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  "body": "\n## Proof\n\nWe concatenate decompositions into transpositions. Products in the [[symmetric group|0RJMNEN3BJ]] $S_n$ are composites of maps, and $n \\geq 2$.\n\n1. Let $\\sigma, \\tau \\in S_n$. By the [[decomposition into transpositions|0XK6FECMW5]], write $\\sigma = \\tau_1 \\cdots \\tau_m$ and $\\tau = \\tau'_1 \\cdots \\tau'_l$, where each $\\tau_i$ and each $\\tau'_j$ is a [[transposition|02DTRYKGE8]].\n2. By the [[definition of the sign|0SK90238BH]], $\\operatorname{sgn}(\\sigma) = (-1)^m$ and $\\operatorname{sgn}(\\tau) = (-1)^l$. This does not depend on the chosen decompositions, because the sign is well defined.\n3. By associativity, $\\sigma \\tau = \\tau_1 \\cdots \\tau_m \\tau'_1 \\cdots \\tau'_l$. This is a product of $m + l$ transpositions.\n4. By the definition of the sign again, $\\operatorname{sgn}(\\sigma \\tau) = (-1)^{m+l} = (-1)^m (-1)^l = \\operatorname{sgn}(\\sigma) \\operatorname{sgn}(\\tau)$.\n5. The set $\\{1, -1\\}$ is a group under multiplication, and step 4 says that $\\operatorname{sgn}$ preserves products. So $\\operatorname{sgn}$ is a [[group homomorphism|0HV9Y7WG5R]].\n6. For surjectivity, the transposition $(1\\ 2)$ is a product of one transposition, so $\\operatorname{sgn}((1\\ 2)) = -1$. This uses $n \\geq 2$.\n7. The identity $e$ equals $(1\\ 2)(1\\ 2)$, a product of two transpositions. So $\\operatorname{sgn}(e) = (-1)^2 = 1$.\n8. By steps 6 and 7, both $1$ and $-1$ are values of $\\operatorname{sgn}$. So $\\operatorname{sgn} \\colon S_n \\to \\{1, -1\\}$ is a surjective group homomorphism.\n\n## Idea\n\nIf $\\sigma$ uses $m$ transpositions and $\\tau$ uses $l$, then writing the two products side by side expresses $\\sigma \\tau$ with $m + l$ transpositions. Since the parity of the count is well defined, the signs multiply. All the depth is in the well-definedness of the sign, not in this step.\n",
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