Why $H\neq N_G(H)$? Let $K$ be a field, $f(x)$ a separable irreducible polynomial in $K[x]$. Let $E$ be the splitting field of $f(x)$ over $K$. Let $\alpha,\beta$ be distinct roots of $f(x)$. Suppose $K(\alpha)=K(\beta)$. Call $G=\mathrm{Gal}(E/K)$ and $H=\mathrm{Gal}(E/K(\alpha))$. How can I prove that $H\neq N_G(H)$?
My idea was to take a $\sigma: E\rightarrow\bar{K}$ with $\sigma_{|K}=id$ and $\sigma(\alpha)=\beta$. Then $\sigma\in G\backslash H$. So if I prove that for every $\tau\in H$ one has $\sigma\tau\sigma^{-1}(\alpha)=\alpha$, then this means $\sigma\in N_G(H)$. But I don't know how to prove it, I even don't know if it is true. Any help?

Since every $\tau\in H$ by definition fixes $\alpha$, one has $\sigma\tau\sigma^{-1}(\beta)=\sigma\tau(\alpha)=\sigma(\alpha)=\beta$, so $\sigma\tau\sigma^{-1}$ fixes $\beta$ and therefore $K(\beta)=K(\alpha)$, whence $\sigma\tau\sigma^{-1}\in H$. So indeed $\sigma\in N_G(H)\setminus H$. This was too easy.
