# Abstract Nonsense

## Review of Group Theory: Sylow’s Theorems

Point of post: In this post we discuss the concept of Sylow’s Theorems.

Note: A considerably slicker, and more coherent version of these proofs is given in this post of mine.

Motivation

The majority of the interesting theorems in basic finite group theory can be proven using Sylow’s Theorem (or their ilk) in one way or another. That said, not much motivation is needed.

Cauchy’s Lemma for Abelian Groups

We begin with a lemma of Cauchy which will help us greatly along the way. In essence Cauchy’s Lemma says that if $|G|$ is a finite abelian group and $p$ a prime such that $p\mid |G|$ then $G$ has an element of order $p$. More precisely:

Remark: In the following proof keep in mind the convention that for an order of an element $|\cdot|$ and $\text{ord}$ are used interchangeably.

Theorem: Let $G$ be a finite abelian group and $p$ a prime such that $p\mid |G|$. Then, $G$ contains an element of order $p$.

Proof: It suffices to prove this for a fixed but arbitrary prime $p$. We induct on abelian groups of order $np$ where $n\in\mathbb{N}$. Clearly the result is true for $n=1$ since $|G|=p$ implies that $G$ is cyclic and thus even non-identity element of $G$ has order $p$. Assume the result is true for $pm,\text{ }m and assume that $|G|=n$. If $G$ is cyclic we’re done since every cyclic group has precisely one subgroup of each order which divides the ambient group (see here for a proof). So, assume that $G$ is not cyclic and let $g\in G$ be such that $\left\langle g\right\rangle \ne G$. If $p\mid |g|$ we’re done since $\left\langle g\right\rangle$ would have an element of order $p$ so assume not. Then, $\displaystyle p\mid \left(G:\left\langle g\right\rangle\right)$ and since $G/\left\langle g\right\rangle$ is an abelian group whose order is of the form $pm,\text{ }m we may apply the induction hypothesis to product some $h\left\langle g\right\rangle$ such that $\text{ord}\left(h\left\langle g\right\rangle\right)=p$. Note though that if $\pi:G\to G/\left\langle g\right\rangle$ is the canonical projection then using the fact that for any homomorphism $\text{ord}\left(\phi(g)\right)\mid \text{ord}(g)$ we may conclude that

$p=\text{ord}\left(h\left\langle g\right\rangle\right)=\text{ord}\left(\pi(h)\right)\mid \text{ord}(h)$

But, this implies by previous theorem that

$\displaystyle \text{ord}\left(h^{\frac{|h|}{p}}\right)=\frac{|h|}{\left(|h|,\frac{|h|}{p}\right)}=\frac{|h|}{\frac{|h|}{p}}=p$

Thus, $h^{\frac{|h|}{p}}$ is an element of $G$ of order $p$. The induction is complete. $\blacksquare$

Sylow’s Theorems

With Cauchy’s Lemma we are now able to prove the Sylow theorems. First though we need a little terminology.

Let $G$ be a finite group and $H\leqslant G$ be such that $|H|=p^m$ where $p$ is prime and $p^{m+1}\nmid |G|$. Then, $H$ is called a Sylow $p$-subgroup. In general a group $G$ is called a $p$-group if for every $g\in G$, $|g|=p^m$ for some $m\in\mathbb{N}$ ($m$ is dependent upon $g$). A subgroup of a group $G$ which is a $p$-group is called a $p$-subgroup of $G$.

Theorem( First Sylow’s Theorem): Let $G$ be a finite group and let $p$ be a prime. If $p^k\mid |G|$ then $G$ has a subgroup of order $p^k$ where $k\in\mathbb{N}$.

Proof: We do this by induction on $n$. Clearly the result is true for $n=1$. Assume then that every group of order $m has the property that if $p^k\mid m$ then any group of order $m$ has a subgroup of order $p^k$ and let $G$ be a group of order $n$. If $p\mid\left|\mathcal{Z}\left(G\right)\right|$ then by Cauchy’s Lemma we have that there exists some $g\in \mathcal{Z}\left(G\right)$ such that $|g|=p$. Note then that $\left\langle g\right\rangle\unlhd G$ and $\left(G:\left\langle g\right\rangle\right)<|G|$. Thus, by the induction hypothesis there exists some subgroup $H'\leqslant G/\left\langle g\right\rangle$ such that $|H'|=p^{k-1}$. Recall though that one of the consequences of the Fourth Isomorphism Theorem is $H'=H/\left\langle g\right\rangle$ for some $H\leqslant G$. Note though that

$|H|=|H'|\left|\left\langle g\right\rangle\right|=p^{k-1}p=p^k$

and thus the conclusion follows.

Suppose now that $p\nmid \left|\mathcal{Z}\left(G\right)\right|$ then evidently by the class equation

$\displaystyle |G|=\left|\mathcal{Z}\left(G\right)+\right|\sum_{r\in\Gamma}\left(G:G_r\right)$

and thus by assumption we must have that there exist some $r_0\in\Gamma$ for which $p\nmid \left(G:N(r_0)\right)$. Note then though that we must have that $p^m\mid \left|N(r_0)\right|$ otherwise we’d see from

$\displaystyle p^m\frac{|G|}{p^m}=\left(G:N(r_0)\right)\left|N(r_0)\right|$

that $p\mid\left|\mathcal{Z}\left(G\right)\right|$ contradictory to our assumption. Thus $p^m\mid N(r_0)$ and since $\left|N(r_0)\right|<|G|$ the induction hypothesis implies that $N(r_0)$ and consequently $G$ has a subgroup of order $p^m$. The induction is complete. $\blacksquare$

Corollary(Cauchy’s Theorem): Let $G$ be a finite group and $p$ a prime such that $p\mid |G|$. Then, $G$ contains an element of order $p$.

Proof: By the First Sylow’s Theorem we know that $G$ has a subgroup $H$ of order $p$ and since this subgroup must be cyclic (see here for proof) we know that $H=\{e,h,\cdots,h^{p-1}\}$ for some $h$ and $\left|h^k\right|=p$ for $k=1,\cdots,p-1$ and thus $G$ contains at least $p-1$ elements of order $p$. Since $p-1>0$ the conclusion follows. $\blacksquare$

Corollary: Let $G$ be a finite group. Then $G$ is a $p$-group if and only if $|G|=p^m$ for some $m\in\mathbb{N}$.

Proof: Evidently if $|G|=p^m$ then clearly $G$ is a $p$-group. Conversely, let $G$ be a $p$-group and suppose that $q$ is a prime other than $p$ such that $q\mid |G|$. Then, by Cauchy’s theorem there exists some $h\in G$ such that $|h|=q$. But, clearly $|h|$ cannot then be a power of $p$ contradicting that $G$ is a $p$-group. It follows if $d\mid |G|$ then $d=p^k$ for some $k\in\mathbb{N}\cup\{0\}$. Thus, $|G|=p^n$ for some $n\in\mathbb{N}$. $\blacksquare$

References:

1. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge: Cambridge UP, 1994. Print.

2. Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

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

January 8, 2011 -

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