Galois group of $f(x^2)$

Question

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?

Answer 1

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 $.