Burnside normal p complement theorem

Given: A finite group $G$, a prime $p$, a Sylow $p$-subgroup $P\le G$ with $P\le Z(N_G(P))$.

Write $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$; then $\mathrm{\mid }P\mathrm{\mid }=p^a$ and the index $n:=[G:P]=m$ is prime to $p$ (Sylow $p$-subgroups of a finite group, Sylow I: every finite group has a Sylow $p$-subgroup, Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).

As $P\le Z(N_G(P))$ and $P\le N_G(P)$, every two elements of $P$ commute: $P$ is abelian (The center $Z(G)$ of a group, The normalizer $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$ of a subgroup).

The transfer $V=V_{\phi }$ of the homomorphism $\phi ={\mathrm{id}}_P:P\to P$ is a homomorphism $G\to P$, explicitly $V(x)={\prod }_{\alpha \in P\mathrm{\backslash }G}\phi (t_{\alpha }x{t}_{\alpha x}^{-1})$ for any transversal, and it agrees with the cycle formula $V(x)={\prod }_i\phi (t_i{x}^{n_i}{t}_i^{-1})$ where $n_i$ are the orbit sizes of $⟨x⟩$ acting on $P\mathrm{\backslash }G$ (Transfer homomorphism for a finite index subgroup, Transfer is a homomorphism, Transfer cycle decomposition formula, Transfer is independent of the transversal).

The orbits of the action of $⟨x⟩$ on the finite set $P\mathrm{\backslash }G$ partition it, so their sizes $n_1,\dots ,n_r$ satisfy $n_1+\cdots +n_r=[G:P]=n$ (The orbits of a group action are the equivalence classes of $x\sim y$ iff $y=g\cdot x$ for some $g$, and hence partition the acted-on set, Left group actions, transitive actions, and faithful actions).

If $P$ is abelian and $u,v\in P$ are conjugate in $G$, then they are conjugate in $N_G(P)$ (Abelian sylow fusion in its normalizer).

If a finite group $H$ and a natural number $k$ with $\gcd (k,\mathrm{\mid }H\mathrm{\mid })=1$ are given, the power map $y\mapsto y^k$ is a bijection $H\to H$: by Bézout there are $k^{\mathrm{\prime }},t$ with $k{k}^{\mathrm{\prime }}=1+t\mathrm{\mid }H\mathrm{\mid }$, and $y^{\mathrm{\mid }H\mathrm{\mid }}=e$, so $(y^{k^{\mathrm{\prime }}}{)}^k=y^{1+t\mathrm{\mid }H\mathrm{\mid }}=y$ (The extended Euclidean algorithm: the same descent produces integers $x,y$ with $ax+by=\gcd (a,b)$, so Bézout coefficients are computed and not merely shown to exist, The order of every element of a finite group divides the order of the group, If $\mathrm{ord}(g)=n$ then $g^k=e$ iff $k$ is an integer multiple of $n$, the powers $g^0,\dots ,g^{n-1}$ are distinct, and $⟨g⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^j=g^k$ only for $j=k$, Exponent laws in a group: $g^{m+n}=g^m{g}^n$ and $(g^m{)}^n=g^{mn}$ for all $m,n\in \mathbb{Z}$, and $(gh{)}^n=g^n{h}^n$ when $g$ and $h$ commute, Powers $g^n$: natural exponents in a monoid and integer exponents in a group, with $g^0=e$).

If a finite group has a Sylow $p$-subgroup $Q$ and an epimorphism $\theta :G\to Q$, then it has a normal $p$-complement (Equivalent forms of having a normal p complement).

Conjugation is a homomorphism: $(z^u{)}^v=z^{vu}$, where $z^u:=uz{u}^{-1}$ (Conjugation $x\mapsto gx{g}^{-1}$ is an automorphism, A group homomorphism automatically satisfies $f(e)=e^{\mathrm{\prime }}$ and $f(g^{-1})=f(g{)}^{-1}$, and $f(g^n)=f(g{)}^n$ for every $n\in \mathbb{Z}$; for monoid homomorphisms preservation of the identity must be assumed, Left group actions, transitive actions, and faithful actions).