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.

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    $\begingroup$ Letting $f(\sigma)=-1$ for all transpositions $\sigma$ is also a viable choice, see this. $\endgroup$ Oct 13, 2022 at 15:49
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    $\begingroup$ 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? $\endgroup$ Oct 13, 2022 at 15:50
  • $\begingroup$ Please use more descriptive titles. $\endgroup$
    – Shaun
    Oct 13, 2022 at 17:34
  • $\begingroup$ $S_n$ 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. $\endgroup$
    – citadel
    Oct 15, 2022 at 10:31

1 Answer 1


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.

Alternatively, we argue in the following way:

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.


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