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

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)

• Are any of $m,n,p$ co-prime to each other? – user458276 Feb 3 at 21:41
• Not necessarily. – koch Feb 3 at 21:41
• 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? – Robert Chamberlain Feb 3 at 21:46
• This rather depends on what $m$, $n$ and $p$ are. – Lord Shark the Unknown Feb 3 at 21:46
• Tensor with $\Bbb{Z}/(p)$. – jgon Feb 3 at 22:34

## 1 Answer

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.

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