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?
Best Answer
There's a simple proof which doesn't use Sylow's theory.
Lemma. Let $G$ a finite group with $n$ elements. If for every $d \mid n$, $\# \{x \in G \mid x^d = 1 \} \leq d$, then $G$ is cyclic.
If $G$ is a finite subgroup of the multiplicative group of a field, then $G$ satisfies the hypothesis because the polynomial $x^d - 1$ has $d$ roots at most.
Proof. Fix $d \mid n$ and consider the set $G_d$ made up of elements of $G$ with order $d$. Suppose that $G_d \neq \varnothing$, so there exists $y \in G_d$; it is clear that $\langle y \rangle \subseteq \{ x \in G \mid x^d = 1 \}$. But the subgroup $\langle y \rangle$ has cardinality $d$, so from the hypothesis we have that $\langle y \rangle = \{ x \in G \mid x^d = 1 \}$. Therefore $G_d$ is the set of generators of the cyclic group $\langle y \rangle$ of order $d$, so $\# G_d = \phi(d)$.
We have proved that $G_d$ is empty or has cardinality $\phi(d)$, for every $d \mid n$. So we have:
$$\begin{align} n &= \# G\\ & = \sum_{d \mid n} \# G_d \\ &\leq \sum_{d \mid n} \phi(d) \\ &= n. \end{align}$$
Therefore $\# G_d = \phi(d)$ for every $d \vert n$. In particular $G_n \neq \varnothing$. This proves that $G$ is cyclic. QED