TITLE: How is $Gal(\mathbb Q(\sqrt{2+\sqrt{2}})/\mathbb Q)\cong \mathbb Z/4\mathbb Z$?
QUESTION [7 upvotes]: $\def\Gal{\operatorname{Gal}}$
I was working on homework, and the problem starts off by saying that I previously showed (I can't find where, though) that with $\def\Q{{\mathbb Q}}\def\Z{{\mathbb Z}}F=\Q$ and $L=\Q(\sqrt{2+\sqrt{2}})$, $G=\Gal(L/\Q)\cong \Z/4\Z$. I can't figure out how this is so.
The min poly for $\sqrt{2+\sqrt{2}}$ over $\Q$ is $x^4-4x^2+2$, whose roots are:
$$
\def\a{{\alpha}}\a_1=\sqrt{2+\sqrt{2}},\,\a_2=-\a_1,\,\a_3=\sqrt{2-\sqrt{2}},\,\a_4=-\a_3
$$
so therefore $\def\s{{\sigma}}\def\t{{\tau}}\s\in G$ can be thought of as a permutation of the $\a_i$. Note that $\a_1\a_3 = \sqrt{2}$. So $G = \{e, \s, \t, \s\t\}$ where:
$$
\s(\a_1)=\a_2,\,\t(\a_1)=\a_3\\
\s(\a_2)=\a_1, \s(\a_3)=\s\left(\frac{\sqrt{2}}{\a_1}\right)=\frac{\sqrt{2}}{\a_2}=\a_4, \s(\a_4)=\a_3\\
\t(\a_3)=\t\left(\frac{\sqrt{2}}{\a_1}\right)=\frac{\sqrt{2}}{\a_3}=\a_1, \t(\a_2)=\t(-\a_1)=\a_4, \t(\a_4)=\a_2,\text{ so}\\
\quad\s\to(12)(34)\\
\quad\t\to(13)(24)\\
\quad\t\s\to(14)(23)
$$
In this group, each element is its own inverse, so this should mean that $G \not\cong \Z/4Z$, but $G\cong V$, the Klein four-group.
Where have I gone wrong?

REPLY [4 votes]: $\def\a{{\alpha}}\def\t{{\tau}}\def\Gal{\operatorname{Gal}}\def\Q{{\mathbb Q}}$Your mistake was the assumption that $\t(\sqrt{2})=\sqrt{2}$; $\t(\sqrt{2})=-\sqrt{2}$:
$$
\a_1 = \sqrt{2+\sqrt{2}}, \a_3 = \sqrt{2-\sqrt{2}}\\
\begin{align}
\t(\a_3)&=\t\left(\frac{\sqrt{2}}{\a_1}\right)\\
&=\t\left(\frac{\color{red}{\a_1^2-2}}{\a_1}\right)\\
&=\frac{\a_3^2-2}{\a_3}\\
&=\frac{-\sqrt{2}}{\a_3}\\
&=-\a_1=\a_2
\end{align}
$$
And since the $\a_1$ then has order $> 2$, we know that $\t(\a_2)=\a_4, \t(\a_4)=\a_1$. Note that we don't need to worry if $\t(2)=2$ since $\t\in G$ so $\t$ fixes $\Q$