Are there two groups $G_1 , G_2 $ of orders $7$ and $8$ respectively & a morphism $f: G_1 \to G_2 $ such that $|\operatorname{Im}(f)| = 4$?

Question

Is is possible to find two groups $G_1, G_2 $ of orders $7$ and $8$ respectively & a morphism $f: G_1 \to G_2$ such that $|\operatorname{Im}(f)| = 4$ ?

Answer 1

Since $G_1/\ker(f) \cong \operatorname{Im}(f)$ per the isomorphism theorem, we have $|G_1/\ker(f)| = |G_1|/|\ker(f)| = |\operatorname{Im}(f)|$. Is this possible?

In greater generality, notice that when $|G| = p$ is prime, any homomorphism $\phi:G \rightarrow G'$ is either trivial or an isomorphism since $|\ker(\phi)| = 1 \text{ or } p$.