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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?

Martin Sleziak
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S.B.
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    Lagrange's theorem is so elementary and so easy to prove that one should ask why would you try to find any proof that does not use it, if you have a proof that does. – Ittay Weiss Mar 11 '15 at 08:39
  • In a book we are using in undergrad Algebra Courses, Lagrange's Theorem is stated a section after these small facts are used. So I wondered that whether it should be before that or there is another way. – S.B. Mar 11 '15 at 08:42
  • As I recall, there is a way to show that if $|G| = n$ then $x^n = 1$ for all $x\in G$ without Lagrange. But it uses a lot of knowledge of permutations, and is probably longer than a proof of Lagrange. – Tobias Kildetoft Mar 11 '15 at 09:08
  • This is my pet question. See http://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory. – lhf Nov 22 '15 at 19:49

2 Answers2

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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 $.

Tobias Kildetoft
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  • This seems to be a different way to explain my orbits proof in http://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory. – lhf Nov 22 '15 at 19:51
  • @lhf I agree, they are indeed very similar. – Tobias Kildetoft Nov 23 '15 at 07:31
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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.

Andreas Caranti
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  • Nice, thanks! Easy to teach undergrads also.. Thank you very much. And for the non-abelian case, maybe, one can not prove this fact without using Lagrange's I think, if someone claims the converse, I would also appreciate it. Thank you very much. – S.B. Mar 11 '15 at 11:32
  • @smyr, you're welcome! – Andreas Caranti Mar 11 '15 at 11:33
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    @smyr As I said in a comment, I recall there being (a rather long) one. But I can't seem to reconstruct it right now. I think it was something about looking at the cycle decomposition of the permutation given by multiplication by the element, and somehow finding a contradiction if one assumed it to have a cycle of length not dividing the order of the group. – Tobias Kildetoft Mar 11 '15 at 11:56
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    So I finally recalled that proof, and it turned out to use the same action as this answer, and just a bit closer examination of the cycle structure (which does mean it becomes a lot less neat than the trick in this answer). – Tobias Kildetoft Apr 13 '15 at 12:48