Injection from the permutation representation of $S_4$ to $\uparrow^{S_4}_{S_2 \times S_2}$? Let $V$ denote the permutation representation of $S_4$. I want to know if there is an injection $\alpha: V \rightarrow \space \uparrow^{S_4}_{S_2 \times S_2} 1$. 
My Answer:
I don't think we can find such an $\alpha$. We have,
$$V \cong k \times k \times k \times k$$ for some algebraically closed field $k$. We also have $$\uparrow^{S_4}_{S_2 \times S_2} 1 \cong k[S_4/(S_2 \times S_2)] = k\bar{1} \oplus k\overline{(1 2)} \oplus k\overline{(1 3)} \oplus k\overline{(2 3)} \oplus k\overline{(1 2 3)}  \oplus k\overline{(1 3 2)}$$ where $\overline{(x_1, ... ,x_m)}$ denotes the equivalence class of $(x_1, ... , x_m)$. 
In order to have an injection, we need to map $V$ isomorphically to a $4$ dimensional subgroup of $\uparrow^{S_4}_{S_2 \times S_2} 1$. But that's impossible, since any subgroup of dimension $4$ is not invariant under the action of $S_4$ (since the action permutes the equivalence classes). For example, if we map $V$ into $k\bar{1} \oplus k\overline{(1 2)} \oplus k\overline{(1 3)} \oplus k\overline{(2 3)}$, we have 
$$(1 2 4 3)(k\bar{1} \oplus k\overline{(1 2)} \oplus k\overline{(1 3)} \oplus k\overline{(2 3)} \\ = k\overline{(1 2 4 3)} \oplus k\overline{(1 4 3)} \oplus k\overline{(2 4 3)} \oplus k\overline{(1 2)(4 3)} \\ = k\overline{(1 4)} \oplus k\overline{(1 3 2)} \oplus k\overline{(1 2 3)} \oplus k\overline{1}$$
and we know that $k\overline{(1 2 3)}$ and $k\overline{(1 3 2)}$ are not contained in $k\bar{1} \oplus k\overline{(1 2)} \oplus k\overline{(1 3)} \oplus k\overline{(2 3)}$. 
Is my answer correct? If not, can anybody give me a hint? 
 A: If you work over a field of characteristic $0$ or a commutative ring that contains $\mathbb Q$ the representation theory of $S_n$ is the same as over $\mathbb C$. Let me assume this.
In this case a lot is known about the induction from $S_a\times S_b$ to $S_{a+b}$. (See e.g. Fulton-Harris p. 58 (4.41), pp.455f (A.7) and (A.8))
I would calculate the decomposition of both $S_4$ representations (using (A.7) for the induced representation).
My solution:

 In particular it can be computed by Pieri's rule (A.7) that
 $ \uparrow_{S_2\times S_2}^{S_4} 1_{S_2\times S_2} \cong V_{(4)}\oplus V_{(3,1)} \oplus V_{(2,2)}.$
 (Here $V_\lambda$ is the irreducible $S_4$-representation corresponding to $\lambda$.)
 It is well-known (and much easier to see) that
 $V^{st} \cong V_{(4)}\oplus V_{(3,1)} .$
 Putting both results together one knows that there is an injection of $V^{st}$ to $\uparrow_{S_2\times S_2}^{S_4} 1_{S_2\times S_2}$. But I have no idea how it looks like.
 (The map described in my comment is therefore not injective.)

