# Problem 15 from Lang's Algebra

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!

• Are you allowing infinite Galois extensions? – Lord Shark the Unknown Apr 3 at 2:58
• 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. – астон вілла олоф мэллбэрг Apr 3 at 2:59
• @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. – ZFR Apr 3 at 13:33
• @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. – ZFR Apr 3 at 15:49
• 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. – астон вілла олоф мэллбэрг Apr 3 at 16:13