Alternating group

The alternating group $A l t 𝑛$ of degree $𝑛$ is the kernel of the alternating character $s g n : S 𝑛 \to S 2$ , #m/def/group and therefore a normal subgroup made up of all even permutations. For $𝑛 \geq 2$ we have the split extension (and hence semidirect product)

1 \to A l t 𝑛 ↪ S 𝑛 ↠ 2 \to 1

Simplicity

An important property of the alternating group $A l t 𝑛$ for $𝑛 \geq 5$ is that it is a simple group. #m/thm/group This is proven using the following lemmata

$A l t 𝑛$ for $𝑛 \geq 3$ is generated by 3-cycles.
If $𝑁 ⊴ A l t 𝑛$ with $𝑛 \geq 3$ contains a 3-cycle, then $𝑁 = A l t 𝑛$ .
Every nontrivial $𝑁 ⊴ A l t 𝑛$ for $𝑛 \geq 5$ contains a 3-cycle.

Proof

Since pairs of transpositions generate $A l t 𝑛$ for $𝑛 \geq 3$ by construction, we need only show that any pair of transpositions can be written as a product of 3-cycles. Noting that $(𝑎, 𝑏) = (𝑏, 𝑎)$ , the following list exhausts any pair of transpositions:

(𝑎, 𝑏) (𝑎, 𝑏) = 𝑒 (𝑎, 𝑏) (𝑐, 𝑑) = (𝑎, 𝑐, 𝑏) (𝑎, 𝑐, 𝑑) (𝑎, 𝑏) (𝑎, 𝑐) = (𝑎, 𝑐, 𝑏)

proving ^S1.

Now $A l t 𝑛$ for $𝑛 \geq 3$ can in fact be generated only from 3-cycles of the form $(𝑖, 𝑗, 𝑘)$ with $𝑖, 𝑗$ fixed in $ℕ 𝑛$ , since every 3-cycle can be expressed as such:

(𝑖, 𝑎, 𝑗) = (𝑖, 𝑗, 𝑎) 2 (𝑖, 𝑎, 𝑏) = (𝑖, 𝑗, 𝑏) (𝑖, 𝑎, 𝑗) (𝑗, 𝑎, 𝑏) = (𝑖, 𝑏, 𝑗) (𝑖, 𝑗, 𝑎) (𝑎, 𝑏, 𝑐) = (𝑗, 𝑐, 𝑎) (𝑗, 𝑎, 𝑏)

Now assume $𝑁 ⊴ A l t 𝑛$ contains a 3-cycle, say $(𝑖, 𝑗, 𝑎)$ . By normality it follows for any $𝑏 \in ℕ 𝑛 ∖ {𝑖, 𝑗, 𝑎}$

(𝑖, 𝑗, 𝑏) = [(𝑖, 𝑗) (𝑎, 𝑏)] (𝑖, 𝑗, 𝑎) 2 [(𝑖, 𝑗) (𝑎, 𝑘)] - 1 \in 𝑁

Hence $𝑁$ contains all 3-cycles of the form $(𝑖, 𝑗, 𝑏)$ and hence all 3-cycles, thus by ^S1 it is $A l t 𝑛$ , proving ^S2.

Now let $𝑁 ⊴ A l t 𝑛$ for $𝑛 \geq 5$ be nontrivial. Then one of the following holds:

Case 1: Suppose there exists $𝜎 \in 𝑁$ with a cycle of length $\geq 4$ , so without loss of generality (by relabelling) $𝜎 = (123 \dots 𝑟) 𝜏$ where $(123 \dots 𝑟)$ and $𝜏$ are disjoint. Since $(123) \in A l t 𝑛$ , it follows

𝜎 - 1 (123) 𝜎 (123) - 1 = 𝜏 - 1 (123 \dots 𝑟) - 1 (123) (123 \dots 𝑟) 𝜏 (123) - 1 = (𝑟 \dots 321) (231 \dots 𝑟) = (13 𝑟) \in 𝑁

so $𝑁$ contains a 3-cycle.

Case 2a: Suppose there exists a $𝜎 \in 𝑁$ which contains two disjoint 3-cycles (and nothing longer). Without loss of generality, $𝜎 = (123) (456) 𝜏$ for disjoint $(123)$ , $(456)$ , and $𝜏$ . Since $(124) \in A l t 𝑛$ , it follows

𝜎 - 1 (124) 𝜎 (124) - 1 = 𝜏 - 1 (456) - 1 (123) - 1 (124) (123) (456) 𝜏 (124) - 1 = (654) (321) (123) (123) (456) (421) = (14263) \in 𝑁

which falls under case 1.

Case 2b: Suppose there exists a $𝜎 \in 𝑁$ containing exactly one 3-cycle and otherwise only transpositions. Without loss of generality $𝜎 = (123) 𝜏$ where $(123)$ and $𝜏$ are disjoint, and $𝜏 - 1 = 𝜏$ . Then $𝜎 2 = (123) 𝜏 (123) 𝜏 = (123) 2 = (132) \in 𝑁$ , so $𝑁$ contains a 3-cycle.

Case 2c: If $𝑁$ contains a 3-cycle we are already done.

Case 3: The only remaining possibility is that there exists a $𝜎 \in 𝑁$ which is a product of disjoint transpositions, and an even number thereof since $𝑁 ⊴ A l t 𝑛$ . Without loss of generality $𝜎 = (12) (34) 𝜏$ with $(12)$ , $(34)$ , $𝜏$ disjoint and $𝜏 - 1 = 𝜏$ . Then

𝜎 - 1 (123) 𝜎 (123) - 1 = 𝜏 - 1 (34) (12) (123) (12) (34) 𝜏 (132) = (34) (12) (123) (12) (34) (132) = (13) (24) \in 𝑁

whence

(13) (24) (135) (13) (24) (135) - 1 = (135) \in 𝑁

thus $𝑁$ contains a 3-cycle.

This proves ^S3 and therewith the simplicity of $A l t 𝑛$ for $𝑛 \geq 5$ .

Note that $A l t 2$ is trivial, $A l t 3 ≅ ℤ 3$ is Abelian and simple, but $A l t 4$ is not simple as ${𝑒, (12) (34), (13) (24), (14) (23)} ◃ A l t 4$ . See Decomposition of S4.

Properties

$A l t 𝑛$ is $(𝑛 - 2)$ -transitive.

#state/tidy | #lang/en | #SemBr