0
$\begingroup$

Does someone have an idea about how to prove the following claim?

If $G_1 $ , and $G_2 $ are two p-groups, of the same order, but $G_2 $ has bigger rank, then $G_2 $ has more normal subgroups than $G_1$ [ rank of a group = minimal number of generators ]

Thanks in advance

  • 0
    This can't possibly be true, any simple group gives a counterexample. Are you missing an assumption, e.g. that they're p-groups?2012-10-16
  • 0
    @NoahSnyder : Yes, you're right... Sorry... I indeed forgot to mention that I'm interested in p-groups Thanks!2012-10-16
  • 0
    "Rank" has many meanings, so what does it mean here?2012-10-16
  • 0
    @DerekHolt: I think I mean here the usual meaning of "rank"- The minimal number of generators(i.e. - the size of a basis)2012-10-16
  • 0
    For reference, although that meaning of rank is "standard" (e.g. it has a wikipedia page) I don't think it's widely known. I certainly didn't know about it.2012-10-16

2 Answers 2