Number of homomorphisms between two arbitrary groups

Suppose $f:A_5 \to S_4$ be a homomorphism. Then $\ker f$ is a normal subgroup of $A_5$. But $A_5$ is simple, so $$\ker f \in \Big\{ \{e\},A_5\Big\}$$

• $\ker f=\{e\}$ implies $$A_5/\{e\} \sim f(A_5)$$ and so $f(A_5)$ is a subgroup of order $60$ in $S_4$, which is not possible in $S_4$.
• $\ker f=A_5$ implies $f$ is trivial

Hence $$\Big\vert\{f \;\vert \;f:A_5 \to S_4 \;\text{is a homomorphism} \}\Big\vert=1$$

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For finding homomorphism $f$ for arbitraay two groups, use the following facts:

• $\vert f(g) \vert$ divides $\vert g \vert$ where $g$ belong to the domain with $\vert g \vert < \infty$ [this is useful for finite groups]
• $f(g^n)=[f(g)]^n$
• List all normal subgroups of domain and use first isomorphism theorem