In Tu's book, Differential Geometry: Connections, Curvature, and Characteristic Classes, he defines a principal G-bundle as (p. 244):
A smooth fiber bundle $\pi:P\rightarrow M$ with fiber $G$ (G is a Lie group) is a smooth principal $G$-bundle if $G$ acts smoothly and freely on $P$ on the right and the fiber-preserving local trivializations $\phi_U:\pi^{-1}(U)\rightarrow U\times G$ are $G$-equivariant, where $G$ acts on $U\times G$ on the right by $(x,h)\cdot g = (x,hg)$.
My question is this: does the fact that the action of $G$ on $P$ is free follow from the other axioms in the definition? It seems like it does by the following argument:
Assume $P$ satisfies all the properties above, except we don't assume the action is free. Suppose that $p\cdot h=p$ for some $p\in P$ and $h\in G$. Fixing a trivialization $\phi_U$, we can write $\phi_U(p)=(x,g)$ for some $x\in M$ and $g\in G$. Then $$(x,g)=\phi_U(p)=\phi_U(p\cdot h)=\phi_U(p)\cdot h =(x,g)\cdot h=(x,gh).$$ Thus, $h$ is the group identity, $h=e$. Doesn't this prove the action of $G$ on $P$ is automatically free?