Let A \mathbf{A} A E \mathbf{E} E A \mathbf{A} A ( x i ) 1 ≤ i ≤ n (x_i)_{1 \leq i \leq n} ( x i ) 1 ≤ i ≤ n E \mathbf{E} E ( i , j ) (i, j) ( i , j ) 1 ≤ j < i ≤ n 1 \leq j < i \leq n 1 ≤ j < i ≤ n
x i x j = λ i j x j x i with λ i j ∈ A .
x_i x_j = \lambda_{ij} x_j x_i \quad \text{with } \lambda_{ij} \in \mathbf{A}.
x i x j = λ ij x j x i with λ ij ∈ A . Then, for every permutation σ ∈ S n \sigma \in \mathfrak{S}_n σ ∈ S n
x σ ( 1 ) x σ ( 2 ) ⋯ x σ ( n ) = ε σ ( λ ) x 1 x 2 ⋯ x n .
x_{\sigma(1)} x_{\sigma(2)} \cdots x_{\sigma(n)} = \varepsilon_\sigma(\lambda) x_1 x_2 \cdots x_n.
x σ ( 1 ) x σ ( 2 ) ⋯ x σ ( n ) = ε σ ( λ ) x 1 x 2 ⋯ x n . The lemma is trivial for n = 1 n = 1 n = 1 n = 2 n = 2 n = 2 n n n n ≥ 3 n \geq 3 n ≥ 3 σ ( n ) = n \sigma(n) = n σ ( n ) = n σ ( n ) = k , k ≠ n \sigma(n) = k, k \neq n σ ( n ) = k , k = n τ \tau τ { 1 , n } \{1, n\} { 1 , n }
{ τ ( i ) = i for i < k τ ( i ) = i + 1 for k ≤ i < n τ ( n ) = k .
\begin{cases}
\tau(i) = i & \text{for } i < k \\
\tau(i) = i + 1 & \text{for } k \leq i < n \\
\tau(n) = k.
\end{cases}
⎩ ⎨ ⎧ τ ( i ) = i τ ( i ) = i + 1 τ ( n ) = k . for i < k for k ≤ i < n Let π = τ − 1 ∘ σ \pi = \tau^{-1} \circ \sigma π = τ − 1 ∘ σ π \pi π n n n σ = τ ∘ π \sigma = \tau \circ \pi σ = τ ∘ π y i = x τ ( i ) , y π ( i ) = x σ ( i ) y_i = x_{\tau(i)},\ y_{\pi(i)} = x_{\sigma(i)} y i = x τ ( i ) , y π ( i ) = x σ ( i ) i ≠ n i \neq n i = n j ≠ n j \neq n j = n π ( i ) > π ( j ) \pi(i) > \pi(j) π ( i ) > π ( j ) σ ( i ) > σ ( j ) \sigma(i) > \sigma(j) σ ( i ) > σ ( j ) τ \tau τ { 1 , n − 1 } \{1, n-1\} { 1 , n − 1 } { 1 , n } \{1, n\} { 1 , n } i ≠ n , j ≠ n i \neq n, j \neq n i = n , j = n σ ( i ) > σ ( j ) \sigma(i) > \sigma(j) σ ( i ) > σ ( j )
y π ( i ) y π ( j ) = x σ ( i ) x σ ( j ) = λ σ ( i ) , σ ( j ) x σ ( j ) x σ ( i ) = λ σ ( i ) , σ ( j ) y π ( j ) y π ( i )
y_{\pi(i)} y_{\pi(j)} = x_{\sigma(i)} x_{\sigma(j)} = \lambda_{\sigma(i),\ \sigma(j)} x_{\sigma(j)} x_{\sigma(i)} = \lambda_{\sigma(i),\ \sigma(j)} y_{\pi(j)} y_{\pi(i)}
y π ( i ) y π ( j ) = x σ ( i ) x σ ( j ) = λ σ ( i ) , σ ( j ) x σ ( j ) x σ ( i ) = λ σ ( i ) , σ ( j ) y π ( j ) y π ( i ) whence, by the induction hypothesis (using the fact that π ( n ) = n \pi(n) = n π ( n ) = n
y π ( 1 ) y π ( 2 ) ⋯ y π ( n ) = ( ∏ i < j < n , σ ( i ) > σ ( j ) λ σ ( i ) , σ ( j ) ) y 1 y 2 ⋯ y n
y_{\pi(1)} y_{\pi(2)} \cdots y_{\pi(n)} = \left( \prod_{i < j < n,\ \sigma(i) > \sigma(j)} \lambda_{\sigma(i),\ \sigma(j)} \right) y_1 y_2 \cdots y_n
y π ( 1 ) y π ( 2 ) ⋯ y π ( n ) = i < j < n , σ ( i ) > σ ( j ) ∏ λ σ ( i ) , σ ( j ) y 1 y 2 ⋯ y n that is
x σ ( 1 ) x σ ( 2 ) ⋯ x σ ( n ) = ( ∏ i < j < n , σ ( i ) > σ ( j ) λ σ ( i ) , σ ( j ) ) x τ ( 1 ) ⋯ x τ ( n ) .
x_{\sigma(1)} x_{\sigma(2)} \cdots x_{\sigma(n)} = \left( \prod_{i < j < n,\ \sigma(i) > \sigma(j)} \lambda_{\sigma(i),\ \sigma(j)} \right) x_{\tau(1)} \cdots x_{\tau(n)}.
x σ ( 1 ) x σ ( 2 ) ⋯ x σ ( n ) = i < j < n , σ ( i ) > σ ( j ) ∏ λ σ ( i ) , σ ( j ) x τ ( 1 ) ⋯ x τ ( n ) . Now
x τ ( 1 ) ⋯ x τ ( n ) = x 1 ⋯ x k − 1 x k + 1 ⋯ x n x k
x_{\tau(1)} \cdots x_{\tau(n)} = x_1 \cdots x_{k-1} x_{k+1} \cdots x_n x_k
x τ ( 1 ) ⋯ x τ ( n ) = x 1 ⋯ x k − 1 x k + 1 ⋯ x n x k and this, by Lemma 1, is equal to
( ∏ j > k λ j k ) x 1 ⋯ x n = ( ∏ σ ( i ) > σ ( n ) λ σ ( i ) , σ ( n ) ) x 1 ⋯ x n .
\left( \prod_{j > k} \lambda_{jk} \right) x_1 \cdots x_n = \left( \prod_{\sigma(i) > \sigma(n)} \lambda_{\sigma(i),\ \sigma(n)} \right) x_1 \cdots x_n.
j > k ∏ λ j k x 1 ⋯ x n = σ ( i ) > σ ( n ) ∏ λ σ ( i ) , σ ( n ) x 1 ⋯ x n . Finally, (20) and (21) give
x σ ( 1 ) ⋯ x σ ( n ) = α ⋅ x 1 ⋯ x n
x_{\sigma(1)} \cdots x_{\sigma(n)} = \alpha \cdot x_1 \cdots x_n
x σ ( 1 ) ⋯ x σ ( n ) = α ⋅ x 1 ⋯ x n with
α = ( ∏ i < j < n , σ ( i ) > σ ( j ) λ σ ( i ) , σ ( j ) ) ⋅ ( ∏ i < n , σ ( i ) > σ ( n ) λ σ ( i ) , σ ( n ) )
\alpha = \left( \prod_{i < j < n, \sigma(i) > \sigma(j)} \lambda_{\sigma(i), \sigma(j)} \right) \cdot \left( \prod_{i < n, \sigma(i) > \sigma(n)} \lambda_{\sigma(i), \sigma(n)} \right)
α = i < j < n , σ ( i ) > σ ( j ) ∏ λ σ ( i ) , σ ( j ) ⋅ i < n , σ ( i ) > σ ( n ) ∏ λ σ ( i ) , σ ( n ) = ∏ i < j , σ ( i ) > σ ( j ) λ σ ( i ) , σ ( j ) = ε σ ( λ )
= \prod_{i < j, \sigma(i) > \sigma(j)} \lambda_{\sigma(i), \sigma(j)} = \varepsilon_\sigma(\lambda)
= i < j , σ ( i ) > σ ( j ) ∏ λ σ ( i ) , σ ( j ) = ε σ ( λ ) which completes the proof of Lemma 3.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.