Question:
The number of group homomorphisms from \( {Z}/47 \) to \( S_4 \) is equal to (answer in integer):
The number of group homomorphisms from \( {Z}/47 \) to \( S_4 \) is equal to (answer in integer):
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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.
Updated On: Feb 1, 2025
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Solution and Explanation
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 \( {16} \).
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