0
$\begingroup$

Only consider the part I of the Sylow theorem.

The statement of the Sylow theorem(part I) in wikipedia is that:

For every prime factor $p$ with multiplicity $n$ of the order of a finite group $G$, there exists a Sylow $p$-subgroup of $G$, of order $p^n$.

I know the general version is also correct. i.e. If the order of $G$ is $p^nm$, $p$ is some prime and $gcd(p,m)=1$, then for each $1\le i \le n$, there exists a subgroup of $G$ with order $p^i$.

I also know a result which says that, if $G$ has a sylow-$p$ group, and $H$ is a subgroup of $G$, then $H$ also has its sylow-$p$ group. This result can be found in the proof of this question:A proof of Sylow theorem

I want to prove the general version of Sylow theorem by using both the version on wiki and the result above.

We can first find a Sylow-$p$ subgroup $P$ of $G$ with order $p^n$. If we can find a subgroup of $P$ with order $p^{n-1}$, then by the version on wiki and the result above, we are done.

But I cannot find such a subgroup of $P$. What can I do best is using the Cauchy theorem to find a subgroup $Q$ of $P$ with order $p$ (Hence cyclic), and |$P:Q|=n-1$. I even cannot prove whether $Q$ is normal or not.

My question is how to find a subgroup of $P$ with order $p^{n-1}$.

Any help will be appreciated.

$\endgroup$
7
  • $\begingroup$ I really am not sure what your question exactly is, but if $\;G\;$ is a group of order $\;p^n\;,\;\;n\in\Bbb N\;$ , then for any $\;0\le k\le n\;$ , the group $\;G\;$ not only has a subgroup of order $\;p^k\;$ but in fact also a normal subgroup of order $\;p^k\;$ . Again, I don't know if this is what you're asking... $\endgroup$
    – DonAntonio
    Apr 1 '18 at 12:29
  • $\begingroup$ Maybe what I wrote is too tedious... My question is how to find a subgroup of $P$ with order $p^{n-1}$, where $P$ is a sylow-p subgroup of $G$. Yeah, I know they are normal, which just follows from the part II of Sylow theorem. Could you help find such a subgroup of $P$? Thanks! $\endgroup$
    – Sam Wong
    Apr 1 '18 at 12:37
  • $\begingroup$ Hmmm, if the number of Sylow-$p$ subgroup is greater than $1$, then I cannot use the part II of Sylow theorem to prove it is normal. So...by the way...how to prove them are normal...? Thanks Don:) $\endgroup$
    – Sam Wong
    Apr 1 '18 at 12:42
  • $\begingroup$ So you are asking how to prove that any group of order $p^n$ with $p$ prime has a subgroup of order $p^{n-1}$? You could have asked that in two lines with no reference to Sylow's theorem. One way to do it is by induction on $n$, using the result that the centre of a nontrivial finite $p$-group is nontrivial. $\endgroup$
    – Derek Holt
    Apr 1 '18 at 12:46
  • $\begingroup$ @Sam (1) No, they are "not normal"...at least not necessarily all of them. For example, out of five subgroups of order $\;2\;$ that the dihedral group $\;D_4\;$ of order $\;8\;$ has, only one is normal. (2) The group $\;P\;$ you're apparently looking for is one of the subgroups the theorem I mentioned in my first comment promises. $\endgroup$
    – DonAntonio
    Apr 1 '18 at 12:47
3
$\begingroup$

First, one needs to know that the center of any finite $\;p\,-$ group isn't trivial , and thus we can proceed by induction on $\;n\;$, when we have $\;|G|=p^n\;$ .

For $\;n=0,1,2,\;$ the result is trivial (why?), so we can assume $\;n\ge3\;$. We look at $\;G/Z(G)\;$ . By the above, this is a $\;p\,-$ group of order $\;p^{n-1}\;$ at most . Use now the inductive hypothesis and then the correspondence theorem to end the business.

$\endgroup$
1
  • 1
    $\begingroup$ I follow it. Thanks Don:) $\endgroup$
    – Sam Wong
    Apr 1 '18 at 13:11

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.