# Finite subgroups of the multiplicative group of a field are cyclic

In Grove’s book Algebra, Proposition 3.7 at page 94 is the following

If $G$ is a finite subgroup of the multiplicative group $F^*$ of a field $F$,
then $G$ is cyclic.

He starts the proof by saying “Since $G$ is the direct product of its Sylow subgroups …”. But this is only true if the Sylow subgroups of $G$ are all normal. How do we know this?

There’s a simple proof which doesn’t use Sylow’s theory.

Lemma. Let $$GG$$ a finite group with $$nn$$ elements. If for every $$d∣nd \mid n$$, $$#{x∈G∣xd=1}≤d\# \{x \in G \mid x^d = 1 \} \leq d$$, then $$GG$$ is cyclic.

If $$GG$$ is a finite subgroup of the multiplicative group of a field, then $$GG$$ satisfies the hypothesis because the polynomial $$xd−1x^d - 1$$ has $$dd$$ roots at most.

Proof. Fix $$d∣nd \mid n$$ and consider the set $$GdG_d$$ made up of elements of $$GG$$ with order $$dd$$. Suppose that $$Gd≠∅G_d \neq \varnothing$$, so there exists $$y∈Gdy \in G_d$$; it is clear that $$⟨y⟩⊆{x∈G∣xd=1}\langle y \rangle \subseteq \{ x \in G \mid x^d = 1 \}$$. But the subgroup $$⟨y⟩\langle y \rangle$$ has cardinality $$dd$$, so from the hypothesis we have that $$⟨y⟩={x∈G∣xd=1}\langle y \rangle = \{ x \in G \mid x^d = 1 \}$$. Therefore $$GdG_d$$ is the set of generators of the cyclic group $$⟨y⟩\langle y \rangle$$ of order $$dd$$, so $$#Gd=ϕ(d)\# G_d = \phi(d)$$.

We have proved that $$GdG_d$$ is empty or has cardinality $$ϕ(d)\phi(d)$$, for every $$d∣nd \mid n$$. So we have:

n=#G=∑d∣n#Gd≤∑d∣nϕ(d)=n.\begin{align} n &= \# G\\ & = \sum_{d \mid n} \# G_d \\ &\leq \sum_{d \mid n} \phi(d) \\ &= n. \end{align}

Therefore $$#Gd=ϕ(d)\# G_d = \phi(d)$$ for every $$d|nd \vert n$$. In particular $$Gn≠∅G_n \neq \varnothing$$. This proves that $$GG$$ is cyclic. QED