Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Alternate Proof of the Sylow Theorems

Point of post: In this post we give some alternate proofs of Sylow’s theorems which use a little more machinery then the “classic” ones.



As remarked the Sylow Theorems are probably the most fundamentally important set of theorems in finite group theory. Consequently, any new proof verifying their validity is welcomed. In this post we give alternate proofs (in some sense) of the Sylow theorems than the “classic” proofs we have already given. They are less constructive and use more machinery. That said, they are very elegant and “cute”.

Sylow’s Theorems


We begin with a simple enough number theoretic lemma:


Lemma: Let p be prime and \alpha\geqslant 0. Then, for all r\in\mathbb{N} one has

\displaystyle {p^\alpha m\choose p^\alpha}\equiv m\text{ mod }p

Proof: We note that since p is prime it’s trivial that


\displaystyle {p\choose k}\equiv 0\text{ mod }p


for 1\leqslant k\leqslant p-1. Thus,


\left(1+x\right)^p\equiv 1+x^p\text{ mod }p


(recalling that two polynomials are equivalent \text{ mod }p if and only if their coefficients are equivalent). It clearly follows using the same argument that


(1+x)^{p^2}=\left((1+x)^p\right)^p\equiv \left(1+x^p\right)^p\equiv 1+x^{p^2}\text{ mod }p


and so proceeding by induction one finds that


\left(1+x\right)^{p^n}\equiv 1+x^{p^n}\text{ mod }p


for every n\in\mathbb{N}. In particular


\left(1+x\right)^{p^\alpha}\equiv 1+x^{p^\alpha}\text{ mod }p


and so


\left(1+x\right)^{p^{\alpha}m}\equiv \left(1+x^{p^\alpha}\right)^m\text{ mod }p


comparing the coefficients of x^{p^\alpha} gives the desired result. \blacksquare


With this we can prove what is the equivalent statement of the first Sylow theorem in our last post.


Theorem: Let G be a group and p a prime such that p\mid |G|. Then, G has a Sylow p-subgroup.

Proof: For notational convenience let |G|=p^\alpha m where p\nmid m. We then let


\mathcal{S}=\left\{S\subseteq G:\#\left(S\right)=p^\alpha\right\}


Note that \mathcal{S} is evidently a G-space under the G-action of left multiplication, namely for each g\in G and s\in S define


g\cdot S=gS


Note by the Orbit Decomposition Theorem that


\displaystyle {p^\alpha m\choose p^\alpha}=\#\left(\mathcal{S}\right)=\sum_{\mathcal{O}_S\in\text{Orb}\left(\mathcal{S}\right)}\#\left(\mathcal{O}_S\right)


and thus we must have, by our lemma, that p\nmid\#\left(\mathcal{O}_S\right) for some S\in\mathcal{S}. It follows then that


\displaystyle p\nmid\#\left(\mathcal{O}_S\right)=\left(G:G_S\right)=\frac{|G|}{|G_S|}=\frac{p^\alpha m}{|G_S|}


from where it follows that p^\alpha\mid |G_S|. Note though that if g\in G_S then S=g\cdot S=gS and thus g\in S. It follows then that G_S\subseteq S and thus |G_S|\leqslant p^\alpha. We may then conclude that |G|=p^\alpha from where the conclusion follows. \blacksquare



1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200

3.Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.


January 11, 2011 - Posted by | Algebra, Group Theory | , ,

1 Comment »

  1. […] Point of post: This is a continuation of this post. […]

    Pingback by Review of Group Theory: Alternate Proof of the Sylow Theorems (Pt. II) « Abstract Nonsense | January 12, 2011 | 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 )

Google+ photo

You are commenting using your Google+ 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 )


Connecting to %s

%d bloggers like this: