Let $L/K$ be a finite Galois extension and $\alpha\in L\setminus K$. Then there exists $h\in G$ with prime power order not fixing $\alpha$. Let $L/K$ be a finite Galois extension of fields and let $G=\text{Gal }L/K$. Let $\alpha\in L$ with $\alpha\notin K$. Show that there exists $h\in G$ with $h$ of prime power order not fixing $\alpha$.
I struggle to get started. An earlier part of the problem asks us to show that there exists $g\in G$ with $g(\alpha)\neq \alpha$, but this is trivial. I am not sure how to use this. One strategy I considered involved raising $g$ to some power $m$ to make $g^m$ have prime power order, but I am not sure how to show that such $g^m$ need not fix $\alpha$. I also considered inducting on the number of prime factors of the order of $g$, but no clear opportunity arises to apply the inductive hypothesis.
 A: Suppose $g$ does not fix $\alpha$ and that $g$ has order $r = p^at$ where $t$ is not divisible by $p$. Choose integers $c, d$ such that $cp^a+dt=1$. If both $g^{p^a}$ and $g^t$ fix $\alpha$, then so do $(g^{p^a})^c$ and $(g^t)^d$. But then $(g^{p^a})^c\cdot (g^t)^d = g^{cp^a+dt} = g$ also fixes $\alpha$. Now use induction on the number of prime factors of $r$.
A: Suppose $|G|=n=p_1^{e_1}\cdots p_k^{e_k}$ where $p_1<\cdots <p_k$ are the prime divisors of $|G|$. Let $P_i$ be the Sylow $p_i$-subgroup of $G$. The elements of $G$ that have prime power order are exactly the elements of the union $\bigcup_{i=1}^k P_i$. The question asked is: Show that if $\alpha\in L\setminus K$, then some $h\in \bigcup P_i$ fails to fix $\alpha$. An equivalent formulation is: Show that if $\alpha\in L$ and $\alpha$ is fixed by all elements of $\bigcup P_i$, then $\alpha\in K$.
Here is an argument for the equivalent formulation: 
If $\alpha\in L$ is fixed by all elements of $\bigcup P_i$, then $\alpha$ is fixed by all elements of the subgroup generated by $\bigcup P_i$, which is $G$. Thus $\alpha\in K$, since $K$ is the fixed field of $G$.
In this argument I am using that any finite group is generated by the union of its Sylow subgroups. The reason for this is: the subgroup $S$ generated by $\bigcup P_i$ contains each $P_i$, hence has order divisible by each $|P_i|=p_i^{e_i}$, hence has order divisible by $\textrm{lcm}(p_1^{e_1},\ldots,p_k^{e_k})=n$. But $S\leq G$ must also have order dividing $|G|=n$. Hence $S$ has order exactly $n$, forcing $S=G$.
