What is the Galois group $Gal(E/F)$ when $F=GF(5^3)$ and $E=GF(5^{24})$ What is the Galois group $Gal(E/F)$ when $F=GF(5^3)$ and $E=GF(5^{24})$. I have attempted to describe the Galois group, but I've become stuck, and it's entirely possible that I've made mistakes as well. Here's what I've done.
I know that for any finite field, there exists irreducible polynomials of any degree $n$ over that field. So then there exists an irreducible polynomial $p$ over $F$ of degree $8$. Since $E$ is a finite extension of $F$, it is a simple extension, so let's say $E = F(\alpha)$ where $\alpha$ is some root of $p$. The basis of $E$ over $F$ is then $\{1, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5, \alpha^6, \alpha^7\}$, and $[E:F] = 8 = |Gal(E/F)|$. So, the Galois group should be isomorphic to either $\mathbb{Z}_8, \mathbb{Z}_4\times\mathbb{Z}_2$ or $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$, unless I'm mistaken.
Assuming that I've not made mistakes so far (which is far from certain), I don't know where to go from here in order to explicitly describe the automorphisms of the Galois group. How do I know what those automorphisms are? Some help would be nice. 
 A: Notice, that we have $x^{p^n} = x$ for all $x\in GF(p^n)$ ($p$ a prime and $n\ge1$ a natural number): this is how $GF(p^n)$ is constructed, namely by adjoining all roots of $X^{p^n}-X$ to $GF(p) = \mathbb Z/p\mathbb Z$.
One checks that the Frobenius $\varphi\colon GF(p^n) \rightarrow GF(p^n)$, $x\mapsto x^p$ is an injective ring homomorphism, and hence an automorphism of fields. Since $\varphi(x) = x$ for all $x\in GF(p)$ (by the little Fermat), $\varphi$ is an element of ${\rm Gal}(GF(p^n)/GF(p))$, of which you know that this is a group of order $n$.
It therefore suffices to show that the powers $\varphi^k:= \varphi\circ\dotsm \circ\varphi$ ($k$ times) are all distinct for for $k$ running through $0,1,\dotsc,n-1$. Assume that $\varphi^j = \varphi^k$ for some $0\le j\le k\le n-1$. It follows that $\varphi^{k-j}= {\rm id}_{GF(p^n)}$, i. e. $\varphi^{k-j}\colon x\mapsto x^{p^{k-j}}$ fixes all points in $GF(p^n)$. In other words, $X^{p^{k-j}}-X$ is a polynomial of degree $p^{k-j}<p^n$ (since $k-j<n$) with $p^n$ many roots. Over a field, this is only possible if $X^{p^{k-j}}-X$ is the zero polynomial, i. e. $k-j=0$. This shows that ${\rm id}_{GF(p^n)} = \varphi^0, \varphi, \varphi^2,\dotsc,\varphi^{n-1}$ are $n$ pairwise distinct elements of ${\rm Gal}(GF(p^n)/GF(p))$. Hence this group is cyclic.
For $k\mid n$, ${\rm Gal}(GF(p^n)/GF(p^k))$ is a subgroup of the cyclic group ${\rm Gal}(GF(p^n)/GF(p))$ and hence itself cyclic (generated by $\varphi^k\colon x\mapsto x^{p^k}$).
