In continuation to my last post:
In class we saw an example that says: $n=[\mathbb{F}_{p^n}:\mathbb{F}_{p}]=|\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p})|$ ; ($p$ is prime).
My thoughts are that if I look at any Galois extension then it is the splitting field of separable polynomials in the field I started with. I know that any automorphism of the extension field that fixes the field I started with sends each root of an irreducible factor (of one of the polynimials that the Galois expenstion is their splitting field), and also that for every such permutation I have an automorphism .
But since if an irreducable factor have $k$ roots then I can permute on them in $k!$ options I deduced that the size of Galois group is of the form $k_1!k_2!\cdots k_t!$, in particular it is either $1$ or devisable by $2$.
But this contradicts what I wrote in my first paragraph if for example we take $p=2,n=5$ since $5$ is odd.
Can someone please point out the mistake ?