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This doesn't seem like it should be true but I'm not entirely sure - I would appreciate anyone looking over the following:

If we have a finite Galois field extension $L/K$ and $M \subset L$ a subfield with $M/K$ that is a cyclotomic extension of $K$ then we have that $M/K$ is a Galois extension and so the Galois group $Gal(L/M)$ is normal and also $Gal(M/K) = Gal(L/K) / Gal(L/M)$ which we know is cyclic (as it's a cyclotomic extension) and hence Abelian.

Then we have a subnormal series ${e} \unlhd Gal(L/M) \unlhd Gal(L/K)$?

Thanks

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  • $\begingroup$ Ah that is what I intended - I'll edit the post $\endgroup$
    – Wooster
    May 15, 2015 at 8:13
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    $\begingroup$ Recall that a solvable group requires a subnormal series in which all factors are abelian. What can you really say about $\mathrm{Gal}(L/M)$? As an extreme case, take any non-solvable extension $L/K$ and choose $M = K$, which trivially is a cyclotomic extension. $\endgroup$
    – k.stm
    May 15, 2015 at 8:17

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No. Every counterexample is with $L/M$ nonsolvable. (A group $G$ is solvable iff every subfactor is solvable, and nonsolvable iff there exists a nonsolvable subfactor. Is this clear to you?)

In particular, let $G=S_5$ act on $F(x_1,x_2,x_3,x_4,x_5)$ in the obvious way. Then we can define the two subfield $M=K=L^G$ so that $M/K$ is cyclotomic (trivial in fact) and $L/M$ is nonsolvable, because we know that ${\rm Gal}(L/M)=S_5$ is not solvable.

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  • $\begingroup$ Great - thanks. The example I had was with $L/M$ a kummer extension and so this is abelian as well - I couldn't see why it wasn't true in general! $\endgroup$
    – Wooster
    May 15, 2015 at 8:24

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