|G|=pq where p and q are distinct prime then G is solvable prove that. 

To prove that a group \( G \) with order \( |G| = pq \), where \( p \) and \( q \) are distinct primes, is solvable, we will rely on several key results from group theory.

### Step-by-Step Proof:

#### 1. **Sylow's Theorems**:
By Sylow's theorems, we know that for a finite group \( G \) of order \( |G| = pq \), where \( p \) and \( q \) are distinct primes, there are Sylow \( p \)-subgroups and Sylow \( q \)-subgroups. Specifically:
- The number of Sylow \( p \)-subgroups, denoted \( n_p \), satisfies \( n_p \equiv 1 \pmod{p} \) and \( n_p \) divides \( q \).
- The number of Sylow \( q \)-subgroups, denoted \( n_q \), satisfies \( n_q \equiv 1 \pmod{q} \) and \( n_q \) divides \( p \).

Since \( p \) and \( q \) are distinct primes, this gives us the following possible values for \( n_p \) and \( n_q \).

#### 2. **Analyzing \( n_p \) and \( n_q \)**:
- \( n_p \equiv 1 \pmod{p} \) and \( n_p \) divides \( q \). Since \( q \) is prime, \( n_p \) can be either \( 1 \) or \( q \).
  - If \( n_p = 1 \), there is a unique Sylow \( p \)-subgroup, and this subgroup is normal in \( G \).
  - If \( n_p = q \), this means there are \( q \) Sylow \( p \)-subgroups.
  
- Similarly, \( n_q \equiv 1 \pmod{q} \) and \( n_q \) divides \( p \). Since \( p \) is prime, \( n_q \) can be either \( 1 \) or \( p \).
  - If \( n_q = 1 \), there is a unique Sylow \( q \)-subgroup, and this subgroup is normal in \( G \).
  - If \( n_q = p \), this means there are \( p \) Sylow \( q \)-subgroups.

#### 3. **At least one Sylow subgroup is normal**:
From Sylow's theorems, it follows that at least one of the Sylow subgroups (either the \( p \)-subgroup or the \( q \)-subgroup) must be normal. This is because \( n_p = 1 \) or \( n_q = 1 \) in most cases, meaning one of these subgroups is unique and normal in \( G \).

#### 4. **Conclusion: \( G \) is solvable**:
Now, consider the structure of \( G \):
- If \( G \) has a normal Sylow \( p \)-subgroup \( P \), then \( G \) can be written as a semi-direct product \( P \rtimes Q \), where \( Q \) is the Sylow \( q \)-subgroup. Since \( |P| = p \) and \( |Q| = q \), both \( P \) and \( Q \) are cyclic groups of prime order, which are abelian and hence solvable.
  
- If both Sylow subgroups are normal, then \( G \) is the direct product of two abelian groups (since cyclic groups are abelian), making \( G \) abelian and therefore solvable.

- Even if only one Sylow subgroup is normal, \( G \) is still a semi-direct product of two abelian groups, making \( G \) a solvable group.

### Conclusion:
A group \( G \) of order \( pq \), where \( p \) and \( q \) are distinct primes, is solvable because:
- At least one Sylow subgroup is normal.
- The group can be expressed as a semi-direct or direct product of abelian groups, and abelian groups are solvable.
- Therefore, \( G \) is solvable.

Would you like a more detailed explanation of any part, or do you have any other questions?

### 5 Related Questions:
1. What are the Sylow theorems and how do they apply to group theory?
2. Can you explain more about semi-direct products and their significance in group theory?
3. How does the concept of solvability relate to the Jordan-Hölder theorem?
4. What are the properties of cyclic groups that make them solvable?
5. What is an example of a non-solvable group, and how does it differ from solvable groups?

### Tip:
Understanding Sylow's theorems is crucial for analyzing groups of various orders. It gives insight into the number and structure of subgroups, which helps in solving more complex group theory problems.

|G|=pq where p and q are distinct primes, prove that G is solvable.

This detailed proof explains how to prove that a group G of order |G|=pq, where p and q are distinct primes, is solvable. Using key concepts from group theory such as Sylow's theorems and the structure of semi-direct products, this solution is suitable for undergraduate mathematics students.

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