Demonstration of a topic via an example - Representation theory Reading chapter 2.4 of linear representations of finite groups by Serre, can someone create an example for me to understand what's going on?

Let the irreducible characters be denoted by $\chi_i$ and their degrees are $n_i$. Let the order of $G$ be $|G|$ and let $V$ be a vector space of dimension $|G|$ with a basis $\left\lbrace e_t \right\rbrace$ for $t \in G$. Define the linear map of $\rho_s$ from $V$ into $V$ as $\rho_s(e_t) = e_{st}$, which is called the regular representation R. Its degree is equal to the order of $G$.
Proposition 5 
The character $r_G$ of the regular representation $R$ is given by
  \begin{align*}
r_G(1) &= |G| \\
r_G(s) &= 0 \qquad \text{if } s\neq 1
\end{align*}
Proof
If $s=1$, we have
  $$
Tr(\rho_s)=Tr(1)=\dim(R)=| G|
$$
  If $s\neq 1$, then $st\neq t$ and $\rho_s(e_t) = e_{st} \neq e_t$ for all $t$ so all diagonal terms of the matrix for $\rho_s$ are zero.
Corollary 6
Every  irreducible representation $W_i$ is contained in the regular representation with multiplicity equal to its degree $n_i$.
Proof
From theorem 2.3.3, this number equates to $(r_G|\chi_i)$
$$
(r_G|\chi_i)=\frac{1}{\,|G|} \sum_{s \in G} r_G(s)\chi_i(s)^* = \chi_i(1)^* = n_i 
$$
Remark: 
  Hence we notice, there is only a finite number of irreducible representations.
Corollary 7
(i) The degrees $n_i$ satisfy $\sum_{i=1} n_i^2 =|G|$
(ii) If $s \in G$ is different from $1$, we have $\sum_{i=1} n_i \chi_i(s) =0$
Proof
By corollary 2.4.2, we have $r_G(s) = \sum n_i \chi_i(s)$ for all $s \in G$. For $(i)$ take $s=1$ and for $(ii)$ take $s\neq1$.
  \end{proof} 
Corollary 8
  The matrix entries of the unitary irreducible representations form an orthogonal basis for the set of all functions on $G$. 
Proof
We know that the matrix entries are all orthogonal as functions. There are $\sum n_i^2=|G|$ of them, and this is the dimension of the vector space of all functions. 

 A: As @Dane suggested, lets consider the group $G = S_3$ the symmetric group on $3$ letters $1,2,3$. We will consider propositions $5,6$ here. Let $\mathsf{k}$ be some field. Consider the six dimensional $\mathsf{k}$-vector space $V$ with basis $e_0$, $e_{(12)}$, $e_{(23)}$, $e_{(13)}$, $e_{(123)}$, $e_{(132)}$. All we have done here is indexed the basis by the elements of $S_3$. We then define a representation $\rho$ of $G$ by $\rho(g)(e_{h}) =e_{gh}$. So for example
$$
\rho((12))e_{(23)} = e_{(132)}, \quad \rho((13))e_{(23)} = e_{(123)}, \quad \rho((12))e_{(12)} = e_0.
$$
Let's explicitly calculate the matrix of $\rho(12)$ with respect to the ordered basis $e_0$, $e_{(12)}$, $e_{(23)}$, $e_{(13)}$, $e_{(123)}$, $e_{(132)}$. Note that
$$
\begin{array}{c | c c }
g & e_0 & e_{(12)} & e_{(23)} & e_{(13)} & e_{(123)} & e_{(132)} \\
\hline \rho((12))e_{g} &  e_{(12)} & e_0 & e_{(132)} & e_{(123)} & e_{(13)} & e_{(23)}
\end{array}
$$
and so
$$
\rho((12)) = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\end{pmatrix}
$$
so note that $\chi((12)) = 0$, where $\chi = \operatorname{trace} \circ \rho$. This is what is called $r_G$ in your post. You might notice that
$$
\chi(g) = \left| \left\{ h \in S_3 \mid gh = h \right\} \right| = \left|\operatorname{Fix}_{g}(S_3) \right|.
$$
Hence indeed
$$
\chi(g) = \left\{ \begin{array}{ll} \left| G \right| = 6, & \text{if} \ g = e_0, \\ 0, & \text{otherwise} \end{array}  \right.
$$
Now we have a representation of $S_3$ given by the sign representation. That is $\rho_{\operatorname{sign}} : S_3 \to \mathsf{k}^{\times}$ such that 
$$
\rho_{\operatorname{sign}}(g) = \left\{ \begin{array}{ll}  1, & \text{if} \ g \ \text{is a product of an even number of transpositions,} \\ -1, & \text{if} \ g \ \text{is a product of an odd number of transpositions.} \end{array} \right.
$$
Now we will find this a subrepresentation of $V$. Consider the vector
$$
v_{\operatorname{sign}} = e_0 - e_{(12)} - e_{(23)} - e_{(13)} + e_{(123)} + e_{(132)} \in V.
$$
and define $V_{\operatorname{sign}} = \mathsf{k} \cdot v_{\operatorname{sign}}$, then I claim that this is a subrepresentation. But it is easy to check that $S_3$ acts on $v_{\operatorname{sign}}$ by $\pm 1$ so this is indeed a subrepresentation and whats more its easy to check that this is indeed the sign representation. It is also easy to check that this is the unique subrepresentation of $V$ isomorphic to the sign representation.
