Galois group of $f(x^2)$

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?

• Do you mean to ask if there is a way that is faster than simply brute force computing it again? – Edward Evans Oct 9 '16 at 14:08
• 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. – IAlreadyHaveAKey Oct 9 '16 at 14:10

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 $H$.
• 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! – IAlreadyHaveAKey Oct 9 '16 at 15:00
• 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 $H$. – Jyrki Lahtonen Oct 10 '16 at 5:38
• (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 $H$. – Jyrki Lahtonen Oct 10 '16 at 5:42