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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?

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    $\begingroup$ Why the down votes with no comments? $\endgroup$
    – Xiao
    Oct 16, 2014 at 3:54
  • $\begingroup$ 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 $|G|$. $\endgroup$
    – user169852
    May 8, 2020 at 18:37

2 Answers 2

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

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  • $\begingroup$ 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 $i=0,...k$. $\endgroup$
    – Xiao
    Oct 16, 2014 at 3:52
  • $\begingroup$ ....er.... what? $\endgroup$ Oct 16, 2014 at 3:57
  • $\begingroup$ @Xiao: Read the page the poster linked to. The given statement is listed explicitly there. $\endgroup$
    – anomaly
    Oct 16, 2014 at 3:58
  • $\begingroup$ 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. $\endgroup$ Oct 16, 2014 at 4:00
  • $\begingroup$ I made an edit, could you make a clarification of the theorem 1 from wiki please. Thanks a lot! $\endgroup$
    – Xiao
    Oct 16, 2014 at 4:20
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Xiao
    Oct 16, 2014 at 4:46
  • $\begingroup$ m is a positive integer $\endgroup$
    – Learnmore
    Oct 16, 2014 at 4:50

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