"Let N be the splitting field of $f(X)=X^4+9$ over $\mathbb{Q}$, and let $G = Gal(N/\mathbb{Q})$.
Using the fact that $α = \sqrt{3}\exp{\frac{πi}{4}}$ is a root of $f(X)$ and $α^{2} = 3i$, show that $|G| = 4 $ and that $G$ contains elements $σ$ and $τ$ such that $σ(α) = iα$ and $τ (α) = −α$.
Find $σ(i)$ and $τ(i)$, and find the orders of $σ$ and $τ$."
Let $N = \mathbb{Q}(\alpha,i)$
so the $deg(N)=[\mathbb{Q}(\alpha,i):\mathbb{Q}]=[\mathbb{Q}(\alpha,i):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]$
We know $\alpha^2=3i$ so $\alpha^2 \in \mathbb{Q}(i)$,so
$deg(N)=[\mathbb{Q}(\alpha,i):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]=2 \times 2=4$
$\lvert G\rvert = deg(N) = 4$
Now this is where I am stuck, why are $σ(α) = iα$ and $τ (α) = −α$?
I understand $σ$ and $τ$ permute the roots of $f(x)$, so we can choose $σ$ and $τ$ to do this, but that isn't enough to SHOW that $σ(α) = iα$ and $τ (α) = −α$
Recall that the Galois' group of the irreducible polynomial $f$ acts transitively on its roots and note that $-\alpha$ and $i\alpha$ are roots of $f$. So there exists such $\sigma$ and $\tau$. Finally, $$ \sigma(i)=\sigma(\frac{\alpha^2}{3})=\frac{\sigma(\alpha)^2}{3}=\frac{(i\alpha)^2}{3}=-i $$ And analogously $$ \tau(i)=\frac{\tau(\alpha)^2}{3}=\frac{(-\alpha)^2}{3}=i $$