well, not quite.

suppose (for example) that F contains all 3rd roots of unity, but no primitive 9th root of unity. well all 3rd roots of unity are also 9th roots of unity, if:

a^{3}= 1 (mod q-1) then:

a^{6}= (a^{3})^{3}= 1^{3}= 1 (mod q-1).

but the subgroup of 9-th roots of unity in F doesn't have order 9, but order 3.

for example F_{25}has a cyclic multiplicative group of order 24, which therefore has a cyclic subgroup of order 3, but no cyclic subgroup of order 9 since 9 does not divide 24

(if the generator of F* is b, then b^{8}is a cube root of 1, so b^{8}is a 9-th root of 1 as well:

(b^{8})^{9}= b^{72}= (b^{24})^{3}= 1^{3}= 1).

so all you can say is 93 and 5^{n}-1 have a common factor > 1. this common factor may be 3,31 or 93.

if F_{5n}is supposed to contain ALL 93-rd roots of unity, THEN 93 would have to divide 5^{n}-1.