Let $f:G \to H$ be a group homomorphism. Which the statement is true? Why?
If $H$ is finite, then so is $G$.
If $H$ is finite cyclic, then so is $G$.
If $H$ is abelian, then so is $G$.
Edit: $f$ isn't trivial group homomorphism.
My attempt:
If $G$ is finite, then so is $H$, but not as converse. The converse is hold iff $f$ is injective.
Same with point 1) above.
Same with point 1) above and let $f(a),f(b)\in H$. Then, $f(a)f(b) = f(b)f(a) \Leftrightarrow f(ab)=f(ba)$. Note that $ab=ba$ is not necessarily true.
Any idea? Thanks in advanced.
Remember that there is a trivial group homomorphism between any two groups, namely the function $f\colon G\to H$ defined by $f(g) = e_H$ for all $g\in G$ (here $e_H$ denotes the identity element of $H$).
With this in mind, is there any information about the group $H$ that would give us information about the group $G$?