# Finding all homomorphisms between $S_3$ and $\mathbb Z_8$

Finding all homomorphisms between $S_3$ and $\mathbb Z_8$

The method I saw on solving this was trying to figure out the homomorphisms based on the possible orders of image and kernel. Another thing was saying that homomorphisms are defined by generators and relations. I didn't quite understand that last bit. Glad if someone could explain, and what's the proper way of approaching such problem.

Thank you for any assistance!

• The possible orders thingy looks promising: the only possibility seems to be that the image of such a homom. has order $\;2\;$...but also because the only non-trivial normal subgroup of $\;S_3\;$ is the alternating group $\;A_3\;$ , of order $\;3\;$ ... – DonAntonio Apr 21 '14 at 16:39

$S_3$ is generated by an element $\tau$ of order three and an element $\sigma$ of oreder two, say $\tau=(1\,2\,3)$ and $\sigma=(1\,2)$, that is every element of $S_3$ can be written as some product of $\tau$s and $\sigma$s, for example $(2\,3)=\tau^2\sigma$. Thus if $f\colon S_3\to G$ is a homomorphism and we know $f(\tau)$ and $f(\sigma)$ then we already know $f$ completely, for example $f(\tau^2\sigma)=f(\tau)f(\tau)f(\sigma)$. However, we cannot pick arbitrary elements of $G$ as images $f(\sigma)$ and $f(\tau)$. First of all, $\sigma^2=1$ implies that $f(\sigma)^2=1$, similarly $f(\tau)^3=1$, i.e. we may pick as $f(\sigma)$, $f(\tau)$ only elements that have the same order as $\sigma$ and $\tau$, respectively, or a divisor of that order. Also, $\sigma\tau=\tau^2\sigma$, so we must also make sure that $f(\sigma)f(\tau)=f(\tau)^2f(\sigma)$, i.e. any relation that holds for $\sigma,\tau$ must also hold for their images.
As it turns out, the relations $\sigma^2=\tau^2=1$ and $\sigma\tau=\tau^2\sigma$ completely determine the group law of $S_3$.
What does this tell us if $G=\mathbb Z/8\mathbb Z$? Here, $G$ has no elements of order $3$, thus $3\cdot f(\tau)=0+8\mathbb Z$ (note that we are using additive notation in $\mathbb Z/8\mathbb Z$) can only be achieved if $f(\tau)=0+8\mathbb Z$. And for $f(\sigma)$ we have only the choices $0+8\mathbb Z$ and $4+8\mathbb Z$ because we need $2\cdot f(\sigma)=0+8\mathbb Z$. Both turn out to be valid choices, which can be verified manually - or one uses the indented paragraph above and checks that the third relation is also respected by (both) $f$.
Hint: consider that, called $\tau$ any homomorphism, $[G,G] \subset Ker(\tau),$ so you have to count homomorphisms from $\mathbb{Z}_2$ to $\mathbb{Z}_8$ and they are two.