📂Abstract Algebra

Alternating Group in Abstract Algebra

Definition ¹

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

Theorem

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

Explanation

The fact that the order of $A_{n}$ is exactly half of $\left| S_{n} \right|$ is quite an interesting property. The alternating group is a very important group, since it is later used to show that equations of degree $5$ or higher do not have a formula for their roots.

Proof

First, we must show that $A_{n}$ is a group:

• (i): Since the composition of even permutations is an even permutation, $A_{n}$ 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 expressed by an even number, $0$, of transpositions, so it is an even permutation, and $A_{n}$ has the identity element $\iota$.
• (iv): For any transposition $(i, j)$, 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 us set $A_{n}^{c} := S_{n} \setminus A_{n}$ for convenience. If a bijection $f : A_{n} \to A_{n}^{c}$ exists, then $$ \left| A_{n} \right| = \left| A_{n}^{c} \right| \\ n! = \left| S_{n} \right| = \left| A_{n} \right| + \left| A_{n}^{c} \right| $$ so $\displaystyle \left| A_{n} \right| = {{ n! } \over {2}}$.

Now, if we concretely show that the function $f ( x ) = (1,2) x$ is a bijection, the proof is complete.

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

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

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Tip

If you look closely at the part in (iii) that establishes the existence of the identity element, you will understand why even permutations, rather than odd permutations, are used.