# Different Applications of Sylow Theorems

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.

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.

• 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. – user138530 Dec 6 '15 at 5:36
• @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... – Qiaochu Yuan Dec 6 '15 at 6:33
• 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." – user138530 Dec 6 '15 at 6:38
• @Christian: well, this is all I got. I've personally never needed the Sylow theorems for anything outside of group theory. – Qiaochu Yuan Dec 6 '15 at 6:45

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.

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.

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 $$p-1$$.