The question in the image asks to prove the following statement:

Problem: "Prove that if $G$ is a finite group and $a \in G$, then the order of $a$ divides the order of $G$."

Approach: This can be solved using Lagrange's Theorem.

Proof Outline:

1. Lagrange's Theorem: This theorem states that for any finite group $G$, the order (i.e., the number of elements) of any subgroup $H$ of $G$ divides the order of $G$.

2. Consider the cyclic subgroup generated by $a$: Let $\langle a \rangle$ denote the cyclic subgroup of $G$ generated by $a$. The order of this cyclic subgroup, $|\langle a \rangle|$, is the same as the order of the element $a$, which is denoted by $o(a)$.

3. Applying Lagrange's Theorem: Since $\langle a \rangle$ is a subgroup of $G$, by Lagrange's theorem, the order of $\langle a \rangle$ (i.e., $o(a)$) must divide the order of $G$, denoted $|G|$.

Thus, the order of any element $a \in G$ divides the order of the group $G$.

Would you like a more detailed explanation on any specific part? Or further clarifications on the theorem?

Here are some additional questions related to this topic:

1. What are some examples of groups where the order of elements divides the order of the group?
2. How does the order of an element help classify the structure of a group?
3. What happens if $G$ is infinite? Does Lagrange's theorem still apply?
4. Can the order of an element in a group ever be larger than the order of the group?
5. How does this theorem relate to cosets in group theory?

Tip: Lagrange’s Theorem is crucial in understanding the structure of finite groups and plays a vital role in proofs involving divisibility and subgroup orders.