I need to find all group homomorphisms of $S_n\to\mathbb{C}^*$ (for $n\ge2$), and also all group homomorphisms of $A_n\to\mathbb{C}^*$ (for $n\ge2$).

Someone on this site, two years ago, seems to have already asked the same question (Finding homomorphisms of $S_n$ → C* and $A_n$ → C*). I understand what the questioner says there, but I don't understand it when he says:

Therefore there is exactly one $g:\mathbb{Z}/2\mathbb{Z}\to\mathbb{C}^*$, so all elements of $a\in\mathbb{C}^*$ give us $\operatorname{ord}(a)\mid g[\mathbb{Z}/2\mathbb{Z}]$, so $a$ can only be $-1$ or $1$. Now we know that $f:S_n\to\{-1,1\}$ …

What is he saying there? And how to go on from there? (I do not understand the answer given either …)

I hope that someone here can help me further. (And if I'm not allowed to ask this question because is has already been asked two years ago, please tell me what to do instead).

Thank you in advance!

  • 3
    $\begingroup$ A transposition $(a,b)$ has order $2$, so $f((a,b))=\pm1$. The transpositions generate $S_n$, so $f$ maps everything to $\pm1$. $\endgroup$ – David C. Ullrich May 2 '18 at 13:50

The question (and comment) you refer to is a sketch of how to solve this problem. The key fact is the following: $C^* = \{ z \in \mathbb{C} \mid z \neq 0 \}$ is an abelian group under the operation of multiplication of complex numbers.

The relevant theorem is the following: if $G$ is a group and $A$ is an abelian group, then any homomorphism $f:G \to A$ must factor through the abelianization of $G$. This means that there is a homomorphism $f_1:G/[G,G] \to A$ such that $f = f_1 \circ \pi$, where $\pi: G \to G/[G,G]$ is the natural map to the abelianization, which we take to be defined as quotient of $G$ by its commutator subgroup $[G,G]$.

In your case, it is well-known that if $G = S_n$ and $n > 2$, then $G/[G,G] \cong C_2 = \{-1,1\}$, the cyclic group of order two. Any homomorphism $f: S_n \to C^*$ will be determined by a homomorphism $f_1:C_2 \to C^*$.

So, this reduces your question to the following: What are all homomorphisms $f_1:C_2 \to C^*$?


Here is another take for the first part.

Let $f: S_n \to \mathbb C^*$ be a group homomorphism. Let $\tau$ be a transposition. Then $\tau^2=1$ implies $f(\tau)=\pm 1$. Since every permutation $\sigma$ is a product of transpositions, we have $f(\sigma)=\pm 1$. Therefore, the image of $f$ has size at most $2$ and so the index of $\ker f$ in $S_n$ is at most $2$. Thus, the only possibilities for $\ker f$ are $S_n$ and $A_n$. Therefore, $f$ is trivial or the sign homomorphism.

  • $\begingroup$ For the last part, I'd like to say that every homomorphism $A_n \to \mathbb C^*$ can be extended to an homomorphism $S_n \to \mathbb C^*$ but I'm not sure this is true. $\endgroup$ – lhf May 2 '18 at 16:05
  • 1
    $\begingroup$ It is not true that every homomorphism $A_n \to \mathbb{C}^*$ extends to $S_n$. Since $A_3 \cong C_3$, it can be mapped isomorphically onto the cube roots of unity; this map does not extend to $S_3$ since it is generated by elements of order 2. The group $A_4$ is non-abelian; its abelianization is $C_3$, with the nontrivial elements having coset representatives $(123)$ and $(132)$; the element $(12)(34)$ belongs to $[A_4,A_4]$. For $n \geq 5$, $A_n$ is simple and so its abelianization is trivial; in this case, every homomorphism $S_n \to \mathbb{C}^*$ can be extended. $\endgroup$ – Robert Bell May 3 '18 at 20:01
  • $\begingroup$ @RobertBell, thanks! $\endgroup$ – lhf May 3 '18 at 20:08

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