Let $G$ denote the group of all automorphisms of the field $F_{3^{100}}$. Then, what is the number of distinct subgroups of $G$?
First of all I have to compute $G$. Now \begin{equation*} F_{3^{100}}(+) \cong Z/3Z \times Z/3Z \times Z/3Z\times....Z/3Z \end{equation*} ($100$ times). Now what do I do compute the possible group automorphisms first?
Should I look at the basis elements $(1,0,0,0,0,....0),(0,1,0,0,0.....0)...$ and where I send them to would determine all automorphisms? or am I wrong?
In general, the automorphism group of $\mathbb{F}_{p^n}$ is generated by the Frobenius automorphism $\phi$ defined where $a \mapsto a^p$. In particular, this implies that the automorphism group is cyclic (and ultimately of order $n$).
One way of going about showing this is by first proving that finite extensions of finite fields are Galois, then recalling that, for any Galois extension $K$ of a base field $F$, we have $| \operatorname{Aut}(K/F) | = [K:F]$. In this case, $K = \mathbb{F}_{p^{n}}$ and $F = \mathbb{F}_p$, and $[K:F] = n$. So it suffices to show that $\operatorname{id}, \phi, \phi^2, \phi^3, ..., \phi^{n-1}$ are all unique automorphisms, and since there are $n$ of them, the list must be exhaustive. To show these are unique automorphisms and none wrap back around to the identity for any $0 \leq k < n$, note that $\phi^k$ maps $a \mapsto a^{p^k}$, and then consider the polynomial $x^{p^k} - x$; you can arrive at a contradiction via a degree of the polynomial / number of roots argument.
At any rate, the first paragraph implies that $\displaystyle \operatorname{Aut}(\mathbb{F}_{3^{100}}) \cong \mathbb{Z}/100\mathbb{Z}$. Now counting the number of subgroups is much easier!
For further reading, see this document by Keith Conrad: http://www.math.uconn.edu/~kconrad/math5211s13/handouts/finitefield.pdf