We are given that a finite group $G$ has exactly two proper nontrivial subgroups, a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$, where $p$ and $q$ are distinct primes. We need to prove that $G$ is cyclic and determine its order.

Step 1: Analyze the structure of $G$

From the problem statement, we know that $G$ has exactly two proper nontrivial subgroups, one of order $p$ and the other of order $q$. We can infer that these two subgroups must be distinct, as $p \neq q$ (since $p$ and $q$ are distinct primes).

Since $H$ and $K$ are both proper subgroups, they must be contained in $G$, and thus the order of $G$ must be a multiple of both $p$ and $q$. Therefore, the order of $G$, denoted $|G|$, must be divisible by both $p$ and $q$. Hence, the least common multiple of $p$ and $q$ gives a candidate for the order of $G$. Since $p$ and $q$ are distinct primes, $\text{lcm}(p, q) = pq$.

Thus, the order of $G$ must be $|G| = pq$.

Step 2: Verify the structure of $G$

We now verify that $G$ is cyclic of order $pq$. If $G$ is cyclic, it must have an element $g$ of order $pq$. The subgroups of $G$ would then be the trivial subgroup, the subgroup generated by $g^p$ (which would have order $p$), and the subgroup generated by $g^q$ (which would have order $q$).

Since we are given that $G$ has exactly two proper nontrivial subgroups—one of order $p$ and one of order $q$—the subgroups $H = \langle g^p \rangle$ and $K = \langle g^q \rangle$ must be exactly these two subgroups. Therefore, $G$ must be cyclic.

Step 3: Conclusion

Since $G$ is cyclic and its order is $pq$, the order of $G$ is exactly $pq$. Thus, we have shown that $G$ is cyclic and that its order is $pq$.

Summary of the steps:

1. We used the fact that the order of $G$ must be divisible by both $p$ and $q$ to conclude that $|G| = pq$.
2. We showed that if $G$ is cyclic of order $pq$, the two given subgroups of order $p$ and $q$ are precisely the subgroups of $G$.
3. Therefore, $G$ is cyclic and its order is $pq$.

Do you want more details about any part of this proof or have any questions?

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Here are some follow-up questions to help deepen understanding:

1. What is the structure of a cyclic group of order $pq$?
2. How can we determine the number of subgroups of a finite group given its order?
3. What happens if $G$ had more than two nontrivial proper subgroups? How would this affect the conclusion?
4. How does the classification of cyclic groups of order $pq$ help us understand the nature of groups of prime order?
5. Why does the fact that $p$ and $q$ are distinct primes matter in this proof?

Tip: When proving properties about finite groups, always use group order and the Sylow theorems to reason about possible subgroups and their relations.