Motivated by this question, can one prove that the order of an element in a finite group divides the order of the group

withoutusing Lagrange’s theorem? (Or, equivalently, that the order of the group is an exponent for every element in the group?)The simplest proof I can think of uses the coset proof of Lagrange’s theorem in disguise and goes like this: take $a \in G$ and consider the map $f\colon G \to G$ given by $f(x)=ax$. Consider now the

orbitsof $f$, that is, the sets $\mathcal{O}(x)=\{ x, f(x), f(f(x)), \dots \}$. Now all orbits have the same number of elements and $|\mathcal{O}(e)| = o(a)$. Hence $o(a)$ divides $|G|$.This proof has perhaps some pedagogical value in introductory courses because it can be generalized in a natural way to non-cyclic subgroups by introducing cosets, leading to the canonical proof of Lagrange’s theorem.

Has anyone seen a different approach to this result that avoids using Lagrange’s theorem? Or is Lagrange’s theorem really the most basic result in finite group theory?

**Answer**

Consider the representation of $\langle a \rangle$ on the free vector space on $G$ induced by left multiplication. Its character is $|G|$ at the identity and $0$ everywhere else. Thus it contains $|G|/|\langle a \rangle|$ copies of the trivial representation. Since this must be an integer, $|\langle a \rangle|$ divides $|G|$. Developing character theory without using Lagrange’s theorem is left as an exercise to the reader.

**Attribution***Source : Link , Question Author : lhf , Answer Author : joriki*