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I have the following question :

Proof/Disproof : is there a surjective homomorphism such that $f:\mathbb{Z}_{20}\rightarrow \mathbb{Z}_{2} \oplus\mathbb{Z}_{2}$

I don't really know how to approach this problem.

I do understand that $\mathbb{Z}_{20}$ cyclic and I think that $\mathbb{Z}_{2} \oplus\mathbb{Z}_{2}$ is not cyclic since $0$ is not a generator $1$ is also not we get the following sub group when we use $1$ $\{0,0\},\{1,1\}$ and $2$ is also not a generator.

Any ideas how approach this?, Is it true is it false?

Thank you.

  • 2
    Every such homomorphism is determined by the image of 1. This implies... – 2017-01-27
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    @LeonSot I'm still don't get it, can you go into more detail? – 2017-01-27
  • 1
    @JaVaPG: See my answer. – 2017-01-27

1 Answers 1

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Let $\varphi:\mathbb Z_{20}\longrightarrow \mathbb Z_2\oplus \mathbb Z_2$ an homomorphism. In particular, $\varphi(k)=k\varphi(1)$, and thus $$\text{Im}(\varphi)=\left<\varphi(1)\right>.$$ Therefore, the range of such homomorphism is cyclic. Since $\mathbb Z_2\oplus\mathbb Z_2$ is not cyclic, such homomorphism can't be surjective.