Shengwei did this in class. She noted that since we are trying to decompose the
module into 1dimensional irreducibles, then the corresponding matrix representation
will have the form
$$X(g) = \left[ \begin{matrix} a&0\\0&d \end{matrix} \right]$$
where $a, d \in {\mathbb C}$
and since $g^4 = e$, then $X(g^4) = X(e) = X(g)^4$ so
$$X(g)^4 = \left[ \begin{matrix} a^4&0\\0&d^4 \end{matrix} \right] =
\left[ \begin{matrix} 1&0\\0&1 \end{matrix} \right]$$
So that means that $a \in \{ 1, 1, i, i \}$. Since an arbitrary element
in $V$ is of the form $r(e  g^2) + s(g  g^3)$, then
$g (r(e  g^2) + s(g  g^3)) = a (r(e  g^2) + s(g  g^3))$.
By comparing coefficients, we see that
if $a=1$ or $1$ then there are are no solutions. If $a=i$ (by choosing $r=1$
arbitrarily), then $r=1, s=i$ produces
$$g ((e  g^2) + i(g  g^3)) = i ((e  g^2) + i(g  g^3))$$
is an element that satisfies the equation. If $a=i$, then
$$g ((e  g^2)  i(g  g^3)) = i ((e  g^2)  i(g  g^3))$$
is an element that satisfies the equation.
Therefore
$$V = {\mathcal L}\{ e  g^2, g  g^3 \} =
{\mathcal L}\{ (e  g^2) + i(g  g^3) \} \oplus
{\mathcal L}\{ (e  g^2)  i(g  g^3) \}~.$$

So we discussed this problem a lot longer than I expected to in class on May 21
and I only outlined the problem. Step 1: Conjecture the answer. Method 1: figure out the graded dimensions for the first few terms and then look the answer up in the "online integer sequence database." Method 2: recall that a basis for the submodule ${\mathbb C}[x_1, x_2]^{S_2} = $ submodule consisting of those elements invariant under the $S_2$ action is $$\{ x_1^a x_2^b + x_1^b x_2^a\hbox{ for }a \geq b\}~.$$ Use this to guess (big leap here) that a basis for ${\mathbb C}[x_1, x_2, x_3]^{S_3}$ is the set $$\{ R^{S_3}( x_1^a x_2^b x_3^c ) \hbox{ for }a \geq b \geq c \geq 0 \}$$ Step 2: Show that the set above is a basis by demonstrating that it spans and is linear independent. Step 3: Show that generating function for the number of elements in the set $$\{ (a,b,c) : a \geq b \geq c \geq 0 \}$$ is equal to $\left( \frac{1}{(1q)(1q^2)} \right) \frac{1}{(1q^3)}$. Conclusion $$dim_q( {\mathbb C}[x_1, x_2, x_3]^{S_3} ) = \frac{1}{(1q)(1q^2)(1q^3)}$$ On May 26 we will have all the tools necessary to prove this without explicitly producing a basis. 