Looking for various proofs there is no faithful representation $\rho:S_4\rightarrow GL(\mathbb{R}^2)$ I stumbled on the question when thinking about a representation $\rho:D_4\rightarrow GL(\mathbb{R}^2)$ as symmetries of the four points $\{(\pm 1, \pm 1)\}$. There didn't seem to be a good geometric picture for how to swap two adjacent vertices without moving the others via linear mapping.
To show $\rho:S_4\rightarrow GL(\mathbb{R}^2)$ is never faithful it is enough to show that the image of the transpositions can not be distinct. It's possible to work this out by calculating all $2\times 2$ matrices of order $2$ and showing there simply aren't enough valid distinct choices for the images.
However, the problem I had in mind came from geometry so I was wondering about other proofs for this problem. Namely:

What are other proofs there does not exist a faithful representation $\rho:S_4\rightarrow GL(\mathbb{R}^2)$?

 A: Here's a proof I like:
Assume (for contradiction) that $\rho$ is a faithful representation.
We note that $S_4$ contains the commuting transpositions $\tau_1 = (1\;2)$ and $\tau_2 = (3\;4)$.  Since neither transposition is in the center of $S_4$, neither transposition can be mapped to a multiple of the identity.  Because $\rho(\tau_i)^2 = I$, we may deduce that each $\rho(\tau_i)$ is diagonalizable with eigenvalues $\pm 1$.  Because the $\rho(\tau_i)$ commute, they are simultaneously diagonalizable.  Because they are distinct and $\rho$ is faithful, they are not mapped to the same matrix.
So, up to an isomorphism of representations (i.e. a change of basis), we have
$$
\rho(\tau_1) = \pmatrix{1\\&-1} \quad
\rho(\tau_2) = \pmatrix{-1\\&1}
$$
This means that
$$
\rho(\tau_1 \tau_2) = \rho(\tau_1) \rho(\tau_2) = -I
$$
So, we have $\rho(\tau_1 \tau_2) \in Z(GL_2)$.  However, $\tau_1\tau_2 \notin Z(S_4)$, which since $\rho$ is faithful would necessarily imply that $\rho(\tau_1 \tau_2) \notin Z(GL_2)$.
So, no faithful representation exists.
Note: It may seem like overkill in this context, but the same proof can be extended to show that there is no faithful representation $\rho: S_n \to GL_{n/2}$ for any even $n \geq 4$.

A more typical approach is as follows: the table of irreducible characters on $S_4$ is given by
$$
\begin{array}{c|rrrrr}
\text{classes:} & 1 & (1\,2) & (1\,2\,3) & (1\,2\,3\,4) & (1\,2)(3\,4)\\
\text{sizes:} & 1&6&8&6&3\\
\hline
\chi_1 & 1&1&1&1&1\\
\chi_2 & 1&-1&1&-1&1\\
\chi_3 & 2&0&-1&0&2\\
\chi_4 & 3&1&0&-1&-1\\
\chi_5 & 3&-1&0&1&1
\end{array}
$$
(copied from Dummit and Foote). Thus, the character induced by any representation $\rho:S_4 \to GL(\Bbb R^2)$ must either have the form $\chi_i + \chi_j$ with $i,j \in \{1,2\}$ or $\chi_3$. Correspondingly, $\rho$ must be isomorphic either to the direct sum $\rho_i \oplus \rho_j$ with $i,j \in \{1,2\}$ or to $\rho_3$ (where $\rho_j$ is a representation inducing character $\chi_j$).
$\rho_1$ is the trivial representation and $\rho_2$ is the sign representation, so $\rho_i \oplus \rho_j$ fails to be faithful for $i,j \in \{1,2\}$ (since $\rho_2$ is not faithful).  Alternatively, since $\chi_1((1\,2\,3)) = \chi_2((1\,2\,3)) = 1$ and the degrees of $\chi_1,\chi_2$ are $1$, it must be that $\rho_1((1\,2\,3)) = \rho_2((1\,2\,3)) = 1$.  Thus, $(1,2,3) \in \ker (\rho_1 \oplus \rho_2) = \ker(\rho_1) \cap \ker(\rho_2) \subset \ker(\rho_i) \cap \ker(\rho_j)$ for $i,j \in \{1,2\}$.
It now suffices to note that $\rho_3$ is also not faithful.  To that end, it suffices to follow our second approach above: note that $\chi_3((1\,2)(3\,4)) = 2$ implies that $\rho_3((1\,2)(3\,4)) = I$, which is to say that $(1\,2)(3\,4) \in \ker(\rho_3)$.
Aside: As this answer notes, the representation $\rho_3$ corresponding to $\chi_3$ is the composition of the canonical surjection $S_4 \to S_4/V \cong S_3$, where $V$ is the subgroup generated by the double transpositions, and the standard 2-dimensional representation $\rho_\mathrm{standard} : S_3 \to \mathrm{GL}(2,\mathbb{C})$.
