What does it mean to be a free $G$-module? This question is very close to one I asked yesterday, but I will reformulate it better, and incorporate the answer I already got.
Define a $G$-module to be an abelian group with an action by some group $G$, that is compatible with its group structure. This notion is equivalent to that of a module over the group ring $\mathbb{Z}[G]$, because since every abelian group is a $\mathbb{Z}$-module, you can $\mathbb{Z}$-linearly extend the $G$-action to get a $\mathbb{Z}[G]$ module structure.
Here is my question: say $A$ is a $G$-module, and that the action of $G$ on $A$ is free. What additional conditions do we need to be able to say that $A$ is a free $\mathbb{Z}[G]$ module? (as in, a direct sum of copies of $\mathbb{Z}[G]$).
The reason I need it, is because it seems like a standard result that if a topological space $X$ has a free $G$-action, then its singular complex $C_\bullet(X)$ is free over $\mathbb{Z}[G]$ (see for example remark 2.3 here). Thanks to this answer, I do see why "free $G$ action on $X$" implies "free $G$ action on $C_\bullet(X)$". But I do not see why that should extend to a free $\mathbb{Z}[G]$ module structure.
Because to be able to say that $C_\bullet(X)$ is a free $\mathbb{Z}[G]$ module, we'd need to show that $\mathbb{Z}[G]$ freely permutes the generators of $C_\bullet(X)$. But given $\sigma \in C_\bullet(X)$, what guarantee do we have that for an arbitrary element $a = \sum a_i g_i \in \mathbb{Z}[G]$, we won't have $a\sigma = \sigma$?

*

*Note: it really does seem like "free $G$ action on an abelian group $A$" does not imply $A$ is a free $\mathbb{Z}[G]$ module. For example, $A = G = \mathbb{Z}/2\mathbb{Z}$.

 A: Let us denote for every set $X$ the free abelian group on $X$ by $ℤ[X]$. More explicitly, $ℤ[X]$ is the free abelian group with basis $(e_x)_{x ∈ X}$.
If $X$ is a $G$-set, then the action of $G$ on $X$ induces a $ℤ[G]$-module structure on $ℤ[X]$ via bilinear extension.
Let us make three observations about this construction from $G$-sets to $ℤ[G]$-modules:

*

*We observe that for any family $(X_λ)_{λ ∈ Λ}$ of sets we have an isomorphism of abelian groups
$$
  ℤ\Biggl[ \coprod_{λ ∈ Λ} X_λ \Biggr]
  ≅
  \bigoplus_{λ ∈ Λ} ℤ[X_λ] \,.
$$
More explicitely, an element $(x, μ)$ of the set $\coprod_{λ ∈ Λ} X_λ$ corresponds on the right hand side to the element $(a_λ)_{λ ∈ Λ}$ with $a_μ = x$ and $a_λ = 0$ for every index $λ$ distinct from $μ$.


*If in the above situation each $X_λ$ is a $G$-set, then this isomorphisms of abelian groups is already an isomorphism of $ℤ[G]$-modules.


*For the $G$-set $X = G$ the resulting $ℤ[G]$-module $ℤ[X]$ is precisely the ring $ℤ[G]$ as a left module over itself.
Suppose now that $X$ is a free $G$-set.
This means that
$$
  X ≅ \coprod_{λ ∈ Λ} G
$$
as $G$-sets for some index set $Λ$.
More explicitly, if $B = (b_λ)_{λ ∈ Λ}$ is a basis of $X$ as a $G$-set, then the map
$$
  \coprod_{λ ∈ Λ} G \to X
  \quad
  (g, λ) \mapsto g · b_λ
$$
is an isomorphisms of $G$-sets.
It follows for the resulting $ℤ[G]$-module $ℤ[X]$ that
$$
  ℤ[X]
  ≅
  ℤ\Biggl[ \coprod_{λ ∈ Λ} G \Biggr]
  ≅
  \bigoplus_{λ ∈ Λ} ℤ[G]
  =
  ℤ[G]^{\oplus Λ}
$$
as $ℤ[G]$-modules.
More explicitly, the element $b_λ$ (of the basis $B$ of $X$) on the left hand side corresponds to the standard basis element $e_λ$ on the right hand side.
In your example, the free action of $G$ on $X$ induces a free action of $G$ on the set $S$ of $n$-simplices in $X$, which in turn results in a free $ℤ[G]$-module structure on $ℤ[S] = \operatorname{C}_n(X)$.
