1
$\begingroup$

Let $G$ be a $p$ group of order $p^n$ and $k\leq n$.

Theorem:Number of the subgroups of $G$ with order $p^k$ is congurent to $1$ modulo $p$.

I have found a proof of this theorem in Rotman's book but I wonder that "Is there any shorter or elegant proof for this theorem" ?

$\endgroup$
  • 2
    $\begingroup$ There is a proof by Wielandt, which applies to any finite group of order divisible by $p^k$. You can find it in my answer to math.stackexchange.com/questions/479839 $\endgroup$ – Derek Holt Feb 15 '15 at 19:33
  • $\begingroup$ @DerekHolt: I am not asking the proof of Sylow's theorem. $G$ is already a $p$ group. Do you mean that when we take $m=1$ then $|G|=p^n$ then the therem can be applied to this problem ? $\endgroup$ – mesel Feb 15 '15 at 20:12
  • $\begingroup$ Yes that's right, it's a more general result than the one you are asking about. The proof may not be shorter than the one in Rotman's book, but it could be said to be elegant. $\endgroup$ – Derek Holt Feb 15 '15 at 20:20
  • $\begingroup$ @DerekHolt: I liked this proof. It also shows existence of sylow-p subgroup. Can this method be used for proving Schur Zasenaus therem ? $N$ is normal subgroup of $G$ such that $(|N|,|G:N|)=1$. Let $\Omega$ be set of all subset of $G$ with cardinality $|G:N|$ and $G$ acts on $\Omega$ by right multiplication. .. $\endgroup$ – mesel Feb 15 '15 at 21:50

Your Answer

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

Browse other questions tagged or ask your own question.