Group Theory, Part 2: The Sylow Theorems

July 6th, 2017

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The Sylow theorems tell us quite a lot of information in general about the subgroups of a group of finite order. In this post, I’ll be covering the basic ideas behind and finally the proofs of the Sylow theorems.

Group Actions

First, we describe group actions. For any group $G$ and a set $S$, we can define a group action $\rho:\ G \rightarrow Perm(S)$ which is a homomorphism from the group to the symmetric group (the set’s permutations group). Alternatively, we could define the group action as operating on pairs in the product $G \times S$, i.e.:

$$ \rho:\ G \times S \rightarrow S $$

With this definition, we also require associativity and also the action of the identity ($\rho(e,s) = s$).

Example

For example, we look at conjugation. Let $C_x:\ G \rightarrow G$ be the map so that $C_x: y \mapsto xyx^{-1}$.

We can indeed think of $C: x \mapsto C_x$ as a homomorphism, from $G$ to $Aut(G)$. The kernel of this homomomorphism is all the elements $x$ so that:

$$ C_x(y) = xyx^{-1} = y \\ xy = yx $$

We denote the kernel of this homomorphism as the center of $G$, denoted $Z(G)$, and it consists of all elements $x \in G$ which commute with the group.

Definition: For a defined group action $\rho:\ G\rightarrow S$, the stabilizer of $s\in S$ consists of all the elements $x\in G$ so that $\rho(x,s) = s$. In other words, the stabilizer of $s$ is all the group elements which send $s$ to itself. Often it is denoted $G_s$.

Definition: In the case where $\rho$ is the conjugation group action, we define the centralizer of $a$. These are all the elements $x\in G$ which commute with $a$. The group of all elements which fix a subgroup $H \subseteq G$ is called the normalizer of $H$.

Definition: We define the orbit of $s \in S$ under $G$ to be the set $\{xs \vert x \in G\}$. In the case of the conjugation action, we refer to the orbits as conjugacy classes and it is clear that they form an equivalence class for the set.

There is a natural bijection between $G/G_s$ and $Gs$, with the explicit bijection given by $hG_s \rightarrow hs$. Furthermore, the order of the orbit $Gs$ is equal to the index $(G:G_s)$.

In particular, the number of conjugate subgroups to $H$ is equal to the index of the normalizer of $H$. Also, the number of elements in the conjugacy class of $x$ is equal to the index of the centralizer, i.e. $G/G_s$.

Example

We will prove that every subgroup of index $2$ is normal. Let $S$ be the set of cosets of $H$ defined in the usual way (there should be two). Then, the group action is defined as a homomorphism from $G$ to $S_2 =Z_2$. Consider the kernel $K$ of this homomorphism; this is all group elements $g \in G$ so that $gH = H$. So, the kernel of this homomorphism is a subgroup of $H$. But then that means $G/K$ is a subgroup of $Z_2$. So $(G:K)$ is either $1$ or $2$. But we know that:

$$ (G:K) = (G:H)(H:K) $$

And $(G:H)=2$, so we must have $(H:K)=1$ so this tells us that $(H:K)=1$ and indeed $H=K$. So that means $H$ is the kernel of the given homomorphism.

But for any homomorphism $\phi$, if we pick an arbitrary $g\in G$ and $k$ in the kernel, we have $\phi(gkg^{-1}) = \phi(g)\ \phi(k)\ \phi(g)^{-1} = e$. So indeed, the kernel is a normal subgroup. Therefore, $H$ from the previous example is a normal subgroup.

We also can write the obvious fact that the order of a set $S$ is the sum of all the orbits in this new notation, in the decomposition formula:

$$ |S| = \sum\limits_{G_s} (G:G_s) $$

We have a special case where the group action is conjugation. In this case, we take $Y$ to be a set of representatives from each conjugation class and write the class formula:

$$ |G| = |Z(G)| + \sum\limits_{y\in Y} (G:G_y) $$

Sylow Subgroups

Definition Let $p$ be a prime number. Then a Sylow $p$-group is a finite group of order $p^n$ for some $n$.

With this definition, we prove an intermediate theorem:

Theorem

Let $G$ be a non-trivial $p$-group. Then $G$ has a non-trivial center. Furthermore, $G$ is solvable.

Proof

From the class formula, we have:

$$ |G| = |Z(G)| + \sum\limits_{y\in Y} (G:G_y) $$

If the center is trivial, then the sum on the RHS of the equation adds up to $|G| - 1 $. Since $G$ is a $p$-group, the prime $p$ divides both sides of the equation. However, this is clearly impossible since $p$ does not divide $|G|-1$. So the center is non-trivial.

We also know that $G/Z(G)$ is a $p$-group as well, and it is strictly smaller than $G$. The rest of the proof proceeds by induction on $n$.

We know for a fact that cyclic groups are solvable (with the shortest normal series which is abelian). Suppose the theorem holds for all $p$-groups with $n \le k-1$. Then, we have that $Z(G)$ and $G/Z(G)$ are both solvable since they are $p$-groups as well. So, we have proved the $n=k$ case.

Theorem: First Sylow Theorem

Suppose that we have $p^n \mid G$ and $p^{n-1} \nmid G$ for some prime $p$. Then, define a Sylow $p$-subgroup of $G$ to be a subgroup of order $p^n$. We will prove that for every $p \mid G$, there exists a $p$-Sylow subgroup.

Lemma: Cauchy’s Theorem

We start with the brief lemma that for any finite abelian group $G$, for any prime $p \mid\ \vert G \vert$, there is a subgroup in $G$ of order $p$. This is known as Cauchy’s lemma and is a special case of Sylow’s First Theorem.

By the fundamental theorem of finite abelian groups, $G$ can be written as the direct product:

$$ G = G(p) \times G' $$

Where $\vert G’ \vert$ is prime to $p$. Take an element $a\in G(p)$ with order $p^k$. Then take:

$$ b = a^{p^{k-1}} $$

We know that $b$ is not the identity. However, $b^p = a^{p}=e$. So, $b$ generates a cyclic group of order $p$, and we are done.

In fact, we can generalize Cauchy’s theorem to arbitrary finite groups using the class equation:

$$ |G| = |Z(G)| + \sum\limits_{y\in Y} (G:G_y) $$

Suppose that $p \mid \vert Z(G) \vert$. Then we can apply the earlier proof to $Z(G)$ to find an element of order $p$ in $Z(G) \subseteq G$; so we are done.

Suppose instead that $p \nmid |Z(G)|$. Then, the sum on the right hand side cannot divide $p$ either, and in particular there is at least one conjugacy class $G_y$ with order prime to $p$. Therefore, $p \nmid \vert \frac{G}{G_y} \vert$. But this means that $G_y$ divides $p$! Applying Cauchy’s lemma to this group, we find an element of order $p$ in $G_y$ and therefore in $G$, and we are done.

Proof of Sylow’s First Theorem

Cauchy’s lemma proves that $G$ certainly has a subgroup of size $p$. With Sylow’s first theorem, we prove that $G$ has a subgroup of size $p^n$, where $p^n$ is the largest power of $p$ dividing the order of $G$. We proceed by induction on the size of $G$.

Suppose that there is a proper subgroup $H \subset G$ so that $(G:H)$ does not divide $p$. Then $p^n \mid |H|$, and so by induction $H$ has a $p$-Sylow subgroup which is also a $p$-Sylow subgroup of $G$. We assume then that every proper subgroup $H \subset G$ has the property that $p \mid (G:H)$. Now, we let $G$ act on itself by conjugation and consider the class formula:

$$ |G| = |Z(G)| + \sum\limits_{y\in Y} (G:G_y) $$

The order of each of the orbits divides the order of the group (since a stabilizer or in this case a conjugacy class, is a subgroup), so $p$ divides the order of $Z(G)$, and $G$ has a non-trivial center; furthermore, by a similar proof to Cauchy’s lemma, we have that $G$ contains a subgroup $H$ of size $p$. Furthermore, since $H$ is a subgroup of the center in particular, it is a normal subgroup.

We consider the group $G/H$. By the inductive hypothesis, this group has a subgroup $K’$ of order $p^{n-1}$. By the third isomorphism theorem, $K’$ is really (isomorphic to) a subgroup of the form $K/H$, where $H \subset K \subset G$. Take this $K$; it has order $p^n$ as desired.

Theorem: Second Sylow Theorem

The second Sylow Theorem states that all $p$-Sylow subgroups are conjugate to each other. Before we get to that, let’s prove a related lemma:

Lemma: Every $p$-subgroup of $G$ is contained in some $p$-Sylow subgroup.

This isn’t hard to prove using the class formula. Let $S$ be the set of $p$-Sylow subgroups of $G$, and $G$ operates on $S$ by conjugation. Let’s take $P$ to be one of those Sylow subgroups. Then we have:

$$ |Orb(P)| = \frac{\vert G \vert}{\vert G_{P} \vert} $$

Where $G_P$ is the normalizer of $P$ which sends $P$ to itself under conjugation. Evidently, $G_P$ contains $P$, so that means $|Orb(P)|$ is prime to $p$. Let’s take a look at the action of a $p$-subgroup $H$ on $Orb(P)$ using the class formula:

$$ \vert Orb(P)\vert = |Z| + \frac{\vert Orb(P) \vert}{\vert H_{Orb(P)} \vert} $$

By Lagrange, each term in the right hand sum divides $ |H| $ so that means there is a non-trivial center $P’$ within $|Orb(P)|$. This means that $H$ is contained within the normalizer of $P’$. Since $H$ is contained in the normalizer, by the second isomorphism theorem we know that $HP’$ is a subgroup with $P’ \trianglelefteq HP’$. The order of the quotient group $HP’/P \cong H/(H \cap P’)$ is a power of $p$, so the order of $HP’$ is also a power of $p$. But then $HP’ = P’$ since $P’$ is a maximal $p$-subgroup of $G$; and therefore $H \subseteq P’$. We have now found a Sylow subgroup which contains $H$.

Proof of Sylow’s Second Theorem

Let H be some $p$-Sylow subgroup of $G$. $H$ is contained in some conjugate $P’$ of $P$ by the above argument. And again by the above construction, because $H$ and $P’$ have equal orders, $H = P’$.

This second theorem implies as a direct corollary that if there is only one $p$-Sylow subgroup, it is normal.

Sylow’s Third Theorem

Sylow’s Third Theorem is arguably the most useful. It says that the number of $p$-Sylow subgroups of $G$ is equal to $1\mod p$. That means by counting arguments, if $|G|/p^{n} < p$, then the Sylow subgroup is normal.

Proof of Sylow’s Third Theorem

Take the action of $H = P$ on the set $S$ consisting of the Sylow $p$-subgroups. Any other orbit cannot have just one element $P’$, or else $H$ is contained in the normalizer of $P’$, and by earlier arguments $H=P=P’$. Take an element $s’$ of any other orbit. It has a stabilizer, which is a proper subgroup of $H$; the stabilizer is a subgroup, so it is divisible by $p$.

As a result, the number of elements in $S’$ is divisible by $p$. So the size of $S$ (the set of all $p$-Sylow subgroups) is equal to 1 (the size of the center of conjugation under $P$) plus the size of all the remaining orbits (all of which have orders which divide p). So we are done.

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