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Question:

Let F be a field of characteristic $0$ such that $|F:\mathbb Q|=2$, and let U be a finite subgroup of F*, the multiplicative group of F. Show that $|U|$ is 1, 2, 3, 4 or 6.

Attempt at solution:

I know all finite subgroups of the multiplicative group of a field are cyclic. I am trying to consider finite subgroups of $\mathbb Q$* (which I think are just {1}, {1,-1}) and then multiplying by the algebraic element which extends $\mathbb Q$ to F, but I can't quite get a solution.

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Hint: what are the cyclic subgroups of $\Bbb{C}^*$? Which are in two-dimensional extensions of $\Bbb{Q}$? – Chris Eagle Feb 8 at 14:59
There are cyclic subgroups of $\mathbb C$* for any integer n. This can be seen by taking the generator to be the $n^{th}$ root of unity. – FBI Feb 8 at 15:01
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Right, and what's the degree over $\Bbb{Q}$ of a primitive $n$th root of $1$? – Chris Eagle Feb 8 at 15:02
The number of co-prime integers with n which are less than or equal to n. So we want values of n so the degree is 2 or less, which are precisely 1, 2, 3, 4, 6. Thanks Chris – FBI Feb 8 at 15:08

1 Answer

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If the order of the group is $n$, then the generator is a root of the $n$th cyclotomic polynomial. The degree of this polynomial is $\phi(n)$ and must be $\le 2$. What can you say about $\phi(n)$ if $p|n$ for some prime $p>3$? And what is $\phi(2^a3^b)$ with $a,b\ge 0$?

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