Happy Peaceful New Year !

In this question, I recalled that if $H$ is a proper subgroup of a *finite* group $G$, such that
$$({\bf A1})\qquad(g\not\in H)\Longrightarrow(g^{-1}Hg\cap H=(1)),$$
then
$$N:=(1)\bigcup\left(G\setminus\bigcup_g g^{-1}Hg\right)$$
is a normal subgroup. The normality is obvious, but the fact that $N$ is a group is proved by character theory (one doesn't know a direct proof). The terminology is that $G$ is a Frobenius group, $H$ is the Frobenius complement of $G$ and $N$ is the Frobenius kernel.

What happens when we allow $G$ to be infinite ? I suppose that the status of this question is well-known. I suspect that there be some counter-example, a pair $(G,H)$ for which $N$ is

nota subgroup. Perhaps the Frobenius Theorem above adapts with some extra assumption; an additional hypothesis could have a topological flavour, in the spirit of functional analysis, where we use topology in order to extend linear algebra in the infinite dimensional context.

**Edit**. After the negative answers, essentially based on the use of free groups or free product, I came to the following strengthen context. Let $K$ be the subgroup generated by the union of the conjugate subgroups $g^{-1}Hg$ (notice that $K$ is a normal subgroup). Then $G$ is the union of $K$ and $N$, where $K\ne(1)$ and $N\ne G$. If the same conclusion as in Frobenius Theorem holds true ($N$ a subgroup), we have easily that $K=G$. This leads me to add the following assumption:

Assume in addition that $$({\bf A2})\qquad\bigcup_g g^{-1}Hg\quad\hbox{generates}\quad G.$$

Then can we say that $N$ is a subgroup ?

Let me discuss a little the case where $H=\langle a\rangle$ is a cyclic subgroup of $\mathbb{L}_2$, the free group with generators $a,b$. Then $K$ is a proper subgroup, the kernel of the morphism $\phi:\mathbb{L}_2\rightarrow\mathbb{Z}$ defined by $\phi(a)=0$ and $\phi(b)=1$. This explains why $N$ cannot be a subgroup of $\mathbb{L}_2$. In a general configuration, suppose that $H$ is a malnormal subgroup of $G_0$, but that $K$ is proper. We might replace $G$ by $G_1=K$, and $H$ is still malnormal. However, the new $K$, which I denote $K(H,G_1)$ is smaller than $K$, because there are less conjugate subgroups $g^{-1}Hg$ (the constraint is now $g\in G_1$). Whence the necessity to define $G_2=K(H,G_1)$. This is the beginning of an induction. This induction is not necessatily finite or denumerable, it can be transfinite, but its length is bounded by the cardinal of $G/H$. Eventually, we reach a subgroup $G^\dagger$ in which $H$ is malnormal and satisfies (A2). It is unclear to me whether $G^\dagger$ is bigger than $H$ or not.

What is $G^\dagger$ in the case described above, where $G=\mathbb{L}_2$ and $H=\langle a\rangle$ ? In particular, can $G^\dagger$ be equal to $H$ ?