Exercise 14.1.4

Proof.

$\text{(⇒)}$ Assume that $G$ is a Frobenius group. Then $G$ acts transitively on a set $X$ such that $1<|X|<|G|$, and for every $(x,y)\in X\times X$ such that $x\ne y$, the identity is the only element of $G$ fixing $x$ and $y$.

First we show that every isotropy group $G_x$ is non trivial, i.e. $G_x\ne \{e\}$ and $G_x\ne G$, for all $x\in G$.

Since $G$ acts transitively on $X$, $X=G\cdot x$ is the orbit of $x$, thus

and since $1<|X|<|G|$, this proves $1<|G_x|<|G|$, so $G_x\ne \{e\},G_x\ne G$. Fix $x_0\in G,x_0\ne e$, and take $H=G_{x_0}$ the isotropy group of this chosen element $x_0$. Then $1<|H|<G$.

Assume that $g\in G,g\notin H$, and $h\in H\cap \mathit{gH}{g}^{-1}$. Then $h$ and $g^{-1}\mathit{hg}$ are both in $H=G_{x_0}$, so that $h\cdot x_0=x_0$, and $(g^{-1}\mathit{hg})\cdot x_0=x_0$, that is

Since $g\notin H=G_{x_0}$, $x_0\ne g\cdot x_0$, thus $h$ fixes two distinct elements of $X$, and this shows that $h=e$. We have proved $H\cap \mathit{gH}{g}^{-1}=\{e\}$ for all $g\notin H$.

$\text{(⇐)}$ Conversely, assume that $G$ has a subgroup $H$ such that $1<|H|<|G|$ and $H\cap \mathit{gH}{g}^{-1}=\{e\}$ for all $g\notin H$.

Take $X$ as the set of left cosets $\mathit{hH},h\in G$ relative to $H$, and consider the action of $G$ on $X$ defined for all $h\in G$ by

$\text{∙}$

This action is transitive: if $\mathit{kH}$ and $\mathit{lH}$ are left cosets, then ${(\mathit{lk})}^{-1}\cdot \mathit{kH}=\mathit{lH}$.

$\text{∙}$

Since $1<|H|<|G|$, then $1<|G|∕|H|<|G|$, thus $1<|X|<|G|$.

$\text{∙}$

Assume that $g$ fixes two distinct left cosets $\mathit{hH}\ne \mathit{kH}$: $$\begin{array}{llll}\hfill g\cdot \mathit{hH}& =\mathit{hH},& \hfill & \\ \hfill g\cdot \mathit{kH}& =\mathit{kH}.& \hfill & \end{array}$$

Then $l=h^{-1}\mathit{gh}\in H,m=k^{-1}\mathit{gk}\in H$, therefore $m=k^{-1}\mathit{gk}=k^{-1}\mathit{hl}{h}^{-1}k\in H$, so that

$$l\in H,{(h^{-1}k)}^{-1}l(h^{-1}k)\in H.$$

This proves $l\in H\cap \mathit{gH}{g}^{-1}$, where $g=h^{-1}k\notin H$ (since $\mathit{hH}\ne \mathit{kH}$), and the hypothesis $H\cap \mathit{gH}{g}^{-1}=\{e\}$ gives $l=e$, and $g=\mathit{hl}{h}^{-1}=e$. The identity is the only element of $G$ fixing $\mathit{hH}$ and $\mathit{kH}$.

Therefore $G$ is a Frobenius group. □