# Abstract Nonsense

## Review of Group Theory: Alternate Proof of the Sylow Theorems (Pt. II)

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

We continue now with a proof what was previously called the third Sylow theorem. Namely, that the number of Sylow $p$-subgroups of a group is congruent to $1$ modulo $p$. Namely:

Theorem: Let $G$ be a finite group with $|G|=p^\alpha m$ where $p\nmid m$. Then, if  $\text{Syl}_p\left(G\right)$ denotes the set of all Sylow $p$-subgroups of $G$ then $\#\left(\text{Syl}_p\left(G\right)\right)\equiv 1\text{ mod }p$.

Proof: Let $P\in\text{Syl}_p\left(G\right)$ (we know that $\text{Syl}_p\left(G\right)$ by our last theorem). Then, consider the $P$-action on $\text{Syl}_p\left(G\right)$ given by

$g\cdot S=gSg^{-1}$

for $g\in P$ and $S\in\text{Syl}_P\left(G\right)$. Evidently $\mathcal{O}_P=\{P\}$, and we claim that this is the only orbit under this action with one element. To see this suppose that $H\in\text{Syl}_p\left(G\right)$, $\mathcal{O}_H=\{H\}$, and $H\ne P$. Consider then that for each $g\in P$ and $h\in H$ we have that $ghg^{-1}\in H$ by assumption. Thus the map

$\displaystyle \varphi:P\longrightarrow \text{Aut}\left(H\right):g\mapsto i_g$

where $i_g$ is the variant of the inner automorphism given by $i_g(h)=ghg^{-1}$ is well-defined. Thus, consider the group

$H\rtimes_\varphi P$

Note then that the map

$\phi:H\rtimes_\varphi P\to G:(h,g)\mapsto hg$

is a homomorphism. Indeed,

\displaystyle \begin{aligned}\phi\left((h,g)(h',g')\right) &= \phi\left(\left(h\varphi_g(h'),gg'\right)\right)\\ &= \phi\left(\left(hgh'g^{-1},gg'\right)\right)\\ &= hgh'g^{-1}gg'\\ &= hgh'g'\\ &= \phi\left(\left(h,g\right)\right)\phi\left(\left(h',g'\right)\right)\end{aligned}

Note though that $\phi\left(\tilde{P}\right)=P$ and $\phi\left(\tilde{H}\right)=H$ and since $H\ne P$ we clearly must have that

$\displaystyle \left|\phi\left(H\rtimes_\varphi P\right)\right|\geqslant \left|H\cap P\right|=|H|+|P|-\left|H\cap P\right|>p^\alpha$

since $H\cap P\ne H$ and $H\cap P\ne P$. Otherwise $H\subseteq P$ or  $P\subseteq H$ and since they are of the same cardinality it would follow that they are equal, contradicting our assumption. Note though that as a corollary of the First Isomorphism Theorem we have that

$\displaystyle \left|\phi\left(H\rtimes_\varphi P\right)\right|\text{ divides }\left|H\rtimes_\varphi P\right|=p^{2\alpha}$

and thus it follows that $\left|\phi\left(H\rtimes_\varphi\right)\right|=p^k$ for some $k\geqslant 0$. But, recalling that $\left|\phi\left(H\rtimes_\varphi\right)\right|>p^\alpha$ we may then conclude that $\left|\phi\left(H\rtimes_\varphi P\right)\right|=p^\beta$ where $\beta>\alpha$. But, by Lagrange’s Theorem we may conclude that

$p^\beta=\left|\phi\left(H\rtimes_\varphi P\right)\right|\text{ divides }|G|=p^\alpha m$

but this clearly implies that $p\mid m$, which is a contradiction. It follows then that for every $\mathcal{O}_H\in\text{Orb}\left(\text{Syl}_p\left(G\right)\right)$ with $H\ne P$ we have that $\#\left(\mathcal{O}_H\right)>1$. But, by the Orbit-Stabilizer Theorem

$\displaystyle \#\left(\mathcal{O}_H\right)=\left(P:G_H\right)=\frac{|P}{|G_H|}=\frac{p^\alpha}{|G_H|}$

And so in particular we have that for every $\mathcal{O}_H\in\text{Orb}\left(\text{Syl}_p\left(G\right)\right)$ we have that $\#\left(\mathcal{O}_H\right)=p^r$ with $r\in\mathbb{N}$. But, from previous comment we know that for $\mathcal{O}_H\in\text{Orb}\left(\text{Syl}_p\left(G\right)\right)$ with $H\ne P$ we have that $\#\left(\mathcal{O}_H\right)>1$ and so in conjunction with the previous sentence we have that $\#\left(\mathcal{O}_H\right)=p^k$ with $k>0$. It follows then from the Orbit Decomposition Theorem that

\displaystyle \begin{aligned}\#\left(\text{Syl}_p\left(G\right)\right) &=\sum_{\mathcal{O}_H\in\text{Orb}\left(\text{Syl}_p\left(G\right)\right)}\#\left(\mathcal{O}_H\right)\\ & \\ &=1+\sum_{\substack{\mathcal{O}_H\in\text{Orb}\left(\text{Syl}_p\left(G\right)\right)\\ H\ne P}}\#\left(\mathcal{O}_H\right)\\ & \\ &\equiv 1\text{ mod }p\end{aligned}

from where the conclusion follows. $\blacksquare$

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.