Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Sylow’s Theorems (Pt. II)


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

We now prove the second Sylow theorem. But first we remark on some notation. If G is a group we can think of H\leqslant G acting on \mathcal{P}\left(G\right) (the power set) by conjugation. In other words, h\cdot S=hSh^{-1}. This is easily verified to be a group action with the orbit of S equal to C(S)=\left\{hSh^{-1}:h\in H\right\} and isotropy subgroups N(S)=\left\{h\in H:hSh^{-1}=S\right\}. Thus, by earlier theorem we know that \left(G:N(S)\right)=\#(C(S)) for every S\in\mathcal{P}\left(G\right). If A\in C(B) we say that A is conjugate to B. So, with one small lemma we are ready to prove the second Sylow theorem:

 

Lemma: Let G be a group and H,K\leqslant G such that H\leqslant N(K). Then, HK\leqslant G.

Proof: By a previous theorem it suffices to show that HK=KH. To do this we merely note that by definition for each h\in H one has hKh^{-1}=K and thus hK=Kh (this is easy to verify, we proved in in the post about normality) and so

\displaystyle \begin{aligned}HK &= \bigcup_{h\in H}hK\\ &= \bigcup_{h\in H}Kh\\ &= KH\end{aligned}

from where the conclusion follows. \blacksquare

 

With this in mind:

Theorem(Second Sylow Theorem): Let G be a finite group. Then:

\displaystyle \begin{aligned}&\mathbf{(1)}\quad \textit{If }P\textit{ is a }p\textit{-subroup of }\textit{, then }P\textit{ is containted in some Sylow }p\textit{-subgroup}\\ &\mathbf{(2)}\quad\textit{All Sylow }p\textit{-subgroups are conjugate}\end{aligned}

Let G be a finite group, and p a prime. Then all Sylow p-subgroups of G are conjugate.

Proof: Let H be a Sylow p-subgroup of G. Let then C(H) denote the orbit of H under the G-action of conjugation on \mathcal{P}\left(G\right). More directly, let

\displaystyle C(H)=\left\{gHg^{-1}g\in G\right\}

Note firstly that each element of C(H) being the image of H under an inner automorphism is itself a Sylow p-subgroup. Note next then that by previous comment

\displaystyle \#\left(C(H)\right)=\left(G:N(H)\right)=\frac{|G|}{N(H)}

and since H\leqslant N(H) it follows that

\displaystyle p\nmid \frac{|G|}{|N(H)|}=\#\left(C(H)\right)

consider P acting on C(H) by p\cdot L=pLp^{-1} for p\in P and L\in C(H). Note by the Orbit Decomposition Theorem we have that

\displaystyle \#\left(C(H)\right)=\sum_{\mathcal{O}_L\in\text{Orb}\left(C(H)\right)}\#\left(\mathcal{O}_L\right)

note thought that from an earlier theorem we have that

\#\left(\mathcal{O}_L\right)=\left(H:H_L\right)\mid |H|=p^k

and thus \#\left(\mathcal{O}_L\right)=p^e,\text{ }e\geqslant 0 for each \mathcal{O}_L\in\text{Orb}\left(C(H)\right). But, if p\mid \#\left(\mathcal{O}_L\right) for each \mathcal{O}_L\in\text{Orb}\left(C(H)\right) then one would have that

\displaystyle p\mid \sum_{\mathcal{O}_L\in\text{Orb}\left(C(H)\right)}\#\left(\mathcal{O}_L\right)=\#\left(C(H)\right)

which is a contradiction. Thus, it follows that \#\left(\mathcal{O}_L\right)=1 for some \mathcal{O}_L\in\text{Orb}\left(C(H)\right) and thus \mathcal{O}_L=\{L\}. Note though that this implies that H\leqslant N(L) and thus by our lemma we know that HL\leqslant G. But, by an earlier theorem we know that

\displaystyle \left|HL\right|=\frac{|H||L|}{|H\cap L|}

but it’s fairly evident that if

\displaystyle \frac{|H|}{|H\cap L|}=p^e>1

then HL would be a subgroup of G whose order is higher power of p than L, but this contradicts that L is a Sylow p-subgroup. Thus, |H|=|H\cap L| and thus H=H\cap L, and so H\subseteq L. The conclusion follows.

\mathbf{(2)}: Note that evidently since “being conjugate to” is an equivalence relation it suffices to show that every Sylow p-subgroup of G is conjugate to a fixed Sylow p-subgroup P. Note though that using the above methodology taking $laetx P$ as our initial Sylow p-subgroup we have that for any Sylow p-subgroup K there exists some conjugate gPg^{-1} of P such that K\subseteq gPg^{-1}. But, since \#\left(K\right)=\#\left(gPg^{-1}\right) we may conclude that K=gPg^{-1}. Thus, every Sylow p-subgroup of G is conjugate to P and thus must be conjugate to one another.

\blacksquare

Lastly, we prove the third Sylow theorem.

Theorem(Third Sylow Theorem): Let G be a finite group. Then, if s_p denotes the number of Sylow p-subgroups of G then s_p\mid |G| and s_p\equiv 1\text{ mod }p.

Proof: Fix a Sylow p-subgroup P of G. By the second part of the second Sylow theorem we know that the set of all Sylow p-subgroups is equal to

\displaystyle \#\left(C(P)\right)=\frac{|G|}{|N(P)|}

and thus s_p\mid |G|.

To prove the other part of the theorem we recall that if P acts on C(P) by conjugation we get that

\displaystyle \#\left(C(P)\right)=\sum_{\mathcal{O}_L\in\text{Orb}\left(C(P)\right)}\#\left(\mathcal{O}_L\right)

but we know that \#\left(\mathcal{O}_P\right)=1 and if L\ne P then \#\left(\mathcal{O}_L\right)>1. Thus, recalling that \#\left(\mathcal{O}_L\right)=p^k,\text{ }k\geqslant 0 for each \mathcal{O}_L\in\text{Orb}\left(C(P)\right) we may conclude that

\displaystyle s_p=\#\left(C(P)\right)=1+\sum_{\substack{\mathcal{O}_L\in\text{Orb}\left(C(P)\right)\\ \#\left(\mathcal{O}_L\right)>1}}\#\left(\mathcal{O}_L\right)\equiv 1\text{ mod }p

 

 

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, 20


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January 8, 2011 - Posted by | Algebra, Group Theory | , , , ,

2 Comments »

  1. […] continue now with a proof what was previously called the third Sylow theorem. Namely, that the number of Sylow -subgroups of a group is congruent […]

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

  2. […] This follows immediately from the fact that all Sylow -subgroups are conjugate to each […]

    Pingback by Groups of Order pq (pt. I) « Abstract Nonsense | April 19, 2011 | Reply


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