Let $K/k$ be a Galois extension, and let $F$ be an intermediate field between $k$ and $K$. Let $H$ be the subgroup of $\text{Gal}(K/k)$ mapping $F$ into itself. Show that $H$ is the normalizer of $\text{Gal}(K/F)$ in $\text{Gal}(K/k)$.

Remark: This problem has appeared in MSE couple of times but all these problems have a bit different condition. More precisely, they consider "mapping $F$ onto itself", i.e. for $\sigma \in H$ it follows that $\sigma(F)=F$.

My solution: Denote this normalizer by $N$. We have to show that $H=N$.

I was able to show that $N\subseteq H$ without any difficulties.

Let show the converse, namely $H\subseteq N$.

Take $\sigma \in H$ then $\sigma \in \text{Gal}(K/k)$, $\sigma(F)\subseteq F$. We have to show that that $\sigma \in N$, i.e. $$\sigma \text{Gal}(K/F)\sigma^{-1}=\text{Gal}(K/F).$$ So basically speaking we have to show double containment.

i) Take $\tau\in \text{Gal}(K/F)$ then we see that $\sigma^{-1}\tau \sigma\in \text{Gal}(K/k)$. For any $x\in F$ we see that $\sigma^{-1}\tau \sigma(x)=\sigma^{-1}(\sigma(x))=x$ here I've used that $\sigma(x)\in F$ and $\tau$ fixes $F$ pointwise. So $\sigma^{-1}\tau \sigma\in \text{Gal}(K/F)$ $\Rightarrow$ $\tau \in \sigma \text{Gal}(K/F)\sigma^{-1}$. Therefore, $\text{Gal}(K/F)\subseteq\sigma \text{Gal}(K/F)\sigma^{-1}$.

ii) Let $\tau \in \sigma \text{Gal}(K/F)\sigma^{-1}$ $\Rightarrow$ $\tau=\sigma\hat{\tau}\sigma^{-1},$ where $\hat{\tau}\in \text{Gal}(K/F)$.

Easy to see that $\sigma\hat{\tau}\sigma^{-1}\in \text{Gal}(K/k)$.

Take any $x\in F$ then $\tau(x)=\sigma\hat{\tau}\sigma^{-1}(x)$.

If $\sigma(F)=F$ then I can easily show this case.

But we have that $\sigma(F)\subseteq F$ and in this case we have some troubles because in this case $\sigma^{-1}(x)$ may not be in $F$.

I would be very grateful if anyone can show how to proceed this!

  • $\begingroup$ Are you allowing infinite Galois extensions? $\endgroup$ – Lord Shark the Unknown Apr 3 at 2:58
  • $\begingroup$ I hope you are working with finite Galois extensions, otherwise you may have to take closures, for example I am not sure the normalizer will be closed as a subgroup of the whole Galois group. $\endgroup$ – астон вілла олоф мэллбэрг Apr 3 at 2:59
  • $\begingroup$ @LordSharktheUnknown, I don't know. Since it says that $K/k$ is Galois extension so it could be finite or even infinite extension. But is my point in the topic is correct? Since we have $\sigma(F)\subseteq F$ then we can have some troubles. $\endgroup$ – ZFR Apr 3 at 13:33
  • $\begingroup$ @астонвіллаолофмэллбэрг, if we have ginite Galois extension then we can consider $\sigma$ as linear mapping from finite-dimensional vector space to itself. Since it is injective then it is onto. But I didn't understand your reasoning in infinite case. $\endgroup$ – ZFR Apr 3 at 15:49
  • $\begingroup$ I am saying that in the infinite case it could potentially be false, because the normalizer need not be a closed subgroup, while FTGT requires closed subgroups in the infinite case. In the finite case what you have done seems ok. $\endgroup$ – астон вілла олоф мэллбэрг Apr 3 at 16:13

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