Prove $\operatorname{Gal}(M_3/M_1)\cong\operatorname{Gal}(M_2/(M_1\cap M_2))$

abstract-algebraautomorphism-groupfield-theorygalois-extensionsgalois-theory

I've been solving problems from my Galois Theory course, and I got the idea to solve this one but I'm unable to prove some details. The problem goes like this:

Being $L/K$ a Galois extension, and being $M_1/K$ and $M_2/K$ Galois
subextensions. Prove that, if $M_3$ is the smallest subfield of $L$
which contains $M_1$ and $M_2$, then $\operatorname{Gal}(M_3/M_1)\cong
\operatorname{Gal}(M_2/(M_1\cap M_2))$.

After thinking about it and drawing the corresponding Hasse diagram, I concluded the isomorphism between those Galois groups is
$$\varphi:\operatorname{Gal}(M_3/M_1)\longrightarrow \operatorname{}Gal(M_2/(M_1\cap M_2))$$
$$ \sigma\in \operatorname{Gal}(M_3/M_1) \implies \varphi(\sigma)=\sigma|_{M_2} $$
Now I need to prove it's an isomorphism, so I need to prove it's well defined, it's a homomorphism and it's a bijection. My problem is that I'm unable to prove some of these. Here's what I already did:

  1. Well defined: Being $\sigma\in Gal(M_3/M_1)$, it was easy to prove that $\sigma|_{M_2}(\lambda)=\lambda$, $\forall\lambda\in (M_1\cap M_2)$. However, I don't really know how to prove $\sigma|_{M_2}$ is an automorphism in $M_2$ (this may be trivial, but I'm not sure how to prove it).
  2. Homomorphism: this one was obvious I think, because it's trivial that, being $\sigma,\tau\in \operatorname{Gal}(M_3/M_1)$, then $$\sigma \circ\tau|_{M_2} = \sigma|_{M_2}\circ\tau|_{M_2}.$$
  3. Bijection: I thought the easiest way to prove this last one was first proving it's injective and then prove $|\operatorname{Gal}(M_3/M_1)|=|\operatorname{Gal}(M_2/(M_1\cap M_2))|$ (I assumed I would be able to prove it by degree transitivity but now I'm not sure if I can do it with just the hypothesis that $M_3$ is the smallest field that contains $M_1$ and $M_2$). However, I couldn't prove any of these statements (maybe surjectivity is easier to prove than $|\operatorname{Gal}(M_3/M_1)|=|\operatorname{Gal}(M_2/(M_1\cap M_2))|$, I don't know).

How can I finish this exercise? Is what I already did correct? Any help or hint will be appreciated, thanks in advance.

Best Answer

This follows directly form your recent post since $M_3=M_1M_2\subseteq L$ (assuming that all extensions are finite; but as you are talking about the order of the groups I think you have this case in mind too).

What you did reiterates the proof of the theorem I linked and is correct so far. The only difficult part left is surjectivity. If this part is not clear to you I can try to formulate it explictly.

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