Motivated by this question, can one prove that the order of an element in a finite group divides the order of the group without using 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 orbits of $f$, 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?
Best 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.