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

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$S_n, n>1,$ 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?

Thanks in advance!

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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 $S_n$.

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$\DeclareMathOperator\GL{GL}$I know this post is old, and probably you know it by now. But in general, $G$ having a $1$-dimensional (irr.) representation (let us say over the field $\mathbb{C}$) means to have a homomorphism $G \to \GL_1(\mathbb{C}) = \mathbb{C}^{\times}$. But $\mathbb{C}^{\times}$ is abelian, which means that the homomorphism factors through $G / [G,G]$. On the contrary, any $1$-dim. representation of $G / [G,G]$, which is just a homomorphism $G / [G,G] \to \mathbb{C}^{\times}$ lifts (one also says inflates) to a 1-dimensional (irr.) representation $G \to \mathbb{C}^{\times}$, and since any irr.representation of $G / [G,G]$ is of this form, there is a natural correspondence: $$ \{1\text{-dim. reps of } G\} \, \longleftrightarrow \, \{\text{irr. reps of } G/ [G,G]\} $$ the $1$-dimensional (irr.) representations of $G$, and the irr.representations of $G / [G,G]$.

Now, since $[S_n,S_n] = A_n$ and $S_n/A_n = \{1, \tau\}$ for some transposition $\tau$, the result follows.