Is there a way to prove that the order of an element in a group divides the order of the Group, WITHOUT USING LAGRANGE'S

Question

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?

Answer 1

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.

Answer 2

As I mentioned, here is a way to show that the order of any element divides the order of the group, which does not use Lagrange. Instead, it uses a bunch of knowledge about permutations, which might sometimes have been introduced before Lagrange.

So let $x\in G $ and note that $a\mapsto xa $ defines a permutation of $G $. So write this permutation as a product of disjoint cycles, with the shortest of length $m $.

Now $x^m $ has a fixed point (any element in the $m $-cycle). But this means that $x^mb = b $ for some $b\in G $ and thus $x^m= 1$. However, if there was also some cycle of length $k>m $ in the decomposition of $x $ then no element in this cycle would be fixed by $x^m $, but since $x^m $ was equal to $1$ this is clearly absurd, so we conclude that all cycles have length $m $.

Now we get the conclusion since the sum of the lengths of the cycles is precisely the order of the group, so $m $ divides this and is clearly the order of $x $.