# Determine the fixed field of a subgroup of $S_3$ and check the Galois correspondence

I came across this homework problem and I'm stumped. This is the third part of the problem, so to give some context, here's what we have so far:

Let $f(x)=x^3-2 \in \mathbb{Q}[x]$. We know that the splitting field is $K=\mathbb{Q}(\sqrt[3]{2},\omega)$, where $\omega$ is the 3rd root of unity. We have $[K:\mathbb{Q}]=6$, which is also the order of the Galois group $G(K,\mathbb{Q})$. Lastly, $G(K,\mathbb{Q})$ is isomorphic to the symmetric group $S_3$.

Here's the part with which I'm having trouble:

For a single nontrivial subgroup $H$ of $S_3$, find $K_H$ (the fixed field of $H$) and check the correspondence given in the fundamental theorem of Galois theory.

I decided to choose $H=\{id, (1,2,3), (1,3,2)\} \subset S_3$. We know that $K_H=\{a\in K\mid \sigma(a)=a \,\,\forall\,\, \sigma\in H\}$.

Let $\sigma \in H$. Then, we have the three following automorphisms

$\sigma_1 = \begin{pmatrix} 1&2&3\\3&1&2\end{pmatrix}$

$\sigma_2 = \begin{pmatrix} 1&2&3\\2&3&1\end{pmatrix}$

$\sigma_3 = \begin{pmatrix} 1&2&3\\1&2&3\end{pmatrix}$

Let $T=\mathbb{Q}(\omega)$ and let $\sigma\in G(K,T)$.

By looking at $\sigma_1$ or $\sigma_2$, we see that none of $r_1,r_2,r_3$ is in the fixed field $K_H$.

For the correspondence, I'm not exactly sure what the question is asking me to do. The fundamental theorem of Galois theory states that for $\mathbb{Q}\subset T \subset K$, we associate to $T$ the subgroup $G(K,T)\subset G(K, \mathbb{Q})$. Conversely, for $H\subset G(K,\mathbb{Q})$, associate to $H$ the fixed field $K_H$.

Under this correspondence, we have:

(1) $T=K_{G(K,T)}$

(2) $H=G(K,K_H)$

(3) $[K:T]=o(G(K,T))$ and $[T:\mathbb{Q}]=\frac{o(G(K,\mathbb{Q}))}{o(G(K,T))}$

(4) $T$ is a normal extension of $\mathbb{Q}$ iff $G(K,T)$ is a normal subgroup of $G(K,\mathbb{Q})$.

(5) If $T$ is normal over $\mathbb{Q}$, then $G(K,T)$ is isomorphic to the quotient group $G(K,\mathbb{Q})/G(K,T)$

This statement is very long, and I'm not sure if the question is asking me to verify that each of the five conditions holds for my specified $H$ and $T$ in $K$, or if it just wants me to verify that a one-to-one correspondence exists between $H$ and $T$. In either case, I'm not sure how to proceed.

I apologize for the long post. Any guidance at this point is greatly appreciated.

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I am sure by $H=(1,3,2)$ you actually mean subgroup generated by $(1~3~2)$... you said $\sigma\in H$ implies $\sigma =(1~3~2)$ which is not actually the case –  Praphulla Koushik Mar 23 '14 at 5:14
It is asking you to verify that any sub field is coming from a subgroup and any subgroup would give a sub field... you have to check this correspondence.. I am not sure if he wants you to check for those normal subgroup cases... It is advised to write down explicitly all elements in the galois group i.e., given an element in galois group where does $\sqrt[3]{2}$ go and where does $\omega$ go.. –  Praphulla Koushik Mar 23 '14 at 5:18
You seem to have misunderstood what fixed field of $\sigma$ means. It consists of solutions of $\sigma(a)=a$. Your definition of $\sigma$ already told you that $\sigma(r_1)=r_3$, $\sigma(r_2)=r_1$ and $\sigma(r_3)=r_2$ (assuming that's what you meant by the permutation $(132)$). But $r_1\neq r_3\neq r_2\neq r_1$, so this very clearly means that none of $r_1,r_2,r_3$ is in the fixed field. See what happens to $$d=(r_1-r_3)(r_3-r_2)(r_2-r_1)$$ when you apply $\sigma$ to it. –  Jyrki Lahtonen Mar 23 '14 at 6:13
And, of course, +1 for explaining your thoughts. Makes it easier to see where the problem is. There is are two different but related concepts of fixed under $\sigma$. One could be called setwise, i.e. a set $S$ is fixed under $\sigma$, iff $\sigma(S)\subseteq S$. The other is elementwise, i.e. $\sigma(a)=a$ for all $a\in S$. Both of these concepts occur in Galois theory, so you need to pay attention to the difference. –  Jyrki Lahtonen Mar 23 '14 at 6:18
@PraphullaKoushik ah yes, I made a mistake. I do indeed mean the subgroup generated by $(1,3,2)$, namely $H=\{id, (1,2,3),(1,3,2)\}$. I'll edit that above. Thanks. –  Hildegarde Mar 23 '14 at 15:06