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The theorems of Sylow are very well known and almost every mathematician learns in his undergraduate course.

The applications of Sylow theorems given in books are of the kind

"If $|G|=....$ then show that $G$ is not simple/ $G$ is solvable/ ..."

I would like to know if there are some other, interesting, applications of this theorem.

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I know exactly one other direct application of the Sylow theorems outside of group theory, which is to proving the fundamental theorem of algebra.

Suppose $K$ is a Galois extension of $\mathbb{R}$. We'll aim to show that either $K = \mathbb{R}$ or $K = \mathbb{C}$. (In particular, $\mathbb{C}$ itself must therefore be algebraically closed.) Let $G$ be its Galois group and let $H$ be the Sylow $2$-subgroup of $G$.

By Galois theory, $K^H$ is an odd extension of $\mathbb{R}$. But $\mathbb{R}$ has no nontrivial odd extensions: any such extension has primitive element something with an odd degree minimal polynomial over $\mathbb{R}$, but any such polynomial has a root by the intermediate value theorem. Hence $K^H = \mathbb{R}$, or equivalently $H = G$, so $G$ has order a power of $2$.

But now $K$ is an iterated quadratic extension of $\mathbb{R}$, and it's easy to explicitly show using the quadratic formula that the only nontrivial quadratic extension of $\mathbb{R}$ is $\mathbb{C}$, which itself has no nontrivial quadratic extensions.

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  • $\begingroup$ This is neat, but the word "application" suggests something else entirely to my mind, namely a result that one would have a hard time proving without the Sylow theorems. $\endgroup$ – user138530 Dec 6 '15 at 5:36
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    $\begingroup$ @Christian: this seems nonstandard to me. It means I can't start any proof of the fundamental theorem with "this is an application of..." because of course given just about any technique you can think of there's a proof which doesn't use that technique... $\endgroup$ – Qiaochu Yuan Dec 6 '15 at 6:33
  • $\begingroup$ The context matters of course. Anyway, I liked your answer, but I was also hoping for something addressing the OP's (and mine, secretly) implied concern "the Sylow theorems, yes, you never need them for anything." $\endgroup$ – user138530 Dec 6 '15 at 6:38
  • $\begingroup$ @Christian: well, this is all I got. I've personally never needed the Sylow theorems for anything outside of group theory. $\endgroup$ – Qiaochu Yuan Dec 6 '15 at 6:45
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The fundamental theorem of algebra is probably the best example, but how about the following proof that a finite subgroup of the multiplicative group of a field is cyclic:

Let $G \subset F^\times$ be finite. Let $H_p$ be a $p$-Sylow subgroup. Then if $|H_p|=p^k$, we claim that $H_p$ is simply the set of all roots in $F$ of the polynomial $x^{p^k}-1$: every element of $H_p$ is a root by LaGrange's theorem, and there can be no more roots since the degree of the polynomial is $p^k$. Now using this, it's easy to see that $H_p$ is cyclic: it's generated by any $y \in H_p$ such that $y^{p^{k-1}}\neq 1$ (such $y$ exist because there are only $p^{k-1}$ roots of $x^{p^{k-1}}-1$). But now since $G$ is abelian, $G$ is the direct product of its Sylow subgroups (this is where a Sylow theorem is used - although you could also use abelian group theory...), which are all cyclic of relatively prime order.

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The Sylow Theorems often play a crucial role in finding all groups of a certain order. For example, all groups of order $pq$, or all groups of order $p^n$, where $p$ and $q$ are primes can be found in this manner. You may find more information in this book by J.S. Milne, chapter 5.

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Let $p$ be a natural number larger than $1$. Then $p$ is prime if and only if $(p-1)! \equiv -1 \mod p$.

This is Wilson's theorem. It can be proved using Sylow's theorem.

Indeed the implication $\Leftarrow$ is rather elementary, for the implication $\Rightarrow$, if $p$ is a prime the symmetric group $G=S_p$ of degree $p$ contains exactly $(p-1)!$ elements of order $p$ ($p$-cycles): this is also elementary. Each of them generates a Sylow $p$-subgroup of $G$, because the order of the Sylow $p$-subgroups of $G$ is exactly $p$.

It follows that the number of Sylow $p$-subgroups of $G$ is $(p-1)!/(p-1)= (p-2)!$, since each Sylow $p$-subgroup of $G$ contains precisely $p-1$ elements of order $p$ and any two Sylow $p$-subgroups of $G$ intersect trivially.

Sylow's theorem then implies that $(p-2)! \equiv 1 \mod p$, and the result follows by multiplying both sides by $p-1$.

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