Alternating Groups

After learning the symmetric group, we now study one of its most important subgroups: the alternating group. The alternating group consists of even permutations. It is a standard finite group that appears throughout abstract algebra, especially in permutation theory, group actions, and the study of solvability. The smallest nontrivial alternating group,

\mathbb{A}_3

, has only three elements, while

\mathbb{A}_n

has

\dfrac{n!}{2}

elements for

n\geq2

. In this lesson, students will verify that alternating groups are finite groups under composition.

Let

\sigma\in\mathbb{S}_n

. The permutation

\sigma

is called an

\textbf{even permutation}

if it can be expressed as a product of an even number of transpositions.

\mathbb{A}_3

, has only three elements, while

\mathbb{A}_n

has

\dfrac{n!}{2}

elements for

n\geq2

. In this lesson, students will verify that alternating groups are finite groups under composition.

Let

\sigma\in\mathbb{S}_n

. The permutation

\sigma

is called an

\textbf{even permutation}

if it can be expressed as a product of an even number of transpositions.

Let

n\in\mathbb{N}

. The

\textbf{alternating group}

\mathbb{A}_n

is the set of all even permutations in

\mathbb{S}_n

. Thus

\begin{equation*} \mathbb{A}_n=\{\sigma\in\mathbb{S}_n:\sigma \text{ is even}\}. \end{equation*}

The operation on

\mathbb{A}_n

is composition of permutations.

The parity of a permutation is well-defined: a permutation cannot be both even and odd. This fact allows us to separate

\mathbb{S}_n

into even and odd permutations. For

n\geq2

, exactly half of the permutations are even and half are odd, so

\begin{equation*} |\mathbb{A}_n|=\frac{n!}{2}. \end{equation*}

The identity permutation is even because it is a product of zero transpositions.

Let

n\geq2

. Then

(\mathbb{A}_n,\circ)

is a finite group under composition of permutations, and

\begin{equation*} |\mathbb{A}_n|=\frac{n!}{2}. \end{equation*}

Given that

n\geq2

and

\mathbb{A}_n

is the set of all even permutations in

\mathbb{S}_n

. To prove that

(\mathbb{A}_n,\circ)

is a finite group. [1] To prove closure. Let

\sigma,\tau\in\mathbb{A}_n

. Then

\sigma

and

\tau

are even permutations. A product of two even permutations is even because the total number of transpositions in the combined expression is even. Therefore

\begin{equation*} \sigma\circ\tau\in\mathbb{A}_n. \end{equation*}

[2] To prove associativity. Let

\sigma,\tau,\rho\in\mathbb{A}_n

. Since composition of functions is associative,

\begin{equation*} (\sigma\circ\tau)\circ\rho=\sigma\circ(\tau\circ\rho). \end{equation*}

[3] To prove the existence of identity element. The identity permutation

I

is even because it is a product of zero transpositions. Hence

I\in\mathbb{A}_n

. Also

\begin{equation*} \sigma\circ I=I\circ\sigma=\sigma \quad \forall\, \sigma\in\mathbb{A}_n. \end{equation*}

Therefore

I

is the identity element. [4] To prove the existence of inverse elements. Let

\sigma\in\mathbb{A}_n

. Then

\sigma

is even. If

\sigma

is expressed as a product of an even number of transpositions, then

\sigma^{-1}

is obtained by reversing the order of those transpositions. The number of transpositions remains even. Therefore

\sigma^{-1}

is even and

\begin{equation*} \sigma^{-1}\in\mathbb{A}_n. \end{equation*}

Also

\begin{equation*} \sigma\circ\sigma^{-1}=\sigma^{-1}\circ\sigma=I. \end{equation*}

Thus every element has an inverse in

\mathbb{A}_n

. Since half of the

n!

permutations in

\mathbb{S}_n

are even for

n\geq2

, we get

\begin{equation*} |\mathbb{A}_n|=\frac{n!}{2}. \end{equation*}

Hence,

(\mathbb{A}_n,\circ)

is a finite group.

\square

The group

\mathbb{A}_3

\begin{equation*} \mathbb{A}_3=\{I,(123),(132)\}. \end{equation*}

It has

\begin{equation*} |\mathbb{A}_3|=\frac{3!}{2}=3 \end{equation*}

elements. The two

3

-cycles are even because each

3

-cycle can be written as a product of two transpositions.

The group

\mathbb{A}_4

has

\begin{equation*} |\mathbb{A}_4|=\frac{4!}{2}=12 \end{equation*}

elements. It contains the identity permutation, all

3

-cycles in

\mathbb{S}_4

, and the products of two disjoint transpositions such as

(12)(34)

Prove that

\mathbb{A}_3

is a finite group under composition.

Given that

\begin{equation*} \mathbb{A}_3=\{I,(123),(132)\}. \end{equation*}

To prove that

(\mathbb{A}_3,\circ)

is a finite group. Since

\mathbb{A}_3

consists of the even permutations in

\mathbb{S}_3

, closure follows because the composition of two even permutations is even. Associativity follows from associativity of function composition. The identity permutation

I

belongs to

\mathbb{A}_3

and satisfies

\begin{equation*} \sigma\circ I=I\circ\sigma=\sigma \quad \forall\, \sigma\in\mathbb{A}_3. \end{equation*}

The inverse of

I

I

, the inverse of

(123)

(132)

, and the inverse of

(132)

(123)

. Thus every element has an inverse in

\mathbb{A}_3

. Since

\mathbb{A}_3

has

3

elements, it is finite. Hence,

(\mathbb{A}_3,\circ)

is a finite group.

\square

Find the order of

(123)

\mathbb{A}_3

Let

\sigma=(123)\in\mathbb{A}_3

. To find the order of

\sigma

. We compute

\begin{equation*} \sigma^1=(123), \end{equation*}

\begin{equation*} \sigma^2=(132), \end{equation*}

and

\begin{equation*} \sigma^3=I. \end{equation*}

Since no smaller positive power of

\sigma

I

, the least positive integer

n

such that

\sigma^n=I

3

. Hence,

\begin{equation*} o((123))=3. \end{equation*}

\square

The alternating group

\mathbb{A}_n

is not merely a subset of

\mathbb{S}_n

; it is a group in its own right under the same operation. The main verification points are closure under composition and closure under inverses. Both rely on parity of permutations. If a proof simply says that

\mathbb{A}_n

is inside

\mathbb{S}_n

, it is incomplete because a subset of a group need not be a group.

Use the calculator below to compute the size of

\mathbb{A}_n

for

n\geq2

. It uses the formula

|\mathbb{A}_n|=n!/2

. This reinforces that alternating groups are finite groups.

[1] Define an even permutation. [2] Define

\mathbb{A}_n

. [3] Find

|\mathbb{A}_5|

. [4] List the elements of

\mathbb{A}_3

. [5] Find the inverse of

(123)

\mathbb{A}_3

[1] An even permutation is a permutation expressible as a product of an even number of transpositions. [2]

\mathbb{A}_n

is the set of all even permutations in

\mathbb{S}_n

. [3]

|\mathbb{A}_5|=5!/2=60

. [4]

\mathbb{A}_3=\{I,(123),(132)\}

. [5] The inverse of

(123)

(132)

It is the group of all even permutations in

\mathbb{S}_n

The operation is composition of permutations.

For

n\geq2

|\mathbb{A}_n|=n!/2

Yes. It has three elements:

I,(123),(132)

BMLabs Mathematics

BMLabs Mathematics

Alternating Groups

Alternating Groups

EVEN PERMUTATION

ALTERNATING GROUP

ALTERNATING GROUP SIZE

KLEIN FOUR GROUP

Coordinates and Locus

Invariants under Orthogonal Transformation