Just to clarify, I'm thinking of finite abelian groups, but it may be true in general.

This feels like it's obvious, but I'm not quite sure why. Maybe someone can justify it?

(Without using that the group is isomorphic to the product of its Sylow p-subgroups, since I feel that there's a more simple reason.)

  • 3
    $\begingroup$ If you try a few examples you will see that in fact not all elements are contained in a Sylow subgroup. $\endgroup$ – Tobias Kildetoft Nov 15 '16 at 20:29

Certainly not.

An element of a Sylow $p$-subgroup as or a power of $p$. Hence any element that is not of prime power order is not in a Sylow subgroup. Example: A generator of the cyclic group of order $6$.

  • $\begingroup$ You have a minor typo in your second sentence. $\endgroup$ – Cameron Williams Nov 15 '16 at 20:33
  • $\begingroup$ Ah right, in that case, is every element a product of elements from a finite number of Sylow p-subgroups? $\endgroup$ – IIM Nov 15 '16 at 20:42
  • $\begingroup$ At least, if we are talking about finite groups. $\endgroup$ – IIM Nov 15 '16 at 20:43

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