The proof proceeds by identifying $A_n$ as the kernel of a homomorphism and then using standard facts about kernels and the counting argument for even versus odd permutations. **Step 1: $A_n$ is a subgroup.** The even permutations form the [set](/page/Set) $A_n = \{\sigma \in S_n : \operatorname{sgn}(\sigma) = 1\} = \ker(\operatorname{sgn})$. Since the kernel of any group homomorphism is a subgroup, $A_n \leq S_n$. **Step 2: $A_n$ is normal.** The kernel of any homomorphism is a normal subgroup. Therefore $A_n \unlhd S_n$. Alternatively: the index $|S_n : A_n| = 2$ (since $\operatorname{sgn}$ is surjective onto $\{+1, -1\}$, the [First Isomorphism Theorem](/theorems/791) gives $S_n / A_n \cong \{+1, -1\}$). Any subgroup of index $2$ is normal. **Step 3: $|A_n| = n!/2$.** [claim:Equal Count of Even and Odd Permutations] For $n \geq 2$, there are exactly as many even permutations in $S_n$ as odd ones. [/claim] [proof] Let $\tau = (1\; 2) \in S_n$. Define $\varphi : A_n \to S_n \setminus A_n$ by $\varphi(\sigma) = \tau\sigma$. This is well-defined since $\operatorname{sgn}(\tau\sigma) = (-1)(+1) = -1$, so $\tau\sigma$ is odd. It is injective: if $\tau\sigma_1 = \tau\sigma_2$, then $\sigma_1 = \sigma_2$ by left cancellation. It is surjective: given any odd $\pi$, the permutation $\tau\pi$ is even (since $\operatorname{sgn}(\tau\pi) = (-1)(-1) = +1$), and $\varphi(\tau\pi) = \tau(\tau\pi) = \pi$. So $|A_n| = |S_n \setminus A_n|$, giving $|A_n| = n!/2$. [/proof]