To determine the number of homomorphisms from the alternating group $A_4$ to $\mathbb{Z}_3$, we follow these steps:

Key Observations:

1. Homomorphism Basics:

   • A homomorphism $\phi: A_4 \to \mathbb{Z}_3$ maps the identity element of $A_4$ to the identity element of $\mathbb{Z}_3$.
   • The image of $\phi$ must be a subgroup of $\mathbb{Z}_3$.

2. Structure of $A_4$ and $\mathbb{Z}_3$:

   • $A_4$ is a non-abelian group of order 12.
   • $\mathbb{Z}_3$ is a cyclic group of order 3, so it has only one proper subgroup, namely the trivial subgroup $\{0\}$.

3. Homomorphism Implies Kernel:

   • The kernel of a homomorphism is a normal subgroup of $A_4$.
   • Since $A_4$ is simple (it has no non-trivial normal subgroups), the kernel must either be $A_4$ itself or $\{e\}$.

Analysis:

• If the kernel is $A_4$, the homomorphism is trivial (maps all elements of $A_4$ to 0 in $\mathbb{Z}_3$).
• If the kernel is $\{e\}$, the homomorphism is injective. However, since $A_4$ cannot be embedded into $\mathbb{Z}_3$ (due to order constraints), there are no injective homomorphisms.

Conclusion:

The only homomorphism from $A_4$ to $\mathbb{Z}_3$ is the trivial homomorphism. Therefore, the number of homomorphisms is 1.

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Would you like a detailed explanation of any specific step or related examples? Here are some related questions to explore:

1. What is the definition of a group homomorphism?
2. Why is $A_4$ a simple group, and how does this affect homomorphisms?
3. How do we determine the kernel of a homomorphism in general?
4. What are the normal subgroups of $A_4$?
5. Can $A_4$ be mapped homomorphically to any other cyclic group?

Tip: Always analyze the structure and normal subgroups of a group when evaluating homomorphisms!