Is there a surjective homomorphism from $\oplus_{i=0}^\infty $ $\Bbb Z_p$ into $\prod_{i=0}^\infty$ $\Bbb Z_p$?

Question

Is there a surjective homomorphism from $\oplus_{i=0}^\infty $ $\Bbb Z_p$ into $\prod_{i=0}^\infty$ $\Bbb Z_p$?

Can this be shown by the order of element?

Can we say this homomorphism is one-one, then if onto, then the order of this element is the same?

I mean all of the elements of $\oplus_{i=0}^\infty $ $\Bbb Z_p$ have finite order but in $\prod_{i=0}^\infty$ $\Bbb Z_p$ there are elements of infinite order. Therefore, one cannot find an onto homomorphism.

Answer 1

Your idea is right:

If $x$ has finite order and your group homomorphism is $\phi$, then $\phi(x)$ has an order that is less than or equal to that of $x$. This is true regardless of whether $\phi$ is 1-1 or not. Thus you can't map onto elements of infinite order. When I read the original post, I misread that the indexing was over $p$! But of course, if you are just looking at many copies of $\Bbb Z_p$ for fixed $p$, elements still have finite order, so your approach will not work.

Another way to see that there is no onto homomorphism is to consider cardinalities. There cannot even be a set mapping of $\oplus \Bbb Z_p$ onto $\prod \Bbb Z_p$, because the former is countable and the latter is uncountable.