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?

**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∣n, #{x∈G∣xd=1}≤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 xd−1 has d roots at most.

*Proof.* Fix d∣n and consider the set Gd made up of elements of G with order d. Suppose that Gd≠∅, so there exists y∈Gd; it is clear that ⟨y⟩⊆{x∈G∣xd=1}. But the subgroup ⟨y⟩ has cardinality d, so from the hypothesis we have that ⟨y⟩={x∈G∣xd=1}. Therefore Gd is the set of generators of the cyclic group ⟨y⟩ of order d, so #Gd=ϕ(d).

We have proved that Gd is empty or has cardinality ϕ(d), for every d∣n. So we have:

n=#G=∑d∣n#Gd≤∑d∣nϕ(d)=n.

Therefore #Gd=ϕ(d) for every d|n. In particular Gn≠∅. This proves that G is cyclic. QED

**Attribution***Source : Link , Question Author : QETU , Answer Author : Shaun*