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Let $F$ be a field. Show that every finite subgroup of $F^\times$ is cyclic.

My attempt:

Let $H$ be a subgroup of $F^\times$. Suppose $p\mid|H|$ with $p$ prime.

Any element in $H$ of order $p$ is a root of the polynomial $X^p-1 \in F[X]$, and such a polynomial can have at most $p$ zeros. Thus, $H$ has at most $p-1$ elements of order $p$ ($1$ is also a root of the polynomial). By Cauchy's theorem, there is also an element of order $p$, and the subgroup this element generates has order $p$ and thus there are precisely $p-1$ elements of order $p$ in $H$.

By the fundamental theorem of finitely generates abelian groups, it follows that the factorisation of $H$ must have precisely one factor $C_p$ (if it has more than one such factor, then there would be more than $p-1$ elements of order $p$. This is because $C_p^n$ contains $p^n-1$ elements of order $p$).

Thus, we have proven

$$H \cong \prod_{p||H|, p \text{ prime}} C_p$$

and by the Chinese Remainder theorem, $H$ is cyclic.

Is this proof ok?

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  • $\begingroup$ Looks like the standard proof more or less. Or see this duplicate. $\endgroup$ – Dietrich Burde Apr 17 at 19:53
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    $\begingroup$ This is roughly the first proof that I was taught. But it’s perfectly possible that higher powers of primes occur: for instance $\Bbb F_9$, the quadratic extension of $\Bbb Z/3\Bbb Z$, has a total multiplicative group of order $8=2^3$. I’m sure you can modify the proof you’ve given above to achieve correctness. $\endgroup$ – Lubin Apr 17 at 20:42
  • $\begingroup$ @Lubin. Thanks for your comment. I see the problem now. It should be that the decomposition contains only one factor of the form $C_{p^n}$ for some n. $\endgroup$ – user661541 Apr 17 at 20:55
  • $\begingroup$ Yes, exactly... $\endgroup$ – Lubin Apr 17 at 21:00

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