p-sylow: Let \( G \) be a finite group, let \( n \) be the p-adic valuation of \( |G| \). Then there exists a subgroup of \( G \) that has order \( p^n \).
A priori, given a finite group \( G \), it is not guaranteed that there exists a subgroup whose order is a specific divisor of the order of \( G \). However, if such a divisor is a power of a prime, there exists at least one subgroup of that order. This was first proven by Peter Ludwig Sylow, a Norwegian mathematician, after whom these theorems are named.
The theorems are 4 in total, and they provide significant information about the p-subgroups of a finite group: Let \( G \) be a finite group of order \( |G|=p^n m \), such that \( (p,m)=1 \)
For every \( i \leq n \), there exists \( H < G \) such that \( |H|=p^i \)
For every \( H < G \) of this form, there exists \( H' > H \) such that \( [H':H]=p \), in particular every p-subgroup is contained in a p-Sylow.
All p-Sylows are conjugate, but this is not necessarily true for smaller p-subgroups.
Let \( n_p \) be the number of p-Sylows, then \( n_p \equiv 1 \pmod{p} \), \( n_p=[G:N(S)] \Rightarrow n_p \mid m \).
Lemma 1: If \( G \) acts on a set \( X \), then the cardinality of \( X \) is equal to the sum of the cardinalities of the orbits:
\[ |X|= \sum_{x \in R} |Orb(x)|=\sum_{x \in R} \frac{|G|}{|stab(x)|} \]where \( R \) is a set of representatives. The first equality trivially follows from the fact that the orbits of \( x \) form a partition of the set. The subsequent equality derives from the bijection between the orbits of an element and the left cosets of its stabilizer.
Definition: Let \( X \) be a set and \( G \) a group acting on it via \( \phi \). We define \( Fix_G(X)= \left\{x \in X \mid \forall g \in G, \phi_g(x)=x\right\} \)
In short, these are the elements that have the entirety of \( G \) as their stabilizer.
Lemma 2: If \( G \) is a p-group, we can use the formula from Lemma 1 to observe that:
\[ |X| \equiv |Fix_G(X)| \pmod{p} \]Indeed, in their case, the stabilizer of the action is the whole group, while in the others it is not; so for the others, their factor evaluates to zero modulo \( p \), whereas the fixed points have a value of one.
Sylow 1 + 2: Let \( H < G \) be a p-group. We know that at least one exists by Cauchy's theorem. Let \( X=G/H \); \( H \) acts on \( X \) by left multiplication.
\[ g \in Fix_H(X) \iff \forall h \in H, hgH=gH \iff \forall h \in H, g^{-1}hg \in H \iff g \in N_G(H) \]this means that \( Fix_H(X)=N_G(H)/H \).
If \( H=p^{\alpha}, \alpha < n, \Rightarrow p \mid [G:H] \), but \( [G:H]=|X| \), so from Lemma 2 \( 0\equiv [G:H]\equiv |X| \equiv |Fix_H(X)| \equiv |N_G(H)/H| \pmod{p} \iff p \mid [N_G(H):H] \). Since \( H \lhd N_G(H) \), the quotient is a group, thus by Cauchy's theorem it contains an element of order \( p \). Taking the subgroup generated by it, the preimage will be a subgroup containing \( H \), of index \( p \), exactly as we wanted. Furthermore, by iterating the process we obtain the p-Sylow.
Sylow 3: Let \( P, P_1 \) be two \( p \)-Sylow subgroups. And let \( X=G/P \). Then \( P_1 \) acts on \( X \) by conjugation. Since \( P_1 \) is a p-group, we can use Lemma 2, which tells us that:
\[ |X| \equiv |Fix_{P_1}(X)| \pmod{p} \]but \( LHS \) is not 0 (modulo \(p\)), therefore \( RHS \) is not 0 either, which implies that it is not empty. So let \( gP \in Fix_{P_1}(X) \). It will hold that:
\[ \forall h \in P_1, hgP=gP \iff g^{-1}hg \in P \iff P_1=gPg^{-1} \]which implies that they are conjugate.
Sylow 4: Let \( S \) be a p-Sylow. Let \( X \) be the set of p-Sylows. \( X \) therefore coincides with the orbit of \( S \) through the conjugation action of the subgroups. So \( n_p=|X|=[G:N_G(S)] \Rightarrow n_p \mid |G| \). Considering instead the conjugation action of \( S \) on \( X \), the only trivial orbit is that of \( S \). Indeed, let \( H \) be a p-Sylow with a trivial orbit. Then
\[ Orb(H)=\left\{gHg^{-1} \mid g \in S\right\}=H \iff S \subset N_G(H) \]this implies that \( SH \) is a group, but its order is a power of \( p \) greater than or equal to \( p^n \), \( |SH|=\frac{|S||H|}{|S\cap H|} \), so \( H=S \). From Lemma 1 we have that:
\[ n_p=|X|=\sum_{x \in R} |Orb(x)|= 1 + \sum_{x \in R\setminus \left\{S\right\}} \frac{|S|}{|stab(x)|} \equiv 1 \pmod{p} \]
© Poisson di Alessandro Avellino
Diego University of Pisa. (n.d.). GitHub repositories. Retrieved October 29, 2024.
Conrad, K. (n.d.). Some Topics in Group Theory. Retrieved October 29, 2024.