I'm reading this expository paper about group theory in the Rubik's cube.
I'm a little confused by the definition of the wreath product in this paper.
Example 3.12 on page 12 states that the elements of the wreath product $(\mathbb{Z}/2\mathbb{Z})^3 \wr S_3$ for the set X={0, 1, 2} are
{$(0,0,0) \rho, (1,0,0) \rho, (0, 1, 0) \rho, (0, 0, 1) \rho, (1, 1, 0) \rho, (0, 1, 1) \rho, (1, 0, 1) \rho, (1, 1, 1) \rho$}
where $\rho$ is an element of $S_3$.
My question is: is $\rho$ just an arbitrary element in $S_3$, or is $\rho$ somehow specially chosen? I don't understand how $S_3$ acts on $(\mathbb{Z}/2\mathbb{Z})^3$ in the same way as $S_3$ acts on the set X.
More generally, I don't understand how, for a direct product $G^t \wr H$, where $H$ acts on a set of size $t$, how $H$ acts on $G^t$. Specifically, which elements of $H$ are acting on $G^t$?
P.s I'd really appreciate it if answers were geared towards a high school student with little formal abstract algebra background. Thank you. :)
$S_3$ acts by permuting the entries of a $3$-tuple in $\newcommand{\Z}{\mathbb{Z}} \DeclareMathOperator{\Aut}{Aut} (\mathbb{Z}/2\mathbb{Z})^3$. In symbols: given $\pi \in S_3$, $$ \pi \cdot (a_1, a_2, a_3) = (a_{\pi^{-1}(1)}, a_{\pi^{-1}(2)}, a_{\pi^{-1}(3)}) \, . $$ (The inverse is just there for bookkeeping, to make sure this defines a left-, rather than right-, action.) So for instance, taking $\pi = (1\ 2)$ and $(a_1, a_2, a_3) = (0,1,0)$, then $\pi^{-1} = \pi$ and $$ \pi \cdot (a_1, a_2, a_3) = (a_2, a_1, a_3) = (1,0,0) \, . $$ This action induces a homomorphism $\varphi: S_3 \to \Aut((\Z/2\Z)^3)$, which allows us to define the semidirect product $(\Z/2\Z)^3 \rtimes_\varphi S_3$. As a set $(\Z/2\Z)^3 \rtimes_\varphi S_3$ is just the cartesian product $(\Z/2\Z)^3 \times S_3$, so as a set $$ (\Z/2\Z)^3 \rtimes_\varphi S_3 = \{((a_1, a_2, a_3), \rho) : a_1, a_2, a_3 \in \Z/2\Z, \rho \in S_3\} \, . $$ However, the multiplication is "twisted" by the action of $S_3$: \begin{align*} ((a_1, a_2, a_3), \pi) \ ((b_1, b_2, b_3), \rho) &= ((a_1, a_2, a_3) + \pi \cdot (b_1, b_2, b_3), \pi \rho)\\ &= ((a_1, a_2, a_3) + (b_{\pi^{-1}(1)}, b_{\pi^{-1}(2)}, b_{\pi^{-1}(3)}), \pi \rho)\\ &= ((a_1 + b_{\pi^{-1}(1)}, a_2 + b_{\pi^{-1}(2)}, a_3 + b_{\pi^{-1}(3)}), \pi \rho) \, . \end{align*}
I only defined the wreath product for $S_t$, but the the same thing works for any subgroup of $S_t$, or more generally for any group $H$ with a homomorphism $H \to S_t$. (This is what it means for a group to act on a set of size $t$.)