The strategy has three parts: (1) show that multiplying any permutation by a transposition changes the cycle count by $\pm 1$, (2) use this to prove that the parity of any transposition decomposition is an invariant (well-definedness), and (3) verify the homomorphism property. **Step 1: Transpositions change the cycle count by $\pm 1$.** Let $c(\sigma)$ denote the number of cycles in the [disjoint cycle decomposition](/theorems/775) of $\sigma$ (including $1$-cycles), so $c(\mathrm{id}) = n$. [claim:Cycle Count Parity Shift] For any $\sigma \in S_n$ and any transposition $\tau = (k\; l)$, $c(\sigma\tau) = c(\sigma) \pm 1$. [/claim] [proof] There are two cases. **Case 1:** $k$ and $l$ lie in different cycles of $\sigma$, say $(k\; a_1\; \ldots\; a_r)$ and $(l\; b_1\; \ldots\; b_s)$. Then: \begin{align*} (k\; a_1\; \ldots\; a_r)(l\; b_1\; \ldots\; b_s)(k\; l) = (k\; b_1\; \ldots\; b_s\; l\; a_1\; \ldots\; a_r). \end{align*} Two cycles merge into one, so $c(\sigma\tau) = c(\sigma) - 1$. **Case 2:** $k$ and $l$ lie in the same cycle of $\sigma$, say $(k\; a_1\; \ldots\; a_r\; l\; b_1\; \ldots\; b_s)$. Then: \begin{align*} (k\; a_1\; \ldots\; a_r\; l\; b_1\; \ldots\; b_s)(k\; l) = (k\; b_1\; \ldots\; b_s)(l\; a_1\; \ldots\; a_r). \end{align*} One cycle splits into two, so $c(\sigma\tau) = c(\sigma) + 1$. [/proof] In both cases, $c(\sigma\tau) \equiv c(\sigma) + 1 \pmod{2}$. **Step 2: Well-definedness of the sign.** Suppose $\sigma = \tau_1 \cdots \tau_a = \tau'_1 \cdots \tau'_b$ where each $\tau_i, \tau'_j$ is a transposition. By induction on the number of transpositions applied to $\mathrm{id}$, Step 1 gives: \begin{align*} c(\sigma) = c(\tau_1 \cdots \tau_a) \equiv n + a \pmod{2}, \end{align*} and similarly $c(\sigma) \equiv n + b \pmod{2}$. Therefore $n + a \equiv n + b \pmod{2}$, which gives $a \equiv b \pmod{2}$, i.e., $(-1)^a = (-1)^b$. So $\operatorname{sgn}(\sigma) = (-1)^a$ is independent of the choice of decomposition. **Step 3: Homomorphism property.** Let $\alpha, \beta \in S_n$ with $\alpha = \tau_1 \cdots \tau_a$ and $\beta = \tau'_1 \cdots \tau'_b$. Then $\alpha\beta = \tau_1 \cdots \tau_a \tau'_1 \cdots \tau'_b$ is a product of $a + b$ transpositions, so: \begin{align*} \operatorname{sgn}(\alpha\beta) = (-1)^{a+b} = (-1)^a(-1)^b = \operatorname{sgn}(\alpha)\operatorname{sgn}(\beta). \end{align*} Surjectivity: $\operatorname{sgn}(\mathrm{id}) = +1$ and $\operatorname{sgn}((1\; 2)) = -1$.