[proofplan] We compute the signature of the given product using the homomorphism property of $\operatorname{sgn}$. Since each factor is a transposition, each factor contributes a multiplicative factor of $-1$. The independence of parity then follows by comparing two decompositions of the same permutation and observing that $(-1)^a=(-1)^b$ exactly when $a$ and $b$ have the same parity. [/proofplan] [step:Apply multiplicativity of the signature to the transposition product] If $m=0$, then by the empty product convention $\sigma=\operatorname{id}_{\{1,\dots,n\}}$. Since $\operatorname{sgn}:S_n\to\{1,-1\}$ is a [group homomorphism](/page/Group%20Homomorphism) by [citetheorem:7877], it sends the identity of $S_n$ to the identity $1$ of $\{1,-1\}$. Hence \begin{align*} \operatorname{sgn}(\sigma)=1=(-1)^0. \end{align*} Assume now that $m\ge 1$. Since $\operatorname{sgn}:S_n\to\{1,-1\}$ is a group homomorphism by [citetheorem:7877], applying multiplicativity successively to the product $\tau_1\tau_2\cdots\tau_m$ gives \begin{align*} \operatorname{sgn}(\sigma)=\operatorname{sgn}(\tau_1)\operatorname{sgn}(\tau_2)\cdots\operatorname{sgn}(\tau_m). \end{align*} [guided] First we separate the possible empty product case. If $m=0$, the hypothesis says that $\sigma$ is the empty product in $S_n$, so $\sigma=\operatorname{id}_{\{1,\dots,n\}}$. The map $\operatorname{sgn}:S_n\to\{1,-1\}$ is a group homomorphism by [citetheorem:7877], so it preserves identity elements. Therefore \begin{align*} \operatorname{sgn}(\sigma)=\operatorname{sgn}(\operatorname{id}_{\{1,\dots,n\}})=1. \end{align*} Since $(-1)^0=1$, the desired formula holds when $m=0$. Now suppose $m\ge 1$. The point of using [citetheorem:7877] is that it converts the signature of a product into the product of the signatures. The theorem applies because every factor $\tau_i$ lies in $S_n$, and the product $\tau_1\tau_2\cdots\tau_m$ is an element of $S_n$. Applying the homomorphism property first to $\tau_1$ and $\tau_2\cdots\tau_m$, and then repeatedly to the remaining product, gives \begin{align*} \operatorname{sgn}(\tau_1\tau_2\cdots\tau_m)=\operatorname{sgn}(\tau_1)\operatorname{sgn}(\tau_2)\cdots\operatorname{sgn}(\tau_m). \end{align*} Since the hypothesis identifies $\sigma$ with this product, we obtain \begin{align*} \operatorname{sgn}(\sigma)=\operatorname{sgn}(\tau_1)\operatorname{sgn}(\tau_2)\cdots\operatorname{sgn}(\tau_m). \end{align*} [/guided] [/step] [step:Replace each transposition signature by $-1$] For each index $i\in\{1,\dots,m\}$, the permutation $\tau_i$ is a transposition by hypothesis. Hence [citetheorem:7875] gives \begin{align*} \operatorname{sgn}(\tau_i)=-1. \end{align*} Substituting this into the product from the previous step yields \begin{align*} \operatorname{sgn}(\sigma)=\prod_{i=1}^m(-1)=(-1)^m. \end{align*} This proves the asserted formula for every $m\in\mathbb{Z}_{\ge 0}$. [/step] [step:Compare two decompositions to prove parity independence] Suppose that $\sigma$ also has a second [transposition decomposition](/theorems/777) \begin{align*} \sigma=\rho_1\rho_2\cdots\rho_\ell, \end{align*} where $\ell\in\mathbb{Z}_{\ge 0}$ and each $\rho_j\in S_n$ is a transposition. Applying the formula already proved to the two decompositions gives \begin{align*} (-1)^m=\operatorname{sgn}(\sigma)=(-1)^\ell. \end{align*} Therefore $(-1)^{m-\ell}=1$, so $m-\ell$ is even. Thus $m$ and $\ell$ have the same parity. This proves that the parity of the length of a transposition decomposition of $\sigma$ is independent of the chosen decomposition. [/step]