Quick question about formal definition of a representation So I know that a linear representation is defined as $\rho : G \to GL(V)$ over some finite group $G$. So if we define the action of some group ring, say F[G] over some representation V, is this representation V, the same thing as the homomorphism $\rho$? 
Taking G to be any group for example, when we say that V is the trivial representation of G over F, do we mean that V is the subspace left invariant by the action of the trivial representation, in this case, just the multiplicative group $F^{*}$? So what exactly does it mean to define the action of G on V in this case then?
 A: So we are given a group $G$ and a field $F.$ Then there is a "natural" correspondence between linear representations of $G$ over $F$ and linear representations of $F[G]$ over $F.$ This correspondence works as follows.
Suppose $V$ is an $F$-vectorspace. Suppose, we are given a linear representation $\rho:G\rightarrow GL(V).$ Then we can extend $\rho$ to a homomorphism $\sigma:F[G]\rightarrow End(V)$ by defining $\sigma(g) = \rho(g)$ for every basis element $g\in G\subseteq F[G]$ and linear extension. Here, $End(V)$ is the $F$-algebra of $F$-linear maps from $V$ to $V$. It's "very" easy to check that $\sigma$ is indeed a homomorphism of $F$-algebras.
Now, suppose we are given a linear representation $\sigma: F[G]\rightarrow End(V).$ Then we get a group homomorphism $\rho:G\rightarrow GL(V)$ by defining $\rho(g) = \sigma(g)$ for every $g\in G\subseteq F[G].$ Since $g\in G$ is invertible in $F[G],$ $\sigma(g)\in End(V)$ is invertible, and thus we have indeed $\rho(g) \in GL(V).$
These correspondences $\rho \leadsto \sigma$ and $\sigma\leadsto\rho$ are inverse to each other in the sense that in the concatenation $\rho \leadsto \sigma \leadsto \rho',$ we have $\rho = \rho'$ and in the concatenation $\sigma \leadsto \rho \leadsto \sigma',$ we have $\sigma = \sigma'.$
About the trivial representation: to me, "the trivial representation" of a group $G$ over a field $F$ is $V = F$ (one-dimensional vector space) and $g\mapsto 1 \in F.$
If we consider an arbitrary linear representation $\rho:G\rightarrow GL(V)$ of $G$ over $F,$ the we have the fixed point space
$$
V^G := \{v \in V\ |\ \forall g\in G:\rho(g)\cdot v = v \}.
$$
One often says that $G$ "acts trivially" on $V^G.$ This makes perfectly sense, since every $g \in G$ acts as the identity on $V^G.$ Of course, $V^G$ is a direct sum of one-dimensional trivial representations as described above.
A: Suppose we are working over a field $F$.
If you have a group homomorphism $\rho:G\to GL(V)$ then it is easy to see that there is a unique action $\alpha:F[G]\times V\to V$, turning $V$ into a $F[G]$-module, such that $\alpha(g,v)=\rho(g)(v)$ for all $g\in G$ and all $v\in V$. 
Conversely, given an action $\alpha:F[G]\times V\to V$ turning $V$ into a $F[G]$-module, there is a unique group homomorphism $\rho:G\to GL(V)$ such that $\rho(g)(v)=\alpha(g,v)$ for all $g\in G$ and all $v\in V$.
Moreover, these correspondences are mutually inverse.
So no, the homomorphism $\rho$ and the action $\alpha$ are most certainly not the same thing, but they determine each other.
A: Suppose that $V$ is a vector space over a field $F$, and $G$ is a group. Other answers have explained that a group homomorphism $\rho: G\to GL(V)$ gives $V$ the structure of a module over the ring $F[G]$ (and vice-versa; an $F[G]$-module is clearly a vector space and it is easy to reconstruct the $\rho: G\to GL(V)$ by, say, choosing a basis of $V$).
Linguistically, sometimes people say that $\rho$ is the representation.  On the other hand, sometimes they say that $V$ is the representation, but in that case they are already thinking of $V$ as a module over the ring $F[G]$. That is, they have in mind a particular action of $G$ on $V$.
So if someone says that "$V$ is the trivial representation of $G$ over $F$," the action of $G$ on $V$ is already defined.  Specifically, it is the trivial action: $g\cdot v = v$ for all $g\in G$ and $v \in V$. 
Indeed, when someone talks about the trivial representation of $G$ over $F$, they almost certainly mean a $1$-dimensional representation, so that $V$ is isomorphic to $F$. The reason being: an $n$-dimensional vector space $V \cong F^n$ which is endowed with the trivial $G$-action may obviously be "decomposed" as a direct sum of $n$ copies of a $1$-dimensional $F[G]$-module, each of which is invariant under the $G$-action, and we normally like to break up a representation into its "indecomposable" summmands.
A: Ok, I'm gonna try to explain this. Please correct me if wrong...
The mapping from the group to a representation of the group is called a homomorphism because it "preserves the structure". A group action is defined as a function of a group element on elements from some set. The representation has nothing to do with any such action but is only a remapping of group elements to matrices and the group operation to matrix multiplication such that they behave in the same way. 
You can view the representation as a kind of "implementation" of a group if you will. Then it may be that any group action you want to define for the group and some set may also be expressible with matrices, but that is something different from the group representation.
