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Let $f$ be a polynomial over a field $K$ of degree $n$ such that $f(x^2)$ is separable. Assume that the Galois group $G$ of (a splitting field of ) $f(x^2)$ is maximal, that is to say, the wreath product $G= C_2 {\rm wr} S_n$. (In particular, the Galois group of $f$ is $S_n$.)

The action of the group $G$ on the roots of $f(x^2)$ coincides with the imprimitive action of the wreath product.

My question is whether there is a natural set with Galois action which corresponds to the primitive action of the wreath product.

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  • $\begingroup$ When you say "the primitive action", which action do you mean? Any transitive, primitive finite group action comes from the action of a group on the cosets of a maximal subgroup. It doesn't seem to me that $C_{2} {\rm wr} S_{n}$ ever has a transitive, primitive action for $n \geq 3$. $\endgroup$ – Jeremy Rouse Jan 28 '18 at 20:49
  • $\begingroup$ I am refereeing to the natural actions (the imprimitive on pairs and the primitive on functions): en.wikipedia.org/wiki/… $\endgroup$ – Lior Bary-Soroker Jan 28 '18 at 20:56
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Yes, there is such a set. First, identify $C_{2}$ with $\{ \pm 1 \}$, and note that the primitive action of $C_{2}~{\rm wr}~S_{n} = C_{2}^{n} \rtimes S_{n}$ is on the set of functions $\theta : \{ 1, 2, \ldots, n \} \to C_{2}$. An element of $C_{2}^{n}$ acts component-wise and an element of $h \in S_{n}$ acts by sending $\theta$ to the function $h \cdot \theta(s) = \theta(h^{-1}(s))$. (I'm following the notation from here.)

Let $\beta_{1}$, $\beta_{2}$, $\ldots$, $\beta_{n}$ be the roots of $f$, and define $S = \{ \pm \sqrt{\beta_{1}} \pm \sqrt{\beta_{2}} \pm \cdots \pm \sqrt{\beta_{n}} \}$. Given an element $s = \sum_{i=1}^{n} \epsilon_{i} \sqrt{\beta_{i}}$, we can identify $s$ with the function $\theta : \{ 1, 2, \ldots, n \} \to C_{2}$, where $\theta(i) = \epsilon_{i}$. (It's easy to see that $\sqrt{\beta_{1}}$, $\sqrt{\beta_{2}}$, $\ldots$, $\sqrt{\beta_{n}}$ are linearly independent over $K$, and so this is well-defined.)

Let $L$ be the splitting field of $f(x)$ and $M$ be the splitting field of $f(x^{2})$. The elements of $C_{2}^{n}$ correspond to those elements in ${\rm Gal}(M/L)$. For $\sigma \in {\rm Gal}(M/L)$, define $\epsilon_{i} = \sigma(\sqrt{\beta_{i}})/\sqrt{\beta_{i}}$. Then, $\sigma$ corresponds to the function $\theta(i) = \epsilon_{i}$.

If $h \in S_{n}$, the element of ${\rm Gal}(M/K)$ corresponding to $h$ is the one for which $\sigma(\sqrt{\beta_{i}}) = \sqrt{\beta_{h(i)}}$. We then have that if $s = \sum \epsilon_{i} \sqrt{\beta_{i}} \in S$, then $$ \sigma(s) = \sum \epsilon_{i} \sqrt{\beta_{\sigma(i)}} = \sum_{i} \epsilon_{\sigma^{-1}(i)} \sqrt{\beta_{i}}. $$

This shows the actions of $C_{2}^{n}$ and $S_{n}$ on the set of functions corresponds to the actions of $C_{2}^{n}$ and $S_{n}$ on $S$.

Addendum for those who are might be confused about the terminology of "primitive" action: In the literature on wreath products, the action of $A~{\rm wr}~B$ on the set of functions is sometimes called the product action. This product action is primitive if and only if $B$ is transitive and $A$ is primitive but not regular. (This is apparently Lemma 2.7A from Dixon and Mortimer's book "Permutation groups".) The action of $C_{2}~{\rm wr}~S_{n}$ is not primitive because $C_{2}$ is primitive and regular.

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  • $\begingroup$ Nice! Exactly what I wanted. Thanks a lot. $\endgroup$ – Lior Bary-Soroker Jan 29 '18 at 0:13

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