Operation of permutations on functions Let $P$ be the additive group of mappings from $\mathbf{Z}^n$ to $\mathbf{Z}$. For $f \in P$ and $\sigma \in \mathfrak{S}_n$ (the symmetric group of degree $n$) let $\sigma f$ be the element of $P$ defined by 
$$\sigma f(z_1,\dots,z_n) = f(z_{\sigma(1)},\dots,z_{\sigma(n)}).$$
Is it not true that if $\sigma, \tau \in \mathfrak{S}_n$, we have 
$$(\tau(\sigma f))(z_1,\dots,z_n) = \sigma f(z_{\tau(1)},\dots,z_{\tau(n)}) = f(z_{\sigma\tau(1)},\dots,z_{\sigma\tau(n)}) = (\sigma\tau)f(z_1,\dots,z_n)?$$
The book says that $\tau(\sigma f) = (\tau\sigma) f$.

One way to dissolve the confusion here is to think of it this way. A point $(z_1,...,z_n)$ is a function $z:\{1,...,n\}\to Z$. Hence, the point $(z_{\sigma(1)},...,z_{\sigma(n)})$, is the composition $z\circ \sigma$. Now we can think of the domain of $f$ as the set of functions
$$Z^n\simeq \{z:\{1,...,n\}\to Z\}\;,$$
and this way you can rewrite the action on $f$, evaluated at a point $z$, via precomposition by $\sigma$.
$$\sigma f (z_1,...,z_n)=(\sigma f)(z):=f(z \circ \sigma).$$
Then 
\begin{eqnarray} \tau(\sigma f)(z_1,...,z_n) &=& \tau(\sigma f)(z)
\\
&:=& (\sigma f)(z \circ \tau)
\\
&:=& f ((z\circ \tau)\circ \sigma)
\\
&=& f (z\circ (\tau\circ \sigma))
\\
&=:& (\tau \sigma)f(z)
\\
&=& (\tau \sigma) f(z_1,...,z_n)
\end{eqnarray}
Remark
Thanks to Malice Vidrine for pointing out a few mistakes on the original post.
Edit by Randy Randerson:
By definition, $\mathbf{Z}^n$ is the set of functions $z = (z_1,\dots,z_n)$ from $\{1,\dots,n\}$ to $\mathbf{Z}$. A permutation $\sigma \in \mathfrak{S}_n$ maps $\{1,\dots,n\}$ onto $\{1,\dots,n\}$. Therefore $z \circ \sigma \in \mathbf{Z}^n$. By definition $(\sigma f)(z) = f(z \circ \sigma)$, so $\tau(\sigma f) = (\tau \sigma)f$ as shown in your answer.
