Question

Let \( \phi : S_3 \to S_1 \) be a non-trivial non-injective group homomorphism. Then the number of elements in the kernel of \( \phi \) is .............

Hint:

For finite groups, use \( |G| = |\ker \phi| \times |\text{Im } \phi| \). Non-injective means \( \ker \phi \) has more than one element.

Answer 1

Correct Answer: 3

Step 1: Order of \( S_3. \)
\(|S_3| = 6.\)

Step 2: Use the Fundamental Theorem of Homomorphisms.
\[ |S_3| = |\ker \phi| \cdot |\text{Im } \phi|. \] Since the homomorphism is non-trivial and non-injective, \(|\text{Im } \phi| > 1\) and \( |\ker \phi| > 1.\) Possible factors of 6 satisfying this: - \(|\ker \phi| = 3, |\text{Im } \phi| = 2.\)

Step 3: Verify subgroup structure.
A normal subgroup of order 3 exists in \( S_3 \) (the cyclic subgroup generated by a 3-cycle). Hence, \(|\ker \phi| = 3.\)

Final Answer: \[ \boxed{3} \]