Burnside's transfer theorem in group theory

"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.

• Firstly, this is a theorem of Burnside, not of Wedderburn, and secondly it is not necessarily true that $G = N \times P$. In general we just have a semidirect product $G = N \rtimes P$. Jul 5, 2018 at 18:20
• No, it should be a semidirect product. I also don't think I have seen this attributed to Wedderburn. I usually see it referred to as Burnside's transfer theorem. Jul 5, 2018 at 18:20
• This is Burnside's Normal Complement Theorem and its proof uses transfer as the the fundamental technique. Note that $P \subseteq Z(N_G(P))$ is equivalent to $C_G(P)=N_G(P)$ and see for example ysharifi.wordpress.com/2011/01/20/… or read I.M. Isaacs Finite Group Theory. Jul 5, 2018 at 22:38

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).
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.