Let G = S3 and let V be the regular F G-module. Find irreducible submodules U1, U2, U3, U4 of V such that V =U1 ⊕U2 ⊕U3 ⊕U4 Let $G = S_3$ and let $V$ be the regular $F$ $G$-module.  
i) Find irreducible submodules $U_1, U_2, U_3, U_4$ of $V$ such that $V =U_1$ ⊕$U_2$ ⊕ $U_3$ ⊕ $U_4$  where dim($U_1$) = dim($U_2$) = 1 and dim($U_3$) = dim($U_4$) = 2.
For 1 ≤ i ≤ 4, you should prove that $U_i$ is a submodule of V. 
ii) Prove that $U_3$ is isomorphic to $U_4$.
iii) Which of the $F$ $G$-modules $U_1$, $U_2$, $U_3$ are faithful?
I am trying to follow an example given to me in lectures but it makes absolutely no sense and I would really appreciate someone to work this example through with me
 A: The problem basically boils down to finding the irreducible representations of $S_3$. Since $S_3$ has three conjugacy classes, there are only three such nonisomorphic representations. Further, we have that if $n_i$ is the dimension of the $i^{\text{th}}$ irreducible representation, then $n_1^2 + n_2^2 + n_3^2 = |S_3| = 6$. The only solution is $n_1 = n_2 = 1$ and $n_3 = 2$. Thus we've found the dimensions of the representations.
Now, we consider first the one-dimensional representations. These are group homomorphisms $\phi: S_3 \to \mathbb{C}^*$. Since $\mathbb{C}^*$ is abelian, any commutator in $S_3$ must map to $1$ in $\mathbb{C}^*$. Since $(12)(13)(12)^{-1}(13)^{-1} = (132)(132) = (123)$, we have that $\phi(123) = 1$. Thus our only choice is in where to send $(12)$. Since $(12)$ is an element of order 2, it maps to a second root of unity, i.e. $\phi(12) = \pm 1$. These two choices determine the two one-dimensional representations of $S_3$.
Finally, we need to find an irreducible two-dimensional representation of $S_3$. There is no general method to find such a representation of an arbitrary group, but in this case we happen to know of one (my knowledge of this representation is due to Lang's Algebra, 3rd edition). Generally, each $S_n$ has an $n-1$-dimensional representation due to the fact that for any $n$ there exists an irreducible polynomial over $\mathbb{Q}$ of degree $n$ whose Galois group is $S_n$. Let $f$ be such a polynomial and let $\alpha_1, \ldots, \alpha_n$ be its (necessarily distinct) roots. Consider the vector space $\mathbb{C}<\alpha_1, \ldots, \alpha_n>$. This is an $n-1$-dimensional vector space (why?) and $S_n$ acts on it by permuting the roots. Thus, this action yields an $n-1$-dimensional representation of $S_n$.
For $S_3$ in particular, this is a $2$-dimensional representation. It is easy to compute the character of this representation, and using the orthogonality relations, one can check that it is an irreducible representation. Thus, we have all of the irreducible representations of $S_3$.
Now that you know the isomorphism classes of irreducible representations of $S_3$, you need to find submodules of $V$ isomorphic to each of these. You should be able to do this by hand just by messing around with $V$ until you find the submodules you want. To get you started, the submodule isomorphic to the trivial representation is $U_1 = \mathbb{C}<g_1 + g_2 + \ldots + g_6>$, where $g_i$ are the elements of $S_3$. Now you just need to find a submodule $V'$ such that $U_1 \oplus V' = V$ and continue.
