# 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?

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## 3 Answers

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?

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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 $K=\{g_1, g_5, g_{13}, g_{17}\}\cong Z_2^2$.

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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...

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Is the subgroup just C2? –  Mathematician_in_Training Nov 7 '12 at 19:39
Well, isomorphic to: going with the isomorphisms I wrote above, it could as well be one of $\,C_2\times\{1\}\times\{1\}\cong \{1\}\times C_2\times\{1\}\cong\{1\}\times\{1\}\times C_2\,$ . Once you have this one, you have to take the quotient of the Galois group by it to get the group of the extension you want, according to Galois Correspondence theorem. –  DonAntonio Nov 7 '12 at 19:42
how would one going about actually taking the quotient of the Galois group? –  Mathematician_in_Training Nov 7 '12 at 20:01