0

I know that for some prime $p$ ,$p$-elements (i.e. order of elements are some power of prime $p$) need not generate a $p$-group as in seen in $S_3$, but is it true in nilpotent groups that $p$-elements generate a $p$-group. It is definitely true in abelian groups but not sure about nilpotent groups.

Departed
  • 1,528

1 Answers1

2

If by $p$-elements you mean elements of order $p$ then yes, they generate a $p$-group in a (finite) nilpotent group, since they are all contained in the unique Sylow $p$-subgroup.

verret
  • 6,691
  • That argument only works for finite nilpotent groups, but the statement is true in general, and the question did not say anything about finiteness. – Derek Holt Mar 30 '16 at 09:34
  • @DerekHolt can you supply a proof for this statement for nilpotent groups (not necessarily finite) – Departed Mar 30 '16 at 21:49
  • 1
    The elements of finite order in a nilpotent group form a subgroup: see http://math.stackexchange.com/questions/79474/ for a proof. So any two $p$-elements generate a finite subgroup and then the result follows from the finite case. – Derek Holt Mar 30 '16 at 22:02