Show that the permutation representation of $S_3$ is induced from a trivial representation of $S_2$ Let $x,y,z$ be 3 independent variables and consider the polynomial vector space $P$ over $\mathbb{R}$ generated by $<<xy,xz,yz>>$.. Show that the permutation representation of $S_3$ is induced from a trivial representation of $S_2$

So when it says a trivial representation of $S_2$, does that mean the trivial map from $S_2 \rightarrow Aut(P)$ where $S_2$ can act on the entirety of $P$ (although it acts trivially)?
I'm just confused because $S_2$ is action on $P$, wouldn't the induced representation when we consider $S_2$ as a subgroup of $S_3$ act on $\oplus_{\sigma} P_{\sigma}$ where the $\sigma$ that index the direct sum span a transversal? I'm just a bit confused in general I guess. If somebody could help me with the original question that'd be rad!
 A: The trivial representation of $S_2$ doesn't have to act on $P$.
Let's say we are looking at the permutation representation $\pi : S_3 \to \operatorname{Aut}(\Bbb{C}^3)$ acting on canonical vectors as $\sigma(e_i) = e_{\sigma^{-1}(i)}$ for $1 \le i \le 3$ and $\sigma \in S_3$.
Now look at the trivial representation $S_2 \to \operatorname{Aut}(\Bbb{C})$ and let's construct the induced representation. We have to select a complete set of representants $\sigma_1, \sigma_2, \sigma_3 \in S_3$ of $S_3/S_2$. We can take
$$\sigma_1 = (1 \ 3), \quad \sigma_2 = (2 \ 3), \quad \sigma_3 = \operatorname{id}$$
where $\sigma_i$ represents the coset of maps $\sigma \in S_3$ such that $\sigma(3) = i$.
The induced representation acts on the space $\sigma_1\Bbb{C} \oplus \sigma_2\Bbb{C} \oplus \sigma_3\Bbb{C} \cong \mathbb{C}^3$ such that for $\sigma \in S_3$ and $1 \le i \le 3$ we find $1 \le j(i) \le 3$ and $\tau_i \in S_2$ such that $$\sigma\sigma_i = \sigma_{j(i)}\tau_i$$
and then define
$$\sigma(z_1,z_2,z_3) = \sigma(\sigma_1z_1+\sigma_2z_2+\sigma_3z_3) = \sigma_{j(1)}(\tau_1z_1) + \sigma_{j(2)}(\tau_2z_2) + \sigma_{j(3)}(\tau_3z_3).$$
In our case we have $$(\sigma\sigma_i)(3) = \sigma(\sigma_i(3)) = \sigma(i)$$
which implies that $\sigma\sigma_i$ is in the coset represented by $\sigma_{\sigma(i)}$, or $\sigma\sigma_i = \sigma_{\sigma(i)} \tau_i$ for some $\tau_i \in S_2$. Since $S_2$ acts trivially, we have
$$\sigma(z_1,z_2,z_3) =  \sigma_{\sigma(1)}(z_1) + \sigma_{\sigma(2)}(z_2) + \sigma_{\sigma(3)}(z_3) =(z_{\sigma^{-1}(1)}, z_{\sigma^{-1}(2)}, z_{\sigma^{-1}(3)})$$
so it acts as $\sigma(e_i) = e_{\sigma(i)}$, i.e. it is precisely the permutation representation $\pi$.
