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I want to show that $S_n$ has only two 1 dimensional represnetations. mainly the trivial and sign represnetations.

Where I assumed that our Field we're working on is with characteristic $\neq 2$.

What I wrote so far, that if I assume that there exists another nonequivalent represnetation to the above representations then (denote it by $\tau$, and $\chi_{\tau} = trace\ \tau $):

$$\sum_{\sigma \in A_n} \chi_{\tau}(\sigma) = \sum_{\sigma \in S_n -A_n} \chi_{\tau} (\sigma) =\sum_{\sigma \in S_n} \chi_{\tau} (\sigma) = 0 $$

Somehow, I want to derive some sort of contradiction, I thought of deriving that $\dim \tau =0$, but don't see exactly how?

I mean I know from above that:

$$ 1=\dim \tau = -\sum_{\sigma \in A_n \ \sigma \neq id} \chi_{\tau} (\sigma)$$

But other than that, I don't see how to derive a contradiction?

Any help is appreciated

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A hint for the complex number case: It suffices to show that every homomorphism $\chi: S_n$ to $\mathbb{C}^*$ is real, which then forces any such homomorphism to be either identity or sign, depending on the value at the cycle (12). To show that it's real, it suffices to show that $\chi(g) = \overline{\chi(g)}$ for any $g$, i.e. $\chi(g) = \chi(g^{-1})$. Try to show that $g$ and $g^{-1}$ are actually conjugates which lead to the result. – user27126 Dec 24 '12 at 15:43
up vote 3 down vote accepted

In general, if you are looking for $1$ dimensional representations of a group $G$ over a (commutative) field $K$, i.e., group homomorphisms $G\to K^\times$, then every commutator $aba^{-1}b^{-1}$ must map to $1$ (because of the commutatitivity of $K^\times$). With the commutator subgroup $G'$ in their kernel, such representations amount essentially to characters of the Abelianisation $G/G'$ of $G$. In the case of the symetric group $G=S_n$ one has $G'=A_n$, and $G/G'$ is a group of order $2$ (if $n\geq 2$). Counting the characters $C\to K^*$ of a cyclic group $C$ of order $2$ is easy: there is only the trivial one if $\operatorname{char} K=2$, and there are two such characters otherwise.

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Hint: $ GL(\mathbb{K})=\mathbb{K}^*$ $\Rightarrow$ $(\phi(\sigma))^3=1, (\phi(\tau))^2=1;$ where $\sigma$ is the 3-cycle and $\tau$ is the 2-cycle.

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How does this uses the fact that char of the field is different than 2? – MathematicalPhysicist Dec 24 '12 at 16:29
Why is this implication correct? – MathematicalPhysicist Dec 24 '12 at 16:30
$\forall \chi: G \rightarrow G'$ omomorphism $o(\chi(x))| o(x)$, so the implication it correct. – Ivan Dec 24 '12 at 17:07
I am not sure I understand why you don't include the other cycles why just 2 and 3? – MathematicalPhysicist Dec 24 '12 at 17:35
OK, I read the part on generator in WIKI on symmetric group and I see how to solve this. – MathematicalPhysicist Dec 24 '12 at 17:45

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