I'm beginning to study group theory and I need some help with this question:

Let $f:S_3 \rightarrow \mathbb{C^x}$ be a group homomorphism, where $\mathbb{C^x}=\mathbb{C} \setminus \{0\}$ denotes the multiplicative complex group.

Show that $f(S_3)$ is a subgroup of $\mu_6=\{z \in \mathbb{C}; z^6=1\}$ and determine all the homomorphisms from $S_3$ to $\mathbb{C^x}$.

My attempted solution:

Let $S_3=<\sigma_1,\sigma_2>$. We know that $f$ is determined by $f(\sigma_1)$ and $f(\sigma_2)$ and we have $\sigma_i^2=1$, so $f(\sigma_i)^6=f(\sigma_i)^2f(\sigma_i)^2f(\sigma_i)^2=f(\sigma_i^2)f(\sigma_i^2)f(\sigma_i^2)=1.1.1=1$

Then, $f(S_3)$ is contained in $\mu_6$ and therefore is a subgroup. (Is that right ?)

Furthermore, it is easy to show that $f(\sigma_1)=f(\sigma_2)$ using that $\sigma_1 \sigma_2 \sigma_1= \sigma_2 \sigma_1 \sigma_2$

Then, if $f(\sigma_1)=a \in \mu_6$, it is straightforward that $a$ has order 2 and it divides the order of $\mu_6$, but how can I describe all the homomorphisms?


It's always true that the image of a group homomorphism is a subgroup, which just follows from the fact that a homomorphism respects the group operations, i.e., that $f(g_1 g_2) = f(g_1) f(g_2)$.

In fact, you can show that $f(S_3)$ is a subgroup of $\mu_6$ quickly: Since $S_3$ has order $6$, the order of any element $s \in S_3$ has order dividing $6$---so what can we say about the order of $f(s)$? (You've managed to prove somewhat more than required here!)

There are a few approaches to describing all of the homomorphisms $f$: One is simply to pick a set of generators for $S_3$, and see to which elements $f$ can map them while still respecting the product rules. You've already observed that $f(\sigma_i)^2 = 1$, so each of $f(\sigma_i)$ must be $\pm 1$, and this leaves just four possible maps (only two turn out to be homomorphisms).

There's another approach that generalizes nicely, but may use some ideas you haven't see quite yet; the commutator subgroup of a group $G$ is the (normal) subgroup $[G, G]$ of $G$ generated by the elements of the form $ghg^{-1}h^{-1}$ (by definition such a product is equal to $1_G$ iff $g$ and $h$ commute). Now, given a homomorphism $f$ from $G$ to an abelian group $H$ (like $\mathbb{C}^{\times}$), any generator $ghg^{-1}h^{-1}$ of $[G, G]$ satisfies $$f(ghg^{-1}h^{-1}) = f(g) f(h) f(g^{-1}) f(h^{-1}) = f(g) f(h) f(g)^{-1} f(h)^{-1} = 1_H,$$ and the same reasoning applies to any element of $[G, G]$. So, the homomorphism $f$ factors to a homomorphism $G / [G, G] \to H$. Computing gives that $[S_3, S_3] = A_3$ and so $$S_3 / [S_3, S_3] = S_3 / A_3 \cong \mathbb{Z}_2,$$ and so any homomorphism $f: S_3 \to H$ is determined by a homomorphism $\mathbb{Z}_2 \to H$. This is determined by what $f$ does to the nonidentity element $z \in \mathbb{Z}_2$. In our case, $H = \mathbb{C}^{\times}$ and $$f(z)^2 = f(z^2) = f(1) = 1,$$ so $f(z)$ is either $1$ or $-1$, and each of these defines a homomorphism.


It seems that you managed to prove more than you were expected to. Good for you!

You know that $f(\sigma_1)=f(\sigma_2)$, and you know that this is a complex number $a$ satisfying $a^2=1$. How many of these do you know...? (I know two such numbers). It means that you have exactly two options - once you determine the image of $\sigma_1$, the image of $\sigma_2$ is the same, and all other images are determined since $\sigma_1,\sigma_2$ generate the whole group.


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