= 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, equivalently
$$
N=\{1\}\cup\left(G\setminus\bigcup_{g\in G}H^g\right).
$$
The <Frobenius kernel theorem> asserts that this set is a <normal subgroup>, acts regularly on the points, and gives $G=N\rtimes H$. 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.
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