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.

Example: Consider $G=S_4$, since $|S_4|=24=2^3\cdot 3$, the Sylow theorem states that there is a subgroup of order $8$.
In fact, we can take

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

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

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

# The Second Sylow Theorem

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

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

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 $D_{2n}$, 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.

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

Example: Consider $D_5$ and $p = 2$. The number of Sylow $2$−subgroups is $5$, which does divide $10/2$ and is congruent
to $1$ modulo $2$.

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 $G$ of order $|G|=15=5\cdot 3$, by Sylow III, for $p=5$, the number of Sylow $5$-subgroups divides $3=15/5$ 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 $H \leq G$ such that $|H| = 5$. Since there is only one subgroup of size 5, Sylow II indicates that $gHg^{-1} = H$.
That is, $H$ is normal: the Sylow theorems indicate automatically that there is a normal subgroup of size 5.
For $p = 3$, 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 $K \unlhd G$ such that $|K| = 3$.

Moreover, $H \cap K = \{1\}$, since $H$ is the cyclic group of order 5 and $K$ is the cyclic group of order 3.
Nontrivial elements of $H$ have order 5, while elements of $K$ 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 $h\in H,k\in K$, $hk=kh$;
• $H\times K\cong G$.

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

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

$$|G|=p_1^{e_1}...p_r^{e_r}.$$

Then we know that we have a Sylow subgroup $H_i$ such that

$$|H_i|=p_i^{e_i}.$$

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

One can verify that $G\cong H_1\times...\times H_r$, where $f:(x_1,...,x_r)\mapsto x_1+...+x_r$ is an isomorphism.

# Proof of Sylow I

Let $S$ be the set of subsets of $G$ of size $p^e$, and $n$ be the order of group $G$. Then by basic combinatorics,

$$|S|=\begin{pmatrix} n\\ p^e \end{pmatrix}.$$

Let $G$ act on $S$ by left translations:

$$g:U\mapsto gU.$$

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

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

$$|S|=\begin{pmatrix} n\\ p^e \end{pmatrix}=m\mod p.$$

考虑展开$(1+x)^n$ 考察系数？这个就是一个数论问题。

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

Proof: Since $H$ stabilizes $U$, for any $h\in H$, $hU=U$. In other words, for any $u\in U$,

$$Hu\subseteq U.$$

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

We know that the orbits partition $S$, so

$$|S|=|O_1|+...+|O_r|.$$

Since $p$ does not divide LHS, there must exist an orbit $\theta$ such that

$$gcd(p,|\theta|)=1.$$

注意，$\theta$ 是$S$ 的子集，是一个 “大小为$p^e$ 的$G$ 的子集的集合”。
Let the size of $\theta$ be $k$.
Now, consider some element $U\in \theta$. By the orbit-stabilizer theorem, we also know that

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

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

# Proof of Sylow II

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

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

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

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

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

for all $k\in K$. Thus $a^{-1}Ka\leq H$.
$\square$