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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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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) –  Fredrik Meyer Mar 25 '11 at 6:19
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For abelian groups the proof is pretty simple, just multiply all the elements in the group and in the image of $f$. –  N. S. Apr 9 '11 at 20:04
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In case you didn't see it: Martin Brandenburg asked this question with essentially the same thrust. –  t.b. Aug 6 '11 at 17:57
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@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. –  Pete L. Clark Aug 6 '11 at 22:39
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@lhf: Also, I like the proof you give above using orbits. –  Pete L. Clark Aug 6 '11 at 23:00

1 Answer 1

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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Wasn't there a book that said "... is left as an exercise for the masochistic reader"? –  Arturo Magidin May 10 '11 at 20:42
    
thanks, though it's hardly in the elementary nature I'm looking for. –  lhf May 10 '11 at 21:03
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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 :-) –  joriki May 11 '11 at 7:23
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I wonder what Linderholm would say (Mathematics made difficult) –  Bill Dubuque May 12 '11 at 3:44
    
@Bill: Ah -- never heard of that -- looks like a book I might want to read :-) –  joriki May 12 '11 at 4:38

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