A Galois Extension over a field of characteristic p 
Hi! I've been having a bit of trouble with this question, namely the very last part. It's from a past paper for a Galois Theory course, in which I have an exam on Monday. Basically, I can show every part of the question, except the final part: it doesn't really make any sense to me. 
I reason that E/F is Galois, and has degree p. Hence, as it's finite, then |Gal(E:F)|=p. Because p is prime, then Gal(E/F) is isomorphic as an abstract group to the cyclic group of order p. But this then makes is solvable trivially. So the statement that E/F is not solvable must be false? 
Can anyone please verify that I've not missed something? I've e-mailed the course lecturer with no response at all. 
Thanks! 
 A: There are several different definitions of solvable (from hereafter, soluble) extensions, as you point out, using your given definition the problem is wrong. However, if we use the definition given below (the two are not in general equivalent) then we can solve this problem using the earlier parts.
We can say $L/K$ is soluble if there is a chain of extensions $K=K_1\subset K_2\subset\dots\subset K_n $ such that $L$ is contained in $K_n$, and each $K_i=K_{i-1}(x_i)$ for some $x_i$ such that $x_i^m \in K_{i-1}$ for some power $m$. 
So, let us take $E/F$ a Galois extension of degree $p$ as given in the question. 
Suppose we have a chain of radical extensions $F=F_1\subset F_2\subset\dots\subset F_n=L $ given as above. Without loss of generality, if $F_i=F_{i-1}(x_i)$ with $x_i^m \in F_{i-1}$, we may assume that $m$ is prime. In particular this means that any element of $F_i\setminus F_{i-1}$ will work for a choice of $x_i$.
But then by $b)$ part $i)$, $x_2 \not\in E$. By part $ii)$ $E(x_2)/F_2$ is Galois of degree $p$, so applying part $i)$ again, $x_3 \not \in E(x_2)$, so $x_3\not\in E$. So none of the $x_i$ lie in $E$ for any $i$, by repeatedly applying this procedure. Since we may choose any element of $F_i\setminus F_{i-1}$ for $x_i$, this means that no element of $F_n\setminus F$ lies in $E$. In particular, $E$ is not contained in $L$ since it is not contained in $F$.
