I am studying linear representation theory for finite groups and came across the claim in title. When $n\geq 5$, $S_n$ does not have an irreducible, $2$- dimensional representation.But I am not sure where to begin with.
Although it seems that this result will follow from this as a special case, I am interested in a solution that is specific to this problem.
The condition $n\geq 5$ seems to suggest that we need to use the fact that $A_n$ is simple for $n\geq 5$.
I would appreciate any hint.
On
It is also possible to prove the following, using elementary methods of the above link:
For $n \geq 5$, the only representations of $S_n$ of dimension $<n$ are direct sums of
(a) the trivial representation
(b) the sign representation
(c) the $n-1$ dimensional representation on $n$-tuples of numbers summing to $0$
(d) the tensor product of (b) and (c).
We see that every $2$-dimensional representation then is reducible. The argument with the eigenvalues for $5$-cycles is already given in the comments by Jyrki Lahtonen at the linked question, and by David Speyer in his answer.
Here's a hint for one fairly elementary proof.
Suppose $\rho:S_n\to\textrm{GL}(2,\mathbb{C})$ is a representation, where $n\geq5$.
Consider the eigenvalues of $\rho(\sigma)$ for $5$-cycles $\sigma$.