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Assume that $K$ is a finite normal extension of a field F with Galois group $G:=G(K/F)$. Assume that $K$ has exactly 2 distinct subfields $E_1$ and $E_2$ such that $F\subsetneq E_1$ and $E_2 \subsetneq K$, and assume $E_1$ and $E_2$ are both normal extensions of $F$. Prove $G$ is cyclic.

How am going to proceed? I think some sylow theorems might help. But I don't know how to start.

Any help would be appreciated.

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  • $\begingroup$ Hint: Think about the Galois-correspondence and see this question : math.stackexchange.com/questions/322849/… $\endgroup$ – Mathematician 42 Apr 20 '17 at 8:42
  • $\begingroup$ I don't think we need both intermediate fields to be normal extensions of the base field. The result holds true without this condition. $\endgroup$ – Paramanand Singh Apr 20 '17 at 12:32
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Be sure you can follow the following providing justifications:

First of all, what can be $\;n:=\;$ the extension's degree? Clearly $\;n\;$ can't be a prime number (why?), and it can't also be that $\;n\;$ has more than two proper non-trivial divisors. Thus, we either have $\;n=pq\;$ , for two different primes $\;p,q\;$ , or else $\;n=p^3\;$ for some prime $\;p\;$ .

In the first first case $\;n=pq\;$: if the extension isn't cyclic then the Galois Group isn't even abelian and is thus the non-trivial semidirect product of two cyclic groups of order $\;p,\,q\;$. This case is possible only if $\;p<q\;$ and $\;p\,\mid\,(q-1)\;$ , say $\;q-1=kp\;$ , which would then give us $\;q>1\;$ different subgroups of order $\;p\;$, a contradiction. Thus, $\;|G|=pq\;$ is the cyclic group of order $\;pq\;$ and indeed has one single subgroup of order $\;p,\,q\;$, and both subgroups correspond with normal extensions of $\;F\;$ since the whole extension is abelian and thus the subgps. are abelian, too.

If $\;[K:F]=|G=\text{Gal}\,(K/F)|=p^3\;$ , the group $\;G\;$ has two subgroups of orders $\;p,\,p^2\;$ . If $\;G\;$ isn't cyclic then it must have at least one subgroup more of at least one of the two orders $\;p,\,p^2\;$, which cannot be true by assumption...

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  • $\begingroup$ Just one quick question. If I choose base field to be $\mathbb{Q} $ and the intermediate fields obtained by adjoining $\sqrt{2},\sqrt{3} $respectively and $K=\mathbb{Q} (\sqrt{2},\sqrt{3})$, then Galois group is of order $4$ and yet only two intermediate fields $\mathbb{Q} (\sqrt{2}), \mathbb{Q} (\sqrt{3}) $ are there. What am I missing here. Sorry if this question appears silly/trivial. $\endgroup$ – Paramanand Singh Apr 20 '17 at 12:49
  • $\begingroup$ @ParamanandSingh I think you may be forgetting the intermediate extension $\;\Bbb Q(\sqrt6)\;$ ...And indeed: this extesion is not cyclic as its Galois group is $\;V\;$ , the Klein group. $\endgroup$ – DonAntonio Apr 20 '17 at 12:51
  • $\begingroup$ Oh yes I did miss the $\sqrt{6}$. And yeah the group is $Z_2\times Z_2$ same as $V$. $\endgroup$ – Paramanand Singh Apr 20 '17 at 13:05

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