# Show that finite subgroup of units of a field is cyclic.

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?

• Looks like the standard proof more or less. Or see this duplicate. – Dietrich Burde Apr 17 at 19:53
• 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. – Lubin Apr 17 at 20:42
• @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. – user661541 Apr 17 at 20:55
• Yes, exactly... – Lubin Apr 17 at 21:00