Converse of first isomorphism theorem

Question

Prove or disprove the following statement :

If the mapping from $G/\ker\phi$ to $\phi(G)$, given by $g\ker\phi\mapsto\phi(g)$, is an isomorphism, then $\phi:G\to\bar G$ is a homomorphism.

Answer 1

This is the closest thing to a converse I can think of.

Let $G, G'$ be groups, let $\phi: G \rightarrow G'$ be a function, and let $K = \{ g \in G : \phi(g) = e_{G'}\}$. Suppose that $K$ is a normal subgroup of $G$, $\phi(G)$ is a subgroup of $G'$, and suppose that the function $\overline{\phi}:G/K \rightarrow \phi(G)$ given by $\overline{\phi}(xK) = \phi(x)$ is well defined and an isomorphism. Then $\phi$ is a group homomorphism.

Answer: Yes, because $\phi(xy) = \overline{\phi}(xyK) = \overline{\phi}(xK) \overline{\phi}(yK) = \phi(x)\phi(y)$.