This is a very easy fact we use in Group Theory,
But somehow, I wondered that whether there may be another way (other than Lagrange's Theorem) to prove that the order of an element divides the order of Group.
I attempted to go on the term "exponent" of the Group (just assume that G is a Finite Group), which we may define the least common multiple of orders of elements in G. But this exponent divides $|G|$, since the order of each element divides $|G|$. So it became a little paradox.
Shall I think that; before teaching this fact in a Course, one should teach about Lagrange's first?
Best Answer
There is a nice, Lagrange-free proof for Abelian groups.
Let $G$ be an Abelian, finite group of order $n$, and let $G = \{a_{1}, \dots, a_{n} \}$. Let $x \in G$.
The map $$ G \to G, \qquad a \mapsto a x $$ is readily seen to be a bijection, so that $G = \{a_{1} x, \dots, a_{n} x \}$. Therefore $$\tag{key} \prod_{i=1}^{n} a_{n} = \prod_{i=1}^{n} (a_{n} x) = \left(\prod_{i=1}^{n} a_{n}\right) x^{n}. $$ Simplifying, we get $x^{n} = 1$
Note that in (key) I have rearranged the terms in the products, taking advantage of the fact that $G$ is abelian.