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In Henri Cohen's book, A Course in Computational Algebraic Number Theory, the following remark about Galois groups of sextic fields is given:

Cohen Comp. ANT

I would like to know exactly how this can be shown; specifically how does one determine the groups $G_{72}$ (in the quadratic subfield case) and $S_4\times C_2$ ( in the cubic subfield case)?

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    What group is $ G_{72} $ here? – 2017-02-21
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    Is the meaning of *a sextic field* = the splitting field of an irreducible polynomial of degree six? I would have thought that a sextic field is one that is a degree six extension, but that doesn't seem right at all. – 2017-02-21
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    @Starfall, it is a transitive group of 72 elements. I think it is the Frobenius group $F_{32}$. However, I do not know very much about Frobenius groups. – 2017-02-21
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    My guess would be that for Cohen $G_{72}$ is the normalizer of a Sylow $3$-subgroup of $S_6$ (isomorphic to $S_3\wr S_2$). But if the Galois group is $S_6$, why wouldn't the fixed field of $A_6$ be quadratic? – 2017-02-21
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    @Jyrki: it is quadratic, but not a subfield of the sextic extension. – 2017-02-21
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    @JyrkiLahtonen Yes, in this case a sextic field is the splitting field of an irreducible sextic polynomial. – 2017-02-21
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    Getting warmer - possibly. My guess now is that a sextic field is a field $K=\Bbb{Q}(a)$ such that $[K:\Bbb{Q}]=6$, but the Galois group $G$ here is $Gal(L/\Bbb{Q})$ where $L$ is the splitting field of the minimal polynomial $m(x)$ of $a$. Then the question is whether there exist intermediate fields between $\Bbb{Q}$ and $K$. When we identify $G$ as a subgroup of permutations of roots of $m(x)$, this amounts to asking whether the point stabilizer $Stab_G(a)$ is maximal or not. In other words, whether $G$ acts primitively on the six roots or not. – 2017-02-21
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    So the first two lists consist of transitive but non-primitive subgroups of $S_6$. The subgroup $S_3\wr S_2$ is certainly non-primitive. Sorry about being dense. – 2017-02-21
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    So if I understood the question, it is about finding transitive non-primitive subgroups of $S_6$. I am fairly sure that A) this is explained in Dixon & Mortimer, and B) Derek Holt or some other resident expert on that topic could give the argument without breaking a sweat. I will add the group-theory tag to attract the right people. – 2017-02-21
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    @JyrkiLahtonen, I did notice that in Butler and McKay's paper on transitive groups, that there was a correspondence between the groups that were imprimitive and those which were listed in Cohen's remark. But I wasn't sure if this had anything to do with it. I guess it does. Also, thanks for the references. – 2017-02-21

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