# Abstract Nonsense

## Review of Group Theory: Alternate Proof to the Sylow Theorems

Point of post: This is a continuation of this post.

We lastly prove that all Sylow $p$-subgroups of a group $G$ are conjugate. This is what was initially the second Sylow theorem.

Theorem: Let $G$ be a group and $p$ a prime such that $p\mid |G|$. Then, for any $H\in\text{Syl}_p\left(G\right)$ one has that $C(H)=\left\{gHg^{-1}:g\in G\right\}=\text{Syl}_p\left(G\right)$.

Proof: Clearly we have that $G$ acts on $\text{Syl}_p\left(G\right)$ by conjugation. Our claim is equivalent to $\mathcal{O}_{G,H}=\text{Syl}_p\left(G\right)$ where $H$ is the orbit representative and $G$ is to keep track that this is the orbit under the $G$-action (there will be more actions to come). So, choose $H_0\in\text{Syl}_p\left(G\right)$ and suppose that $\text{Syl}_p\left(G\right)$ admits more orbits than $\mathcal{O}_{G,H_0}$, suppose they are $\mathcal{O}_{G,H_1},\cdots,$. We know that $H_0$ acts on $\text{Syl}_p\left(G\right)$ by conjugation though and that this action induces its own orbits; denote such orbits in general as $\mathcal{O}_{H_0,L}$ (where $L$ is the orbit representative and $H_0$ is to remind us that these are orbits under the $H_0$-action). We claim that for each $\mathcal{O}_{G,H_k}$ if we define

$\Omega_{H_k}=\left\{\mathcal{O}_{H_0,L}:\mathcal{O}_{H_0,L}\cap\mathcal{O}_{G,H_k}\ne\varnothing\right\}$

then

$\displaystyle \mathcal{O}_{G,H_k}=\bigcup_{\mathcal{O}_{H_0,L}\in\Omega_{H_k}}\mathcal{O}_{H_0,L}$

Indeed, it’s clear that

$\displaystyle \mathcal{O}_{G,H_k}\subseteq\bigcup_{\mathcal{O}_{H_0,L}\in\Omega_{H_k}}\mathcal{O}_{H_0,L}$

Conversely though, if $h_0Lh_0^{-1}\in\mathcal{O}_{G,H_k}\cap\mathcal{O}_{H_0,L}$ then we have in particular that $h_0Lh_0^{-1}\in\mathcal{O}_{G,H_k}$. But, since the orbits are equivalence classes it follows that

$\mathcal{O}_{G,H_k}=\mathcal{O}_{G,h_0Lh_0^{-1}}=\mathcal{O}_{G,L}\supseteq \mathcal{O}_{H_0,L}$

It clearly follows then using the same methodology as in our last proof and noticing that $\mathcal{O}_{H_0,H_0}\subseteq\mathcal{O}_{G,H_0}$ (it contains the only trivial orbit, and $\mathcal{O}_{G,H_1}$ being distinct from $\mathcal{O}_{G,H_0}$ doesn’t) that

$\#\left(\mathcal{O}_{G,H_0}\right)\equiv 1\text{ mod }p\;\;\text{ and }\;\;\#\left(\mathcal{O}_{G,H_2}\right)\equiv 0\text{ mod }\cdots$

But, notice that there was nothing special about using the $H_0$ action of conjugation on $\text{Syl}_p\left(G\right)$. We could easily repeat the above argument for $\mathcal{O}_{G,H_2}$ to arrive at

$\#\left(\mathcal{O}_{G,H_0}\right)\equiv 0\text{ mod }p\;\;\text{ and }\;\;\#\left(\mathcal{O}_{G,H_1}\right)\equiv 1\cdots$

which is evidently a contradiction. It follows that there can be at most one distinct orbit, from where by prior discussion 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.