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.
- We first consider the case that any two
-
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
.