Question

The number of group homomorphisms from \( {Z}/47 \) to \( S_4 \) is equal to (answer in integer):

Hint:

When counting homomorphisms from a cyclic group, focus on the orders of elements in the target group that divide the order of the source group.

Answer 1

Step 1: Properties of \( {Z}/47 \). The group \( {Z}/47 \) is cyclic of order 47. For a group homomorphism \( \phi: {Z}/47 \to S_4 \), the image of \( \phi \) is completely determined by the image of the generator of \( {Z}/47 \). Step 2: Constraints on homomorphisms. The generator of \( {Z}/47 \) can be mapped to any element of \( S_4 \). However, for \( \phi \) to be a homomorphism, the order of the image element must divide 47 (the order of the source group). Since 47 is prime, the possible orders of the image are 1 and 47. Step 3: Valid elements in \( S_4 \). - The identity element of \( S_4 \) has order 1. - There are 15 elements of \( S_4 \) that have order dividing 47 (all elements of \( S_4 \), except those whose orders are incompatible with 47). Step 4: Counting homomorphisms. Thus, there are \( 1 + 15 = 16 \) valid mappings. Step 5: Conclusion. The number of group homomorphisms from \( {Z}/47 \) to \( S_4 \) is \( {16} \).