Unitary G-module I'm not sure if I understand this sentence correctly:
"By a unitary G-module we will mean a Hilbert space W on which G acts by means of a strongly continuous unitary representation".
$G$ is a locally compact group. 
I know that an action of a group $G$ on $W$ is a map $G\times W\rightarrow W$ and a unitary representation of $G$ is a homomorphism $U: G\rightarrow U(W)$ where $U(W)$ is the set of unitary operators on W. Is "unitary G-module" just another word for "A Hilbert space with a unitary representation of G which is strongly continuous"? I'm not sure about the "on which G acts by means of" part.
Thank you for your time.
 A: Yes, you are right.
In the following I will try to explain how actions of $G$ and representations of $G$ are related, as I believe this was your main problem here.
An action of $G$ on $W$ is a map $G \times W \rightarrow W, (g,w) \mapsto g.w$ which has the two properties $1.w = w$ and $(gh).w = g.(h.w)$ for all $g,h \in G$ and $w\in W$. Thus the maps $w \mapsto g.w$ are all bijections and from the properties above it follows that
$$ G \rightarrow \text{Sym}(W), g \mapsto (w \mapsto g.w) $$
is a homomrphism of groups (where $\text{Sym}(W)$ denotes the symmetric groups of the set $W$).
Conversely, every homomorphism $\rho: G \rightarrow \text{Sym}(W)$ defines an action $G \times W \rightarrow W$given by $ g.w = \rho(g)(w)$ for $g \in G$ and $w \in W$.
Now let's say we want the maps $w \mapsto g.w$ not only to be bijections but unitary operators, then from this assumption on the action $G \times W \rightarrow W$, the corresponding homomorphism into the symmetric group has its image in $U(W)$, so in fact we get a homomorphism into $U(W)$, thus a unitary representation $G \rightarrow U(W)$.
So having a unitary representation $G \rightarrow U(W)$ is equivalent to having an action $G \times W \rightarrow W$ with $w \mapsto g.w$ being unitary for all $g$.  
