
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!!