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Is there a functor $F$ from the category of abelian groups to itself such that for every non trivial group $G$, $F(G)$ can not be embedded in $G$?

Edit: According to the comment by Prof. Goodwillie I change the question as follows:

Is there a functor $F$ on the category of infinite abelian groups which does not increase the cardinality of groups but for every infinite group $G$, the group $F(G)$ can not be embedded in $G$? By "Does not increase the cardinality" we mean $\text {Card}(F(G)) \leq \text{Card}( G)$

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    $\begingroup$ Hint: First solve the problem for sets instead of abelian groups. $\endgroup$
    – Tom Goodwillie
    Commented Apr 15, 2018 at 2:43
  • $\begingroup$ @TomGoodwillie Thank you very much for your comment and your hint. I am sorry if I posted a non research question. According to your comment I try to improve it as follows. "Is there a functor $F$ on the category of infinite abelian groups which does not increase the cardinality but $F(G)$ can not be embedded in $G$? $\endgroup$
    – Ali Taghavi
    Commented Apr 15, 2018 at 17:45
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    $\begingroup$ It's again not research level, because for every infinite cardinal $\kappa$ there's an abelian group in which every abelian group of cardinal $\le\kappa$ embeds. $\endgroup$
    – YCor
    Commented Apr 15, 2018 at 18:09
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    $\begingroup$ Sure it's countable (countable direct sum of countable groups). $A^{(B)}$ denotes $\bigoplus_{b\in B}A$. $\endgroup$
    – YCor
    Commented Apr 17, 2018 at 7:21
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    $\begingroup$ Possibly. Actually I don't know for uncountable groups: it a first step it might be better to first separately ask whether for some infinite cardinal $\kappa$ there exists a group of cardinal $\kappa$ in which every group of cardinal $\le\kappa$ embeds as a subgroup. (I just did an unsuccessful Google search). $\endgroup$
    – YCor
    Commented Apr 17, 2018 at 8:14

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