splitting field of separable and irreducible polynomial: does isomorphism of Galois subgroups imply isomorphism of subfields? The original problem is here: Galois extension: does isomorphism in subgroups implies isomorphism of the subfield?. My question was, given $L/Q$ finite Galois extension, and suppose for $S$, $H$ Galois subgroups and $S \cong H $, is it necessarily true that $L^H \cong L^S$. This is not true as Hagen von Eitzen pointed a counterexample with $Q(\sqrt{2}, \sqrt{3})$. However, I noticed that part of the reason is that $\sqrt{2}, \sqrt{3}$ are not the roots of the same irreducible polynomials. So $x^2 -3$ has a solution in one field but doesn't have in the other. Now suppose $L$ is the splitting field of $Q$ with respect to some irreducible, separable polynomial $p(x) \in Q[x]$, does the statement hold in this case? This seems correct to me, because $Q(\alpha) \cong Q[x]/p(x) \cong Q(\beta)$ for $\alpha, \beta$ roots of $p$, so intuitively these roots are indistinguishable, and if the permutation of these roots are isomorphic, it seems we can just swap the roots to get an isomorphism of the subfields. Is this statement correct?
 A: Your new restriction doesn’t really buy you anything new. The old example still fits.
The Primitive Element Theorem tells you that any finite extension of $\mathbb{Q}$ is simple, and hence is isomorphic to $\mathbb{Q}(\alpha)$ for some root $\alpha$ of an irreducible polynomial. In particular, if $F$ is a finite Galois extension of $\mathbb{Q}$, then there exists $\alpha\in F$ such that $F=\mathbb{Q}(\alpha)$ and $F$ is the splitting field of the irreducible polynomial of $\alpha$.
In the case of the example you already have: it is well known that $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$. So if you take $p(x)$ to be the irreducible of $\sqrt{2}+\sqrt{3}$, namely
$$p(x) = x^4-10x^2+1$$
then $\mathrm{Gal}(p(x))\cong C_2\times C_2$, with the fixed fields of the three subgroups of order $2$ being $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{6})$, no two of which are isomorphic as fields.
While it is true that adjoining any of the four roots of $p(x)$ gives you the same field, it just gives you the splitting field in all cases.

For a more involved example, take a polynomial with Galois group isomorphic to $S_4$. Such polynomials exist. Saying “has Galois group isomorphic to $S_4$” is just a way of saying the splitting field $L$ has a Galois group $S_4$. In $S_4$, you have two subgroups isomorphic to the Klein $4$-group, $N$ and $H$, with $N\triangleleft S_4$ and $H$ not normal, namely:
$$\begin{align*}
N &= \{e, (12)(34), (13)(24), (14)(23)\}\\
H &= \{e, (12), (34), (12)(34)\}.
\end{align*}$$
But that means that the fixed field of $N$ is Galois over $\mathbb{Q}$, while the fixed field of $H$ is not. Hence, they two fields cannot be isomorphic.
