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?

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 dn, #{xGxd=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 xd1 has d roots at most.

Proof. Fix dn and consider the set Gd made up of elements of G with order d. Suppose that Gd, so there exists yGd; it is clear that y{xGxd=1}. But the subgroup y has cardinality d, so from the hypothesis we have that y={xGxd=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 dn. So we have:

n=#G=dn#Gddnϕ(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

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