I am trying to solve the following exercise (this is the exercise 248 from John Rose's A Course on Group Theory):

Let $G$ be a finite group. We make the following definitions:

(a) Suppose that $G$ acts on the set $X$. The action is said to be a Frobenius action if it is transitive and there exists at least an element $x\in X$ with a nontrivial stabiliser, $|X|>1$, and whenever $x_1, x_2$ are distinct elements of $X$, $\operatorname{Stab}_G(x_1)\cap \operatorname{Stab}_G(x_2)=1$.

(b) $G$ is said to be a Frobenius group if it has a nontrivial proper subgroup $H$ such that $N_G(H)=H$ and whenever $H^{g_1}, H^{g_2}$ are distinct conjugates of $H$ in $G$ (with $g_1, g_2\in G$), $H^{g_1}\cap H^{g_2}=1$. Any such subgroup is called a Frobenius complement in $G$.

Prove the following statements:

(i) $G$ has a Frobenius action on some set if and only if $G$ is a Frobenius group.

(ii) If $G$ is a Frobenius group and $H$ is a Frobenius complement in $G$ then $|G:H|\equiv 1 \mod |H|$.

(iii) If $G$ is a Frobenius group then $Z(G)=1$.

(iv) Let $n$ be a positive integer, and consider the natural action of $S_n$ on the set $\{1,...,n\}$. This is a Frobenius action if and only if $n=3$.

(v) Let $n$ be an integer, $n\geq 3$. Then $D_{2n}$ is a Frobenius group if and only if $n$ is odd.

For the first question, I think I need to use the following fact: let $X$ be a transitive $G$-space, $Y$ a transitive $H$-space. If $(G,X)\cong (H,Y)$ then for any $x\in X, y\in Y$, $\operatorname{Stab}_G(x)\cong \operatorname{Stab}_H(y)$.

My attempts at solution:

(i) $\Rightarrow$ Suppose that $G$ has a Frobenius action on the set $X$. Let $x\in X$ be an element with a nontrivial stabiliser, this is my candidate for a Frobenius complement. Denote $H=\operatorname{Stab}_G(x)$, now consider two distinct conjugates $H^{g_1}, H^{g_2}$. Note that $H^{g_1}=\operatorname{Stab}_G(g_1\cdot x)$, this is due to the fact that every element in $H^{g_1}$ has the form $g_1 h g^{-1}$ for $h\in H$ which stabilises $x$, now consider the action $$ g_1 h g^{-1}_1\cdot (g_1 \cdot x)=g_1 h\cdot (g_{1}^{-1}g\cdot x)=g_1 \cdot (h\cdot x)=g_1 \cdot x $$ Similarly, $H^{g_2}=\operatorname{Stab}_G(g_2\cdot x)$. From there I think I need to show that $g_1\cdot x\neq g_2\cdot x$, so that the stabilisers intersect trivially and therefore $H^{g_1}\cap H^{g_2}=1$.

For the normaliser, we always have that $H\subseteq N_G(H)$, if $n\in N_G(H)$ then $nhn^{-1}\in H$ for every $h\in H$. Now, $nhn^{-1}\in \operatorname{Stab}_G(n\cdot x)$. We either get that $\operatorname{Stab}_G(x)=\operatorname{Stab}_G(n\cdot x)$ or $\operatorname{Stab}_G(x)\cap \operatorname{Stab}_G(n\cdot x)=1$, I think that the second case is not possible, so we only get the first case. Not sure where to from here.

$\Leftarrow$ Now suppose that $G$ is a Frobenius group with a Frobenius complement $H$. Consider the action of $G$ on the set $G/H$, it is transitive and $H$ is an element with nontrivial stabiliser. I am not sure how to show the part about stabilisers. From then, I want to show that every Frobenius action is equivalent to the action defined above, so it is enough to consider only this action.

(ii) I am not sure what to do here, maybe count the orbits under the action?

(iii) I think the previous exercise needs to be used here, but I am not sure how exactly.

(iv) In case $n=4$, we get that $\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 3 & 4 \end{pmatrix}\in \operatorname{Stab}_{S_4}(3)\cap\operatorname{Stab}_{S_4}(4)$, this counterexample works for $n\geq 4$ essentially. If $n=3$, then this can be done by computing stabilisers for all three elements.

(v) I know that $Z(D_{2n})=1$ only for $n$ odd, but I am not so sure how to go about the other direction. If $n$ is odd, how do I show that $D_{2n}$ is Frobenius?


You have asked a few too many questions at once. For now, I will just help you finish the proof of (i), since you have nearly done that anyway.

For the $\Rightarrow$ part, if $g_1\cdot x = g_2 \cdot x$ then $H_1^{g_1} = {\rm Stab}_G(g_1 \cdot x) = {\rm Stab}_G(g_2 \cdot x) = H_2^{g_2}$, but you assumed that they were distinct conjugates, so that's not possible.

If $n \in N_G(H)$ then $H^n=H$ so ${\rm Stab}_G(n \cdot x) = {\rm Stab}_G(x)$, and hence $n \cdot x = x$. That is, $n \in H$.

For $\Leftarrow$, the elements of $X$ are the left cosets $gH$, so let $g_1H$, $g_2H$ be two distinct cosets. Then ${\rm stab}_G(g_iH) = H^{g_i}$. If $H^{g_1} = H^{g_2}$ then $g_1^{-1}g_2 \in N_G(H) = H$, so $g_1H=g_2H$ contrary to assumption. Hence $H^{g_1} \ne H^{g_2}$, so they intersect trivially, and hence so do the corresponding stabilizers of $g_1H$ and $g_2H$.

For (ii), let $H = {\rm stab}_G(x_1)$ in the Frobenius action. So by (i) $H$ is a Frobenius complement. Since ${\rm stab}_G(x_1) \cap {\rm stab}_G(x_2) = 1$ for all $x_1 \ne x_2$, the orbits of $H$ on $X \setminus \{x_1\}$ all have length $|H|$ and so $|X| = |G:H| \equiv 1 \bmod |H|$.

  • $\begingroup$ How can I assume that the elements of $X$ are the left cosets $gH$? $\endgroup$ – Jimmy R Jan 21 '14 at 22:35
  • $\begingroup$ Because that is exactly what they are! You wrote "consider the action of $G$ on the set $G/H$", which (since you are using left actions) is the set of left cosets of $H$ in $G$. $\endgroup$ – Derek Holt Jan 21 '14 at 22:47
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    $\begingroup$ But you are assuming this in $\Rightarrow$ part, where we start with Frobenius action of a group $G$ on some set $X$, or am I not understanding your proof? $\endgroup$ – Jimmy R Jan 21 '14 at 22:57
  • $\begingroup$ Oh, I am sorry!!! I had my left and right arrows the wrong way round. I hope it is more comprehensible now. I was just filling in the gaps in your proof. I will also supply a hint for (ii). $\endgroup$ – Derek Holt Jan 22 '14 at 8:52

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