For every integer $n \geq 2$ , there is a unique non-trivial homomorphism $f: {S_n} \to \mathbb{C^*}$.

Question

For every integer $n \geq 2$ , there is a unique non-trivial homomorphism $f: {S_n} \to \mathbb{C^*}$.

How can I prove this statement? I know there will be a homomorphism as $S_n \over A_n$ is isomorphic to ${(1 ,-1).}$ But how can I prove that there is no other homomorphism than that?

Can anyone please help me?

Answer 1

Since $\mathbb{C}^*$ is commutative, any homomorphism from $S_n$ into $\mathbb{C}^*$ is such that its restriction to the derived subgroup of $S_n$ is the trivial homomorphism. And, if $n\geqslant2$, $S_n'=A_n$.

Answer 2

Because normal subgroup equals kernel of homomorphism^[1] I will first recap the normal subgroups of $S_{n}$. More concretely, we seek proper normal subgroups because $f$ is required to be non-trivial.

For $n\geqslant 2$, $A_{n}\triangleleft S_{n}$. For $n\ne 4$, $A_{n}$ is the only proper normal subgroup of $S_{n}$. For $A_{4}$, the only other proper normal subgroup is isomorphic to $K_{4}$. ^[2]

For $n\geqslant 2$ and $n\ne4$, because $\operatorname{ker}{f}\triangleleft S_{n}$ and $f$ is non-trivial, $\operatorname{ker}{f}$ could only be $A_{n}$. Therefore by the first isomorphism theorem, $\operatorname{Im}{f}\cong S_{n}/A_{n}\cong\mathbb{Z}_{2}$. The only subgroup with order $2$ of $\mathbb{C}^{*}$ is $\left\{\pm 1\right\}$. Therefore, there is a unique choice for $f$.

For $n=4$, we wish to exclude the case of $\operatorname{ker}{f}\cong K_{4}$. We attempt a proof by contradiction. Assume that $\operatorname{ker}{f}\cong K_{4}$, then $S_{4}/\operatorname{ker}{f}\cong S_{3}$ ^[3] and by the first isomorphism theorem, $S_{4}/\operatorname{ker}{f}$ is isomorphic to a subgroup of $\mathbb{C}^{*}$. But this is impossible because $\mathbb{C}^{*}$ is Abelian while $S_{3}$ is not, so no $f$ can exist with $K_{4}$ as the kernel.

Answer 3

I think there's a proof that uses fewer advanced facts than the ones that have been posted.

$S_n$ is generated by a 2-cycle $a=(1\ 2)$ and an $n$-cycle $b=(1\ 2\ \cdots\ n)$. Also, $ab=(1\ 3\ 4\ \cdots\ n)$, an $n-1$-cycle.

Now let $f$ be a homomorphism. Then $1=f(1)=f(a^2)=(f(a))^2$, so $f(a)=1$ or $f(a)=-1$. Similarly, $(f(b))^n=1$, and $(f(a))^{n-1}(f(b))^{n-1}=1$. It follows that $f(b)=(f(a))^{n-1}$. So there are two choices for $f(a)$, and once you've made that choice, that determines the value of $f(b)$ – and once you have $f(a)$ and $f(b)$, you have $f$. Hence, there are only the two homomorphisms.