I am reading Braid Groups by Christian Kassel and Vladimir Turaev (GTM 247). Exercise 1.1.6 of the book is:
Prove that each element $\sigma_i\sigma_j^{-1}$ with $1\leq i<j\leq n-1$ belongs to $[B_n,B_n]$ and generates $[B_n,B_n]$ as a normal subgroup provided either $j\neq i+2$ or $n\neq 4$. (Hint: Consider first the case $j=i+1$)
My question is that why the case $j=i+2$ and $n=4$ needs to be treated seperately. If $j=i+2$ and $n=4$, we have $i=1$ and $j=3$. In this case,
$$\sigma_1\sigma_3^{-1}=(\sigma_1\sigma_2^{-1})(\sigma_2\sigma_3^{-1})$$
But $$[\sigma_{i+1},\sigma_i]=\sigma_{i+1}^{-1}\sigma_i^{-1}\sigma_{i+1}\sigma_i=\sigma_i\sigma_{i+1}^{-1}\sigma_i^{-1}\sigma_i=\sigma_i\sigma_{i+1}^{-1}$$
Thus $$\sigma_1\sigma_3^{-1}=[\sigma_2,\sigma_1][\sigma_3,\sigma_2],$$ which is in $[B_4,B_4]$ anyway. Why does the question emphasizes the condition that "either $j\neq i+2$ or $n\neq 4$"?