Frobenius kernel closure is the content of the theorem

Let $G$ be a finite Frobenius group with complement $H$ and let

$N=(G\setminus {\bigcup }_{x\in G}xH{x}^{-1})\cup \{1\}$

be the associated candidate kernel set (Frobenius kernel set, Frobenius complement and frobenius group). Then $N$ is defined by a membership condition on individual elements: $g\in N$ means that $g=1$ or that $g$ lies in no conjugate of $H$. Nothing in the definition asserts that $N$ is closed under products, and the counting statement $\mathrm{\mid }N\mathrm{\mid }=[G:H]$, $N\cap H=\{1\}$ of Frobenius kernel cardinality likewise says nothing about products of elements of $N$.

The remark. The passage from this candidate set to a subgroup is not formal; it is the content of the Frobenius kernel theorem (Frobenius kernel theorem), which proves that $N$ is a normal subgroup of $G$ by character-theoretic means. In particular, the theorem is not a consequence of the cardinality computation, and a proof of the kernel theorem must somewhere use more than the definition of $N$ and the orbit-counting identities.

The reason no product closure is available for free is that $N$ is described by a negative condition on conjugation, while a product $g{g}^{\mathrm{\prime }}$ of two elements avoiding every conjugate of $H$ need not avoid them; the subgroups $H^x=xH{x}^{-1}$ are numerous and the description of $N$ quantifies over all of them. This is visible already in the smallest case $G=S_3$ with $H=⟨(01)⟩$, where the kernel is $A_3$ ($S_3$ as a Frobenius group): the two nonidentity elements of $A_3$ are $3$-cycles, and the product of either one with itself is the other, while their mutual product is the identity, so closure holds there — but this is a computation in one group, not a general structural reason.

Conversely, the reverse implication is immediate: if a normal subgroup $K$ with $K\le N$, $G=KH$ and $K\cap H=\{1\}$ is exhibited by other means, then $K$ is the Frobenius kernel by the uniqueness statement of Frobenius semidirect product decomposition. The character-theoretic theorem is exactly the tool that produces such a $K$ from the Frobenius condition alone.