佐武線型のパフィアンのところ
前提
$n = 2p$次交代行列のパフィアン$P_{n} = P_{n}([x_{ij}]_{1\leq i,j \leq n})$について，関係式
$$
P_{n}([x_{ij}]) = x_{12}^{2-p}P_{n-2}([x_{12}x_{i+2,j+2}-x_{1,i+2}x_{2,j+2}+x_{1,j+2}x_{2,i+2}])$$
が得られている．このとき
$$ P_{n} = \frac{1}{2^{p}p!} \sum_{\sigma\in\mathfrak{S}_{n}} \sgn(\sigma) x_{\sigma(1)\sigma(2)} x_{\sigma(3)\sigma(4)} \cdots x_{\sigma(n-1)\sigma(n)}$$
が成り立つことを数学的帰納法で示そうとしている．以下では，その induction step の計算を追っていく（cf. satakep.82）．
Step 1.
まづ，上述の関係式と帰納法の仮定より，
$$
P_{n} = \frac{x_{12}^{2-p}}{2^{p-1}(p-1)!} \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) (x_{12}x_{\sigma(3)\sigma(4)}-x_{1,\sigma(3)}x_{2,\sigma(4)}+x_{1,\sigma(4)}x_{2,\sigma(3)} ) \cdots (x_{12}x_{\sigma(n-1)\sigma(n)}-x_{1,\sigma(n-1)}x_{2,\sigma(n)}+x_{1,\sigma(n)}x_{2,\sigma(n-1)})$$
を得る．ただし
$$
\mathfrak{S}_{n-2} = \{\sigma \in \mathfrak{S}_{n} \mid \sigma(1) = 1,\ \sigma(2) = 2\}$$
と見做している．
Step 2.
上式右辺の総和記号の中身を展開したときに出てくる項は，各因子からの$x_{12}$を含む項の寄与によって次の3種類に分けられる：
- $x_{12}$を含む項の寄与が$p-1$：
$$ x_{12}^{p-1} \cdot x_{\sigma(3)\sigma(4)}\cdots x_{\sigma(n-1)\sigma(n)};$$ - $x_{12}$を含む項の寄与が$p-2$：
\begin{align} -x_{12}^{p-2} \cdot x_{\sigma(3)\sigma(4)} \cdots x_{1,\sigma(2i-1)}x_{2,\sigma(2i)} \cdots x_{\sigma(n-1)\sigma(n)}&,\ 2 \leq i \leq p;\\ x_{12}^{p-2} \cdot x_{\sigma(3)\sigma(4)} \cdots x_{1,\sigma(2j)}x_{2,\sigma(2j-1)} \cdots x_{\sigma(n-1)\sigma(n)}&,\ 2 \leq j \leq p; \end{align} - $x_{12}$を含む項の寄与が$p-3$以下．
\begin{align} \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{\sigma(3)\sigma(4)} \cdots x_{1,\sigma(2i-1)}x_{2,\sigma(2i)} \cdots x_{\sigma(n-1)\sigma(n)} &= \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)};\\ \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{\sigma(3)\sigma(4)} \cdots x_{1,\sigma(2j)}x_{2,\sigma(2j-1)} \cdots x_{\sigma(n-1)\sigma(n)} &= - \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}. \end{align}
- $\tau \coloneqq (3,2i-1)\circ(4,2i)$とおくと，$\sgn(\tau)=1$であり，写像
$$ \mathfrak{S}_{n-2} \to \mathfrak{S}_{n-2};\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align} \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{\sigma(3)\sigma(4)} \cdots x_{1,\sigma(2i-1)}x_{2,\sigma(2i)} \cdots x_{\sigma(n-1)\sigma(n)} &= \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(2i-1)\sigma_{\tau}(2i)} \cdots x_{1,\sigma_{\tau}(3)}x_{2,\sigma_{\tau}(4)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\ &= \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)} \end{align}
が成り立つ． - $\rho \coloneqq (2j-1,2j)$とおくと，$\sgn(\rho)=-1$であり，写像
$$ \mathfrak{S}_{n-2} \to \mathfrak{S}_{n-2};\ \sigma \mapsto \sigma\circ\rho \eqqcolon \sigma_{\rho}$$
は全単射である．よって，前段と合わせて，
\begin{align} \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{\sigma(3)\sigma(4)} \cdots x_{1,\sigma(2j)}x_{2,\sigma(2j-1)} \cdots x_{\sigma(n-1)\sigma(n)} &= -\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma_{\rho}) x_{\sigma_{\rho}(3)\sigma_{\rho}(4)} \cdots x_{1,\sigma_{\rho}(2j-1)}x_{2,\sigma_{\rho}(2j)} \cdots x_{\sigma_{\rho}(n-1)\sigma_{\rho}(n)} \\ &= - \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)} \end{align}
を得る．
$$ \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) \text{(3)型の項} = 0.$$
(3)型の項が，たとえば$x_{1,\sigma(2i-1)}x_{2,\sigma(2i)}x_{1,\sigma(2j-1)}x_{2,\sigma(2j)},\,i \neq j,\,$という因子を持つ場合，
\begin{align}
\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) \cdots x_{1,\sigma(2i-1)}x_{2,\sigma(2i)} \cdots x_{1,\sigma(2j-1)}x_{2,\sigma(2j)} \cdots
&= \sum_{\sigma\in\mathfrak{A}_{n-2}} \sgn(\sigma) \cdots x_{1,\sigma(2i-1)}x_{2,\sigma(2i)} \cdots x_{1,\sigma(2j-1)}x_{2,\sigma(2j)} \cdots + \sum_{\sigma\in\mathfrak{A}_{n-2}} \sgn(\sigma\circ(2i-1,2j-1)) \cdots x_{1,\sigma(2j-1)}x_{2,\sigma(2i)} \cdots x_{1,\sigma(2i-1)}x_{2,\sigma(2j)} \cdots \\
&= \sum_{\sigma\in\mathfrak{A}_{n-2}} \sgn(\sigma) \cdots x_{1,\sigma(2i-1)}x_{2,\sigma(2i)} \cdots x_{1,\sigma(2j-1)}x_{2,\sigma(2j)} \cdots - \sum_{\sigma\in\mathfrak{A}_{n-2}}\sgn(\sigma) \cdots x_{1,\sigma(2i-1)}x_{2,\sigma(2i)} \cdots x_{1,\sigma(2j-1)}x_{2,\sigma(2j)} \cdots \\
&= 0
\end{align}
が成り立つ．他の場合も同様である．
よって
\begin{align}
P_{n}
&= \frac{1}{2^{p-1}(p-1)!} \left(\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{12}x_{\sigma(3)\sigma(4)}\cdots x_{\sigma(n-1)\sigma(n)} - 2(p-1)\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}\right)\\
&= \frac{1}{2^{p}p!} \left(2p\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{12}x_{\sigma(3)\sigma(4)}\cdots x_{\sigma(n-1)\sigma(n)} - 4p(p-1)\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}\right)
\end{align}
が成り立つ．
Step 3.
あとは
$$
\sum_{\sigma\in\mathfrak{S}_{n}} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)} = 2p\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{12}x_{\sigma(3)\sigma(4)}\cdots x_{\sigma(n-1)\sigma(n)} - 4p(p-1)\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}$$
が成り立つことを示せばよい．
Step 3-0.
$x_{\bullet\bullet}$の添字における$1,2$が現れる位置によって$G \coloneqq \mathfrak{S}_{n}$を以下のように分割する：
\begin{align}
G_{12} &\coloneqq \{\sigma\in G \mid \exists\,i,\ \sigma(2i-1)=1,\ \sigma(2i)=2\};\\
G_{21} &\coloneqq \{\sigma\in G \mid \exists\,i,\ \sigma(2i)=1,\ \sigma(2i-1)=2\};\\
&\\
G_{10,20} &\coloneqq \{\sigma\in G \mid \exists\,i \neq j,\ \sigma(2i-1)=1,\ \sigma(2j-1)=2\};\\
G_{01,02} &\coloneqq \{\sigma\in G \mid \exists\,i \neq j,\ \sigma(2i)=1,\ \sigma(2j)=2\};\\
G_{10,02} &\coloneqq \{\sigma\in G \mid \exists\,i \neq j,\ \sigma(2i-1)=1,\ \sigma(2j)=2\};\\
G_{01,20} &\coloneqq \{\sigma\in G \mid \exists\,i \neq j,\ \sigma(2i)=1,\ \sigma(2j-1)=2\}.
\end{align}
Step 3-1.
各$i \in \{1,\ldots,p\}$に対して
$$
G_{12}(i) \coloneqq \{\sigma\in G_{12} \mid \sigma(2i-1)=1,\ \sigma(2i)=2\}$$
とおく．このとき，$G_{12} = \bigsqcup_{i} G_{12}(i)$であり，
$$
\sum_{\sigma\in G_{12}(i)} \sgn(\sigma)x_{\sigma(1)\sigma(2)}\cdots x_{\sigma(n-1)\sigma(n)} = \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{12}x_{\sigma(3)\sigma(4)}\cdots x_{\sigma(n-1)\sigma(n)}$$
が成り立つ．
$\tau \coloneqq (1,2i-1)\circ(2,2i)$とおくと，$\sgn(\tau)=1$であり，写像
$$
G_{12}(i) \to \mathfrak{S}_{n-2};\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align}
\sum_{\sigma\in G_{12}(i)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(2i-1)\sigma(2i)} \cdots x_{\sigma(n-1)\sigma(n)}
&= \sum_{\sigma\in G_{12}(i)} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(2i-1)\sigma_{\tau}(2i)} \cdots x_{\sigma_{\tau}(1)\sigma_{\tau}(2)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{12}x_{\sigma(3)\sigma(4)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
各$i \in \{1,\ldots,p\}$に対して
$$
G_{21}(i) \coloneqq \{\sigma\in G_{21} \mid \sigma(2i)=1,\ \sigma(2i-1)=2\}$$
とおく．このとき，$G_{21} = \bigsqcup_{i} G_{21}(i)$であり，
$$
\sum_{\sigma\in G_{21}(i)} \sgn(\sigma)x_{\sigma(1)\sigma(2)}\cdots x_{\sigma(n-1)\sigma(n)} = \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{12}x_{\sigma(3)\sigma(4)}\cdots x_{\sigma(n-1)\sigma(n)}$$
が成り立つ．
$\tau \coloneqq (2i-1,2i)$とおくと，$\sgn(\tau)=-1$であり，写像
$$
G_{21}(i) \to G_{12}(i);\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align}
\sum_{\sigma\in G_{21}(i)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(2i-1)\sigma(2i)} \cdots x_{\sigma(n-1)\sigma(n)}
&= -\sum_{\sigma\in G_{21}(i)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(2i)\sigma(2i-1)} \cdots x_{\sigma(n-1)\sigma(n)} \\
&= \sum_{\sigma\in G_{21}(i)} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(1)\sigma_{\tau}(2)} \cdots x_{\sigma_{\tau}(2i-1)\sigma_{\tau}(2i)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= \sum_{\sigma\in G_{12}(i)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)} \\
&= \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{12}x_{\sigma(3)\sigma(4)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
Step 3-2.
各$i,j \in \{1,\ldots,p\},\, i \neq j,\,$に対して
$$
G_{10,20}(i,j) \coloneqq \{\sigma\in G_{10,20} \mid \sigma(2i-1)=1,\ \sigma(2j-1)=2\}$$
とおく．このとき，$G_{10,20} = \bigsqcup_{i \neq j} G_{10,20}(i,j)$であり，
$$
\sum_{\sigma\in G_{10,20}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)} = -\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}$$
が成り立つ．
$i=1$のとき
$\tau \coloneqq (2,2j-1,3)\circ(4,2j)$とおくと，$\sgn(\tau)=-1$であり，写像
$$
G_{10,20}(1,j) \to \mathfrak{S}_{n-2};\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である（ただし$j=2$のときは$\tau=(2,3)$と見做す）．よって
\begin{align}
\sum_{\sigma\in G_{10,20}(1,j)} \sgn(\sigma) x_{1,\sigma(2)}x_{\sigma(3)\sigma(4)} \cdots x_{2,\sigma(2j)} \cdots x_{\sigma(n-1)\sigma(n)}
&= -\sum_{\sigma\in G_{10,20}(1,j)} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(1)\sigma_{\tau}(3)} x_{\sigma_{\tau}(2j-1)\sigma_{\tau}(2j)} \cdots x_{\sigma_{\tau}(2)\sigma_{\tau}(4)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= -\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)} x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
$j=1$のとき
$\tau \coloneqq (1,2i-1,3,2i,4,2)$とおくと，$\sgn(\tau)=-1$であり，写像
$$
G_{10,20}(i,1) \to \mathfrak{S}_{n-2};\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align}
\sum_{\sigma\in G_{10,20}(i,1)} \sgn(\sigma) x_{2,\sigma(2)}x_{\sigma(3)\sigma(4)} \cdots x_{1,\sigma(2i)} \cdots x_{\sigma(n-1)\sigma(n)}
&= -\sum_{\sigma\in G_{10,20}(i,1)} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(2)\sigma_{\tau}(4)}x_{\sigma_{\tau}(2i-1)\sigma_{\tau}(2i)} \cdots x_{\sigma_{\tau}(1)\sigma_{\tau}(3)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= -\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
$i=2$のとき
$\tau \coloneqq (1,3,4,2j,2,2j-1)$とおくと，$\sgn(\tau)=-1$であり，写像
$$
G_{10,20}(2,j) \to \mathfrak{S}_{n-2};\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align}
\sum_{\sigma\in G_{10,20}(2,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)}x_{1,\sigma(4)} \cdots x_{2,\sigma(2j)} \cdots x_{\sigma(n-1)\sigma(n)}
&= - \sum_{\sigma\in G_{10,20}(2,j)} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(2j-1)\sigma_{\tau}(2j)}x_{\sigma_{\tau}(1)\sigma_{\tau}(3)} \cdots x_{\sigma_{\tau}(2)\sigma_{\tau}(4)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= - \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
$j=2$のとき
$\tau \coloneqq (1,2i-1)\circ(2,3,2i)$とおくと，$\sgn(\tau)=-1$であり，写像
$$
G_{10,20}(i,2) \to \mathfrak{S}_{n-2};\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align}
\sum_{\sigma\in G_{10,20}(i,2)} \sgn(\sigma) x_{\sigma(1)\sigma(2)}x_{2,\sigma(4)} \cdots x_{1,\sigma(2i)} \cdots x_{\sigma(n-1)\sigma(n)}
&= - \sum_{\sigma\in G_{10,20}(i,2)} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(2i-1)\sigma_{\tau}(2i)}x_{\sigma_{\tau}(2)\sigma_{\tau}(4)} \cdots x_{\sigma_{\tau}(1)\sigma_{\tau}(3)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= - \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
$i,j > 2$のとき
$\tau \coloneqq (1,2i-1,3,2i,4,2j,2,2j-1)$とおくと，$\sgn(\tau)=-1$であり，写像
$$
G_{10,20}(i,j) \to \mathfrak{S}_{n-2};\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align}
\sum_{\sigma\in G_{10,20}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)}x_{\sigma(3)\sigma(4)} \cdots x_{1,\sigma(2i)} \cdots x_{2,\sigma(2j)} \cdots x_{\sigma(n-1)\sigma(n)}
&= - \sum_{\sigma\in G_{10,20}(i,j)} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(2j-1)\sigma_{\tau}(2j)}x_{\sigma_{\tau}(2i-1)\sigma_{\tau}(2i)} \cdots x_{\sigma_{\tau}(1)\sigma_{\tau}(3)} \cdots x_{\sigma_{\tau}(2)\sigma_{\tau}(4)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= - \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
各$i,j\in\{1,\ldots,p\},\,i\neq j,\,$に対して
$$
G_{01,02}(i,j) \coloneqq \{\sigma\in G_{01,02} \mid \sigma(2i)=1,\ \sigma(2j)=2\}$$
とおく．このとき，$G_{01,02} = \bigsqcup_{i \neq j} G_{01,02}(i,j)$であり，
$$
\sum_{\sigma \in G_{01,02}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)} = -\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}$$
が成り立つ．
$\tau \coloneqq (2i-1,2i)\circ(2j-1,2j)$とおくと，$\sgn(\tau)=1$であり，写像
$$
G_{01,02}(i,j) \to G_{10,20}(i,j);\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align}
\sum_{\sigma\in G_{01,02}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(2i-1),1} \cdots x_{\sigma(2j-1),2} \cdots x_{\sigma(n-1)\sigma(n)}
&= \sum_{\sigma\in G_{01,02}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{1,\sigma(2i-1)} \cdots x_{2,\sigma(2j-1)} \cdots x_{\sigma(n-1)\sigma(n)} \\
&= \sum_{\sigma\in G_{01,02}(i,j)} \sgn(\sigma) x_{\sigma_{\tau}(1)\sigma_{\tau}(2)} \cdots x_{\sigma_{\tau}(2i-1)\sigma_{\tau}(2i)} \cdots x_{\sigma_{\tau}(2j-1)\sigma_{\tau}(2j)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= \sum_{\sigma\in G_{10,20}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)} \\
&= - \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
各$i,j\in\{1,\ldots,p\},\,i\neq j,\,$に対して
$$
G_{10,02}(i,j) \coloneqq \{\sigma\in G_{10,02} \mid \sigma(2i-1)=1,\ \sigma(2j)=2\}$$
とおく．このとき，$G_{10,02} = \bigsqcup_{i \neq j} G_{10,02}(i,j)$であり，
$$
\sum_{\sigma \in G_{10,02}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)} = -\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}$$
が成り立つ．
$\tau \coloneqq (2j-1,2j)$とおくと，$\sgn(\tau)=-1$であり，写像
$$
G_{10,02}(i,j) \to G_{10,20}(i,j);\ \sigma \mapsto \sigma\circ\tau \eqqcolon \sigma_{\tau}$$
は全単射である．よって
\begin{align}
\sum_{\sigma\in G_{10,02}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{1,\sigma(2i)} \cdots x_{\sigma(2j-1),2} \cdots x_{\sigma(n-1)\sigma(n)}
&= - \sum_{\sigma\in G_{10,02}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{1,\sigma(2i)} \cdots x_{2,\sigma(2j-1)} \cdots x_{\sigma(n-1)\sigma(n)} \\
&= \sum_{\sigma\in G_{10,02}(i,j)} \sgn(\sigma_{\tau}) x_{\sigma_{\tau}(1)\sigma_{\tau}(2)} \cdots x_{\sigma_{\tau}(2i-1)\sigma_{\tau}(2i)} \cdots x_{\sigma_{\tau}(2j-1)\sigma_{\tau}(2j)} \cdots x_{\sigma_{\tau}(n-1)\sigma_{\tau}(n)} \\
&= \sum_{\sigma\in G_{10,20}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)} \\
&= -\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
各$i,j\in\{1,\ldots,p\},\,i\neq j,\,$に対して
$$
G_{01,20}(i,j) \coloneqq \{\sigma\in G_{01,20} \mid \sigma(2i)=1,\ \sigma(2j-1)=2\}$$
とおく．このとき，$G_{01,20} = \bigsqcup_{i \neq j} G_{01,20}(i,j)$であり，
$$
\sum_{\sigma \in G_{01,20}(i,j)} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)} = -\sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)}x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}$$
が成り立つ．
上と同様．
Step 3-3.
以上により
\begin{align}
\sum_{\sigma\in\mathfrak{S}_{n}} \sgn(\sigma) x_{\sigma(1)\sigma(2)} \cdots x_{\sigma(n-1)\sigma(n)}
&= \sum_{i}\left(\sum_{G_{12}(i)} + \sum_{G_{21}(i)}\right) + \sum_{i \neq j}\left(\sum_{G_{10,20}(i,j)} + \sum_{G_{01,02}(i,j)} + \sum_{G_{10,02}(i,j)} + \sum_{G_{01,20}(i,j)} \right) \\
&= 2p \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{12}x_{\sigma(3)\sigma(4)} \cdots x_{\sigma(n-1)\sigma(n)} - 4p(p-1) \sum_{\sigma\in\mathfrak{S}_{n-2}} \sgn(\sigma) x_{1,\sigma(3)}x_{2,\sigma(4)} x_{\sigma(5)\sigma(6)} \cdots x_{\sigma(n-1)\sigma(n)}
\end{align}
が成り立つ．
