How would I characterize ${\rm Gal}(\Bbb Q(ζ_{24})/\Bbb Q(i))$ up to isomorphism? How would I characterize ${\rm Gal}(\Bbb Q(ζ_{24})/\Bbb Q(i))$ up to isomorphism where $ζ_{24}$ is a twenty fourth root of unity and ${\rm Gal}(\Bbb Q(ζ_{24})/\Bbb Q(i))$ is the corresponding Galois group?
 A: Well $\operatorname{Gal}(\mathbb Q(\zeta_{24})/\mathbb Q)$ is isomorphic to the multiplicative group $(\mathbb Z/24\mathbb Z)^\times$, which is isomorphic to $(\mathbb Z/2 \mathbb Z)^3$.
Therefore, the group $\operatorname{Gal}(\mathbb Q(\zeta_{24})/\mathbb Q(i))$ is a subgroup of $(\mathbb Z/2 \mathbb Z)^3$ of order 4.
What does a subgroup like this have to look like?
A: Let $G=Gal(Q(\zeta_{24}/Q)=Z_{24}^*\cong Z_2^3$. 
The subgroup of $G$ of order 4 you want is obtained as the kernel $K$ of a homomorphism to $M=Gal(Q(i)/Q)\cong Z_2$ . 
The non-trivial element of $M$ is complex conjugation. 
The group $G$ has elements $g_k: \zeta_{24}\rightarrow \zeta_{24}^k$ for $k$ relatively prime to 24; $g_{23}$ is afforded by complex conjugation.  The kernel $K$ has the elements which act trivially on $i=\zeta_{24}^6$. So these are multiples of 6, say $6k$, which are 6 mod 24; that is $k=1\mod 4$, 
so $K=\{g_1, g_5, g_{13}, g_{17}\}\cong Z_2^2$.
A: Since
$$\operatorname{Gal}\left(\Bbb Q(\zeta_{24})/\Bbb Q\right)\cong\left(\Bbb Z/24\Bbb Z\right)^*\cong \left(\Bbb Z/8\Bbb Z\right)^*\times \left(\Bbb Z/3\Bbb Z\right)^*\cong (C_2\times C_2)\times C_2\cong (C_2)^3$$
and
$$8=[\Bbb Q(\zeta_{24}):\Bbb Q]=[\Bbb Q(\zeta_{24}):\Bbb Q(i)][\Bbb Q(i):\Bbb Q]=$$
we thus need a subgroup of order 2 in the above...
