6
$\begingroup$

How to show that $G=\mathrm{Gal}[\Bbb Q(\xi_{2^n})/\Bbb Q(\xi_{2^m})]$ cyclic? $n>m>1$ are natural numbers. $\xi_{2^n}$ and $\xi_{2^m}$ are cyclotomic roots.

I know that the order of $ G $ is $|G|=2^{n-m}$

$\endgroup$
2
  • $\begingroup$ The only thing I succeed at it is to show that $G \cong Z_2 \times Z_{2^{n-2}}/Z_2 \times Z_{2^{m-2}}$ but why this is cyclic? $\endgroup$ Jun 21, 2018 at 16:36
  • 3
    $\begingroup$ The Galois groups are generated by the automorphisms $\xi \mapsto \xi^{-1}$ and $\xi \mapsto \xi^{5}$. $\endgroup$
    – sharding4
    Jun 21, 2018 at 16:58

0

You must log in to answer this question.

Browse other questions tagged .