Question

The number of surjective (onto) group homomorphisms from \( S_4 \) to \( \mathbb{Z}_6 \) is equal to ................

Hint:

For surjective homomorphisms from \( S_n \) to \( \mathbb{Z}_m \), focus on the possible kernels and the properties of the group orders. The number of surjections can often be found by examining the structure of normal subgroups.

Answer 1

Step 1: Understand the problem.
We are asked to find the number of surjective (onto) group homomorphisms from the symmetric group \( S_4 \) to the cyclic group \( \mathbb{Z}_6 \). A homomorphism from a group \( G \) to a group \( H \) is a function that preserves the group operation, and a surjective homomorphism is one that maps onto all of \( H \). Step 2: Consider the properties of \( S_4 \) and \( \mathbb{Z}_6 \).
The group \( S_4 \) is the symmetric group on 4 elements, which has order 24. The group \( \mathbb{Z}_6 \) is a cyclic group of order 6. The structure of homomorphisms from \( S_4 \) to \( \mathbb{Z}_6 \) depends on the possible kernel sizes and the image of the homomorphism. Step 3: Identify possible kernels.
By the First Isomorphism Theorem, the number of surjective homomorphisms is equal to the number of distinct homomorphisms where the kernel is a normal subgroup of \( S_4 \). The homomorphisms must map elements in \( S_4 \) to elements in \( \mathbb{Z}_6 \). It is known that there are exactly **2** surjective homomorphisms from \( S_4 \) to \( \mathbb{Z}_6 \), one corresponding to the trivial homomorphism and the other corresponding to a specific non-trivial homomorphism. Final Answer: \[ \boxed{2}. \]