# Generalization of the Sylow theorem(part I)

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.

• 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... Apr 1 '18 at 12:29
• 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! Apr 1 '18 at 12:37
• 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:) Apr 1 '18 at 12:42
• 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. Apr 1 '18 at 12:46
• @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. Apr 1 '18 at 12:47

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.