I more or less copied the following notes from my professor:

Find the Galois group of $x^4-2$ over $\mathbb{Q}$.

Solution: $x^4-2$ has roots $\pm\sqrt[4]{2},\pm i\sqrt[4]{2}$. Note that since $G$ acts on the four roots, $G$ is isomorphic to a subgroup of $S_4$. The splitting field is $\mathbb{Q}(\sqrt[4]{2},i)$, so $|G|=8$. By Sylow, any two subgroups of $S_4$ of order $8$ must be isomorphic. So $G\cong D_8$.
Everything about this proof makes perfect sense to me except the part in bold: Why is $G$ a subgroup (isomorphic) of $S_4$? It cannot be simply because $G$ acts on the four roots (because for instance any group of any order acts on any set trivially). All I can tell from that is that there is a homomorphism from $G$ into $S_4$. So, how do I conclude that $G$ is a subgroup (isomorphic) of $S_4$?

Also, I'm a bit confused as to how it can be that $|G|=8$. For each $\alpha\in G$ is completely determined by $\alpha(i)$ and $\alpha(\sqrt[4]{2})$. Now, $1=\alpha(1)=\alpha(-i^2)=-\alpha(i)^2$, which seems to imply $\alpha(i)=i$. But then $\alpha(\sqrt[4]{2})$ must be one of the four roots. So $|G|\leq 4$, a contradiction. Where did I go wrong, I wonder?

Any help would be much appreciated!

2. Originally Posted by hatsoff
I more or less copied the following notes from my professor:

Everything about this proof makes perfect sense to me except the part in bold: Why is $G$ a subgroup (isomorphic) of $S_4$? It cannot be simply because $G$ acts on the four roots (because for instance any group of any order acts on any set trivially). All I can tell from that is that there is a homomorphism from $G$ into $S_4$. So, how do I conclude that $G$ is a subgroup (isomorphic) of $S_4$?

Any help would be much appreciated!

As you remark the part in bold is not completely correct. For it to be it should be that the Galois group acts transitively on

the four roots, and then it'd follow that the group is isomorphic to a subgroup of $S_4$.

There's another way to show that $G\cong D_8$: first, it's easy to see that $G$ cannot be abelian and that

its order is divisible by four and divides 8, so it is either $D_8\,\,or\,\,Q_8$. Nevertheless, this extension has a non-normal

subextension, $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}$ , from where it follows that there is a non-normal subgroup of $G$

of index 4, and since all the subgroups $Q_8$ are normal then we're done.

Tonio

3. Originally Posted by hatsoff
I more or less copied the following notes from my professor:

Everything about this proof makes perfect sense to me except the part in bold: Why is $G$ a subgroup (isomorphic) of $S_4$? It cannot be simply because $G$ acts on the four roots (because for instance any group of any order acts on any set trivially). All I can tell from that is that there is a homomorphism from $G$ into $S_4$. So, how do I conclude that $G$ is a subgroup (isomorphic) of $S_4$?

Also, I'm a bit confused as to how it can be that $|G|=8$. For each $\alpha\in G$ is completely determined by $\alpha(i)$ and $\alpha(\sqrt[4]{2})$. Now, $1=\alpha(1)=\alpha(-i^2)=-\alpha(i)^2$, which seems to imply $\alpha(i)=i$. But then $\alpha(\sqrt[4]{2})$ must be one of the four roots. So $|G|\leq 4$, a contradiction. Where did I go wrong, I wonder?

Any help would be much appreciated!

$\alpha(i)^2=-1\Longrightarrow \alpha(i)=\pm i$ , since $i^2=(-i)^2=-1$

Tonio

4. Okay, I see now what is going on. Thanks a bunch for the help!