3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ what is $p$‌‌‌‌‌‌‌‌‌‌? $\endgroup$ – user795571 Feb 16 '15 at 18:45
  • 4
    $\begingroup$ What is the downvote for? This question shows a lot of thought compared to many PSQs... $\endgroup$ – rschwieb Feb 16 '15 at 18:49
  • 2
    $\begingroup$ @rschwieb The voting behaviors on this website are sometimes incomprehensible. The question even got a close vote! Seriously... $\endgroup$ – Najib Idrissi Feb 16 '15 at 20:53
7
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ It seems to me that all the elements in $\prod \mathbb{Z}_p$ actually have finite order dividing $p$... If $x = (x_i)_i \in \prod \mathbb{Z}_p$, then $p \cdot x = (p \cdot x_i)_i = (0)_i$ (here I assume that $\mathbb{Z}_p$ is the cyclic group with $p$ elements -- the $p$-adic integers all have infinite order anyway). $\endgroup$ – Najib Idrissi Feb 16 '15 at 20:56
  • $\begingroup$ Well no, that's my point... Since all the elements of $\prod \mathbb{Z}_p$ have finite order too, you reach no contradiction (in this way) by assuming that they are all in the image. Your argument is that elements of infinite order cannot be in the image, but there aren't any. (The countability argument works though). $\endgroup$ – Najib Idrissi Feb 17 '15 at 8:10
  • $\begingroup$ @NajibIdrissi Hmm... I initially misread your comment apparently, so I removed my old one and here is a new one: what do you believe the order of the all $1$'s element is? $\endgroup$ – rschwieb Feb 17 '15 at 11:00
  • $\begingroup$ Isn't it just $p$? $(1,1,\dots) + (1,1,\dots) + \dots + (1,1,\dots) = (p,p,\dots) = (0,0,\dots)$. $\endgroup$ – Najib Idrissi Feb 17 '15 at 12:07
  • $\begingroup$ @NajibIdrissi Oh, that's where we were looking at things differently. When I read the post, I misread that the index $p$ ranging over primes, and not many copies of $\Bbb Z_p$ for fixed $p$! While reading the user's approach, I thought they were using this angle, and I read carelessly. Thanks for alerting me to this: I'll definitely make changes. $\endgroup$ – rschwieb Feb 17 '15 at 13:20

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.