Abstract Nonsense

Crushing one theorem at a time

Sylow’s Theorems Revisited

Point of Post: In this post we give a more refined proof of Sylow’s Theorems.

\text{ }


On this blog I have given a proof of Sylow’s theorems and an alternate proof of Sylow’s first theorem. A year or so later, with more time, and more finesse I’d like to give the simplest, most coherent proof of Sylow’s theorems.

\text{ }

Sylow’s Theorems

\text{ }

We assume all the notations such as \text{Syl}_p(G) being the set of Sylow p-subgroups of G and n_p=\#(\text{Syl}_p(G)),etc. are known to the reader:

Theorem (Sylow): Let G be a finite group with |G|=p^kn where (n,p)=1, then:

\text{ }

\begin{aligned}&\;\;\;(I)\quad \textit{If }p^m\mid |G|\textit{ for some prime }p\textit{ then }G\textit{ contains a subgroup }H\textit{ of order }p^m\\ &\;(II)\quad \textit{If }Q\textit{ is a }p\textit{-subgroup of }G\textit{ and }P\in\text{Syl}_p(G)\textit{ then }Q\leqslant gPg^{-1}\textit{ for some }P\in\text{Syl}_p(G),\\ &\quad\quad\quad\textit{in particular, any two elements of }\text{Syl}_p(G)\textit{ are conjugate}\\ &(III)\quad n_p=\left[G:N_G(P)\right]\textit{ for any }P\in\text{Syl}_p(G)\textit{ and }n_p\equiv 1\text{ mod }p\end{aligned}


\text{ }

(I): We construct such a group inductively. Evidently if m=0 then we may take the trivial group. Suppose then that we have found a subgroup H of |G| with |H|=p^m for m<k. We know then that p\mid \left[G:H\right] and so by a previous theorem we have that p\mid \left[N_G(H):H\right]. Thus, by Cauchy’s Theorem there exists xH\in N_G(H)/H with |\langle xH\rangle|=p. By the fourth isomorphism theorem we know that \langle xH\rangle=K/H for some H\leqslant K\leqslant N_G(H) and so, in particular, we have that |K|=|\langle xH\rangle||H|=p^{m+1}. As stated, from this we may construct subgroups H_0,\cdots,H_k of G with |H_k|=p^k.

\text{ }

(II): Let Q act on X=G/P by left multiplication. Since Q is a p-group we have by the ‘fundamental theorem’ that \#(X)\equiv \#(X^Q)\text{ mod }p. That said, since p\nmid \#(X) we have in particular that p\nmid \#(X^Q) and so X^Q is non-empty. So, let xP\in X^Q then qxP=xP for all q\in Q and so x^{-1}qx\in P for all q\in Q, or equivalently x^{-1}Qx\subseteq P. Thus, Q\subseteq xPx^{-1}. In particular, if Q\in\text{Syl}_p(G) we have that |Q|=|xPx^{-1}| and so Q=xPx^{-1}.

\text{ }

\text{ }

(III): Let G act on \text{Syl}_p(G) by conjugation. By (II) we know that this action is transitive. In particular, we have that n_p=\left[G:\text{stab}(P)\right] for any P\in\text{Syl}_p(G). That said, we note by definition that \text{stab}(P)=N_G(P) and so n_p=\left[G:N_G(P)\right].

\text{ }

For the second part of the statement we let P\in\text{Syl}_p(G) act on \text{Syl}_p(G)=X by conjugation. By the ‘fundamental theorem’ we know that \#(X)\equiv \#(X^P)\text{ mod }P. What we claim though is that X^P=\{P\}. Indeed, it’s evident that P\in X^P, and so now assume that Q\in X^P. We have then that P\subseteq N_G(Q) and so PQ\leqslant G, but we know that

\text{ }

\displaystyle |PQ|=\frac{|P||Q|}{|P\cap Q|}=\frac{p^{2k}}{|P\cap Q|}

\text{ }

and since |PQ|\mid |G| and p^k is the maximal power of p dividing |G| we may conclude that |P\cap Q|=p^k and so P=Q. Thus, \#(X)\equiv \#(X^Q)=1\text{ mod }p. Noting then that \#(X)=n_p completes the proof.

\text{ }

\text{ }


1. Isaacs, I. Martin. Finite Group Theory. Providence, RI: American Mathematical Society, 2008. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.


September 20, 2011 - Posted by | Algebra, Group Theory | , , , , , , ,


  1. […] Our first proof makes clever use of Sylow’s theorems: […]

    Pingback by Unit Group of a Finite Field is Cyclic « Abstract Nonsense | September 21, 2011 | Reply

  2. […] how much information about a group is contained in its actions on objects (e.g. Sylow’s theorems). Consequently, it makes sense that to understand rings we should perhaps look at the actions of […]

    Pingback by Modules (Pt. I) « Abstract Nonsense | October 27, 2011 | Reply

  3. […] A considerably slicker, and more coherent version of these proofs is given in this post of […]

    Pingback by Review of Group Theory: Sylow’s Theorems « Abstract Nonsense | December 14, 2011 | Reply

  4. […] and is prime such that . By Bezout’s lemma we can find integers such that . Now, by Sylows’ theorems we can find with . Note then (since and since the field is characteristic ), which is a […]

    Pingback by Finite Fields « Abstract Nonsense | April 12, 2012 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: