There were two question of group theory posted recently to prove something without Sylow theorem (see 1, and 2) . Both questions have some answer which use Cauchy's theorem. But it seems that any one of these theorem can be proved using another.

In the proof of Sylow theorem given by J. B. Fraleigh (Abstract Algebra), he uses Cauchy's theorem as first step, and after existence of Sylow subgroups (or $p$-subgroups), other parts of Sylow theorem (their conjugacy, their cardinality etc.) can be proved easily.

While Cauchy's theorem can be considered as a special case of Sylow theorem.

Question: Can we say that both these theorems are equivalent?

Sylow theorem:

1) If $G$ is a finite group and $p^n$ divides $|G|$ but $p^{n+1}$ does not, then $G$ has subgroups of order $p^i$ for ($1\leq i\leq n$),

2) any two subgroups of order $p^n$ are conjugate,

3) number of subgroups of order $p^n$ is $1$(mod $ p$).

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    $\begingroup$ Any two true theorems are equivalent. $\endgroup$ – Tobias Kildetoft Sep 29 '11 at 7:57

As mentioned by Tobias, (formally) in a certain sense any two theorems are equivalent.

More informally, both Cauchy's theorem and the first Sylow theorem take about the same amount of "effort" to prove and once one is proven the other follows "quickly". In this informal sense, they are equivalent. However, "effort" and "quickly" are subjective terms here.

This same language is used when speaking of the implicit and inverse function theorems. Both are difficult to establish, but once one is known the other follows quickly. So in some loose sense, they are equivalent.

Reverse mathematics tries to make these ideas more formal. It is the study of which mathematical axioms are needed to prove various statements. In terms of the appropriate reverse math axiom system, I imagine Cauchy and Sylow are equivalent (but Lagrange's theorem is probably equivalent to Cauchy and Sylow as well). http://en.wikipedia.org/wiki/Reverse_mathematics

By the way, both Sylow's and Cauchy's theorems have dozens of different and quite interesting proofs. If you're interested in a particularly neat proof of Cauchy's theorem, check out Joseph Rotman's An Introduction to the Theory of Groups (I have the 4th edition -- see page 74 midpage there is a proof by McKay that is really different and quite slick). Rotman gives multiple proofs of quite a few of the main theorems.

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    $\begingroup$ McKay's proof of Cauchy's theorem is very slick, and does not use Sylow's theorems. In the other direction, Wielandt's proof of Sylow's (existence) theorem is also very slick, and does not use Cauchy's theorem. $\endgroup$ – user641 Sep 29 '11 at 21:27
  • $\begingroup$ To bring an almost dead horse back, there seems to be only one proof of a special case of Lagrange's theorem: math.stackexchange.com/questions/28332/… $\endgroup$ – lhf Sep 29 '11 at 21:54
  • $\begingroup$ @Steve D: of course there are different independent proofs of both theorems... $\endgroup$ – Beginner Sep 30 '11 at 10:17

As others have pointed out, it is not really clear what a formal, non-tautologous meaning of the equivalence of two theorems (i.e., true mathematical statements) would be. Often when two theorems are said to be "equivalent" it means under set-theoretic axioms weaker than the full set of standard ones: e.g. in this sense the Well-Ordering Theorem is equivalent to the Axiom of Choice whereas the theorem that every filter can be extended to an ultrafilter is strictly weaker than the Axiom of Choice. But here we have two theorems concerning only finite structures, so it is hard for me to believe that there are any set-theoretic issues or distinctions arising here.

Nevertheless there is a common informal meaning of the equivalence of Theorem A with Theorem B. Usually when people say this they mean that is easier to prove "Theorem A $\implies$ Theorem B" and "Theorem B $\implies$ Theorem A" than it is to prove either Theorem A or Theorem B. For instance, in this sense the Prime Number Theorem is equivalent to the theorem that the $n$th prime $p_n$ is asymptotic to $n \log n$, whereas the Prime Number Theorem is not equivalent to the assertion $\sum_{n=1}^{\infty} \frac{1}{p_n} = \infty$: it's stronger.

In the sense of the above paragraph, I am tempted to say that Cauchy's Theorem is not equivalent to Sylow's Theorem. Note that Sylow's Theorem as you have written it immediately implies Cauchy's Theorem. For some reason it is more common to state "Sylow's First Theorem" as the assertion that if $p^a \ | \ \# G$ and $p^{a+1} \nmid \# G$ then there is a subgroup of order $p^a$. In this form the result does not immediately imply that there are subgroups of all orders $p^k$ for $p^k$ a prime power dividing $\# G$ although the deduction of this is still rather easy compared to the proof of Sylow's Theorem: it follows easily from the fact the $p$-groups have nontrivial centers that the converse of Lagrange's Theorem holds for $p$-groups. Taking $a = 1$ we recover Cauchy's Theorem.

On the other hand, knowing Cauchy's Theorem does not seem to be particularly helpful in proving Sylow's Theorem. There are some proofs which begin by establishing Cauchy's Theorem (perhaps just for finite commutative groups), but to me they don't seem appreciably shorter or simpler than the ones which establish Sylow's Theorem "from scratch" (e.g. Wielandt's proof using group actions). I know of at least one truly easy proof of Cauchy's Theorem: the famous proof of James McKay. In order for the two theorems to be equivalent, there would need to be a proof of Sylow's Theorem granting Cauchy's Theorem which is no harder than McKay's proof, and I have never seen that.

However, this recent paper of Geoff Robinson comes closer than any other I have seen to giving a "McKay level" proof of Sylow's Theorem. He uses a clever induction argument, but granting some (truly easy, in my opinion) preliminary group-theoretic results it comes out quite short and sweet. It seems to be of some significance that he proves the "stronger" form of Sylow's First Theorem mentioned above: i.e., existence of subgroups of all prime power orders dividing $\# G$, not just the largest such order. Putting this into the "strong" induction hypothesis is a big part of the proof. (Ultimately Robinson's argument reduces to the statement to the Sylow Theorem in a finite cyclic group, where it is of course obvious.) It makes me think that it could be a "mistake" that the standard statement of Sylow's Theorem is the slightly weaker one, as the stronger one is not only stronger but seems easier to prove!


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