The question asks to prove that if $G$ is a finite group with order $n$, then for every element $a \in G$, it holds that $a^n = e$, where $e$ is the identity element of the group. The hint suggests using Lagrange's Theorem to establish this result.

Proof using Lagrange's Theorem:

Step 1: Subgroup generated by $a$

For any element $a \in G$, the set of powers of $a$, denoted by: $\langle a \rangle = \{a^k \mid k \in \mathbb{Z}\}$ is a cyclic subgroup of $G$ generated by $a$. The order of this subgroup, denoted as $|\langle a \rangle|$, is the smallest positive integer $m$ such that $a^m = e$. This $m$ is called the order of the element $a$.

Step 2: Application of Lagrange's Theorem

Lagrange's Theorem states that the order of any subgroup of a finite group divides the order of the group. Since $\langle a \rangle$ is a subgroup of $G$, the order of $\langle a \rangle$ (denoted $m$) must divide $n$, the order of $G$. Hence, there exists some integer $k$ such that: $n = km$ where $m$ is the order of $a$ (i.e., the smallest integer such that $a^m = e$).

Step 3: Conclusion

Since $a^m = e$ by the definition of the order of $a$, and $m$ divides $n$, we can write: $a^n = a^{km} = (a^m)^k = e^k = e$ Thus, $a^n = e$ for every $a \in G$.

Conclusion:

We have proved that for any finite group $G$ with order $n$, for every element $a \in G$, $a^n = e$, using Lagrange’s Theorem.

Would you like further clarification or details on any part of the proof? Here are some related questions you might explore:

1. What are the conditions for Lagrange’s Theorem to hold?
2. How can this result be applied to cyclic groups specifically?
3. What is an example of a group where this theorem applies?
4. How does this proof differ if $G$ is infinite?
5. What is the significance of the identity element $e$ in group theory?

Tip: Lagrange's Theorem is extremely useful in determining the possible orders of elements in a group, especially when working with finite groups.