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How to show that there is no surjection between $Z \oplus Z$ to $Z_m \times Z_n \times Z_p$?.

Intuitively, $Z_m \times Z_n \times Z_p$, if there is such a surjection then $Z_m \times Z_n \times Z_p$ can be generated by two elements, but $Z_m \times Z_n \times Z_p$ "has to be generated by at least three elements". How to formalize this idea?

EDIT: The original question is that if $G = \oplus_{i=1}^{n} Z_{p^{k_i}}$, $k_1 \geq k_2 \geq \ldots \geq k_n$, and there are two elements in $G$ that generates $G$, then $n$ can at most be $2$.($p$ here is a prime)

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  • $\begingroup$ Are any of $m,n,p$ co-prime to each other? $\endgroup$ – user458276 Feb 3 at 21:41
  • $\begingroup$ Not necessarily. $\endgroup$ – koch Feb 3 at 21:41
  • $\begingroup$ I assume you mean surjective homomorphism? Also, there is such a homomorphism for some triples $(m,n,p)$, are there other conditions or do you just want to find find some case for which no such homomorphism exists? $\endgroup$ – Robert Chamberlain Feb 3 at 21:46
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    $\begingroup$ This rather depends on what $m$, $n$ and $p$ are. $\endgroup$ – Lord Shark the Unknown Feb 3 at 21:46
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    $\begingroup$ Tensor with $\Bbb{Z}/(p)$. $\endgroup$ – jgon Feb 3 at 22:34
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If $m$ and $n$ are coprime a way to state the Chinese Remainder Theorem is that the natural map $$ \Bbb Z\longrightarrow\Bbb Z_m\times\Bbb Z_n $$ is surjective. Using this you can construct surjective maps as in the question for many triples $(m,n,p)$.

This was written BEFORE the question was edited.

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  • $\begingroup$ I am sorry, but I was not fast enough to edit before you posted the answer. $\endgroup$ – koch Feb 3 at 22:00

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