# 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.

## 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.

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 $A_5$.