$S_n, n>1,$ has precisely two $1$-dimensional irreducible representations

$S_n, n>1,$ has precisely two $1$-dimensional irreducible representations

My attempt/thoughts: I have proven that for any $1$-dimensional representation $X$ of $G$, it is constant over conjugacy classes. In fact this follows from the equality of the character $\chi$ and the representation $X$ and from the fact that $\chi$ is constant over conjugacy classes.

This result seems helpful in order to prove the claim about $S_n$. Well, every transposition has the same cycle type, so they are always conjugate to each other. Write $\tau_1,\cdots,\tau_k$ for all transpositions. They generate $G$, so any $g\in G$ may be written as $g = \tau_{i(1)}\cdots \tau_{i(s)}$. Since $X$ is constant over conjugacy classes, let $X(\tau_i) =c$. So we get that $X(g) = c^s$.

How to proceed with this? Or should I expose two $1$-dimensional representation (trivial and sgn) and then prove by contradiction that cannot be more than these ones?

• $X$ is a representation, so what are the possible values of $c$? End of story. – ancientmathematician May 17 '18 at 11:38
• @ancientmathematician $1$ and $-1$? Since $X(\tau_i)X(\tau_i) = X(\tau_i^2) = X(e) = 1?$ – math.h May 17 '18 at 11:41
• For $n>1$, yes, but $S_1$ has only one. – Somos May 17 '18 at 12:44
Since all the transpositions $\tau_i$ generate $S_n$, it is sufficient to find all the possible values of $X(\tau_i$) . Furthermore, this value shall be equal to every other transposition $\tau_j$, as they are in the same conjugacy class.
Now, $X^2(\tau_i)=X(\tau_i^2) = X(e) = 1$ implies that $X(\tau_i)= \pm1$.This shows that there are only two possible values of $X$ in the group generators, which ensures only two possible $1$-dimensional representations of $S_n$.