Abstract Nonsense

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 -

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 […]

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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