How can I deduce (3.6) from (3.5)? 
My attempt:-

I could verify the above formula for $f\in A_2(V) $ and $g\in A_1(V)$
  and vise versa. I know the fact that number of even permutation in
  $S_{k+l}$ is $\frac{(k+l)!}{2}$. So, sgn($\sigma$)=$1$ for
  $\frac{(k+l)!}{2}$ terms and sgn($\sigma$)=$-1$ for
  $\frac{(k+l)!}{2}$. I couldn't able to deduce for general case.

 A: From @Jaroslaw Matlak 's Comment I typed my understanding of the section in the textbook.
Consider any $\sigma \in S_{k+l}$ and $\tau \in S_k$.$$(sgn\sigma \tau)f(v_{\sigma \tau(1)}, v_{\sigma \tau(2)},..., v_{\sigma \tau(k)})=(sgn\sigma \tau)(sgn\tau)f(v_{\sigma ((1))}, v_{\sigma ((2))},..., v_{\sigma ((k))})=(sgn\sigma)f(v_{\sigma ((1))}, v_{\sigma ((2))},..., v_{\sigma ((k))}).$$  There exists one $\tau'\in S_k$ such that $\sigma \tau'(1)<\sigma \tau'(2)<...<\sigma \tau'(k). $ We have $\sigma \tau \in S_{k+l},$ fixes the argument of $g$. T
herefore $k!$ number of equal quantities exists in the ($3.5$).
Consider any $\sigma \in S_{k+l}$ and $\rho \in S_l$.$$(sgn\sigma \rho)g(v_{\sigma \rho(k+1)}, v_{\sigma \rho(k+2)},..., v_{\sigma \rho(k+l)})=(sgn\sigma \rho)(sgn\rho)f(v_{\sigma ((k+1))}, v_{\sigma ((k+2))},..., v_{\sigma ((k+l))})=(sgn\sigma)g(v_{\sigma ((k+1))}, v_{\sigma ((k+2))},..., v_{\sigma ((k+l))}).$$  There exists one $\rho'\in S_l$ such that $\sigma \rho'(k+1)<\sigma \rho'(k+2)<...<\sigma \rho'(k+l). $ We have $\sigma \rho \in S_{k+l},$ fixes the argument of $f$.  Applying in ($3.5$) We get ($3.6$).
