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If the Galois group of some irreducible polynomial $f(x)$ over some field $F$ is known, is there some method for calculating the Galois group of the polynomial $f(x^2)$ over the same field?

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  • $\begingroup$ Do you mean to ask if there is a way that is faster than simply brute force computing it again? $\endgroup$ – Edward Evans Oct 9 '16 at 14:08
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    $\begingroup$ Well yes, somehow using the Galois group of $f(x)$. It seems to me that the groups should be related somehow, but I have no evidence to base that on, it just seems like they should. $\endgroup$ – IAlreadyHaveAKey Oct 9 '16 at 14:10
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The two groups are related in the following sense: assuming that $ f \in F[x] $ is separable with $ \operatorname{char} F \neq 2 $, if $ L/F $ is the splitting field of $ f(x) $ and $ M/F $ is the splitting field of $ f(x^2) $ in some fixed algebraic closure $ \bar{F} $, then $ L $ is a subfield of $ M $.

Identifying $ H = \textrm{Gal}(L/F) $ and $ G = \textrm{Gal}(M/F) $, we observe that $ H $ is a quotient of $ G $, that is, there is a surjective homomorphism $ G \to H $ given by restriction to $ L $. Furthermore, the extension $ M/L $ is obtained by adjoining square roots of elements to $ L $, therefore the Galois group $ \textrm{Gal}(M/L) $ admits a particularly simple decomposition: it is the direct sum of finitely many copies of $ \mathbf Z/2\mathbf Z $. Thus, we see that the Galois groups $ G $ and $ H $ fit into a short exact sequence

$$ 0 \to (\mathbf Z/2\mathbf Z)^n \to G \to H \to 0 $$

where $ [M:L] = 2^n $. I do not think more can be inferred, in general, about the group $ G $ from the structure of the group $ H $.

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  • $\begingroup$ So this relationship can be used to find the order of $G$ given $H$ and $[M : L]$. I guess other techniques for determining $G$ are required? The order does help a fair bit though, especially for relatively low values. Thanks a lot! $\endgroup$ – IAlreadyHaveAKey Oct 9 '16 at 15:00
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    $\begingroup$ A good answer! I am just wondering whether something more can be squeezed out of the following. Assume that $\deg f=m$. Then $H$ can be viewed as a subgroup of $S_m$ as its permutation action on the zeros of $f$ is faithful. Then $G$ could be seen as a group of "signed" permutations on the set of roots (with signs determined according to a fixed choice of signs for the square roots of the zeros of $f$). This would then realize $G$ as a subgroup of the wreath product $({\bf Z}/2{\bf Z})\wr H$. Clearly $m\ge n$, and the subgroup of "pure sign chages in $G$" needs to be stable under $H$. $\endgroup$ – Jyrki Lahtonen Oct 10 '16 at 5:38
  • $\begingroup$ (cont'd) Anyway, this should give some information about the group operation of $G$. Obviously it won't be fully determined, but I think it gives a bit more: I think we can realize $G$ as a subgroup of $({\bf Z}/2{\bf Z})\wr S_m$ such that the projection $p:G\to S_m$ equals the projection $G\to H$, and the extra comes from the fact that $\rm{ker}\,p$ must be stable under the action of $H$. $\endgroup$ – Jyrki Lahtonen Oct 10 '16 at 5:42
  • $\begingroup$ That's a nice idea - yes, I think it gives a little more information. $\endgroup$ – Ege Erdil Oct 10 '16 at 5:55

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