Sylow’s Theorems Revisited
Point of Post: In this post we give a more refined proof of Sylow’s Theorems.
On this blog I have given a proof of Sylow’s theorems and an alternate proof of Sylow’s first theorem. A year or so later, with more time, and more finesse I’d like to give the simplest, most coherent proof of Sylow’s theorems.
We assume all the notations such as being the set of Sylow -subgroups of and ,etc. are known to the reader:
Theorem (Sylow): Let be a finite group with where , then:
We construct such a group inductively. Evidently if then we may take the trivial group. Suppose then that we have found a subgroup of with for . We know then that and so by a previous theorem we have that . Thus, by Cauchy’s Theorem there exists with . By the fourth isomorphism theorem we know that for some and so, in particular, we have that . As stated, from this we may construct subgroups of with .
Let act on by left multiplication. Since is a -group we have by the ‘fundamental theorem’ that . That said, since we have in particular that and so is non-empty. So, let then for all and so for all , or equivalently . Thus, . In particular, if we have that and so .
Let act on by conjugation. By we know that this action is transitive. In particular, we have that for any . That said, we note by definition that and so .
For the second part of the statement we let act on by conjugation. By the ‘fundamental theorem’ we know that . What we claim though is that . Indeed, it’s evident that , and so now assume that . We have then that and so , but we know that
and since and is the maximal power of dividing we may conclude that and so . Thus, . Noting then that completes the proof.
1. Isaacs, I. Martin. Finite Group Theory. Providence, RI: American Mathematical Society, 2008. Print.
2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.