Let $L$ be a field, let $U$ be a subgroup of $Aut(L)$ and let $K$ be the fixed field of $U$. Show that an element $\alpha \in L$ is algebraic over K iff the set $\{\sigma(\alpha)|\sigma \in U\}$ is finite.
I have done the part that if the set $\{\sigma(\alpha)|\sigma \in U\}$ is finite then $\alpha \in L$ is algebraic over K.
If $\alpha$ is algebraic over $K$, take a polynomial $f$ in $K[X]$ such that $f(\alpha)=0$ (it can be the minimal polynomial). Since the coefficients of $f$ are fixed by $U$, we have
$\sigma (f(\alpha)) = f (\sigma(\alpha))=0$ for every $\sigma\in U$.
This means that the set $\{\sigma(\alpha)\}_{\sigma\in U}$ is contained in the set of roots of $f$. In particular, it is finite.