The only surjective homomorphism from $S_n$ to $\{1,-1\}$ is $\text{sign}$

Question

I want to show that if there is a surjective homomorphism $\phi: S_n \rightarrow \{1,-1\}$ for $n\geqslant 2$, then that homomorphism is $\text{sign}$, where $$\text{sign}\ \sigma = (-1)^{\text{no. of transpositions composed to produce} \ \sigma}.$$

My idea is this: as $\phi$ is a surjective function, then there is a $\sigma \in S_n$ such that $\phi (\sigma) = -1$. Now suppose $\sigma$ can be represented as a product of transpositions $t_1, t_2, ..., t_k$, so have $\phi(t_1)\phi(t_2)...\phi(t_k)=-1$, so at least one of $\phi(t_i)$ is $-1$.

My aim is to show that every transposition maps to $-1$, and then the result follows as any permutation can be represented as a product of transpositions. But so far, I only have that there must be a single transposition that maps to $-1$, any hints on where to go from here?

Answer 1

Hint Any two transpositions are conjugate in $S_n$, that is, for any transpositions $t, t' \in S_n$, there is a permutation $g \in S_n$ such that $$t' = g t g^{-1} .$$ What happens if you apply a homomorphism $\phi : S_n \to \{\pm 1\}$ to both sides of this equation?

Additional hint Use that $\{\pm 1\}$ is abelian.

Answer 2

The kernel must be a normal subgroup of index $2$. The only normal subgroup of index $2$ is $A_n$. For $n\geq 5$ this follows from simplicity; for $n=4$, if $N$ were a normal subgroup of index $2$ in $S_4$, different from $A_4$, then $N\cap A_4$ would be a normal subgroup of index $2$ in $A_4$... but $A_4$ does not have subgroups of order $6$. For $n=2,3$, this follows by inspection.

So all of $A_n$ maps to $1$; and everything not in $A_n$ must map to $-1$. So the map is indistinguishable from the sign map, regardless of how it is defined.