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Conjecture

My Conjecture: Suppose that $A = \{1,2,3,...,n\}$ where $n \in \mathbb{Z}^+$. Then there always exists $E/ \mathbb{Q}$ such that $\mathcal{P}(A) \cong \text{Gal}(E/\mathbb{Q})$. Moreover, since $\mathcal{P}(A) \cong (\mathbb{Z}_2)^n$ ($\color{Blue} 1$), we have that $\text{Gal}(E/\mathbb{Q}) \cong (\mathbb{Z}_2)^n$.

$\color{Red}{Note}$: Here, $\mathcal{P}(A)$ is the power set, an abelian group under the symmetric difference operator ($\triangle$). Also, $E/ \mathbb{Q}$ denotes that $E$ is the extension field of $\mathbb{Q}$ (in this case, is as well a Galois extension) and $\text{Gal}(K/F)$ the Galois group.

Example:

Suppose that $A = \{1,2\}$, then $\mathcal{P}(A) = \{ \phi, \{1\}, \{2\}, A \}$. Hence, $(\mathcal{P}(A), \triangle) \cong (\mathbb{Z}_2 \times \mathbb{Z}_2, +)$ by $$\alpha: \mathcal{P}(A) \to \mathbb{Z}_2 \times \mathbb{Z}_2$$ such that $$B \mapsto (1_B(1), 1_B(2)) (\color{Red}3)$$ Now take $E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ be an extension field of $\mathbb{Q}$. Therefore, one can show that $\text{Gal}(E/\mathbb{Q}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \cong \mathcal{P}(A)$. Thus, $\text{Gal}(E/\mathbb{Q}) \cong \mathcal{P}(A)$ (by the transitivity of isomorphisms).

Question

Is this true for any set $A$ defined above? ($\color{Brown} 2$) That is, $|A|$ can be any finite size as defined above and still work?

Appendix:

$\color{Blue} 1. $ The reason is that if $f: \mathcal{P}(A) \to (\mathbb{Z}_2)^n$ such that $f(B)=(1_B(1), 1_B(2), ..., 1_B(n))$ is a homorphism with $1_{B \triangle C}(m) = 1_B(m) + 1_C(m) \pmod 2$ and is bijective. This is very easy to prove. However, I'm not going to prove this here. It is a good exercise to show why this is true, though.

$\color{Brown} 2.$ Actually, as long as $X$ is a set such that $|X| = |A|$ it will work as well.

$\color{Red} 3.$ $1_B(m)$ is the indicator function.

Comment:

Thank you for reading. Hope you guys help me. This does not come from any site (I think) that I've seen so far. If you guys like this, please consider giving it a thumbs up!

Arturo Magidin
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Joshua Ortiz
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    It is well-known that every finite abelian group occurs as Galois group of some extension of $\mathbb Q$. See https://en.wikipedia.org/wiki/Inverse_Galois_problem and e.g. https://math.stackexchange.com/q/131376/96384. – Torsten Schoeneberg Dec 21 '21 at 06:19
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    For reference: the problem of finding, given a finite group $G$, a Galois extension $E$ of $\mathbb{Q}$ such that $\mathrm{Gal}(E/\mathbb{Q})\cong G$ is called the "inverse Galois problem." It's known that the answer is affirmative for any solvable group, in particular for the elementary abelian groups you posit. But you don't need to invoke such a big theorem; you can show that the extension given by adjoining the square roots of $n$ distinct primes will do. See e.g. here. – Arturo Magidin Dec 21 '21 at 06:19
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    For square roots in particular see this thread and its answers (the structure of the Galois group follows easily when you have the dimension of the field extension). For more general elementary abelian $p$-groups, see the thread linked to by Torsten. This may also help, particularly as complementing Arturo's fine answer. – Jyrki Lahtonen Dec 21 '21 at 10:58

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