Application of First Fundamental theorem on Isomorphism.

Question

I am studying Homomorphism and came across its first fundamental theorem, I am curious about how it can be used to solve the below given problem and shall be thankful to know if there is some other way to approach this sort of a problem.

Let $f:S_3 \to S^1$ be a non-trivial non-injective group homomorphism. Then, the number of elements in the Kernel of $f$ is?

Here $S_3$ is group of symmetries on $3$ symbols and $S^1 = \{z\in \mathbb C ~\mid~ |z|=1\}$.

Answer 1

Look at the commutator subgroup $[S_3,S_3]$. This is the subgroup generated by all of the commutators. The commutator of $a,b\in S_3$ is defined as $[a,b]=aba^{-1}b^{-1}$. Why care? Because for any $a,b\in S_3$: $$f([a,b])=f(a)f(b)f(a)^{-1}f(b)^{-1}$$ Because $S^1$ is albelian/communative we can simplify this to: $$f(a)f(b)f(a)^{-1}f(b)^{-1}=f(a)f(a)^{-1}f(b)f(b)^{-1}=1$$ So $[S_3,S_3]\subset\ker{f}$. This is a common technique. Whenever you have a homomorphsim from a group to an albelian group, the kernel must contain the commutator-subgroup, which really constrains what $\ker{f}$ can be.

It turns out this subgroup is exactly $A_3$, but why is this the case? Here's a proof from my algebra book: $A_3$ is generated by all of the 3-cycles, and any 3-cycle is a commutator because of the following formula: $$[(x\ y),(x\ z)]=(x\ y)(x\ z)(x\ y)(x\ z)=(x\ y\ z)$$ So $A_3\subseteq [S_3,S_3]$. Then let $\varepsilon:S_3\longrightarrow \{1,-1\}$ be the signature image. Notice that for any $a,b\in S_3$: $$\varepsilon([a,b])=\varepsilon(a)\varepsilon(b)\varepsilon(a)^{-1}\varepsilon(b)^{-1} =\varepsilon(a)\varepsilon(a)^{-1}\varepsilon(b)\varepsilon(b)^{-1}=1$$ So $[S_3,S_3]\subseteq A_3$, so $[S_3,S_3]=A_3$.

Using that $A_3\subseteq\ker{f}$ and that $f$ cannot be trivial, I think you can prove yourself that $A_3=\ker{f}$, so $\#\ker{f}=3$.

Answer 2

More generally, let's consider the possible group homomorphisms $f: S_n \to \mathbb C^*$.

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 $2$. (Use the isomorphism theorems here!) Thus, the only possibilities for $\ker f$ are $S_n$ and $A_n$.

Therefore, $f$ is trivial or the sign homomorphism.