📂Abstract Algebra

Alternating groups in Abstract Algebra

Definition ¹

The group formed by the even permutations of the symmetric group $S_{n}$ is called the Alternating group and is denoted by $A_{n}$.

Theorem

For $n \ge 2$, $$\left| A_{n} \right| = {{\left| S_{n} \right|} \over {2}} = {{ n! } \over {2}}$$

Description

It is quite interesting that the order of $A_{n}$ is exactly half of $\left| S_{n} \right|$. The alternating group is considered to be very important, as it is used later to show that equations of degree higher than $5$ do not have a formula for roots.

Proof

First, we have to show that $A_{n}$ forms a group:

• (i): The composition of two even permutations is an even permutation, hence $A_{n}$ is closed under the operation $\circ$.
• (ii): Since $A_{n} \subset S_{n}$, the associative law holds.
• (iii): The identity function $\iota = \begin{bmatrix} 1 & 2 & \cdots & n \\ 1 & 2 & \cdots & n \end{bmatrix}$ is represented by an even number $0$ of transpositions, making it an even permutation, and thus $A_{n}$ has an identity element $\iota$.
• (iv): For any permutation $(i, j)$, since $(i, j) (i , j) = \iota$, the inverse of an even permutation is also an even permutation.

Now to show that $\displaystyle \left| A_{n} \right| = {{ n! } \over {2}}$, let’s conveniently say $A_{n}^{c} := S_{n} \setminus A_{n}$. If a bijective $f : A_{n} \to A_{n}^{c}$ exists, $$\left| A_{n} \right| = \left| A_{n}^{c} \right| \\ n! = \left| S_{n} \right| = \left| A_{n} \right| + \left| A_{n}^{c} \right|$$ then $\displaystyle \left| A_{n} \right| = {{ n! } \over {2}}$ would hold.

Finally, to specifically prove that the function $f ( x ) = (1,2) x$ is bijective, the proof concludes.

• If $f ( \sigma ) = f ( \tau )$, then $(1,2) \sigma = (1,2) \tau$ and multiplying both sides by $(1,2)$ yields $\sigma = \tau$, so $f$ is injective.
• For any $\xi \in A_{n}^{c}$, since $(1,2) \xi$ is an even permutation and $f( (1,2) \xi ) = (1,2)(1,2) \xi = \xi$, $f$ is surjective.

Thus, the following holds. $$\left| A_{n} \right| = {{\left| S_{n} \right|} \over {2}} = {{ n! } \over {2}}$$

■

Tip

Part (iii) on the existence of an identity element illustrates why even permutations are used instead of odd ones.