12
$\begingroup$

There are many proofs for Cauchy's Theorem from group theory, which states that if a prime $p$ divides the order of a finite group $G$, then $\exists g\in G$ of order $p$.

Recently I've encountered an elegant combinatorial proof for this theorem in the abelian case. It is self-contained, and appears in Hecke's "Lectures on the Theory of Algebraic Numbers". It has 3 steps:

  • Assume $p \mid |G|$, where $G=\{g_1,\cdots, g_n\}$. Consider all the product $\prod_{i=1}^{n} g_i^{a_i}$ where $0 \le a_i < \text{ord}(g_i)$.
  • Each element of $G$ appears in this multiset of size $\prod_{i=1}^{n} \text{ord}(g_i)$ an equal amount of times. This is the (only) point where the fact that $G$ is abelian is used.
  • This implies $\prod_{i=1}^{n}\text{ord}(g_i)$ is divisibly by $p$. As $p$ is prime, we deduce $p \mid \text{ord}(g_i)$ for some $1\le i \le n$. Now we take a suitable power of $g_i$. $\blacksquare$

My question is: Can the second step be generalized to the general case? Is it true that the multiset of products $\{ \prod_{i=1}^{n} g_i^{a_i} \mid 0 \le a_i < \text{ord}(g_i) \}$ contains each element of $G$ the same amount of times? Note that there's no "canonical" product $\prod_{i=1}^{n} g_i^{a_i}$ in the non-abelian case - there are several ways to interpret this product (maybe the $g_i$'s come before the $g_j$'s for $i<j$; maybe they interlace), and I don't want to specify the way.

I've read a bit about equations in groups, and haven't seen this kind of exponential problem discussed.

(I am also interested in the origins of this proof, by the way. If some one knows, please drop a comment.)

$\endgroup$
  • 1
    $\begingroup$ I don't think it is realistic to expect a proof exploiting commutativity to carry over to the non-abelian case. When I first learned about groups, they were all abelian (unit groups mod $m$). For the results (i) order of each $g \in G$ divides $|G|$ and (ii) $g^{|G|} = e$ for all $g \in G$, I had learned a proof of (ii) that exploited commutativity and derived (i) from (ii). To prove the same two results for general finite groups, you prove (i) first via cosets and then derive (ii) from (i). At the end of the day you have the same two theorems, but without commutativity new ideas are needed. $\endgroup$ – KConrad May 18 '15 at 20:43
  • 1
    $\begingroup$ As a silly comment, an induction on $\lvert G\rvert$ allows one to use this argument to reduce the proof of Cauchy's theorem to the case where $G = [G, G]$. That sort of condition gladdens the heart of a Lie theorist, but I'm not sure that it's so useful for a finite-group theorist. $\endgroup$ – LSpice May 18 '15 at 20:48
  • $\begingroup$ @KConrad I agree that my question is mathematically awkward, and that there are beautiful proofs for the non-abelian case with new insights. If, by chance, the abelian proof still works, it will be a) a small miracle, and b) deeper than Cauchy's Theorem. I don't want this problem to be solved so we'll have another proof of Cauchy's Thm, I want it to be solved to satisfy my curiosity and my understanding of non-abelian groups. $\endgroup$ – Ofir Gorodetsky May 18 '15 at 20:56
  • 14
    $\begingroup$ I presume that you know the proof of Cauchy's Theorem by McKay, which works much the same for Abelian and non-Abelian cases, and is rather similar in spirit: look at ordered p-tuples of group elements $(g_{1},g_{2},\ldots,g_{p}) : g_{1}g_{2}\ldots g_{p} = 1_{G}$. There are $|G|^{p-1}$ such $p$-tuples so a multiple of $p$ when $p$ divides $|G|$. But these are invariant under cycling ( ie the obvious action of $\mathbb{Z}/p\mathbb{Z}$). The fixed points have all "coordinates" equal, but the number is divisible by $p$. But $(1,1,\ldots,1)$ is fixed. So there is another $g$ with $g^{p}=1.$ $\endgroup$ – Geoff Robinson May 18 '15 at 21:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.