# Abstract Nonsense

## 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.

Motivation

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. $\blacksquare$

References:

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.