# How is the generator used in Feldman's Verifiable Secret Sharing scheme determined? [duplicate]

According to the Wikipedia description of Feldman's VSS scheme

First, a cyclic group G of prime order p, along with a generator g of G, is chosen publicly as a system parameter. (Typically, one takes a subgroup of (Zq)*, where q is a prime such that p divides q-1.)

Why is it necessary to use a generator g in this process?
What is the procedure to be followed to implement a method which produces this generator for a given prime order group Zp?

The multiplicative group $$\mathbf{Z}_p^*$$ of non-zero integers modulo $$p$$ is cyclic of order $$p-1$$, so it has exactly one subgroup of order $$k$$ for each divisor $$k$$ of $$p-1$$.
To obtain an element $$g$$ of $$G$$, take any element $$a$$ of $$\mathbf{Z}_p^*$$ and let $$g = a^{(p-1)/q} \bmod p$$. Then $$g$$ is an element of $$G$$ because $$g^q = \left(a^{(p-1)/q}\right)^q = a^{p-1} = 1 \pmod p.$$
The order of $$g$$ is a divisor of the order of $$G$$, so it is a divisor of $$q$$, and since $$q$$ is prime, it equals either $$1$$ or $$q$$.