I was solving some questions about homomorphism and isomorphism but then I made up a question. Here is the question: "If $\dfrac{G}{Ker(\phi)} \cong H$ then can we say there must be a homomorphism between $G$ and $H$?
$\underline{\text{Where does this question come from?}}$
I encountered a question on Bhattacharya's book:
Show that a cyclic group of order 8 is homomorphic to a cyclic group of order 4
Solution:
Since these cyclic groups are finite I use the theorem saying every finite cyclic group is isomorphic to $\mathbb{Z}_n$. Consider the normal subgroup $\langle 4 \rangle$ of $\mathbb{Z}_8$. Hence we can consider Quotient group $\mathbb{Z}_8/\langle4\rangle$. A clear consequence of that is: $$\mathbb{Z}_8/\langle4\rangle \cong \mathbb{Z_4}$$
So by First Isomorphism Theorem we can say there must be exist a homomorphism between $\mathbb{Z}_8$ and $\mathbb{Z}_4$
Firstly, as was mentioned in the comments, the statement "$G/\text{ker} \phi \cong H$" has no meaning without first defining $\phi$. The proper question to ask would be if $G/N\cong H$ for some normal subgroup $N\trianglelefteq G$, does this imply that there is a homomorphism $\phi: G\rightarrow H$? The answer to this question is of course yes since we can compose with the canonical map $\pi:G \rightarrow G/N$ to obtain a map $$ G\xrightarrow{\pi} G/N \xrightarrow{\cong} H$$
To determine all of the homomorphic images of a group $G$, we need only look at all the normal subgroups since these will correspond to the possible kernels of maps out of $G$.