Proof:

Let $G$ be a group that does not have proper subgroups. We want to prove that $G$ is either the identity group (a group consisting solely of the identity element) or a group of prime order.

Case 1: $G$ is the identity group

If $G = \{e\}$, where $e$ is the identity element, then $G$ has only one element. In this case, $G$ is trivially a group, and since it only contains the identity element, it has no proper nontrivial subgroups. Thus, this case satisfies the condition that $G$ has no proper subgroups.

Case 2: $G$ has more than one element

Now, assume that $G$ contains more than one element. We claim that in this case, the order of $G$, denoted by $|G|$, must be a prime number.

Let $|G| = n$. By Lagrange's theorem, the order of any subgroup $H$ of $G$ must divide $n$. This means the order of any possible subgroup of $G$ must be a divisor of $n$.

Since we are given that $G$ does not have any proper subgroups, the only subgroups of $G$ are $G$ itself and the trivial subgroup $\{e\}$. This implies that no divisor of $n$ other than $1$ (corresponding to $\{e\}$) and $n$ (corresponding to $G$) can exist. Therefore, $n$ must have no divisors other than 1 and itself, which is the definition of a prime number.

Hence, $|G|$ must be prime.

Conclusion:

Thus, we have proven that if a group $G$ does not have proper subgroups, then $G$ is either the identity group (when $|G| = 1$) or a group of prime order (when $|G| = p$, where $p$ is a prime number).

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Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. What is Lagrange's theorem in group theory?
2. Can a group of composite order have no proper subgroups?
3. What is an example of a group of prime order?
4. How does the concept of normal subgroups relate to this proof?
5. What is the significance of prime numbers in group theory?

Tip: Understanding the divisors of the group's order is key in many group-theoretic results, especially when using Lagrange’s theorem.