Let $K\subset L$ a finite field extension and $G=G(L/K)$ a Galois group of $L$ over $K$ then $|G|\leq[L:K]$.
I was trying to show this.
By the Primitive Element Theorem $\exists\alpha\in L$ such that $L=K(\alpha)$. Take $p(x)=irr(\alpha,K)$ (irreducible polynomial of $\alpha$ in $K$), $\forall \rho\in G$ we have $\rho(\alpha)\in L$ and $p(\rho(\alpha))=0$. Set: $$\begin{array}{lccc} \phi:&G&\longrightarrow&R_p\cap L\\ &\rho&\longrightarrow&\rho(\alpha) \end{array}$$
But I can't show that $\phi$ is a homomorphism. (My ideia is to use the first theorem of isomorphism). Somebody have any hints?
Let $L=K(\alpha)$, as you noted. Any automorphism of $L/K$ is completely determined by the image of $\alpha$. Since the minimal polynomial of $\alpha$ has degree $n=[L:K]$, this polynomial has at most $n$ roots in a splitting field, and thus $\alpha$ has at most $n$ possible images. Thus, the automorphism group has order at most $n$.