We know the fact that: If a module $M$ is finitely cogenerated, then every module that cogenerates $M$ finitely cogenerates $M$. Conversely, it is not true. I find an example in the book "Rings and Categories of modules" written by Frank W. Anderson and Kent R. Fuller.

Example The abelian group $\bigoplus_{\mathbb{P}}{\mathbb{Z}_p}$, ($p$ is a prime) is not finitely cogenerated yet every group that cogenerates it finitely cogenerates it.

I am at a loss for this example. Any help will be appreciated. Clearly, we can regard the abelian group $\bigoplus_{\mathbb{P}}{\mathbb{Z}_p}$ as a module over $\mathbb Z$.

I post my effort. (1) $\mathbb{Z}_p$ is simple. (2) $\bigoplus_{\mathbb{P}}{\mathbb{Z}_p}$ is semisimple and since $\mathbb{P}$ can be $\infty$, $\bigoplus_{\mathbb{P}}{\mathbb{Z}_p}$ is not finitely generated, then it is not finitely cogenerated.


I think what you've written so far is in the right direction, but probably could be expressed more directly this way:

$\bigoplus_{p\in P}\mathbb Z_p\hookrightarrow \prod_{p\in P}\mathbb Z_p$ obviously cannot be pruned down to a finite subset $F\subseteq P$, so $\bigoplus_{p\in P}\mathbb Z_p$ is not finitely cogenerated. (Eliminating the position in the product for $p$ would remove all possible nonzero images for an element of order $p$ in $\bigoplus_{p\in P}\mathbb Z_p$.)

Then for the reverse direction, the point is to establish that if $\bigoplus_{p\in P}\mathbb Z_p\hookrightarrow\prod_{i\in I}G$, you can embed it into finitely many copies of $G$. In this case, actually, I think it's true that $\bigoplus_{p\in P}\mathbb Z_p\hookrightarrow G$.

Let $x_p$ generate $\mathbb Z_p$ and $\phi$ be the embedding. $\phi(x_p)$ is nonzero on some position in $\prod _{i\in I}G$, and we can project on that coordinate to get a nonzero element of $G$, necessarily of order $p$. So $G$ contains a copy of $\mathbb Z_p$. This is true for every prime $p$, and obviously owing to the prime orders of these copies, their sum is direct. So $G$ contains a copy of $\bigoplus_{p\in P}\mathbb Z_p$.

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    $\begingroup$ Thank you very much, what is the $G$, is it any abelian group? $\endgroup$ – Daisy Apr 19 '17 at 14:11
  • $\begingroup$ @ rschwieb I know the fact that: Every finitely generated abelian group $G\neq \langle e \rangle $ in which every element (except e) has order $p$ is isomorphic to $\mathbb{Z}_p \bigoplus \mathbb{Z}_p \bigoplus ...\bigoplus \mathbb{Z}_p$ (n summands) for some $n\geq 1$. So maybe we can not embed it , it is epimorphism. $\endgroup$ – Daisy Apr 19 '17 at 14:26
  • $\begingroup$ @Daisy Yes, $G$ is an arbitrary abelian group. I do not understand the second comment. If you are considering using an elementary $p$-group for $G$, then you should see immediately that it does not cogenerate $\bigoplus_{p\in P} \mathbb Z_p$ right away, finitely or otherwise. I'm not sure why you are mentioning epimorphisms either. $\endgroup$ – rschwieb Apr 19 '17 at 16:14
  • $\begingroup$ I understand it, thank you very much. $\endgroup$ – Daisy Apr 20 '17 at 1:47

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