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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 $\text {Card}(F(G)) \leq \text{Card}( G)$

Ali Taghavi
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    Hint: First solve the problem for sets instead of abelian groups. – Tom Goodwillie Apr 15 '18 at 02:43
  • @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 $G$? – Ali Taghavi Apr 15 '18 at 17:45
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    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. – YCor Apr 15 '18 at 18:09
  • @YCor what is that group for the countable case? – Ali Taghavi Apr 16 '18 at 04:36
  • @YCor According to your argument, do not we need that the universal group you pointed out has the same cardinality $k$? And is it always possible? Please see https://mathoverflow.net/questions/28999/is-there-a-universal-countable-group-a-countable-group-containing-every-counta – Ali Taghavi Apr 17 '18 at 06:57
  • Yes forgot to say "of the same cardinal". Such a group is $(\mathbf{Q}\oplus\mathbf{Q}/\mathbf{Z})^{(\kappa)}$. – YCor Apr 17 '18 at 07:08
  • @YCor but this group is not countable when k is countable. – Ali Taghavi Apr 17 '18 at 07:17
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    Sure it's countable (countable direct sum of countable groups). $A^{(B)}$ denotes $\bigoplus_{b\in B}A$. – YCor Apr 17 '18 at 07:21
  • @YCor yes. I was thinking to direct product. – Ali Taghavi Apr 17 '18 at 07:23
  • @YCor but please read the linked question and its answer. – Ali Taghavi Apr 17 '18 at 07:30
  • I know this standard fact: for countable groups instead of countable abelian groups, the result fails. Have you got any particular reason to point me to this link? – YCor Apr 17 '18 at 07:45
  • @YCor I did not pay attention that "Abelian" is not included in the linked question. So I wondered "is there a contradictory situation"?. So i sent you that link because i was surprised by that contradiction!!Any way sorry for my mistake. – Ali Taghavi Apr 17 '18 at 07:51
  • @YCor so it seems that if I remove the word "Abelian"from my question then the question is nontrivial. Yes? – Ali Taghavi Apr 17 '18 at 07:53
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    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). – YCor Apr 17 '18 at 08:14

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