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.