Free group actions for singleton group or for empty set

Question

My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

I refer to Section 27.1.

Let $M$ be a set, possibly empty. Let $G$ be a group, possibly a singleton. Let $G$ act right on $M$ by the action $\mu: M \times G \to M$. For each $x \in M$, let $\text{Stab}(x):=\{g \in G | \mu(x,g) = x\}$ denote a stabilizer subgroup of $G$. Let $1_G$ be the identity of $G$.

I understand Wikipedia's second definition of $\mu$ to be free as follows:

$\mu$ is free if for all $g \in G$, if there exists $x \in M$ such that $\mu(x,g)=x$, then we have that $g=1_G$.

Later Wikipedia says

if $\mu$ is free and $M$ is a non-empty set, then $\mu$ is faithful. $\tag{A}$

Question: Given the idea that we can have $M$ as an empty set, I would like to clarify, similar to what I did here: Are these correct?

1. If $G$ is a singleton, then every action $\mu$ is free, whether or not $M$ is empty.

2. If $M$ is empty, then every action $\mu$ is free, whether or not $G$ is a singleton.

3. By (1), if $M$ is empty, $G$ is a singleton and $\mu$ is free, then $\mu$ is faithful

4. The reason for the assumption for $M$ non-empty in $(A)$ is that if $M$ is empty and $G$ is not a singleton, then by this every action $\mu$ is free while every action $\mu$ is not faithful.

5. If there exists a free action $\sigma: M \times G \to M$ and if every free action $\mu$ is faithful, then either $M$ is non-empty or $G$ is a singleton. (The existence of $\sigma$ is to avoid cases where no free actions exist, if that's even possible)

---

Update: Based on this answer, I think (2) is correct.

Answer 1

You are correct. Regarding your fifth point, there do exist pairs $(G, M)$ such that no action of $G$ on $M$ is free. Consider $G$ a finite cyclic group of prime order $p$ and $M$ a finite set. If $G$ acts freely, then by the orbit-stabilizer theorem every orbit has cardinality $p$. Since $M$ is the union of its orbits, if $\# M$ is not divisible by $p$ then no action of $G$ on $M$ can be free.