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Supposing we know the Galois groups of $f(x)$ and $g(x)$ over $K$, what can be said about the Galois group of $f(g(x))$?

I suppose we can restrict the question to normal polynomials over $\mathbb{Q}$, though the general case would be interesting also.

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Not much, I suppose. With fixed $g$ cubic and varying $f$ even just linear, you can obtain $S_3$, $C_3$ and even $S_2$ (i.e. produce a reducible cubic). – Hagen von Eitzen Nov 29 '12 at 7:29
up vote 5 down vote accepted

If $f(x)$ is a separable polynomial of degree $n$ then its Galois group is a subgroup of $S_n$ (via the action on the roots). And in fact any subgroup $H$ of $S_n$ occurs as a Galois group of a separable polynomial of degree $n$ over some field.

From this we get that if $f,g$ are polynomials of respective degrees $n,m$ then the Galois group $G$ of $f(g(x)$ is a subgroup of $S_{nm}$. However here not every subgroup can be achieved, because the action of $G$ on the roots of $f(g(x))$ is imprimitive (provided $n,m>1$).

The group that should replace $S_{nm}$ is the permutational wreath product $S_m\wr_n S_n$. Then one can show that $G$ always embeds into $S_m\wr_n S_n$, via the action on the roots, and that for any subgroup $H$ of $S_m\wr_n S_n$, there are a field $K$ and $f,g\in K[X]$ of respective degrees $n,m$ such that $Gal(f(g(x)),K)\cong H$.

I will end with the definition of the permutational wreath product: it is the semi-direct product $S_m^n \rtimes S_n$ where the action of $S_n$ on $S_m^{n}$ is the translation action, that is $$ (\tau_1, \ldots, \tau_n)^\sigma := (\tau_{\sigma(1)}, \ldots, \tau_{\sigma(n)}) $$ for all $\tau_i\in S_m$ and $\sigma\in S_n$.

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