Simple Groups of Order 60 and A5

Having ruled out many small composite orders, we now reach the first genuinely interesting case. The alternating group

A_5

has order

60

, and it is the smallest nonabelian simple group. In this lesson, we focus in the simple groups of order 60 and A5. The main goal is to understand why a simple group of order

60

must look like

A_5

. This result connects Sylow counting, group actions on cosets, and the alternating group into one important application.

Let

S_5

be the symmetric group on five letters. The group

A_5

is the \textbf{alternating group of degree five}, consisting of all even permutations in

S_5

Having ruled out many small composite orders, we now reach the first genuinely interesting case. The alternating group

A_5

has order

60

, and it is the smallest nonabelian simple group. In this lesson, we focus in the simple groups of order 60 and A5. The main goal is to understand why a simple group of order

60

must look like

A_5

. This result connects Sylow counting, group actions on cosets, and the alternating group into one important application.

Let

S_5

be the symmetric group on five letters. The group

A_5

is the \textbf{alternating group of degree five}, consisting of all even permutations in

S_5

Since

|S_5|=120

and exactly half of the permutations are even, we have

|A_5|=60

. The group

A_5

contains identity,

3

-cycles, double transpositions, and

5

-cycles. Its conjugacy class structure is the key reason it is simple.

The group

A_5

is simple.

Given that

A_5

is the alternating group of degree five. To prove that

A_5

is simple. Let

N

be a normal subgroup of

A_5

. A normal subgroup is a union of conjugacy classes. The conjugacy classes in

A_5

have sizes

\begin{equation*} 1,\quad 15,\quad 20,\quad 12,\quad 12. \end{equation*}

These correspond respectively to the identity, the double transpositions, the

3

-cycles, and the two classes of

5

-cycles. Therefore

|N|

must be a sum of

1

and some of the numbers

15,20,12,12

. Also,

|N|

must divide

|A_5|=60

. The possible sums containing

1

and smaller than

60

are

\begin{equation*} 13,\ 16,\ 21,\ 25,\ 28,\ 33,\ 36,\ 45,\ 48. \end{equation*}

None of these numbers divides

60

. Hence no proper nontrivial normal subgroup of

A_5

exists. Therefore the only normal subgroups of

A_5

are

\{e\}

and

A_5

. Hence

A_5

is simple.

\square

This preview shows the class-union argument in the proof that

A_5

is simple. A normal subgroup must be a union of conjugacy classes and must contain the identity. Select conjugacy classes and compare the resulting size with the divisors of

60

. The key observation is that no selected proper nontrivial class union has an order that divides

60

The identity class must always be included in a subgroup. The class-sum test eliminates possible normal subgroups because subgroup orders must divide the group order.

Let

G

be a simple group of order

60

. Then

G

contains a subgroup of order

12

Given that

G

is a simple group of order

60=2^2\cdot 3\cdot 5

. To prove that

G

contains a subgroup of order

12

. Let

n_2

be the number of Sylow

2

-subgroups of

G

. By Sylow's third theorem,

\begin{align*} n_2 &\equiv 1 \pmod 2, \\ n_2 &\mid 15. \end{align*}

Thus

n_2\in \{1,3,5,15\}

. Since

G

is simple,

n_2\neq 1

. If

n_2=3

, then conjugation on the three Sylow

2

-subgroups gives a homomorphism

G\to S_3

. Since

G

is simple and the action is nontrivial,

G

would embed in

S_3

, impossible because

60\nmid 6

. Thus

n_2\neq 3

. If

n_2=5

, then the normalizer of a Sylow

2

-subgroup

P

has order

\begin{equation*} |N_G(P)|=\frac{60}{5}=12, \end{equation*}

G

contains a subgroup of order

12

. If

n_2=15

, then a standard Sylow intersection and normalizer argument gives a subgroup of order

12

. In detail, the Sylow

2

-subgroups have order

4

. If all pairwise intersections were trivial, they would contribute too many distinct elements when combined with the Sylow

3

- and Sylow

5

-counts. Hence two Sylow

2

-subgroups intersect in a subgroup

C

of order

2

. Its normalizer contains more than one Sylow

2

-subgroup, so

|N_G(C)|

is divisible by

4

and greater than

4

. The possible orders then force a subgroup of order

12

. Hence

G

contains a subgroup of order

12

\square

This calculator connects the Sylow and coset-action steps used in the order

60

argument. Enter a group order, a prime, and a candidate number of Sylow subgroups. The calculator computes the normalizer order when the candidate is consistent. It also checks a subgroup order and index, showing how a subgroup of order

12

gives an action on

5

cosets and therefore a homomorphism into

S_5

The normalizer calculation comes from counting conjugates of a Sylow subgroup. The coset action turns a subgroup of index

5

into a permutation representation, which is the bridge from an abstract simple group of order

60

A_5

Any simple group of order

60

is isomorphic to

A_5

Given that

G

is a simple group of order

60

. To prove that

G\cong A_5

. By the preceding lemma,

G

contains a subgroup

H

of order

12

. Hence

\begin{equation*} [G:H]=\frac{60}{12}=5. \end{equation*}

Let

G

act on the set of left cosets of

H

by left multiplication. This action gives a homomorphism

\begin{equation*} \varphi:G\to S_5. \end{equation*}

The action is nontrivial, so

\ker\varphi\neq G

. Since

G

is simple, we must have

\begin{equation*} \ker\varphi=\{e\}. \end{equation*}

Therefore

\varphi

is injective, and

G

is isomorphic to a subgroup

T

S_5

with

\begin{equation*} |T|=60. \end{equation*}

Consider the sign homomorphism restricted to

T

. If

T

were not contained in

A_5

, then

T\cap A_5

would be a subgroup of

T

of index

2

. A subgroup of index

2

is normal, so this would give a nontrivial proper normal subgroup of

T

, contradicting the simplicity of

T

. Thus

T\subseteq A_5

. Since

\begin{equation*} |T|=60=|A_5|, \end{equation*}

we get

T=A_5

. Hence

G\cong A_5

\square

The group

A_5

is the smallest nonabelian simple group.

Given that every nonprime composite order less than

60

does not yield a simple group and that

A_5

is simple of order

60

. To prove that

A_5

is the smallest nonabelian simple group. Groups of prime order are cyclic and abelian. The previous non-simplicity tests rule out all nonprime orders below

60

for nonabelian simple groups. Since

A_5

is nonabelian, simple, and has order

60

, no smaller nonabelian simple group exists. Hence

A_5

is the smallest nonabelian simple group.

\square

Let

G

be a simple group of order

168

. Show that

G

has eight Sylow

7

-subgroups and that the normalizer of a Sylow

7

-subgroup has order

21

Let

|G|=168=2^3\cdot 3\cdot 7

. Let

n_7

be the number of Sylow

7

-subgroups. By Sylow's third theorem,

\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 24. \end{align*}

Thus

n_7=1

n_7=8

. Since

G

is simple,

n_7\neq 1

. Hence

\begin{equation*} n_7=8. \end{equation*}

Let

P

be a Sylow

7

-subgroup of

G

. The Sylow

7

-subgroups are precisely the conjugates of

P

, so their number is

\begin{equation*} [G:N_G(P)]. \end{equation*}

Therefore

\begin{equation*} [G:N_G(P)]=8. \end{equation*}

It follows that

\begin{equation*} |N_G(P)|=\frac{168}{8}=21. \end{equation*}

Hence

G

has eight Sylow

7

-subgroups and

|N_G(P)|=21

\square

1. Compute

|A_5|

from

|S_5|

. 2. Explain why a subgroup of index

2

is normal. 3. If a simple group

G

of order

60

has a subgroup of order

12

, explain why it gives an embedding

G\to S_5

1. Since

|S_5|=5!=120

and the even permutations form a subgroup of index

2

|A_5|=60

. 2. A subgroup of index

2

has exactly two left cosets and two right cosets, so the left and right cosets agree. 3. The action on the five left cosets gives a homomorphism

G\to S_5

. Since

G

is simple and the action is nontrivial, the kernel is trivial.

It is the first nonabelian simple group and shows that Sylow methods do not always force non-simplicity.

A subgroup of index

5

creates a natural permutation representation into

S_5

If the image had odd permutations, its intersection with

A_5

would be a normal subgroup of index

2

, contradicting simplicity.

BMLabs Mathematics

BMLabs Mathematics

Simple Groups of Order 60 and A5

Simple Groups of Order 60 and A5

ALTERNATING GROUP A5

EXPLORE A5 CLASS SUMS

A5 CLASS SUM EXPLORER

INTERPRET THE CLASS SUM

CALCULATE ORDER SIXTY SYLOW DATA

ORDER SIXTY SYLOW AND COSET ACTION CALCULATOR

CONNECT THE CALCULATOR

FREQUENTLY ASKED QUESTIONS

Coordinates and Locus

Invariants under Orthogonal Transformation