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 not a 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 $H$ ?

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 not a 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 $H$ ?

Happy Peaceful New Year !

In this question, I recalled that if $H$ is a subgroup of a finite group $G$, such that $$(g\not\in H)\Longrightarrow(g^{-1}Hg\cap H=(1)),$$ 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 not a 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.

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 not a 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 $H$ ?

Happy Peaceful New Year !

In this question, I recalled that if $H$ is a subgroup of a finite group $G$, such that $$(g\not\in H)\Longrightarrow(g^{-1}Hg\cap H=(1)),$$ 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).

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 not a 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.

Happy Peaceful New Year !

In this question, I recalled that if $H$ is a subgroup of a finite group $G$, such that $$(g\not\in H)\Longrightarrow(g^{-1}Hg\cap H=(1)),$$ 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 not a 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.