Say there is an epimorphism $g$ from $G_1$ to $G_2$ , where $G_2$ is a cyclic group. Is $G_1$ also cyclic ? Prove or disprove.

Question

The converse holds so if $G_1$ is cyclic it is shown that also $G_2$ is cyclic. I can not take the inverse of the function $g$ since the function is not injective ( i wanted to show that the generator of $G_2$ is mapped to the generator of $G_1$). Since the identity element of $G_2$ is mapped to $Kerg$ , then the generator of $G_2$ is mapped to one of the partitioned classes of $G_1$ and because of structure perserverance $G_1/Kerg$ isomorphic to $G_2$ and hence is cyclic. From this point on I am stuck since I can not prove that if $G_1/Kerg$ is cyclic then $G$ is cyclic?
Thank you,

Answer 1

This is false. Map $S_n \to \mathbb{Z}_2$ ($n\geq 3$) using the sign of the permutation. The target is cyclic (and in fact, $\mathbb{Z}_2$ is isomorphic to a subgroup of $S_n$ for $n\geq 2$, just take $\langle (1,2) \rangle$ as the subgroup - just in case you need an endomorphism specifically), but the source is decidedly not.

To put in an edit quickly, if $f$ is an injection, then this is true. See my comment.

Answer 2

If by epimorphism of groups one understands a surjective homomorphism the the existence is sometimes possible, sometimes not. A counterexample has already been given in another answer. An example where it exists is given by $G_1 = C_{2m}$, a cyclic group of order $2m$ and $G_2 = C_m$, a subgroup of $G_1$, the epimorphsm given by the projection $g \mapsto g^2$.