$G$ is a group: $|G|=20$.
Is there such a group G, for which the homomorphism $\tau :G-->Z_{10}$ exist?$$$$ The same question for: $\tau :G-->Z_{15}$
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I think that I should use here the fact $o\left(\tau \left(g\right)\right)| o\left(g\right)$
What I know is: In $Z_{10}$ the order of it's elements is: 1,2,5,10, and in $G$ the order of it's elements can be (1,2,4,5,10,20).
And $\tau$ is homomorphism so: $\tau \left(g_1g_2\right)=\tau \left(g_1\right)+\tau \left(g_2\right)\left(modn\right)$
Now, from here how do I correlate between the fact about the order and the formula of homomorphism, I can try to find a group like that manually trying to find each element and build such a group (or fail bulding it), but it is hard work and not so smart defintly.
I'll suppose that you want a surjective homomorphism or the answer is trivial.
For the first question you just can choose $\mathbb{Z}_2 \times \mathbb{Z}_{10}$.
For the second answer you can't have a surjective homomorphism because 15 doesn't divide 20, but you know that $\mathbb{Z}_{15}$ contains a subgroup of order five so you can still have a non trivial morphism by choosing $$G=\mathbb{Z}_4 \times \mathbb{Z}_{5}$$