0
$\begingroup$

Let G be any group of order n.Also assume p be the largest prime dividing n.Let n(p) be the maximum no of sylow subgroups a group of order n can have.Is it possible to sa anything definite about the limit n(p)/n as n tends to infinity?

$\endgroup$
2
  • 1
    $\begingroup$ Choosing $\;n=p\,,\,p^2\,,\,p^3\,,\ldots\;$ , we get a subsequence of the above that converges to zero, so if the limit exists it is zero. $\endgroup$
    – Timbuc
    Oct 30 '14 at 19:02
  • $\begingroup$ Did you look at symmetric groups? $\endgroup$
    – Pedro Tamaroff
    Oct 30 '14 at 20:35
1
$\begingroup$

The limit does not exist, but it is easy to see that there are subsequences of $1,2,3,\ldots$ (such as prime powers) on which $n(p)/n=1/n$, which tends to $0$, and other subsequences (such as $n=2^{2k}3$) with $n(p)/n=1/3$, which I think is the largest possible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.