# Isomorphic Galois subgroups generate isomorphic fixed fields

Let $$\mathbb{L}/\mathbb{K}$$ be a finite normal extension with Galois group $$G = \operatorname{Gal}(\mathbb{L}/\mathbb{K})$$. Let $$\mathbb{L}^H := \left\lbrace~ x \in \mathbb{L} \mid \sigma(x) = x \quad \forall \sigma \in H ~\right\rbrace$$ denote the fixed field of $$H$$.

Let $$H, K$$ be two subgroups of $$G$$. Does $$H \cong K$$ imply $$\mathbb{L}^H \cong \mathbb{L}^K$$?

I tried going through an example:

Let $$\mathbb{L} := \mathbb{Q}(\sqrt[3]{7}, \zeta_3)$$ (where $$\zeta_3$$ is the third root of unity) be the splitting field of $$x^3 - 7 \in \mathbb{Q}[x]$$. Clearly, the Galois group is $$S_3$$. Take two subgroups of $$S_3$$ in the isomorphism class of $$S_2$$; $$H_1 := \big\lbrace~\operatorname{Id}, (12)~\big\rbrace$$ and $$H_2 := \big\lbrace~\operatorname{Id}, (23)~\big\rbrace$$. Now, $$\mathbb{L}^{H_1} = \mathbb{Q}(\sqrt[3]{7}\zeta_3^2) \cong \mathbb{Q}(\sqrt[3]{7}) = \mathbb{L}^{H_2}$$. So it seems to work for this case; though I'm not sure if I just got lucky.

• Consider the field $\mathbb Q(\sqrt 2, \sqrt 3)$. The Galois group has one subgroup of order $2$ that fixes $\sqrt 2$ and another that fixes $\sqrt 3$, but the fixed subfields are not the same.
– lulu
Commented Dec 29, 2022 at 20:39
• It is conjugate subgroups that yield isomorphic fixed fields. Commented Dec 29, 2022 at 20:42

There is only one group of order $$2$$ (up to isomorphism), but there are often a huge number of degree-$$2$$ extensions of a field -- you can adjoin the square root of any element!
So, we look for a field $$F$$ with two non-isomorphic quadratic extensions. The easiest example, as lulu points out, is $$\mathbb{Q}$$ with $$\mathbb{Q}(\sqrt{2})$$ and $$\mathbb{Q}(\sqrt{3})$$. We can tell that these extensions are non-isomorphic because $$\mathbb{Q}(\sqrt{2})$$ does not contain a square root of $$3$$: if $$(a + b\sqrt{2})^2 = 3$$ then $$a^2 + 2b^2 = 3$$ and $$2ab = 0$$, which forces $$a = 0$$ or $$b = 0$$, which implies that $$3$$ is either a square or twice a square, which is false.
Now we simply take the composite of these extensions, which is $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$. Since quadratic extensions are finite and normal, this composite is also finite and normal. Since $$\mathbb{Q}$$ has characteristic $$0$$, this extension is also separable, so it is Galois. Then the fundamental theorem of Galois theory tells us that there are subgroups $$H,K$$ of order $$2$$ such that $$\mathbb{Q}(\sqrt{2},\sqrt{3})^H = \mathbb{Q}(\sqrt{2})$$ and $$\mathbb{Q}(\sqrt{2},\sqrt{3})^K = \mathbb{Q}(\sqrt{3})$$.