generating prime p and q, with alpha of element of order q
There are generated the primes $p$ and $q$ such that $q | (p-1)$


What that means is that we search for two prime numbers $p$ and $q$, with the constraint that $q$ is a divisor of $p-1$.  That is, $kq = p-1$ for some integer $k$.

In practice, this is easy to do; we first pick a prime $q$ of an appropriate size, and then search for the value $k$ values of the appropriate size until we find a value $p = kq + 1$ which is also prime.


  and $\alpha \in \mathbb{Z}_p^*$ ...


$\mathbb{Z}_p^*$ is the multiplicative group modulo $p$; its elements have an obvious identification to the between 1 and p that are relatively prime to $p$ (and since $p$ is prime, that means the values $[1, p-1]$), and


  an element of order $q$.


What this means is that the smallest positive value $\lambda$ for which $\alpha^\lambda \equiv 1 \pmod p$ is $\lambda = q$.

It turns out to be a lot easier to find such a value $\alpha$ than you'd expect; you pick an arbitrary value $\beta \in [2, p-2]$ and set $\alpha = \beta ^ {(p-1)/q} \pmod p$.  As long as $\beta \ne 1$, then $\alpha$ will have an order precisely $q$.


We know that $\alpha^q = (\beta ^ {(p-1)/q})^q = \beta ^ {p-1} = 1$ by Fermat's Little Theorem (and because $p$ is prime)
We know that $\alpha^r \ne 1$ for any $0 &lt; r &lt; q$, because if $\alpha^r = 1$, then we could show that $\alpha^{\gcd(q, r)} = 1$, and as $q$ is prime, $\gcd(q, r) = 1$, and so we'd have $\alpha^1 = 1$, which we had verified is not the case