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# 14. Sylow Theorems
In Lesson 12, we introduced Cauchy's theorem, which states that, for any prime number $p$ dividing the order of a group $G$, a subgroup of order $p$ exists. Sylow theorems (named after the Norwegian mathematician Sylow, pronounced "Sülov") extends Cauchy's theorem to subgroups of prime power order, i.e., subgroups of order $p^m$. Such subgroups are called **$p$-subgroups**, as groups of prime power order are often called **$p$-groups**.
There are 3 Sylow theorems. Roughly speaking, they tell us that
1. $p$-subgroups exist and are (sort of) nested, from subgroups of order $p$ all the way up to $p^n$, the largest power of $p$ that divides $|G|$. (The latter is called a **Sylow $p$-subgroup**.)
2. Sylow $p$-subgroups are conjugate to each other.
3. There are strict restrictions on the number of Sylow $p$-subgroups.
## 14.1 The concept of a "normalizer"
As we will soon see, Sylow theorems are proved using cleverly constructed group actions. But before that, it is necessary to introduce the concept of a "normalizer".
Consider a group $G$ and a subgroup $H$ of $G$. The **normalizer of $H$**, denoted $N_H$, is the largest "intermediate subgroup" between $G$ and $H$ that has $H$ as its *normal* subgroup. More precisely, $N_H$ consists of elements $g\in G$ that satisfy the following condition:
For any $h \in H$, $g^{-1}hg\in H$.
It is easy to check that all such elements form a group. And it is also rather obvious that, if $H$ is itself a normal subgroup of $G$, then $N_H = G$.
There is another way of viewing $N_H$. As we know, there is a one-to-one correspondence between the left-cosets of $H$ and the right-cosets of $H$ (i.e., $gH$ corresponds to $Hg$). But unless $H$ is a normal subgroup of $G$, the corresponding pair of left- and right-cosets are not necessarily the same as each other. $N_H$ can be seen as the union of those left-cosets that are identical to their corresponding right-cosets.
## 14.2 Sylow theorem 1
Let us play with with normalizers a bit using group actions, and see if we can arrive at the first Sylow theorem. Let us say that $H$ is a subgroup of $G$, and we construct the following group action:
$H$ acts on *the set of all left-cosets of $H$*: $h$ sends $gH$ to $hgH$.
What are the fixed points of this action? If $gH$ is to be fixed by all $h\in H$, then $gH=hgH$, i.e., $g^{-1}hg\in H$. This is exactly the condition for $g$ to be in $N_H$. That is, the fixed points of this action are exactly the cosets of $H$ that combine to form $N_H$. Otherwise, if some coset is *not* fixed under this action, it must belong to an orbit whose size ($\gt 1$) is a factor of $|H|$ (by the orbit-stabilizer lemma). Counting the number of all cosets, we have
$[G:H] = [N_H:H] + \sum_{\text{0 or more terms}} (\text{something that divides}~|H|~\text{and is >}~1).$
How is this counting useful? Well, let's imagine that $|G| = p^n r$ where $r$ is not divisible by $p$, and $H$ is some subgroup of order $p^k$ with $1\leq k\lt n$. (Cauchy's theorem ensures such $H$ exists for $k=1$). In such case, both the left-hand side and the second term on the right-hand side are divisible by $p$. Consequently, $[N_H:H]$ is divisible by $p$, and by Cauchy's theorem, we can find an element of order $p$ in the quotient group $N_H/H$. (Note that $H$ is normal in $N_H$ so we *can* take the quotient.) This element will generate a cyclic subgroup of order $p$ in $N_H/H$, and its inverse image under the natural homomorphism $N_H \to N_H/H$ is a subgroup of order $p^{k+1}$. We have essentially proven the following:
<span style="color:darkorchid">**Theorem 14.1 (First Sylow theorem):** Let $|G| = p^n r$ where $r$ is not divisible by $p$, then there exist subgroups $H_1, H_2,\cdots, H_n$ of order $p, p^2,\cdots, p^n$ respectively, where $H_1 \subset H_2 \subset\cdots\subset H_n$ and every $H_i$ is normal in $H_{i+1}$.
## 14.3 Sylow theorem 2
In the previous section, we looked at cases when $H$ is a $p$-subgroup, but not a Sylow $p$-subgroup. What if $H$ *is* a Sylow $p$-subgroup? In this case, the formula above does not seem to yield much useful information.
The clever idea here is, rather than looking at the action of $H$ on its own cosets, we instead let $H$ act on the cosets of another Sylow $p$-subgroup, say $K$. That is,
$H$ acts on *the set of all left-cosets of $K$*: $h$ sends $gK$ to $hgK$.
Under this action, we have a similar formula:
$[G:K] = \text{# fixed points} + \sum_{\text{0 or more terms}} (\text{something that divides}~|H|~\text{and is >}~1).$
The key observation here is that the number of fixed points cannot be zero, as the left-hand side is not indivisible by $p$, but the second term on the right-hand side is still divisible by $p$. This means that there exists some $g\in G$ such that $hgK=gK$ holds for all $h\in H$. This is exactly the same as saying that $g^{-1}Hg=K$. In other words, $H$ and $K$ are conjugate to each other. Thus, we have the following
<span style="color:darkorchid">**Theorem 14.2 (Second Sylow theorem):** Let $H$ and $K$ be two Sylow $p$-subgroups of a finite group $G$. Then there must exist some $g\in G$ such that $K=g^{-1}Hg$.
An immediate corollary from this theorem is
<span style="color:darkorchid">**Corollary 14.3:** Given any finite group $G$. A Sylow $p$-subgroup $K$ is normal, if and only if it is the unique Sylow $p$-subgroup of $G$.
## 14.4 Sylow theorem 3
Since all Sylow $p$-subgroups are all conjugate among each other, we can now let the entire group $G$ act on the set of all Sylow $p$-subgroups by conjugation and get only one orbit. By the orbit-stabilizer lemma, the size of the orbit, i.e., the number of Sylow $p$-subgroups, must divide $|G|$.
Next, we let one Sylow $p$-subgroup, say $H$, act on the set of all Sylow $p$-subgroups by conjugation. What are the fixed points of this action? Certainly, $H$ itself is a fixed point, but what about others? Let us assume that some other Sylow subgroup $K$ is also fixed under conjugation by elements of $H$. Then $H$ must be contained in $N_K$, the normalizer of $K$. That is to say, both $H$ and $K$ are subgroups of $N_K$. Not only that, they are both Sylow $p$-subgroups of $N_K$, and one of them, $K$, is normal! This is in direct contradiction with Corollary 14.3. Hence the only fixed point of this action is $H$ itself. All other Sylow $p$-subgroups must belong to orbits of size greater than $1$ and dividing $|H|=p^n$. Therefore, the number of Sylow $p$-subgroups is congruent to $1$ ($\mathrm{mod}~p$).
To summarize,
<span style="color:darkorchid">**Theorem 14.4 (Third Sylow theorem):** Given any finite group $G$, with $|G|=p^n r$ where $r$ is not divisible by $p$. The number of Sylow $p$-subgroups divides $r$, and is congruent to $1$ $\mathrm{mod}~p$.
Note that the second Sylow theorem only applies to $p$-subgroups that are Sylow. Otherwise it does not necessarily hold. One example is $D_8$, the dihedral group describing the symmetry of a square. It has one $2$-subgroup isomorphic to $\mathbb{Z}_4$, generated by a rotation of $\pi/2$ radians. It also has a $2$-subgroup isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$, generated by the two "flip" operations along the two diagonals of the square. These two subgroups are clearly not isomorphic (hence not conjugate) to each other.