43
$\begingroup$

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?

  • 19
    Multiplication is commutative. So $G$ is abelian and every subgroup is normal.2011-08-26
  • 0
    For a finite group, $G$ is nilpotent if and only if it is the direct product of its Sylow subgroups.2011-08-26
  • 0
    A slight generalization of the lemma/theorem you are wondering about is topic of [this question](http://math.stackexchange.com/questions/59665/any-periodic-abelian-group-is-the-direct-sum-of-its-maximal-p-subgroups) (in the moment there is no answer, but a good comment by Geoff).2011-08-26
  • 0
    See also : https://mathoverflow.net/questions/54735/2017-11-11
  • 0
    See also Stroppel, Locally compact groups, Theorem 6.32 and corollary 6.33.2017-11-11

3 Answers 3