Abstract Nonsense

Crushing one theorem at a time

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.


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January 12, 2011 - Posted by | Algebra, Group Theory | , ,

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