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Let $L|K$ be a finite Galois extension. Let $G = \text {Gal} \left (L|K \right ).$ Let $H$ be a normal subgroup of $G$ and $M = \text {Fix}_H L.$ Can we say that for every $\tau \in \text {Gal} \left (M|K \right )$ $\exists$ $\sigma \in G$ such that $\sigma|_M = \tau$?

If we drop the condition $H$ a normal subgroup of $G$ and instead if we take any subgroup what happens to the above case?

Any help in this regard will be highly appreciated. Thanks for reading.

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  • $\begingroup$ I think that it can be done by using primitive element theorem for finite Galois extensions. Because $L|M$ is a galois extension whenever $L|K$ is a Galois extension. So it has a primitive element say $\zeta$ i.e. $L=M[\zeta].$ $\endgroup$ Nov 25, 2019 at 16:51
  • $\begingroup$ Now if $\tau \in \text {Gal} \left ( M|K \right )$ then if allow $\tau$ to send $\zeta$ to some $\zeta'$ where $\zeta'$ is some zero of the minimal polynomial $\mu_{\zeta,M}$ of $\zeta$ over $M$ then $\tau$ extends to an automorphism of $\text {Gal} \left (L|K \right ).$ $\endgroup$ Nov 25, 2019 at 17:01
  • $\begingroup$ But then every $\tau \in \text {Gal} \left (M|K \right )$ extends to $[L:M]$ many automorphisms of $ \text {Gal} \left (L|K \right ).$ $\endgroup$ Nov 25, 2019 at 17:04
  • $\begingroup$ That will imply $\#\ \text {Gal} \left (M|K \right ) = \frac {[L:K]} {[L:M]} = [M:K].$ But this is not always true unless $\text {Gal} \left (L|M \right )$ is a normal subgroup of $ \text {Gal} \left (L|K \right ).$ Where am I going wrong then? $\endgroup$ Nov 25, 2019 at 17:10

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Let $\tau\in Gal(M/K)$. Then it can be viewed as a $K$-embedding $M\to K_{alg}$. The theorem of extension of embeddings implies that there exists $\sigma:L\to K_{alg}$ such that $\sigma_M=\tau$ (you can use the fact that $L=K(\alpha)$ for some $\alpha\in L$ if you like). But $L/K$ is a Galois, hence normal, so the image of $\sigma$ is contained in $L$ and we have in fact $\sigma\in Gal(L/K)$.

Notice this is true even if $H$ is not normal. There is no contradiction, because the equality $\sharp Gal(L/K)=\dfrac{[L:K]}{[L:M]}$ is false: ok, any $K$-automorphism of $M$ extends to a $K$-automorphism of $L$, but conversely, a $K$-automorphism of $L$ does not restrict necessarily to a $K$-automorphism of $M$, but only to a $K$-embedding. It would be true only if $H$ is normal!! So an element of $Gal(L/K)$ is not necessarily an extension of an element of $Gal(M/K)$, and you cannot count like you do.

For a specific example, think about $L=\mathbb{Q}(j,\sqrt[3]{2})$ and $M=\mathbb{Q}(\sqrt[3]{2})$. Take $\sigma:L\to L$ which sends $j$ to $j$ and $\sqrt[3]{2}$ to $j\sqrt[3]{2}$. Its restriction to $M$ is NOT a $\mathbb{Q}$-automorphism of $M$.

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  • $\begingroup$ What do you mean by "... your last implication is false..."? Which of my implication is not correct. Can you please help me so that I can find my mistake? $\endgroup$ Nov 25, 2019 at 19:10
  • $\begingroup$ The equality $\sharp Gal(L/K)=\dfrac{[L:K]}{[L:M]}$ is not true. An element of $Gal(L/K)$ is not necessarily an extension of an element of $Gal(M/K)$, so the way you count is not correct (I(ve modified my answer to be more clear) $\endgroup$
    – GreginGre
    Nov 26, 2019 at 9:00

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