Representation theory notation doubt So I've been reading a little bit about representation theory from the book "Linear Representations of finite groups" and i have a question.
When the author writes , 'Let n be the order of G, and let V be a vector space of dimension n,
with a basis $(e_t)_ {t \in G}$ indexed by the elements t of G." Im not quite sure how to look at this , i dont think this is supposed to be $\mathbb{R} ^n$ , i guess my doubt is that i dont understand what he means with that basis notation. So any help is appreciated thanks.
 A: I'll consider complex representations.
When I learned representation theory for the first time, it took me a while to realize why that definition makes sense, so I'll try to give an intuitive explanation.
I guess the sentence is from the definition of the regular representation of a group in section 1.2 (b) in Serre. Abstractly the vector space defined there is isomorphic to $\mathbb{C}^n$ and (in principle) $G$ just serves as the indexing set. Take some bijection from $G$ to $\{1, \ldots, n\}$ and you have the usual basis indexing. However, there is a good reason to index the basis by elements of the group: 
As you have seen, the idea of a representation is to consider the group $G$ as acting on a finite dimensional vector space i.e. to consider the elements as matrices. Quite early on, one might ask if every group has such a representation and what restrictions there are and if there is a natural choice for a vector space and action etc. Indeed, there are two (somewhat) natural choices. 
The first choice is the trivial representation (defined just in 1.2 (a) in Serre's book) where one represents every $g \in G$ by the $0$-matrix in $\mathbb{C}^{1 \times 1}$. The second one is more interesting:
Every group acts on itself by left multiplication $G \times G \rightarrow G: a.g \mapsto ag$. Now consider $\mathbb{C}^n =: V$ and associate to each element $g \in G$ a basis vector $e_g$ and define for any element $a \in G: a.e_g = e_{ag}$. This gives an action on $\mathbb{C}^n$ that is inherited by the groups action on itself. The elements $V$ are of the form $\sum_{g \in G} \lambda_g e_g$ and the explicit matrix representing an element $g \in G$ is entirely given by how $g$ acts on the other elements in $G$.
Notice now that we have a "natural" way of interpreting $G$ as a group of matrices. However, matrices can sensibly made into an algebra over $\mathbb{C}$ i.e. it does not only make sense to consider the action of $a \in G$ on an element $\sum_{g \in G} \lambda_g e_g$ but also to consider the action of $\sum_{a \in G} \lambda_a e_a$ on an element $\sum_{g \in G} \lambda_g e_g$. This action is uniquely determined by the action og $G$ on itself and the fact that that action on an element of $V$ needs to be linear. Ultimately this gives 
$$ (\sum_{g \in G} \lambda_g e_g) \cdot (\sum_{h \in G} \mu_h e_h) = \sum_{g \in G} (\sum_{h \in G} \lambda_{gh^{-1}} \mu_h) e_g = \sum_{g,h \in G} (\lambda_g \mu_h) (e_{gh})$$
Compare this now to discrete convolution and notice why indexing by the group elements is sensible. Also explicitely look at the example where $G:=\mathbb{Z}/p \mathbb{Z}$!
Considering the vector space $V$ together with the above defined multiplication makes it into a $\mathbb{C}$ algebra - the group algebra over $G$.
Also, if you are just starting out in representation theory, Serre's book may be a bit heavy/dry (at least it was for my taste). Maybe have a look at James & Liebeck. They explicitely compute many examples and give a more concrete introduction. If you are looking for something more geared towards Lie Theory, consider Fulton & Harris (PDF available online).
A: Just to add to G. Chiusole’s excellent answer, I wanted to give a (tiny) example.
Let $G = \mathbb{Z}/3\mathbb{Z} = \{0,1,2\}$. Note that $G$ has three elements, so we have three vectors in $\mathbb{C}^3$:
$$e_0 = \begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}, \qquad e_1 = \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}, \qquad e_2 = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}.$$
I know that $0$ is the identity element of $G$, so my representation of $G$ should send $0$ to
$$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}.$$
Now, since 
$$\begin{cases} 1+0=1 \\ 1+1=2 \\ 1+2=0\end{cases}$$
my representation should send $1$ to a matrix $A$ such that
$$\begin{cases} Ae_0 = e_1 \\ Ae_1 = e_2 \\ Ae_2 = e_0\end{cases}$$
Thus, we need that
$$A = \begin{bmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{bmatrix}.$$
Lastly, since 
$$\begin{cases} 2+0=2 \\ 2+1=0 \\ 2+2=1\end{cases}$$
my representation should send $2$ to a matrix $B$ such that
$$\begin{cases} Be_0 = e_2 \\ Be_1 = e_0 \\ Be_2 = e_1\end{cases}$$
Thus, we need that
$$B = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{bmatrix}.$$
