How do I find all homomorphisms between $S_3 \to \mathbb{Z}_{15}$?

Question

How do I find all homomorphisms between $S_3 \to \mathbb{Z}_{15}$?

I know that since $S_3 \cong \langle x,y: x^3 = y^2 = e;yx = x^2y\rangle$, the homomorphism is dependent on $\phi{(x)}$ and $\phi{(y)}$.

$\phi{(x)}$ must have order of $1$ or $3$ and $\mathbb{Z}_{15}$ has elements of both $1$ and $3$. How do I proceed from here?

Answer 1

Let $\phi$ be a homomorphism from $S_3$ to $\mathbb{Z}_{15}$. Let $\sigma$ be a transposition in $S_3$. Then $\phi(\sigma) = \phi(\sigma^{15}) = 15\cdot \phi(\sigma)= \bar{0}$. So every transposition gets sent to the identity, and since every element of $S_3$ is a product of transpositions, everything gets sent to the identity. Hence $\phi$ is the trivial homomorphism.

Answer 2

You expect to find several homomorphisms.

The image of $x$, $\varphi(x)$, must have order dividing the order of $x$, so must have order $1$ or $3$. Every such choice might yield a homomorphism.

The image of $y$, $\varphi(y)$ must have order dividing the order of $y$, so must have order $1$ or $2$. Every such choice might yield a homomorphism.

In this particular problem, there are no elements of $\Bbb{Z}/15\Bbb{Z}$ of order $2$, so $\varphi(y) = 0$, since that is the only element of this additive group of order $1$. Therefore, there is only one option for $\varphi(y)$.

The elements of $\Bbb{Z}/15\Bbb{Z}$ of order $3$ are $5 \pmod{15}$ and $10 \pmod{15}$. (You can find this by constructing a table of elements and computing their orders, using the $\gcd$ of $15$ and the element.) So $\varphi(x)$ is sent to one of the congruence classes represented by $0$, $5$, and $10$.

So there are three possible images for $x$ and one possible image for $y$. That gives three potential homomorphisms. Fortunately, the image is cyclic, $\langle \varphi(x)\rangle$, so that we do not have to verify that the multiplication table in $S_3$ is carried homomorphically along $\phi$. (For instance, in a more general problem, we would need to verify $\varphi(xy) = \varphi(x)\varphi(y)$ and (if either group is nonabelian) $\varphi(yx) = \varphi(y) \varphi(x)$, as well as check the other $23$ relations in the multiplication table, although many of those can be reduced to others. Happily, the image of a cycle is cyclic, guaranteed by the requirement that the images of $x$ and $y$ have orders dividing the orders of $x$ and of $y$, respectively, so the multiplication table is guaranteed to be carried by $\varphi$.)

(If we didn't impose the image order divides the preimage order, then we would have to verify the entire multiplication table is carried by the map. Since we do impose that requirement, we know that the powers of $x$ are carried to a subgroup and the powers of $y$ are carried to a subgroup. We don't know that powers of other elements, various products of $x$ and $y$, are carried to something sensible, that respects the multiplication tables, so that is something we have to check.)