Given, a prime $q$ what is the probability of obtaining a prime $p$ such that $ a^{(p-1)/q} \equiv 1 \mod p$. Here, $a$ is any given integer and by $ a^{(p-1)/q} \equiv 1 \mod p$ we mean that the remainder on dividing $a^{(p-1)/q}$ by $p$ is $1$
In a paper I found out that the probability is $\frac{1}{q}$. But, I do not know how this was obtained. Kindly help me out.
Thanks in advance !