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.
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.