show that $K/k$ is galois Let $K$ be a finite separable extension of a field $k$ of prime degree $p$. Let $\theta$ in K be such that $K = k(\theta)$ and $\theta_1 ..., \theta_p$ be the conjugates of $\theta$ over $k$ in some algebraic closure $k^a$. Let $\theta=\theta_1$ and if $\theta_2 \in k(\theta)$ Then show that $K/k$ Galois.
As far as I can see $Aut(K/k) \leq p$, nothing else other than this. Can one please give some hint ?
 A: Here is another approach: The galois group is a subgroup of $S_p$ and contains an element of order $p$ by Cauchy's Theorem, let us call it $\sigma$. This is a $p$-cycle. After possibly replacing $\sigma$ by a suitable power, we may assume $\sigma(\theta)=\theta_2$, in particular the restriction of $\sigma$ to $K$ is a map $K \to K$, hence so are all powers of $\sigma$. We conclude by noting that for each $i$ there is some power of $\sigma$ mapping $\theta$ to $\theta_i$, in particular $\theta_i \in K$.
A: Let $|Aut(k(\theta):k|=a$. Consider $E=k(\theta_1,\theta_2,..,\theta_p)$. Clearly $E/k$ is Galois. Now as $E/k$ is separable, so we can let $[E:k(\theta)]_{s}=[E:k(\theta)]=n$.  
Now consider the set $S \subset G$ such that $S=\{\sigma \in G \mid \sigma_{k(\theta)}\in Aut(k(\theta)/k)\}$. Clearly, $S$ is subgroup of $G$. Now for each $\sigma \in Aut(k(\theta)/k)$ there exist an extension of $\sigma$ into $Gal(E/k)$. So, for $a$ number of elements we get total $na$ elements in $G=Gal(E/k)$ and hence $|S|=na$. 
But on other hand $Gal(E/k)=[E:k]=[E:k(\theta)][k(\theta):k]=np$. So, $na \mid np \implies a \mid p \implies |Aut(k(\theta):k)|=p$. But that implies all of conjugates of $\theta$ are in $k(\theta) \implies k(\theta)/k$ is Galois and also is cyclic of order $p$
