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The external direct product of groups $G$ and $H$ is the same group $$ G \times H = \{ (g,h)\| g \in G , h \in H\} $$ whose operation is defined componentwise, that is,

$(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$ for all $g_1,g_2 \in G $

and for all $h_1,h_2 \in H$. Is there a homomorphism from $Z_4$ to $Z_2 \times Z_2$ that is surjective (i.e onto)? If so, specify it.


attempt 1] know that $Z_4$ is cyclic and $Z_2xZ_2$ is not so they wont be isomorphic something is failing either not homorphic, into or bijective at the least

trying to play with something like $f:Z_4 \to Z_2 \times Z_2$ by $f(z)=([z]_{k_1},[z]_{k_2})$ not sure what the external direct product has to do with it.

read from similar post that the order of the kernel is two so i dont think there is one that is onto

Question about method of finding homomorphisms from $\mathbb Z_4$ to $\mathbb Z_2 \times \mathbb Z_2$


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    $\begingroup$ Reference, recall, or prove (!!) the almost-trivial statement that the homomorphic image of a cyclic group is cyclic. Then you have your answer, since you said you know already the right-had side group is not cyclic. $\endgroup$ – mathguy Oct 4 '17 at 21:57
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Both groups are finite and of the same size, hence surjectivity implies injectivity using the pigeonhole principle. Hence, requiring a homomorphism that is surjective actually requires an isomorphism. Contradict that with your reasoning that the groups won't be isomorphic.

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A homomorphic image of a cyclic group is always cyclic but $\mathbb Z_2 \times \mathbb Z_2$ is not cyclic.

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