Representations of the symmetric group $S_3$ I'm considering the following example/application. Let $k$ be a field, such that $char(k)\nmid\mid G\mid$.Let $G=S_3$ then we have the following representation on it:


*

*TRIVIAL REPRESENTATION: $V_0=k$, $G\rightarrow GL_1(k)=k^\times,\sigma\mapsto1$

*SIGN REPRESENTATION: $V_1=k$, $G\rightarrow GL_1(k)=k^\times,\sigma\mapsto sgn(\sigma)\cdot 1$

*STANDARD REPRESENTATION: $V=\{(x_1,...,x_n)\in k^n\mid\sum_ix_i=0\}\subset k^n$, $G\rightarrow GL_n(k),\sigma\mapsto P_\sigma$


CLAIM: $V_0,V_1,V$ are all the irreducible representations up to isomorphism.
To prove this we want to show that:

The regular representation of $G$ is isomorph to $V_0\oplus V_1\oplus V^2$

I have a theorem which states that every finite dimensional representation can be written as direct sum of irreducible representations (and since $\mid G\mid<\infty$ then the regular representation is finite dimensional). I also have that, up to isomorphism, there are finitely many irreducible representations and they all occur as subrepresentations of the regular representations. But this states only the existence of such a decomposition, it does not tell that my direct sum has to be irreducible. Which is the argument used to state this?
The prove states that it is easy to find copies of $V_0$ and $V_1$ in the regular representation ($k\cdot\sum_{\sigma\in S_3}{\sigma}$ resp. $k\sum_{\sigma\in S_3}{\sigma}$).
I have absolut no idea of what is going on here. why we want or need copies of $V_0,V_1$? How one comes to such copies?
Any help would be very appreciated. thaks

Actually now I'm seeing that the first two representations are irreducible since there are no non-trivial subrepresentation. And about $V$ I found a note stating that it is irreducible since $Char(k)\nmid \mid G\mid$. But why should this hold?
 A: It is a remarkable and beautiful fact that the irreducible representations of the symmetric group $S_n$ are in correspondence with the partitions of $\lambda \vdash n$. For example, in the case of $S_3$, the irreducible partitions correspond to all the partitions of 3, namely
$$(3) \quad (1,1,1) \quad (2,1)$$
The full story is too long for this post, but details can be found in Chapter 4 of 'Representation Theory: A first course' by Fulton and Harris.
To see the decomposition of the regular representation into its irreducible components is most easily done via character theory. Let $\chi$ be the character of the left regular representation (that is, let $S_3$ act on $K[S_3]$ on the left); $\chi(\sigma)$ is the number of fixed points of the action of $\sigma$ on $K[S_3]$ (as these contribute to the trace of this action). It is plain to see that the only element in $S_3$ which fixes anything in $K[S_3]$ is the identity element $e$, and moreover, this fixes every element in $K[S_3]$. Thus we have
$$\chi(e) = |S_3|=6 \qquad \chi(\sigma)=0 \quad \forall \ \sigma \in S_3 \backslash \{e\}$$
Let $\chi_\lambda$ be the character of the irreducible representation of $S_3$ corresponding to the partition $\lambda \vdash 3$. It is a fact from character theory that the inner product of characters of a representation $A$ and an irreducible representation $B$, defined by,
$$\langle \chi_A, \chi_B \rangle = \frac{1}{|G|}\left( \sum_{g \in G} \chi_A(g)\chi_B(g) \right)$$
gives the multiplicity of $B$ in $A$. For your question then, we need to compute this inner product with the character $\chi_\lambda$. For this we only need to know one more fact: that $\chi_\lambda(e)$ is the dimension of the corresponding irreducible representation of $S_3$ corresponding to $\lambda$. We can now compute
$$\langle \chi, \chi_\lambda \rangle = \frac{1}{|S_3|}\left(\chi(e)\chi_\lambda(e) \right) = \frac{1}{6}(6\cdot \chi_\lambda(e)) = \chi_\lambda(e)$$
We see that the irreducible representation corresponding to $\lambda$ appears in the decomposition of the regular representation exactly the 'dimension of representation' number of times.
Here are the correspondences in your case:
$$\lambda = (3) \rightarrow V_0, \dim = 1$$
$$\lambda = (1,1,1) \rightarrow V_1, \dim = 1$$
$$\lambda = (2,1) \rightarrow V, \dim = 2$$
Therefore
$$k[S_3] = V_0 \oplus V_1 \oplus V^{\oplus 2}$$
as desired.
If this is new to you, then there are a lot of details to check here, all of which can be found in Fulton Harris. You should know that this story works for any $n$ and we have in general that
$$K[S_n] = \bigoplus_{\lambda \vdash n} V_\lambda^{\oplus \dim V_\lambda}$$
where $V_\lambda$ is the irreducible representation of $S_n$ corresponding to the partition $\lambda$ of $n$.
A: I will try to give some insight by working over $\mathbb{C}$, since that is most standard, but most of what I say should hold in general.

Write the regular $\mathbb{C}G$-module as a direct sum of irreducible
  $\mathbb{C}G$-submodules
  $$ \mathbb{C}G = U_{1} \oplus \cdots \oplus U_{r},$$
  then every irreducible $\mathbb{C}G$-module is isomorphic to one of the $U_{i}$.

(I have this in James & Liebeck, though it is a standard result, and holds in other fields as well).
The regular $\mathbb{C}G$-module is $6$-dimensional.  Consider 
$G = \langle a, b \mid a^{3} = b^{2} = 1, \ b a b = a^{-1} \rangle$.
Okay, what you want to do is decompose the regular $\mathbb{C}G$-module.  You can write a basis for it, and then find ways to embed each of the representations you have found.  You have accounted for $4$ of the $6$ dimensions (once you embed), so you have to find what is left (another copy of $V$, or two more copies of $V_{1}$, etc.).
For example, the trivial representation:
$$U_{1} = \langle 1+a+a^{2} + b + ab + a^{2}b \rangle$$
The sign representation:
$$U_{2} = \langle 1+a+a^{2} - b - ab - a^{2}b \rangle$$
