# What are the group-homomorphisms $f: S_n\rightarrow \mathbb{C}^*$?

Denote by $$S_n$$ the symmetric group of order $$n$$ i.e. elements are bijective maps $$f :\{1,2,...,n\}\rightarrow \{1,2,...,n\}$$. What are the group-homomorphisms $$f: S_n\rightarrow \mathbb{C}^*$$?

It is clear to see that the image of a transposition $$(i,j)$$ is either 1 or -1. Thus for for any $$\sigma\in S_n$$ we have necessarly $$f(\sigma)=\pm 1$$. It seems that $$f$$ is constant and equal to 1.

Any help on that would be greatly appreciated.

• Letting $f(\sigma)=-1$ for all transpositions $\sigma$ is also a viable choice, see this. Oct 13, 2022 at 15:49
• One question you might ask yourself is "What can the Kernel of this homomorphism be?" Also note that $\mathbb C^*$ is abelian so the image of $f$ must be abelian. Are there any exceptional cases to consider? Oct 13, 2022 at 15:50
• Please use more descriptive titles. Oct 13, 2022 at 17:34
• $S_n$ hasn't got "order $n$", but $n!$ ($n$ is rather the degree of $S_n$). Moreover, the same symbol "$f$" is used with two different meanings. Oct 15, 2022 at 10:31

Let $$h:S_n\to \mathbb{C}^{\times}$$ be a group homomorphism. As you have noted, $$h$$ must map the transpositions to elements of order $$1$$ or $$2$$, i.e. $$1$$ or $$-1$$.
Since every permutation may be decomposed into a product of transpositions, then $$h$$ takes values in the discrete set $${-1,1}$$. This places a restriction on the size of the image of $$h$$, i.e., $$1\leq |\text{im}\,h|\leq 2$$.
By the first isomorphism theorem, $$S_n/\ker h\cong \text{im}\,h$$ hence $$\ker h$$ has index $$1$$ or $$2$$ in $$S_n$$. If $$\ker h$$ has index $$1$$, then $$h$$ is the trivial homomorphism. Otherwise, $$\ker h$$ has index two in $$S_n$$, so $$\ker h = A_n$$ and $$h$$ is the sign homomorphism.
In the case where $$|\text{im}\,h| = 1$$, then $$h$$ is the trivial homomorphism, mapping each element to $$1$$.
In the case where $$|\text{im}\, h|=2$$, then at least one transposition in $$S_n$$ must be mapped to $$-1$$ by $$h$$. Since all the transpositions are conjugate, then every transposition must be mapped to $$-1$$ by $$h$$. It follows that $$\ker h = A_n$$ and that $$h$$ is the sign homomorphism.