To prove that a group of order 3 must be cyclic, let's proceed step by step.

Proof:

Let $G$ be a group with $|G| = 3$, meaning the group $G$ has exactly 3 elements. By Lagrange's Theorem, the order of any element in $G$ must divide the order of the group. So the order of any element in $G$ must divide 3. The divisors of 3 are 1 and 3.

Now, consider the elements of $G$. Let's say the identity element of the group is denoted by $e$, and there are two other elements in the group, say $a$ and $b$, such that $G = \{e, a, b\}$.

Case 1: If every non-identity element has order 1.

This is not possible because in a group, the identity element is the only element of order 1. Therefore, the non-identity elements must have an order greater than 1. This means that the orders of $a$ and $b$ must divide 3.

Case 2: If the non-identity elements have order 3.

Suppose $a$ has order 3. This means $a^3 = e$, and the cyclic subgroup generated by $a$, i.e., $\langle a \rangle$, contains three distinct elements: $\{e, a, a^2\}$.

Since the group $G$ has exactly three elements and we already have three elements $e, a, a^2$ generated by $a$, it follows that $G = \langle a \rangle$. This means that $G$ is a cyclic group generated by $a$.

Conclusion:

Thus, in a group of order 3, at least one non-identity element must have order 3, and hence the group must be cyclic, generated by this element. Therefore, any group of order 3 is cyclic.

Do you have any further questions or would you like additional details on this?

Here are 5 related questions for further exploration:

1. What is Lagrange’s Theorem and how does it apply to group theory?
2. Can you provide examples of cyclic groups of different orders?
3. How does the structure of cyclic groups change as the order increases?
4. What are the properties of cyclic groups in general?
5. How do non-cyclic groups differ from cyclic ones?

Tip: A cyclic group is always abelian, meaning that the group operation is commutative (i.e., $ab = ba$ for all $a, b \in G$).