Group action where all stabilizers are equal

Question

Let $\cdot:G\times X \rightarrow X$ be a left group action. We know that:
$(i)$ The action is faithful, if $g\cdot x=x, \forall x\in X$ implies $g=e$;
$(ii)$ The action is free, if $g\cdot x=x$ for some $x\in X$ implies $g=e$;
$(iii)$ Every free action is faithful.

Thinking about the converse of $(iii)$, I realized that a faithful action, in order to be free, needs to have this property $(P)$: all stabilizer subgroups are equal (to the kernel of the action, of course). Thus, we have the equivalence: $\text{free} \Leftrightarrow \text{faithful}+(P)$.

Since I haven't found in literature any info about such actions with property $(P)$, I decided to call them quasi-free.

I am willing to find out any other properties of quasi-free actions. I have posted two answers, sharing my thoughts on this topic.

Answer 1

Definition. The action $\cdot:G\times M \rightarrow M$ is called quasi-free, if $g\cdot x=x$ for some $x\in M$ implies $g\cdot x=x,\forall x\in M$.
The quasi-free set of $\alpha$ is $QF(\alpha):=\{g\in G:g\cdot x=x$ for some $x\in X$ implies $g\cdot x=x, \forall x\in X\}\subseteq G.$
An element $x\in M$ is called a fixed point for $\alpha$, if it is a fixed point for every $\alpha_a$. Equivalently, $\alpha(a,x)=x, \forall a \in G$, or ${\rm Stab}_x=G$. Let us denote the set of fixed points of $\alpha$ by $\mathcal{F}(\alpha)\subseteq M$.

Let $\alpha:G\times M \rightarrow M$ be an action, and $\mathcal{D}(M)$ be the set of derrangements of $M$, i.e. the bijections with no fixed points. Then:
$(i)$ $QF(\alpha)=\{a\in G:\alpha_a=id_M \lor \alpha_a \in \mathcal{D}(M)\}\supseteq\ker(\alpha)$;
$(ii)$ $QF(\alpha)$ is closed under taking conjugations and inverses;
$(iii)$ $QF(\alpha)$ is a subgroup of $G$ if and only if it is closed under group multiplication. In this case, $QF(\alpha)$ is even a normal subgroup of $G$;
$(iv)$ If $\mathcal{F}(\alpha)$ is nonempty, $QF(\alpha)=\ker(\alpha)$, a subgroup of $G$.

Let $G$ be a group, and $M$ a nonempty set. Other standard group actions are:
$(i)$ $\alpha:G\times G \rightarrow G, \alpha(a,x):=ax, \forall a,x\in G$, is called the canonical/regular action of $G$ on itself. This action is always free.
$(ii)$ $\alpha:G\times G \rightarrow G, \alpha(a,x):=axa^{-1}, \forall a,x\in G$, is called the adjoint action of $G$ on itself. This means that the group acts on itself by conjugation. We have that $\mathcal{F}(\alpha)=Z(G)$, therefore $QF(\alpha)=\text{Ker}(\alpha)$. This action is faithful if and only if $Z(G)=\{e\}$, i.e. the center of $G$ is trivial. This action if quasi-free if and only if $G$ is abelian (in this case, the action degenerates to the action given by $(iv)$). Thus, the adjoint action is free if and only if $G=\{e\}$.
$(iii)$ $\alpha:{\rm Sim}(M)\times M \rightarrow M, \alpha(f,x):=f(x), \forall f\in{\rm Sim}(M),\forall x\in M$, is called the canonical action on $M$. It is always faithful, but free only when $|M|<3$.
$(iv)$ $\alpha:G\times M \rightarrow M, \alpha(a,x):=x, \forall a\in G,\forall x\in M$, is called the trivial action of $G$ on $M$. An action is trivial if and only if $\mathcal{F}(\alpha)=M$. This action is always quasi-free, and it is faithful if and only if $G=\{e\}$.
$(v)$ Let $\alpha:G\times M \rightarrow M$ be an action, and $H\subseteq {\rm Ker}(\alpha)$ a normal subgroup of $G$. Then $\beta:G/H\times M \rightarrow M, \beta(\widehat{a},x):=\alpha(a,x), \forall a\in G,\forall x\in M$, is well defined, and ${\rm Ker}(\beta)={\rm Ker}(\alpha)/H$. In the special case $H={\rm Ker}(\alpha)$, $\beta$ is the effective action induced by $\alpha$, therefore, $\beta$ is free if and only if $\beta$ is quasi-free. This is another motivation for introducing our new concept.\

$(vi)$ Let $G$ be a group, and $H$ a subgroup of $G$. There is an action of $G$ on the left $H$-cosets of $G$, denoted $G/H:=\{aH:a\in G\}$, as follows: $a\cdot bH:=abH,\forall a,b\in G$. It is named the left $H$-coset action of $G$. The stabilizer of a given element is ${\rm Stab}_{aH}=\{b\in G:b\cdot aH=aH\}=\{b\in G:H=a^{-1}baH\}=\{b\in G:b\in aHa^{-1}\}=aHa^{-1}$, the conjugate subgroup of $H$ by $a$. Thus, the kernel is the intersection of all those conjugates, i.e. the normal core of $H$ in $G$.

This action is always transitive, and it is free if and only if $H$ is trivial. The action is faithful if and only if the normal core of $H$ in $G$ is trivial. The action is quasi-free if and only if all subgroup conjugates of $H$ are equal, that is, $H$ is a normal subgroup of $G$, or $H$ equals its normal core. In this case, all stabilizers are equal to $H$.

Answer 2

$(i)$ There exists actions which are neither quasi-free nor faithful. For example, we can take the adjoint action for the quaternion group $G:=Q_8$.
$(ii)$ There exists actions which are quasi-free, but not faithful. For example, we can take the adjoint action for the cyclic group of order $n$ $G:=Z_n, n\geq 2$, or (more generally) the trivial action for any nontrivial group.
$(iii)$ There exists actions which are not quasi-free, but are faithful. For example, we can take the adjoint action for the permutation group of order $n$ $G:=S_n, n\geq 3$.

It is natural to ask whether $QF(\alpha)$ is always a subgroup of $G$ or not. This has to deal with the fixed points of a composition of two functions. The intuition tells us that we can have two functions, each without fixed points, but their composition can have fixed points. This means that, to find counter-examples, we need to look after a "big" group (of functions, eventually). The following theorem clarifies the situation. We shall need the following two lemmas:

Lemma $1$. (see here) Let $M$ be a finite set, with $n\geq 1$ elements. Then we have the following formula for the number of derrangements of $M$: $$|\mathcal{D}(M)|:\stackrel{not}{=}!n=n!\sum_{k=0}^n \frac{(-1)^k}{k!}.$$

Lemma $2$. (see here) Let $M$ be a set with at least two elements. Then there exists a permutation of $M$ with no fixed points. Equivalently, $\mathcal{D}(M)\neq \phi$.

Theorem. Let $\alpha:{\rm Sim}(M)\times M \rightarrow M$ be the canonical action on $M$. Then: $QF(\alpha)\leq {\rm Sim}(M)$ if and only if $|M|<4$.
Proof. Since $\alpha_f(x)=\alpha(f,x)=f(x), \forall x\in M$, we have $\alpha_f=f, \forall f\in {\rm Sim}(M)$. Thus, $QF(\alpha)=\{f\in {\rm Sim}(M):f=id_M \lor f\in \mathcal{D}(M)\}=\mathcal{D}(M)\cup\{id_M\}$, an explicit description.

For $|M|<3$, we saw that $\alpha$ is free, thus, quasi-free; so $QF(\alpha)={\rm Sim}(M)$. For $|M|=3$, we can compute $QF(\alpha)=\{id_M,\sigma,\sigma^2\}$, where $\sigma$ is a circular permutation. It is indeed a (proper) subgroup of ${\rm Sim}(M)$.

For $|M|=n\geq 4$, we can compute: $|QF(\alpha)|=1+|\mathcal{D}(M)|=1+!n$. If we really want $QF(\alpha)\leq{\rm Sim}(M)$, from Lagrange's Theorem, we must have $1+!n|n!$. Using Lemma $1$, $!4=9, !5=45$, so for $n\in \{4,5\}$, this is not the case. In fact, for $n=4$, the problem is that $QF(\alpha)$ is not closed under the composition of functions. Let $M=\{x_1,x_2,x_3,x_4\}$, then we can give this example (will be useful in the last part of the proof): $$ f:=\left( \begin{array}{cccc} x_1 & x_2 & x_3 & x_4 \\ x_4 & x_1 & x_2 & x_3 \end{array} \right); g:=\left( \begin{array}{cccc} x_1 & x_2 & x_3 & x_4 \\ x_3 & x_4 & x_2 & x_1 \end{array} \right); f\circ g:=\left( \begin{array}{cccc} x_1 & x_2 & x_3 & x_4 \\ x_2 & x_3 & x_1 & x_4 \end{array} \right),\\ $$ where $f,g\in QF(\alpha)$, but $f\circ g\notin QF(\alpha)$.

Now, we are left to discuss the case $|M|\geq 6$. We can write $M=\{x_1,x_2,x_3,x_4\}\cup \overline{M}$, where $|\overline{M}|\geq 2$. From Lemma $2$, there is a bijection $h\in \mathcal{D}(\overline{M})$. We define $F,G:M\rightarrow M$, based on the above $f,g,h$ as follows: $$ F(x):=\begin{cases} f(x),x=x_i, i\in \overline{1,4}\\ h(x), x\in \overline{M}\end{cases}; G(x):=\begin{cases} g(x),x=x_i, i\in \overline{1,4}\\ h(x), x\in \overline{M}\end{cases}.\\ $$ Clearly, $F,G\in QF(\alpha)$, but $F\circ G\notin QF(\alpha)$, because $(F\circ G)(x_4)=x_4$. This shows that $QF(\alpha)\nleq {\rm Sim}(M)$ , which completes the whole proof. \end{proof}

In fact, the divisibility that we encountered: $1+!n|n!$ can be further studied, and we obtain a second proof for the case $|M|=n\geq 6$ which makes no use of Lemma $2$:

Lemma $3$. There is no $n\in\mathbb{N},n\geq 6$ such that $1+!n|n!$.
Proof. Let us denote $a_n:=\frac{n!}{1+!n},\forall n\geq 6$. From analysis, one can deduce that $\lim_{n\to\infty}a_n=e$. Motivated by this, our idea is to show that $2.5<a_n<3,\forall n\geq 6$, hence the conclusion. These inequalities are equivalent to: $$b_n:=\frac{1}{n!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{(-1)^n}{n!}\in\left(\frac{1}{3},\frac{2}{5}\right),\forall n\geq 6.$$ It can be easily seen that $b_5=3/8<2/5$, and $\lim_{n\to\infty}b_n=e^{-1}>1/3$. Moreover, $$b_n-b_{n+1}=\frac{n+(-1)^n}{(n+1)!}>0,\forall n\geq 5,$$ which shows that $(b_n)_{n\geq 5}$ is a strictly decreasing sequence, and we are done.