My question is about the roots of unity in finite fields. It goes like this:
Suppose we have two primes p and q, both greater than 3, which satisfy $q|(p-1)$. Then there exists a $q$-the root of unity (say $x, x \neq 1)$ in the field F with p elements. Suppose y is the inverse of 3 in F and consider the group $\displaystyle G = F^{*} \times F^{*} \times \ldots \times F^{*},$ a direct product of $\displaystyle \frac{q-1}{2}$ copies of the cyclic group $\displaystyle F^{*}.$ Clearly, $G$ is a group of exponent $p-1.$ Then can we say that the order of the element $ z = (y(1+x+x^{-1}), y(1+x^2+x^{-2}), \ldots, y(1+x^{\frac{q-1}{2}}+x^{- \frac{q-1}{2}}))$ is always of order $(p-1)$ in $G?$