There is the following problem from my text book. I just wanna know if I understand the solution correctly.

For which of the following values of n does the finite field F_{5n}with 5^{n}elements contain a non trivial 93 rd root of unity? 1. 92 2. 30 3. 15 4. 6

The theorem I used was as follows

"Let F be a field and H , the group of nth roots of unity in F, be a subgroup of the multiplicative group FFirst up, the multiplicative group^{X}. Then H is cyclic of some order m such that m divides n. If, in addition, F is finite with order q, |H| = (n, q-1)"

Fhas |F| -1 elements. Since H is a subgroup of^{X }F, o(H) should divide |F| -1. Since H is cyclic, there is a generator whose order is the same as the order of H. So if 'a' is a generator of H, o(a) should divide |F| -1.^{X }

Since H is a group of the nth roots of unity in F, the nth power of every element is 1, the identity element with respect to H andF. So, a^{X }^{93}=1. This means 93 should divide |F| -1.

To sum it up, 93 should divide 5^{n }- 1.

Have I been right so far?

Can someone help me find the right option from here on?

Thanks,