What is the number of distinct subgroups of the automorphism group of $\mathbf{F}_{3^{100}}$?

Question

Let $G$ denote the group of all the automorphisms of the field $\mathbf{F}_{3^{100}}$ that consists of $3^{100}$ elements. What is the number of distinct subgroups of $G$?

Answer 1

Rhetorical questions:

• What is the order of $G(\Bbb F_{3^{100}}/\Bbb F_3)$?
• What kind of group is it?
• What are the subgroups of these kinds of groups?
• So, how many are there in this case?
• Are there any formulas for arithmetic functions of help here?

Answer 2

Answer of question 1) $100$ 2) It is a cyclic group 3) Corresponding each divisor of $100$ there is a unique subgroup of $G(\Bbb F_{3^{100}} / \Bbb F_3)$ 4) There are $9$ like this. 5) Its a sigma function. Now what?

Answer 3

Hint:

$\mathbf F_{3^{100}}=F_3[\alpha]$ for some primitive element $\alpha$. It is a root of an irreducible polynomial in $F_3[x]$ of degree $100$ and its conjugate are $\;\alpha,\alpha^3,\alpha^{3^2},\dots,\alpha^{3^{99}}$, which are all distinct since the polynomial is separable. Hence the Galois group $\;\operatorname{Gal}_{F_{3^{100}}/F_{3^{\phantom1}}}$ is cyclic of order $100$, generated by the Frobenius morphism $\;\varphi\colon x\mapsto x^3$. Its subgroups, as all subgroups of a cyclic group, are cyclic, generated by the $\varphi^n$, for all the divisors $n$ of $100$.