Z-transform of a downsampler
This derivation is a tricky one. The approach suggested before has a flaw. Let me demonstrate this first; then I will give the correct solution.
We wish to relate the $\mathcal{Z}$-transform of the downsampled signal, $Y_D(z) = \mathcal{Z}\{x[Mn]\}$, to the $\mathcal{Z}$-transform of the original signal $X(z) = \mathcal{Z}\{x[n]\}$.
The wrong way
One could think of simply plugging the expression for the downsampled signal into the expression of the $\mathcal{Z}$-transform:
$Y_D(z) = \sum\limits_{n=-\infty}^{+\infty}{x[Mn] z^{-n}}$
A change of variable $n'=Mn$ seems obvious:
$Y_D(z) = \sum\limits_{n' \in M\mathbb{Z}}{x[n'] z^{-n'/M}}$
However, it is important to realize that even though the new summation index $n'$ still runs from $-\infty$ to $\infty$, the sum is now over 1 out of M integer numbers. In other words,
$n' \in M\mathbb{Z}=\{...,-2M,-M,0,M,2M,...\}$,
while the definition of the $\mathcal{Z}$-transform requires
$n \in \{...,-2,-1,0,1,2,...\}$.
Since this is no longer a $\mathcal{Z}$-transform, we cannot write:
$Y_D(z) = X(z^{1/M})$
The right way
Let us first define a 'helper' impulse train signal $t_M[n]$ as:
$\begin{align*}
t_M[n]
&amp; = \sum\limits_{k=-\infty}^{+\infty}{\delta[n-kM]} \\
&amp; =
\left\{
     \begin{array}{lr}
       1 &amp; : n \in M\mathbb{Z}\\
       0 &amp; : n \notin M\mathbb{Z}
     \end{array}
   \right.
\end{align*}
$
This function is $1$ at one out of every $M$ samples, and zero everywhere else.
Equivalently, the pulse train function can be written as:
$ t_M[n] = \frac{1}{M} \sum\limits_{k=0}^{M-1}{e^{j2\pi k n / M}} $
Proof: We need to consider separately the cases $n \in M\mathbb{Z}$ and $n \notin M\mathbb{Z}$:
$$\begin{align*}
 t_M[n] 
&amp; = \frac{1}{M} \sum\limits_{k=0}^{M-1}{e^{j2\pi k n / M}} \\
&amp; = 
\left\{
     \begin{array}{lr}
       \frac{1}{M} \sum\limits_{k=0}^{M-1}{1} &amp;&amp; : n \in M\mathbb{Z}\\
       \frac{1}{M} \frac{1-e^{j2\pi n}}{1-e^{j2\pi n / M}} &amp;&amp; : n \notin M\mathbb{Z}
     \end{array}
   \right. \\
&amp; = 
\left\{
     \begin{array}{lr}
       \frac{1}{M} M &amp;&amp; : n \in M\mathbb{Z}\\
       \frac{1}{M} \frac{1-1}{1-e^{j2\pi n / M}} &amp;&amp; : n \notin M\mathbb{Z}
     \end{array}
   \right. \\
&amp; = 
\left\{
     \begin{array}{lr}
       1 &amp;&amp; : n \in M\mathbb{Z}\\
       0 &amp;&amp; : n \notin M\mathbb{Z}
     \end{array}
   \right. 
\end{align*}$$
In the case $n \notin M\mathbb{Z}$, we used the expression for the finite sum of a geometric series.
Now let's go back to our original problem of finding the $\mathcal{Z}$-transform of a downsampler:
$Y_D(z) = \sum\limits_{n=-\infty}^{+\infty}{x[Mn] z^{-n}}$
We apply the substitution $n'=Mn$, keeping in mind that this makes the summation run only over integer multiples of M:
$Y_D(z) = \sum\limits_{n' \in M\mathbb{Z}}{x[n'] z^{-n'/M}}$
We can now use the above impulse train function to safely rewrite this as a summation over all $n \in \mathbb{Z}$:
$Y_D(z) = \sum\limits_{n=-\infty}^{+\infty}{t_M[n] x[n] z^{-n/M}}$
Using the above formulation for the impulse train function as a finite sum of exponentials, we get:
$\begin{align*}
Y_D(z) 
&amp;= \sum\limits_{n=-\infty}^{+\infty}{(\frac{1}{M} \sum\limits_{k=0}^{M-1}{e^{j2\pi k n / M}}) x[n] z^{-n/M}} \\
&amp;= \frac{1}{M} \sum\limits_{k=0}^{M-1}{\sum\limits_{n=-\infty}^{+\infty}{e^{j2\pi k n / M} x[n] z^{-n/M}}} \\
&amp;= \frac{1}{M} \sum\limits_{k=0}^{M-1}{\sum\limits_{n=-\infty}^{+\infty}{x[n] (e^{-j2\pi k / M} z^{1/M})^{-n}}}
\end{align*}$
The summation on the right is a summation over all integers, and is therefore a valid $\mathcal{Z}$-transform in terms of $z'=e^{-j2\pi k / M} z^{1/M}$. Therefore, we can write:
$
Y_D(z) = \frac{1}{M} \sum\limits_{k=0}^{M-1}{X(e^{-j2\pi k / M} z^{1/M})}
$
This is the formula for the $\mathcal{Z}$-transform of a downsampler.