 Solution

A Frobenius group is a finite transitive group action on more than one point in which a nonidentity element fixes at most one point and a point stabilizer is nontrivial. Equivalently it has a nontrivial proper subgroup $H$ with $H\cap H^g=1$ for every $g\notin H$. The Frobenius complement is such a point stabilizer $H$.

The Frobenius kernel is the identity together with all fixed-point-free elements in this action, equivalentlyThe Frobenius kernel theorem asserts that this set is a normal subgroup, acts regularly on the points, and gives $G=N⋊H$. In particular $C_N(h)=1$ for $1\ne h\in H$.

Brauer's permutation lemma states that an automorphism of a finite group fixes equally many conjugacy classes and irreducible characters. More generally a group of automorphisms has equal permutation characters on those two sets. Indeed, the character table is invertible and intertwines the two permutation matrices: simultaneous permutation of rows and columns preserves character evaluation. Similar matrices have equal traces, and a permutation-matrix trace counts its fixed points. The fixed objects on the group side are classes, not individual elements.