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I'm trying to an exam of last year, to practice. I find this exercise quite difficult:

Determine the number of elements of $\text{Hom}(\mathbb{Z}/5 \mathbb{Z}, D_5)$ and of $\text{Hom} (\mathbb{Z}/6 \mathbb{Z} ,D_6)$.

I don't know where to start. A hint would be appreciated to guide me in the right direction.

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Hint: in both cases the domain of definition is a cyclic group and thus any homomorphism from it to any other group has cyclic image. – DonAntonio May 26 '13 at 17:27

Suppose $\phi: \mathbb{Z}_5 \to D_5$ is a group homomorphism. Then $\phi$ is determined by $\phi(1)$ since $\phi(3) = \phi(1)^3$ etc. So we just need to decide where $\phi(1)$ goes. Where can it go? Well we must have $\phi(1)^5=1_{D_5}$ so $\phi(1)$ is either rotation by $0, 2\pi/5, 4\pi/5, 6\pi/5$ or $8\pi/5.$ It can be any of those, so there are $5$ elements in the first Hom group. Use similar ideas for the second one.

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