Fusion control forces trivial Sylow intersection with the p residual

Given: A finite group $G$, a prime $p$, a Sylow $p$-subgroup $P\le G$ controlling fusion in $P$ with respect to $G$, and $K:=O^p(G)$, $Q:=P\cap K\mathrm{\ne }\{1\}$.

$K⊴G$, $G\mathrm{/}K$ is a finite $p$-group, $KP=G$, and $Q=P\cap K$ is a Sylow $p$-subgroup of $K$; in particular $K$ is finite and $[K:Q]=\mathrm{\mid }K\mathrm{\mid }\mathrm{/}\mathrm{\mid }Q\mathrm{\mid }$ is prime to $p$, since $\mathrm{\mid }Q\mathrm{\mid }$ is the exact power of $p$ dividing $\mathrm{\mid }K\mathrm{\mid }$ (P residual of a finite group, Sylow subgroups of a normal subgroup are intersections with Sylow subgroups, Sylow $p$-subgroups of a finite group, If $[G:N]$ is finite then $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]$; for finite $G$ this equals $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }N\mathrm{\mid }$, Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).

$Q⊴P$: for $u\in P$ one has $uQ{u}^{-1}=u(P\cap K)u^{-1}=uP{u}^{-1}\cap uK{u}^{-1}=P\cap K=Q$, because $uP{u}^{-1}=P$ and $K⊴G$ (Normal subgroup: invariance under conjugation, Subgroup, Conjugation $x\mapsto gx{g}^{-1}$ is an automorphism).

Commutators: $[a,b]=ab{a}^{-1}{b}^{-1}$, $[A,B]=⟨[a,b]:a\in A,b\in B⟩$, and $[P,Q]=[Q,P]$ as subgroups because $[p,q]=[q,p{]}^{-1}$; if $A\le B$ and $D\le C$ with $A,B\le C$ then $[A,D]\le [B,D]$ (Subgroup commutators and the lower central series, Commutators $[g,h]=gh{g}^{-1}{h}^{-1}$ and the commutator subgroup $[G,G]$, The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups, Subgroup, In a group $e^{-1}=e$, $(g^{-1}{)}^{-1}=g$ and $(gh{)}^{-1}=h^{-1}{g}^{-1}$, the order of the last product being essential).

Both $P$ and $Q$ are finite $p$-groups, $Q\mathrm{\ne }\{1\}$ and $Q⊴P$; so by Normal p subgroup has proper commutator in a p group the commutator $[Q,P]$ satisfies $[Q,P]<Q$ and $[Q,P]⊴Q$ (A finite $p$-group has order $p^n$ for a prime $p$ and some $n\in \mathbb{N}$, Every subgroup of a finite $p$-group has order a power of $p$).

$[Q,Q]\le [Q,P]$ because $Q\le P$ by [F3], so the quotient $A:=Q\mathrm{/}[Q,P]$ is a finite abelian group by $G\mathrm{/}N$ is abelian if and only if $[G,G]\subseteq N$ and The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$; it is nontrivial by [F4] and a $p$-group because $\mathrm{\mid }A\mathrm{\mid }$ divides $\mathrm{\mid }Q\mathrm{\mid }$ (If $[G:N]$ is finite then $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]$; for finite $G$ this equals $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }N\mathrm{\mid }$, Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$, A finite $p$-group has order $p^n$ for a prime $p$ and some $n\in \mathbb{N}$).

The quotient map $\phi \text{\hspace{0.17em}⁣}:Q\to A$, $\phi (q)=q[Q,P]$, is a surjective group homomorphism (The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$, Monoid homomorphism and group homomorphism, 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); $\ker \phi =[Q,P]$, so $\phi (q)\mathrm{\ne }1$ exactly when $q\mathrm{\notin }[Q,P]$.

For $q\in Q$ and $u\in P$ one has $q^u{q}^{-1}=[u,q]\in [Q,P]$ and therefore $\phi (q^u)=\phi (q)$ (The conjugacy class ${\mathrm{Cl}}_G(x)$ and centralizer $C_G(x)$ of an element, Commutators $[g,h]=gh{g}^{-1}{h}^{-1}$ and the commutator subgroup $[G,G]$, [F3], [F6]).

The transfer of the homomorphism $\phi \text{\hspace{0.17em}⁣}:Q\to A$ is a group homomorphism $V:=V_{\phi }\text{\hspace{0.17em}⁣}:K\to A$, independent of the transversal (Transfer homomorphism for a finite index subgroup, Transfer is a homomorphism, Transfer is independent of the transversal), and it is computed by the cycle decomposition of Transfer cycle decomposition formula: for $x\in K$ the right cosets $Q\mathrm{\backslash }K$ split into orbits $C_1,\dots ,C_r$ of right multiplication by $⟨x⟩$ of lengths $n_i$, with representatives $t_i$, and $V(x)={\prod }_{i=1}^r\phi \left(t_i{x}^{n_i}{t}_i^{-1}\right)$, where $t_i{x}^{n_i}{t}_i^{-1}\in Q$. The orbits partition the finite set $Q\mathrm{\backslash }K$, which has $[K:Q]$ elements, so $n_1+\cdots +n_r=[K:Q]$ (The coset set $G\mathrm{/}H$ and the index $[G:H]$ of a subgroup, Left and right cosets $gH$ and $Hg$ of a subgroup, Left group actions, transitive actions, and faithful actions, 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, The orbit $G\cdot x$ and stabilizer $G_x$ of a point in a group action).

Order and coprimality: if $a\in A$ and $a^m=1$, then $\mathrm{ord}(a)$ divides $m$ and $\mathrm{ord}(a)$ divides $\mathrm{\mid }A\mathrm{\mid }$; and $\gcd (p^j,[K:Q])=1$ for every $j\ge 0$ because $p\nmid [K:Q]$, so an element of $A$ whose order divides both $\mathrm{\mid }A\mathrm{\mid }$ and $[K:Q]$ is trivial (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 $j=k$, The order of every element of a finite group divides the order of the group, Divisibility is reflexive and transitive on $\mathbb{Z}$, and is linear: if $d\mid a$ and $d\mid b$ then $d\mid ax+by$ for all integers $x,y$; also $d\mid a$ implies $d\mid ac$, $-d\mid a$ and $d\mid -a$, A finite $p$-group has order $p^n$ for a prime $p$ and some $n\in \mathbb{N}$, [F1]).

Every homomorphism from $K$ to a finite $p$-group has $O^p(K)$ in its kernel, and $O^p(K)=K$ when $K=O^p(G)$ (P residual of a finite group, P residual is generated by p prime elements and idempotent).