# Free group actions for singleton group or for empty set [duplicate]

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.

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.