Abstract Algebra – Sylow Theorems and Its applications

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In abstract algebra, finite group theory is in its core. Now from Lagrange theorem, we have

Lemma 1 Let $H$ be a subgroup of a finite group $G$ . Then the order of $H$ is a divisor of the order of $G$ .

This theorem states the necessary condition for a finite group $H$ to be a subgroup of a finite group $G$ . But it does not give us the sufficient condition for a finite group to be a subgroup of another finite group. i.e. the converse is false.

For example, given alternating group $A_4$ of order $12$ , there does not exist a subgroup of order $6$ . (The proof of this part will be remained to the reader.)

Suppose that the order of a finite group is $n$ , how do we know whether there exists a subgroup of order a divisor of $n$ ? Sylow Theorems give the answer.

There are in fact three theorems in the set of Sylow Theorems. Let us state them one by one

Theorem 1 (First Sylow Theorem) Let $G$ be a finite group and let $\vert G\vert=p^nm$ where $n\geq 1$ and where $p$ does not divide $m$ . Then

$G$ contains a subgroup of order $p^i$ for each $i$ where $1\leq i\leq n$ .
Every subgroup $H$ of $G$ of order $p^i$ is a normal subgroup of a subgroup of order $p^{i+1}$ for $1\leq i<n$ .

Before stating the Second Sylow Theorem, we require some terminology right now.

Definition 1 A $p$ -group is a group such that each element in the group has order a power of $p$ .

Definition 2 A Sylow $p$ –subgroup $P$ of a group $G$ is a maximal $p$ -subgroup of G, that is, a $p$ -subgroup contained in no larger $p$ -subgroup.

Now we continue with the Second Sylow Theorem.

Theorem 2 (Second Sylow Theorem) Let $P_1$ and $P_2$ be Sylow $p$ -subgroups of a finite group $G$ . Then $P_1$ and $P_2$ are conjugate subgroups of $G$ .

Theorem 3 (Third Sylow Theorem) If $G$ is a finite group and $p$ divides $\vert G\vert$ then the number of Sylow $p$ -subgroups is congruent to $1$ modulo $p$ and divides $\vert G\vert$ .

As the Second and Third Sylow Theorem are quite complex, and our main focus would be finding the converse of Lagrange theorem, so we will only delve into the First Sylow Theorem.

Proof and idea. We use induction in proving this theorem. As by Cauchy theorem, there exists a subgroup of order $p$ . If we can use induction on the power of $p$ , then we are done.

As Sylow Theorem is highly related to the concept of cosets and normalizer group, we can consider the group $G_H =\{g\in G \mid gHg^{−1} = H\}$ , which we denote as normalizer of $H$ in $G$ .

Lemma 2 Let $H$ be a $p$ -subgroup of a finite group $G$ . Then

$(N[H] : H) \equiv (G : H) \pmod p$

Continuing our proof, as $H$ is a normal subgroup of $N[H]$ , we can form the factor group $N[H]/H$ . Suppose that this factor is of order power of $p$ , then by Cauchy theorem again, there exists a subgroup $K$ of order $p$ , and using canonical homomorphism we could proceed our induction.

Let $H$ be a subgroup of order $p^i$ , where $i<n$ , so $p$ divides $(G:H)$ . By Lemma 2, we derive that $p$ divides $(N[H]:H)$ . Therefore the factor group $N[H]/H$ has a subgroup $K$ of order $p$ .

Consider the canonical homomorphism

$\gamma : N[H]\rightarrow N[H]/H$

Then $\gamma^{-1}[K]=\{x\in N[H]\mid \gamma(x) \in K\}$ has order $p^{i+1}$ . Thus our proof is complete. The second part follows immediately from part 1.

Applications

Proposition 1 No group of order $p^r$ for $r > 1$ is simple.

Proposition 2 No group $G$ of order $20$ is simple.

Proof. By the First Sylow Theorem, $G$ has a Sylow $5$ -subgroup. Also, by the Third Sylow Theorem, the number of such Sylow $5$ -subgroups of $G$ is $1 \pmod 5$ and is a divisor of
$\vert G\vert = 20$ . So the number of such subgroups is $1$ . As the only Sylow $5$ -subgroup, it must be a normal subgroup of $G$ . Thus $G$ is not simple.

Abstract Algebra – Sylow Theorems and Its applications

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