Find irreducible representations of semidirect product $(S_2 \times S_2) \rtimes S_2$ I'm looking at a very specific action of the semidirect product (wreath product) $(S_2 \times S_2) \rtimes S_2 = S_2 \wr  S_2$ on $\mathbb{Q}^3$. 
Namely, the generators acts as follows on a basis $x_1, x_2, x_3$ of $\mathbb{Q}^3$ (write $S_2 = \{ 1, \sigma \}$):
\begin{align*}
  ((1,1), \sigma): \hspace{2mm} &x_1 \mapsto x_2, \hspace{2mm} x_2 \mapsto x_1, \hspace{2mm} x_3 \mapsto -x_3 \\
  ((\sigma,1), 1): \hspace{2mm} &x_1 \mapsto -x_1, \hspace{2mm} x_2 \mapsto x_2, \hspace{2mm} x_3 \mapsto x_3 \\
  ((1, \sigma), 1): \hspace{2mm} &x_1 \mapsto x_1, \hspace{2mm} x_2 \mapsto -x_2, \hspace{2mm} x_3 \mapsto x_3
\end{align*}

Question: How do I find the irreducible subrepresentations of this representation?

What I did so far was to find the conjugacy classes of $(S_2 \times S_2) \rtimes S_2$ and calculate the character of the above representation. But how does this help me?
If we would only look at the direct product rather than the semidirect one, the above action on the basis seems to imply that the representation we're looking at is simply the sum of three times the sign irrep of $S_2$. But I don't understand how to go on in the semidirect case...
Thanks for any help!
 A: I give an ad hoc argument specific to this case.
By perusing the listed action of the generators we see that the subspaces
$$
V_1=\{(x_1,x_2,x_3)\in\Bbb{Q}^3\mid x_3=0\}
$$
and
$$
V_2=\{(x_1,x_2,x_3)\in\Bbb{Q}^3\mid x_1=x_2=0\}
$$
are both stable under the action of $G=S_2\wr S_2$.
Furthermore, if we identify $V_1$ with the $xy$-plane via $(x,y)\leftrightarrow (x,y,0)$, we see that:


*

*The element $((1,1),\sigma)$ acts as a reflection w.r.t. the line $x=y$.

*The element $((\sigma,1),1)$ acts as a reflection w.r.t. the $y$-axis.

*The element $((1,\sigma),1)$ acts as a reflection w.r.t. the $x$-axis.


Those reflections generate the group $D_4$ of symmetries of a plane square
with center at the origin and sides parallel to the coordinate axes. This already implies that the subspace $V_1$ is an irreducible representation of $G$,
because this defining representation of $D_4$ is known to be irreducible. Even without this bit of information we can deduce the irreducibility of $V_2$ as follows. A putative non-trivial subrepresentation $W$ will necessarily be $1$-dimensional. But the vectors of a 1-dimensional representation of any group
must be eigenvectors for all the elements of the group. Here there are no such vectors, because the eigenvectors of a plane reflection are exactly the line of reflectio (eigenvalue $\lambda=+1$) and the vectors perpendicular to it (eigenvalue $\lambda=-1$). But the line $x=y$ makes a 45 degree angle with the others, so no common eigenvectors exist.
As a 1-dimensional representation $V_2$ is obviously also irreducible.
