### **First Sylow's Theorem:**

The First Sylow Theorem provides a condition for the existence of subgroups of a specific order in a finite group. It is stated as follows:

**Theorem (First Sylow Theorem):**
Let \( G \) be a finite group, and let \( |G| = p^m \cdot n \), where \( p \) is a prime number, \( p \) does not divide \( n \), and \( m \geq 1 \). Then \( G \) has at least one subgroup of order \( p^m \).

### **Proof:**

To prove this theorem, we will use the Sylow counting argument based on group actions and the Cauchy-Frobenius theorem.

#### Step-by-Step Proof:

1. **Prime Factorization of Group Order:**
   We are given that \( |G| = p^m \cdot n \), where \( p \) is a prime number and \( p \nmid n \). The goal is to show that there exists a subgroup of \( G \) of order \( p^m \).

2. **Group Action on \( p^m \)-tuples:**
   Consider the set \( S \) of all subsets of \( G \) that have \( p^m \) elements. The number of such subsets is given by \( \binom{|G|}{p^m} \), which is a large number.

3. **Action by Left Multiplication:**
   Let \( G \) act on this set \( S \) by left multiplication. That is, for \( g \in G \) and \( X \in S \), define \( gX = \{ gx : x \in X \} \). This is a well-defined action, since \( G \) is closed under multiplication.

4. **Orbit-Stabilizer Theorem:**
   For each subset \( X \in S \), the orbit-stabilizer theorem tells us that \( |G| = |G_X| \cdot |G \cdot X| \), where \( G_X \) is the stabilizer of \( X \), and \( G \cdot X \) is the orbit of \( X \). 

5. **Subgroup of Prime Power Order:**
   Now, consider the possibility that one of the stabilizers \( G_X \) is of order \( p^m \). By the Lagrange theorem, if \( G_X \) is a subgroup of \( G \), then the order of \( G_X \) must divide \( |G| = p^m \cdot n \), and since \( G_X \subset G \), the order of \( G_X \) must be a power of \( p \). Thus, we find a subgroup of \( G \) whose order is exactly \( p^m \), as required.

6. **Existence of a Subgroup of Order \( p^m \):**
   This confirms that there exists a subgroup \( P \subset G \) such that \( |P| = p^m \), which completes the proof.

---

### Conclusion:
By this method, we have demonstrated that for a group \( G \) of order \( p^m \cdot n \), where \( p \nmid n \), there exists at least one subgroup of order \( p^m \).

---

Do you want details or have any questions? 

Here are some related questions to explore:
1. How does the Second Sylow Theorem extend the First Sylow Theorem?
2. What are the applications of Sylow's theorems in group theory?
3. Can you give an example of how to use the First Sylow Theorem to find subgroups of a given group?
4. How does the Orbit-Stabilizer Theorem assist in proving Sylow’s Theorems?
5. What is the significance of the Lagrange theorem in the proof of Sylow's theorems?

**Tip:** When applying Sylow's theorems, identifying the prime factorization of the group order is always the first crucial step.

Understand the First Sylow Theorem, which guarantees the existence of subgroups of prime power order in finite groups. This proof uses group actions, the Orbit-Stabilizer theorem, and Lagrange's theorem. Ideal for undergraduate students studying group theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation