10

Specifically: If $p$ is a prime divisor of the order of a finite group $G$, then there exists an element of order $p$ in $G$

So I'm looking for a little intuition behind this idea. I understand how to prove it, but I don't understand the idea behind it. I.e, if I hadn't already seen a proof, I wouldn't know why or if the statement of the theorem was correct. As stated, I'm looking for the core idea behind the proof.

Any insights on how you think of it would be appreciated! Thanks!

  • Could you put up your understanding of what the theorem is and what proof you understand? Different things get different names is different places sometimes, especially if it's with someone as prolific as Cauchy. – Tom Oldfield Nov 16 '12 at 00:17
  • I'm sorry, I thought Cauchy did most of his work in analysis and that the "for Groups" would suffice. My apologies –  Nov 16 '12 at 00:19
  • I'm sure it would, just want to make sure everyone is one the same page! – Tom Oldfield Nov 16 '12 at 00:20
  • 1
    @Tom: I didn't have any problem recognizing that the title corresponds to the theorem that the OP has stated. I'm pretty sure that his title is specific enough. – Haskell Curry Nov 16 '12 at 00:26
  • @Carl: The proof that you have seen - is it the one by McKay that uses a group action? – Haskell Curry Nov 16 '12 at 00:29
  • @Haskell Curry Good point, it was probably fine, I just thought it was good style to include exactly what it is that was being asked about and which proof being used was certainly important. – Tom Oldfield Nov 16 '12 at 00:38
  • 1
    The proof I've seen counts the number of strings of elements $x_1x_2\dots x_p=\epsilon$ two different ways and shows there exists such a ring of elements $u_1u_2\dots u_p=\epsilon$ such that we can rotate the ring by less than $p$ elements and end up with the same ring. I.e for $0<k<p$ we have $u_{1+k}u_{2+k}\dots {p+k}=\epsilon$ (where the indicies are taken mod $p$. and we have $u_1 = u{1+k},u_2 = u_{2+k}, \dots u_p = u_{p+k}$ by primality of $p$ is follows $u_1=u_2=\dots=u_p$ and so $u^p=\epsilon$ –  Nov 16 '12 at 00:39
  • 2
    @Carl , not only that prook (McKay's) is one of the most beautiful, cunning and compact proofs I know: it is all this and more specially if you compare it with the usual, old proof, which first takes on abelian groups and then uses a rather not-so-pretty inductive stuff to finally reach the end. It is lengthy, cumbersome and not so clear (although any proof can teach us always new things...) – DonAntonio Nov 16 '12 at 00:44
  • @DonAntonio I checked out McKay's paper (never knew it existed until named) and really loved his intro about generalizing from the multiplication table of a finite group. Thats the kind of thing I was hoping for. McKay's proof also seems to be a stronger proof than Cauchy's original theorem, requiring that the number of solutions of $x^p=1$ be divisable by $p$. One quick question though: in his argument, he says if 2 components of a p-tuple within an equivalence class are distinct, then the equivielence class must contain $p$ distinct tuples. I don't see how to make that claim right away –  Nov 16 '12 at 01:37
  • @Carl, i'm not quite sure what you're asking but I think it'd be a good idea if you better open a new thread with this question, including a link to the proof you're referring to (is it the original paper by McKay?). In the version of McKay's proof I know there are no equivalence classes...unless you're talking of orbits under the cyclic group's action. – DonAntonio Nov 16 '12 at 03:01
  • @DonAntonio this is what I found http://www.cs.toronto.edu/~yuvalf/McKay%20Another%20Proof%20of%20Cauchy's%20Group%20Theorem.pdf. and I already opened a new thread here: http://math.stackexchange.com/questions/238385/clarification-of-mckays-proof-of-cauchys-theorem-for-groups :) –  Nov 16 '12 at 03:04

3 Answers3

9

The idea behind it is that it is a partial converse to Lagrange's Theorem (that the order of any subgroup divides the order of the group). We seeks theorems of this kind because it means we can get information about the elements of a group just from knowing it's order, and then we can start to classify groups. i.e. we can then say that if a group has a certain size, it will be of a certain form. This gives an intuitive reason why we want theorems of this kind, but not of why it is true.

In fact, I don't think the theorem is intuitive(which I know is not what you want to hear!), and the proof that I've seen involving having the $\mathbb{Z_p}$ act on the set of p-tuples of the group is something that seems totally unnatural (although once understood it is very pretty, not to mention clever!) In fact, I don't think that there is a core idea behind the proof to be help understand why it's true. I think it is just an exercise in clever manipulation, not giving us much insight into the problem, in the same way that inductive proofs often give us answers but may not help us understand the question better.

I think that theorems like this one (and also the Sylow theorems) aren't obvious, but they are very, very useful. As such, when groups were first discovered, people were desperately trying to find theorems of this sort, so spent a long time thinking about them and discovered them in some manner or another, and so they were born for utility, not because they were intuitively true.

Tom Oldfield
  • 13,034
  • 1
  • 39
  • 77
  • I find Cauchy Theorem really intuitive. When I was solving a problem about groups, I actually assumed Cauchy's Theorem to be trivial, only later realizing it's a theorem. – Divide1918 Nov 23 '19 at 17:13
6

What is the "intuition: behind the theorem "a function differentiable at some point is continuous at that point"? I'm not sure, but perhaps it'd be that the function is "smooth" enough at that point as to be continuous...or something like that.

What's the intuition behind Cauchy's Theorem? That a finite group having order a multiple of a prime $\,p\,$ has "to pay the price", i.e.: it must have at least one element of order that prime $\,p\,$...or something like this.

I can't say what the core behind Cauchy's idea was, but perhaps it stemmed from checking many examples and seeing there was an apparent common pattern to all of them.

So I'm not sure, but perhaps it was Cauchy's Theorem what gave Sylow some ideas or inspiration to make some research on this and eventually to come up with some of the most important and basic theorems in finite group theory.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
0

The uniform distribution of the elements of a finite group $G$ in its Cayley table hints at some structure in the invariants attached to elements of $G$, and in particular, their individual orders.

MarcFRed Tch
  • 1
  • 1