TITLE: Count the number group homomorphisms from $S_3$ to $\mathbb{Z}/6\mathbb{Z}$?
QUESTION [9 upvotes]: I have to count the number of group homomorphisms from $S_3$ to  $\mathbb{Z}/6\mathbb{Z}$ ?

1
2
3
6

I aware of the formula for calculating group homomorphisms defined on cyclic groups, but here $S_3$ is not cyclic. Please suggest how to proceed.

Thank you so much Egbert for the much detailed and patient reply.

REPLY [5 votes]: So we consider $\phi(S_3).$ Note that $|\phi(S_3)|$ divides $6$. So the possiblities are $1,2,3,6$.
Suppose $|\phi(S_3)| = 6.$ Then it maps to whole of $\mathbb Z_6,$ which is not possible as $\mathbb Z_6$ is abelian and cyclic.
Suppose $|\phi(S_3)| = 3.$ Then $ |\ker\phi| = 2$, as there is no normal subgroup of order $2$ in $S_3$ the case gets violated.
Finally  $|\phi(S_3)| = 2,$ where $\ker\phi$ is $A_3$  and when $|\phi(S_3)| = 1,$ $\ker \phi = S_3.$ So there are two homomorphisms.