Finding G- submodules Let G be cyclic group of orfer 3, the generator being $\alpha$. Let V=$k^3$. Let $\alpha e_1=e_2$, $\alpha e_2=e_3$, $\alpha e_3=e_1$. How to find all $G$-submodules of V when 
a) $K=\mathbb R$
b) $K=\mathbb C$
Try: I have found the all one dimensional $G$-submodules, the problem arises for 2-dimensional $G$-submodules.
 A: For part (b) the field is algebraically closed so there are no non-one-dimensional irreducible representations since G is abelian so all the G-submodules will be directs sums of the one-dimensional ones.
A: Over the reals you will have a one-dimensional module spanned by $e_1+ e_2+ e_3$.
Then there is an irreducible two-dimensional module $U$, spanned by $f_1 = e_1 - e_2$, and $f_2 = e_2 - e_3$. The action is
$$
\alpha f_1 = \alpha(e_1 - e_2) = e_2 - e_3 = f_2,
\qquad
\alpha f_2 = \alpha(e_2 - e_3) = e_3 - e_1 = - f_1 - f_2.
$$
So the matrix of the action of $\alpha$ on $U$, with respect to the basis $f_1, f_2$, is 
$$
B =
\begin{bmatrix}
0 & -1\\
1 & -1\\
\end{bmatrix}.
$$
The submodule $U$ is irreducible, since the primitive $3$rd roots of unity $\omega, \omega^{2}$, which are the eigenvalues of $B$, are not real. Over the complex numbers $U$ splits as the sum of the two one-dimensional modules spanned respectively by
$$
e_1 + \omega e_2 + \omega^{2} e_3,
\qquad
e_1 + \omega^{2} e_2 + \omega e_3.
$$
It all boils down to the factorization of the characteristic polynomial $x^{3} - 1$ of the matrix
$$
A =
\begin{bmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0
\end{bmatrix}
$$
of the action of $\alpha$ on $V$ with respect to the basis of the $e_i$. Over the reals it is
$$
x^{3} - 1 = (x - 1) (x^{2} + x + 1),
$$
over the complex numbers it is
$$
x^{3} - 1 = (x - 1) (x - \omega) (x - \omega^{2}).
$$
This is related to the fact that the action of $G$ on $V$ similar is the regular representation of $G$ on the group algebra
$$
k[G] \cong \frac{k[x]}{(x^3 - 1)}.
$$
A: Sean, with Jim's help, has answered part (b).  For part (a), where the field is $\mathbb R$, think of $V$ as 3-dimensional Euclidean space, and consider the orthogonal complement of the $1$-dimensional submodule.
