Question

Let \( C_n \) denote a cyclic group having \( n \) elements. If there is a surjective group homomorphism from \( C_n \) to \( C_{30} \), then the total number of such distinct surjective homomorphisms is .

Hint:

The number of surjective homomorphisms from a cyclic group onto \(C_m\) is equal to the number of generators of \(C_m\), which is \(\phi(m)\), whenever such a map exists.

Answer 1

Correct Answer: 8

Step 1: Use what fixes a map.
A homomorphism from a cyclic group is decided fully by where it sends one generator.

Step 2: What surjective needs.
If $\phi:C_n\to C_{30}$ is onto, then the image of a generator of $C_n$ must itself be a generator of $C_{30}$.

Step 3: Count generators of $C_{30}$.
The number of generators of a cyclic group of order $30$ is Euler's $\phi(30)$.

Step 4: Compute $\phi(30)$.
\[ \phi(30)=30\Big(1-\tfrac12\Big)\Big(1-\tfrac13\Big)\Big(1-\tfrac15\Big)=8 \]

Step 5: Conclude.
So there are $8$ choices for the image, giving $8$ surjective homomorphisms.
\[ \boxed{8} \]