As said, you have misread the statement. This is a special case of the fact that for an odd prime $\rm\:p\:,\:$ $\rm\: \phi(2\:p)\ =\ \phi(p)\ =\ p-1\:,\ $ i.e. there are $\rm\: p-1\: $ naturals below $\rm\:2\:p\:$ that are coprime to $\rm\:2\:p\:,\:$ namely all $\rm\:p\:$ odd numbers below $\rm\:2\:p\ $ excepting $\rm\:p\:.\ $ Hence, modulo $\rm\:2\:p\:,\:$ an odd prime $\rm\ne p\:$ must lie in one of these congruence classes (else it has a nontrivial gcd with $\rm\:2\:p\:,\:$ so it is composite). $\:$ Hence if $\rm\:q\:$ is prime then $\rm\ q\equiv 1,5\ \ (mod\ 6)\:;\ \ q\equiv 1,3,7,9\ \ (mod\ 10)\:;\ \ q\equiv 1,3,5,9,11,13\ \ (mod\ 14)\ $ etc, assuming that $\rm\:q\:$ is coprime to the modulus. Exploiting reflection symmetry we can state this more succinctly: $\rm\ \ q\equiv \pm 1\ \ (mod\ 6)\:;\ \ q\equiv \pm\{1,3\}\ \ (mod\ 10)\:;\ \ q\equiv\pm \{1,3,5\}\ \ (mod\ 14)\ $ and, more generally, $\rm\:\ \ \ q\equiv \pm\{1,3,5,\cdots,p-2\}\ \ (mod\ 2\:p)\ $