望远镜无助于限定灵魂。

# The First Sylow Theorem

The first Sylow theorem can be viewed as a specific converse of the Lagrange theorem.

(Sylow I) Given a group GG such that

G=n=pem,|G|=n=p^em,

where pep^e is the largest power of pp (that is, gcd(p,m)=1gcd(p,m)=1), then there exists a subgroup HGH\leq G such that

H=pe.|H|=p^e.

Such a subgroup is called a Sylow pp-subgroup. It has maximal prime power order within GG.

Example: Consider G=S4G=S_4, since S4=24=233|S_4|=24=2^3\cdot 3, the Sylow theorem states that there is a subgroup of order 88.
In fact, we can take

H=(12),(34),(13)(24).H=\langle (12),(34),(13)(24)\rangle.

Example: For G=D5G=D_5 is the dihedral group, we have D5=10=25|D_5|=10=2\cdot 5. So there must be subgroups of size 5 and 2. A subgroup generated
by a rotation ρ2π/5\langle \rho_{2\pi/5}\rangle has order 5. A subgroup generated by any reflection has order 2.

(Corollary of Sylow I) If prime number pp divides G|G|, then there exists an element xGx\in G with order pp.

Proof: Using Sylow I, there exists a subgroup HGH\leq G such that H=pe|H|=p^e. Pick any element yHy\in H, one knows yH\langle y\rangle\leq H and thus the order of yy is y=pk|\langle y\rangle|=p^k by Lagrange Theorem. Then x=ypk1x=y^{p^{k-1}} is an element of order pp.
\square

# The Second Sylow Theorem

The first Sylow theorem states that a Sylow pp-subgroup, a subgroup of maximal size pep^e dividing G|G|, exists. In fact, we can say a lot more about what these subgroups look like.

(Sylow II) There are two parts; part one is what is usually referred to as the second Sylow theorem.

  • Given HGH\leq G, where HH is a Sylow pp-subgroup, any other Sylow pp-subgroup HGH'\leq G is conjugate to HH; i.e., there exists gGg\in G such that H=gHg1H'=gHg^{-1}.
  • Given any subgroup KGK\leq G such that K=pd|K|=p^d, for any Sylow subgroup HH, there exists gGg\in G such that gKg1HgKg^{-1}\leq H.
    Notice that K|K| does not have to be the maximal prime power, and can have order smaller than H|H|. Every prime power order subgroup, up to conjugation, sits inside a Sylow subgroup.

Evidently, conjugating a Sylow subgroup will result in a Sylow subgroup (since they have the same size), and Sylow II states that all the Sylow subgroups arise in this way.

Note that the second part is stronger, since K|K| can be a prime power smaller than H|H|, and implies the first part by applying to K=HK = H.
The second part states that given any prime power subgroup KGK \leq G and any Sylow subgroup HGH \leq G, it is possible to conjugate KK to make it land in HH.

Since the conjugation of a group is also a group, with the same size, thus the converse of the first part is true.

Example: For D2nD_{2n}, every subgroup of size 2 is generated by a refection, and Sylow II indicates that all the refections are conjugate.

# The Third Sylow Theorem

The last Sylow theorem indicates the number of these (conjugate) subgroups.

(Sylow III) The number of Sylow pp-subgroups of GG divides

m=npe,m=\frac{n}{p^e},

and is congruent to 1 modulo pp.

This theorem seems kind of weird, but is actually very useful.

Example: Consider D5D_5 and p=2p = 2. The number of Sylow 22−subgroups is 55, which does divide 10/210/2 and is congruent
to 11 modulo 22.

Summary: The first Sylow theorem indicates existence of Sylow subgroups, the second Sylow theorem indicates that all Sylow subgroups are related by conjugation, and the third provides strong (and kind of funky) constraints on the number of such subgroups.

# Applications of the Sylow Theorems

Example: Consider any group GG of order G=15=53|G|=15=5\cdot 3, by Sylow III, for p=5p=5, the number of Sylow 55-subgroups divides 3=15/53=15/5 and is equal to 1 modulo 5. Therefore the only possibility is that the number of Sylow 5-subgroup is 1.
So there is a unique HGH \leq G such that H=5|H| = 5. Since there is only one subgroup of size 5, Sylow II indicates that gHg1=HgHg^{-1} = H.
That is, HH is normal: the Sylow theorems indicate automatically that there is a normal subgroup of size 5.
For p=3p = 3, Sylow III states that the number of Sylow 3−subgroups divides 5 and is 1 mod 3, so it is also 1.
Thus, there exists some unique KGK \unlhd G such that K=3|K| = 3.

Moreover, HK={1}H \cap K = \{1\}, since HH is the cyclic group of order 5 and KK is the cyclic group of order 3.
Nontrivial elements of HH have order 5, while elements of KK have order 3, so they intersect only at the identity.
So the Sylow theorems give, for free, nonintersecting normal subgroups of size 5 and 3 for every single group of size 15. That is quite impressive!

Moreover, we can prove that

  • for hH,kKh\in H,k\in K, hk=khhk=kh;
  • H×KGH\times K\cong G.

Proof: For the first statement, since KK is normal, thus hkh1Khkh^{-1}\in K, and therefore hkh1k1Khkh^{-1}k^{-1}\in K.
Similarly, since HH is normal, hkh1k1Hhkh^{-1}k^{-1}\in H. Because HK={1}H\cap K=\{1\}, thus hkh1k1=1hkh^{-1}k^{-1}=1 and hk=khhk=kh.
The second part follows from the first one, one can verify that (h,k)hk(h,k)\to hk is an isomorphism.
\square

Another application of the Sylow theorems is the decomposition of finite abelian groups.
Consider an abelian group GG such that

G=p1e1...prer.|G|=p_1^{e_1}...p_r^{e_r}.

Then we know that we have a Sylow subgroup HiH_i such that

Hi=piei.|H_i|=p_i^{e_i}.

Since GG is abelian, conjugating a group produces the same group, so by Sylow II, these (abelian) subgroups HiH_i are unique for each prime.

Every finite abelian group GG is isomorphic to a product of groups of prime power order.

One can verify that GH1×...×HrG\cong H_1\times...\times H_r, where f:(x1,...,xr)x1+...+xrf:(x_1,...,x_r)\mapsto x_1+...+x_r is an isomorphism.

# Proof of Sylow I

Let SS be the set of subsets of GG of size pep^e, and nn be the order of group GG. Then by basic combinatorics,

S=(npe).|S|=\begin{pmatrix} n\\ p^e \end{pmatrix}.

Let GG act on SS by left translations:

g:UgU.g:U\mapsto gU.

Our eventual goal is to fnd a subgroup of GG of size pp by looking at stabilizers, as they are always subgroups of GG.
We find the size of a stabilizer by trying to find an orbit of size mm (where G=mpe|G|=m\cdot p^e), as we then know that the stabilizer will be order pep^e by orbit-stabilizer theorem.

Lemma: For G=n=mpe|G|=n=m\cdot p^e such that gcd(m,p)=1gcd(m,p)=1, one has

S=(npe)=mmodp.|S|=\begin{pmatrix} n\\ p^e \end{pmatrix}=m\mod p.

考虑展开(1+x)n(1+x)^n 考察系数?这个就是一个数论问题。

Lemma: Suppose we have a subset USU\in S, which is a subset of GG. Also, let HH be a subgroup of GG that stabilizes UU. Then, H|H| divides U|U|.

Proof: Since HH stabilizes UU, for any hHh\in H, hU=UhU=U. In other words, for any uUu\in U,

HuU.Hu\subseteq U.

This implies that the right cosets partition UU. Since the cosets have the same size, we know that H|H| divides U|U|.
\square

We know that the orbits partition SS, so

S=O1+...+Or.|S|=|O_1|+...+|O_r|.

Since pp does not divide LHS, there must exist an orbit θ\theta such that

gcd(p,θ)=1.gcd(p,|\theta|)=1.

注意,θ\thetaSS 的子集,是一个 “大小为pep^eGG 的子集的集合”。
Let the size of θ\theta be kk.
Now, consider some element UθU\in \theta. By the orbit-stabilizer theorem, we also know that

G=θStab(U).|G|=|\theta|\cdot |Stab(U)|.

And so pem=kStab(U)p^em=k|Stab(U)| and peStab(U)p^e\mid |Stab(U)| since gcd(k,p)=1gcd(k,p)=1. By the second lemma, Stab(U)|Stab(U)| divides U=pe|U|=p^e, thus Stab(U)=pe|Stab(U)|=p^e.
\square

# Proof of Sylow II

Fix HH to be a Sylow subgroup. Consider a set X=G/HX = G/H, the left cosets of HH. The index of HH is the same as X|X|, so X=G/H=mpepe=m|X| = |G|/|H|=\frac{mp^e}{p^e}=m.

Let KK be the subgroup we want to show is a subgroup of HH up to conjugation, where K=pf|K| = p^f.
We will look at how KK acts on XX by left translation, the mapping:

k(aH)(ka)H.k(aH)\mapsto (ka)H.

We decompose into orbits, X=O1+...+Or|X|=|O_1|+...+|O_r|. Note that these orbits are with respect to the action of KK, not the action of GG, as that would be transitive and we’d only have one orbit. That is, Oi=KxiO_i=K\cdot x_i for xiHx_i\in H.
We have that Oi|O_i| divides K=pf|K|=p^f (因为 orbit-stabilizer theorem, K=OStab|K|=|O|\cdot |Stab|), but pp does not divide mm. Thus this orbit decomposition can only work if some orbit OO has size 1.
In other words, there exists some coset aHaH that is fixed by all kKk \in K. Then,

kaH=aH    a1kaH,kaH=aH\implies a^{-1}ka\in H,

for all kKk\in K. Thus a1KaHa^{-1}Ka\leq H.
\square