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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.

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    If $\alpha$ is algebraic over $K$, then it has a minimal polynomial $m(x)$. Can you show that $\sigma(\alpha)$ is then also a zero of $m(x)$? How many zeros can $m(x)$ have? – 2017-01-26

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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.