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 $\mathbb Z_2 \times \mathbb Z_2$

  • 2
    $\begingroup$ 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

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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.