The Alternating Group

Let $\sigma \in S_n$, the number of inversions of $\sigma$ is $\vert \{(i,j); 0 \leq i < j < n; \sigma(j) < \sigma(i)\} \vert$ = inv($\sigma$). For example (1,2) has 1 inversion, $j > i, (i,j)$ has $2(j-i-1)+1$ inversions, (1,2,3) has 2 inversions.

Let $\sigma \in S_n$ we define the sign of $\sigma$ to be $(-1)^{inv(\sigma)} = \varepsilon(\sigma) \in {1, -1}$. For example $\varepsilon((i j)) = -1, \varepsilon((1 2 3)) = 1, \varepsilon(1) = 1$.

Proposition: Let $\Delta = \prod_{i < j}(x_i-x_j)$. Let $\sigma \in S_n$ and let $\sigma(\Delta) = \prod_{i<j}(x_{\sigma(i)}-x_{\sigma(j)})$, then $\sigma(\Delta) = \varepsilon(\sigma) \cdot \Delta$

Proof: $\sigma(\Delta) = \prod_{i < j}(x_{\sigma(i)} - x_{\sigma(j)}) = \prod_{\sigma(i)<\sigma(j)}((x_{\sigma(i)}-x_{\sigma(j)})(-1)^{\varepsilon_{ij}})$ where $\varepsilon_{ij} = 0$ if there is no inversion or 1 if there is inversion. This expression is therefore equal to $(-1)^{inv(\sigma)}\prod_{k <l}(x_k-x_l) = \varepsilon(\sigma)\cdot \Delta$

Proposition: $\sigma : S_n \rightarrow \{1, -1\}$ is a group morphism.

Proof: let $\tau, \sigma \in S_n$. Then $\tau \sigma(\Delta) = \varepsilon(\tau \sigma)\Delta =$ $\prod_{i<j}(x_{\tau(\sigma(i))}-x_{\tau(\sigma(j))}) =$ $\prod_{\sigma(i)<\sigma(j)}(x_{\tau(\sigma(i))}-x_{\tau(\sigma(j))}) =$ $(-1)^{inv(\sigma)} \prod_{k<l}(x_{\tau(k)}-x_{\tau(l)}) =$ $\varepsilon(\sigma)\tau(\Delta) =$ $\varepsilon(\sigma)\varepsilon(\tau)\Delta$ so $\varepsilon(\tau \sigma) = \varepsilon(\tau)\varepsilon(\sigma)$

Alternating group ($A_n$): the alternating group is defined as $A_n =$ ker($\varepsilon) \unlhd S_n$. For $n \geq 2$ by the first isomorphism theorem $\{1, -1\} \cong S_n/A_n$ and by Lagrange $\vert A_n \vert = \frac{\vert S_n \vert}{2} = \frac{n!}{2}$. For example: $A_1 = S_1 = \{1\}, A_2 = \{1\}, A_3 = <(1 2 3)>$

Proposition: if $\gamma \in S_n$ is a k cycle, then $\varepsilon(\gamma) = (-1)^{k-1}$

Proof: let’s consider $\gamma = (0 1 ... k-1) = (k-1 0)(k-2 0)...(2 0)(1 0)$ $\varepsilon(\gamma) = (-1)^{k-1}, \gamma = (a_0 ... a_{k-1})$, where $\sigma \in S_n$. Claim: $\sigma \gamma \sigma^{-1} = (\sigma(a_0)...\sigma(a_{k-1})), 0 < i < k-1$ $\sigma \gamma \sigma^{-1}(\sigma(a_i)) = \sigma \gamma(a_i) = \sigma(a_{i+1})$ $\sigma \gamma \sigma^{-1} (\sigma(a_{k-1})) = \sigma \gamma (a_{k-1}) = \sigma(a_0)$ Let $x \not\in \{\sigma(a_i), 0 \leq i < k\}$ then $\sigma \gamma \sigma^{-1}(x) = \sigma(\sigma^{-1}(x)) = x$ so$\sigma^{-1}(x) \not\in \{a_i, 0 \leq i < k\}$. Let $\sigma(i) = a_i, 0 \leq i < k$. By induction $l \geq k =$ min$\{0 ... n-1\}\backslash \{\sigma(i) j < l\},$ by construction $\sigma$ is injective so $\sigma \in S_n$ $\sigma(0 ... k-1)\sigma^{-1} = (\sigma(0)... \sigma(k-1)) = (a_0 ... a_{k-1}) = \gamma$ $\varepsilon(\gamma) = \varepsilon(\sigma(0 ... k-1)\sigma^{-1}) = \varepsilon(\sigma)\varepsilon((0 ... k-1))\epsilon(\sigma) = (-1)^{k-1}$

From above it follows that $A_4 = \{$1, (1 2 3), (1 3 2), (1 2)(3 4), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3), (1 3)(2 4), (1 4)(2 3)$\}$. Note that $A_4$ is not simple.

Theorem: $A_n$ is simple if $n \geq 5$. $A_n$ is generated by 3 cycles.