While reading this I read the following (slightly rephrased, and edited according to comments) :
"Burnside's transfer theorem : If a $p$-Sylow subgroup $P$ of a finite group $G$ is included in its normalizer's center, i.e. $P \leq Z(N_G(P))$, then there is a normal subgroup $N$ of order $|G|/|P|$ such that $P \cap N = 1$, and $G = N \rtimes P$"
What is the proof and/or applications (within pure math) of the above theorem other than classification of group of order 30? You are welcome to just provide a link. I wasn't able to find one online.
Best Answer
This is the first in a long line of theorems guaranteeing a normal $p$-complement (the technical term for $N$ in the question). This includes Frobenius (if $N_G(H)/C_G(H)$ is a $p$-group for every $p$-subgroup $H$ of $G$ then $G$ has a normal $p$-complement), Thompson (if $p$ is odd and $C_G(Z(P))$ and $N_G(J(P))$ have normal p-complements, so does G -- here $J(P)$ is the Thompson subgroup...the subgroup of $P$ generated by all elementary abelian subgroups of order $p^n$ where $n$ is the largest number such that such subgroups exist), on to Glauberman's normal $p$-complement theorem (for $p$ odd it is enough to $N_G(Z(J(P)))$ to have a normal $p$-complement to guarantee that $G$ does).
Interesting applications of the Burnside theorem include the result that non-abelian simple groups must have order divisible by 12 or by the cube of the smallest prime dividing the order (in particular, non-abelian simple groups of even order must have order divisble by 8 or 12). Another application is a relatively simple proof of the theorem that any finite group with all Sylow subgroups cyclic is a semidirect product of two cyclic groups of coprime orders. A key application of the Thompson p-complement theorem is the proof that Frobenius kernels are nilpotent.
You can find much of this material in chapters 12-14 of Passman's Permutation Groups.
These theorems are useful for recreational group theory if you are trying to show that there is no simple group of a certain order. For example, to show there is no group of order $552=2^3\cdot3\cdot23$, one easily sees that, since the number of 23-Sylows must be congruent to 1 mod 23 and divide 552, it is either 1 (impossible as then the 23-Sylow is normal) or 24. But if it is 24, then a 23-Sylow is its own normalizer and, thus, being abelian, is in the center of its normalizer, so Burnside's theorem guarantees the existence of a normal 23-complement (i.e., in this case, a normal subgroup of order 24). Thus, every group of order 552 either has a normal subgroup of order 23 or a normal subgroup of order 24.