. Lagrange's theorem forces that $S$ has index either three or five in $G$. $S$ cannot have index three, by fact (5.3). Thus, $S$ must have index five.
4. $G$ has a subgroup $S$ of index five: Note that steps (2) and (3) show that for both possible values of $n_2$, $G$ has a subgroup of index five.
5. $G$ is isomorphic to a subgroup $H$ of $A_5$: This follows fron fact (5.1).
6. $G$ is isomorphic to $A_5$: By order considerations, the order of $H$ equals that of $A_5$, so $H = A_5$. Thus, $G$ is isomorphic to $A_5$.