# Sylow's First Theorm in GRE book (typo?)

In my Gre book, the Sylow's First Theorem is stated as

Let $G$ be a finite group of order $n$, and let $n= p^k m$, where $p$ is a prime that does not divide $m$. Then $G$ has at least one subgroup of order $p^i$ for every integer $i$ from $0$ to $k$.

Is this correct? Or is this a corollary? I remember Sylow's theorem only guarantees the existence of a $p$-Sylow subgroup, not for each $p^i$ with $i=0,...k$.

Thanks a lot!

Edit: From Wiki Page

"Theorem 1: For any prime factor $p$ with multiplicity $n$ of the order of a finite group $G$, there exists a Sylow $p$-subgroup of $G$, of order $p^n$."

This Theorem 1 only says the existence of a Sylow $p$-subgroup, not a $p$-subgroup.

I checked with Dummit & Foote book. The first theorem there also only says the existence of a Sylow $p$-subgroup, not a $p$-subgroup.

• $p$-subgroup is order $p^\alpha$ such that $p^\alpha$ divides $|G|$.

• Sylow $p$-subgroup is order $p^\alpha$ such that $|G| = p^\alpha m$ where $p$ does not divide $m$.

For example $|G| = 24 = 2^3 3$, then a subgroup of order $4$ is a $p$-subgroup but not a Sylow $p$-subgroup. Since $24 = 2^2 6$ which means $m=6$.

So can I really say that there exists a subgroup of order $4$ using Sylow's theorem?

• Why the down votes with no comments?
– Xiao
Oct 16, 2014 at 3:54
• Every finite $p$-group contains subgroups of all possible orders, so it's no loss of generality to state the Sylow existence theorem only for the maximal order of $p$ dividing $|G|$.
– user169852
May 8, 2020 at 18:37

## 2 Answers

This is usually known as Sylow's First Theorem. Check out the summary at http://en.wikipedia.org/wiki/Sylow_theorems

• It says there exists a subgroup of order $p^k$. But it doesn't say that there exist subgroups of order $p^i$ for each $i=0,...k$.
– Xiao
Oct 16, 2014 at 3:52
• ....er.... what? Oct 16, 2014 at 3:57
• @Xiao: Read the page the poster linked to. The given statement is listed explicitly there. Oct 16, 2014 at 3:58
• Looking at the wikipedia page gives Theorem 1: A finite group G whose order |G| is divisible by a prime power $p^k$ has a subgroup of order $p^k$. If that doesn't help I can't say anymore. Oct 16, 2014 at 4:00
• I made an edit, could you make a clarification of the theorem 1 from wiki please. Thanks a lot!
– Xiao
Oct 16, 2014 at 4:20

Dont be confused.Sylow's theorem guaranteed only the existence of a subgroup of order p^m if p^m divides order of G.However a much more generalized version which can be shown to be true is that if order of G is of the form p^k*q gcd(p,q)=1 then for each i 1<=i<=k; G has a subgroup of order p^i. and this corollary is sometimes taken as the theorem nowadays.

• I made a 2nd edit. And for your first statement, what is the restriction on $m$? Because I think your first statement (there exists subgroup of order $p^m$ whenever $p^m$ divides $|G|$) is actually more general than your generalized version of the theorem.
– Xiao
Oct 16, 2014 at 4:46
• m is a positive integer Oct 16, 2014 at 4:50