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.

  • 4
    $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Dec 29, 2022 at 20:39
  • 5
    $\begingroup$ It is conjugate subgroups that yield isomorphic fixed fields. $\endgroup$ Commented Dec 29, 2022 at 20:42

1 Answer 1


lulu's comment is a perfect concise answer. To expand on the situation a bit:

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 $\mathbb{Q}(\sqrt{2},\sqrt{3})^K = \mathbb{Q}(\sqrt{3})$.


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