Originally Posted by

**matt.qmar** I take it that an epimorphism is a surjective homomorphism?

is the hint you are giving me, that the order of an image under the homomorphism must divide the order of the group being mapped into?

ie, if $\displaystyle \phi : G \rightarrow H$, then $\displaystyle |\phi(g)|$ divides $\displaystyle |H|$

is that true? I think so!

So there are no surjective homomorphisms, because there would need to be an element of order 8 in $\displaystyle Z_{20}$ to map to an element of order 8 in $\displaystyle Z_{8}$ to get the whole group? Is there some homomorphism property which dictates that?

Sure! if $\displaystyle \Phi:G\to H$ is a group hom. and $\displaystyle ord(g)=k\,,\,g\in G$ , then $\displaystyle \Phi(g)^k=\Phi(g^k)=\Phi(1)=1\Longrightarrow ord(\Phi(g))\mid k$

and there is 3 homomorphisms to $\displaystyle Z_{8}$, because elements of $\displaystyle Z_{20}$ have order 1,2,4,5,10,20 and elements of $\displaystyle Z_{8}$ have orders 1,2,4, and 8, and those have 3 possible orders in common?!

Yeppers

Coincidentially?, Euler's phi function of 8 is 3....?

Not, it is $\displaystyle \phi(8)=4$

Tonio

Is any/some/all of this reasoning valid?

Thank you!!