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?

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    $\begingroup$ How do you define "the most basic result"? Certainly, uniqueness of inverses (et cetera) is a basic result, even though it is a triviality. Anyhow - your question is a good one: I am surprised how often I apply Lagrange's theorem. (even though I hardly think of it under that name. It's just a fact of life) $\endgroup$ Mar 25, 2011 at 6:19
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    $\begingroup$ @Fredrik: I mean, non-trivial result. $\endgroup$
    – lhf
    Mar 25, 2011 at 11:34
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    $\begingroup$ For abelian groups the proof is pretty simple, just multiply all the elements in the group and in the image of $f$. $\endgroup$
    – N. S.
    Apr 9, 2011 at 20:04
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    $\begingroup$ @lhf: I just saw elsewhere the link you provided to a paper of Pengelley which reproduces Cayley's first paper on group theory with insightful footnotes. In particular (as you know...) Cayley states the theorem in question and says only "it can be shown": neither cosets nor Lagrange's Theorem are anywhere in sight. I think this link would make a nice addition to your question. $\endgroup$ Aug 6, 2011 at 22:39
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    $\begingroup$ @lhf: Also, I like the proof you give above using orbits. $\endgroup$ Aug 6, 2011 at 23:00

2 Answers 2


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.

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    $\begingroup$ Wasn't there a book that said "... is left as an exercise for the masochistic reader"? $\endgroup$ May 10, 2011 at 20:42
  • $\begingroup$ thanks, though it's hardly in the elementary nature I'm looking for. $\endgroup$
    – lhf
    May 10, 2011 at 21:03
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    $\begingroup$ Hmm -- I'm wondering whether the humourous aspect of this answer was apparent enough -- perhaps the last sentence should have been followed by a smiley :-) $\endgroup$
    – joriki
    May 11, 2011 at 7:23
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    $\begingroup$ I wonder what Linderholm would say (Mathematics made difficult) $\endgroup$ May 12, 2011 at 3:44
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    $\begingroup$ While I am aware this remark comes many years later - Spivak in his Comprehensive Introduction to Differential Geometry vol 1 makes this remark in an appendix to the chapter on tangent bundles where he is proving their uniqueness. The argument proceeds in two main steps, and after carrying out the first, and many pages of diagram chasing, he makes the remark Arturo mentions. $\endgroup$ Jan 3, 2018 at 1:52

I am late... Here is a proposal, probably not far from Ihf's answer.
For $a \in G$ of order $p$, define the binary relation $x\cal Ry$ : $\exists k\in \mathbb{N} ; k<p$ such that $y=a^kx$
$\cal R$ is an equivalence relation on $G$ and sets up a partition of $G$.
A class is defined by $C_x=\left\{a^kx|k=0,1,\ldots,p-1 \right\}$
All the classes have $p$ elements, then $n=|G|$ is a multiple of $p$ ; $(n=pm)$ and $a^n=a^{pm}=e$

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    $\begingroup$ This is exactly the coset proof for the cyclic group generated by $a$. $\endgroup$
    – lhf
    Aug 2, 2019 at 10:20

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