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Let $A$ and $B$ be free abelian groups and

$0\to A\to B\to\Bbb{Z}/p\to 0$

a short exact sequence.

Are $A$ and $B$ necessarily isomorphic?

The only examples I can come up with are of the form

$0\to \Bbb{Z}\xrightarrow{\cdot p}\Bbb{Z}\to\Bbb{Z}/p\to 0$

or very similar examples with groups of higher rank. I'm not asking whether the map $A\to B$ is an isomorphism, but if there exists some isomorphism $A\cong B$. I've heard that counterexamples can be built using large cardinals, but I don't know anything about cardinal theory. Any alternative or a detailed explanation with references about cardinals would be appreciated.

Javi
  • 6,541
  • Related. – Shaun Oct 03 '19 at 15:51
  • $A$ and $B$ must have the same rank, so yes. – Derek Holt Oct 03 '19 at 16:09
  • @DerekHolt why and what about groups of infinite rank? – Javi Oct 03 '19 at 16:29
  • There is a standard result that, for any subgroup $A$ of a finitely generated abelian group $B$, there is a free basis $b_1,\ldots,b_n$ of $B$ and positive integers $d_1,\ldots,d_r$ for some $r < n$ such that $A$ is generated by $d_1b_1,\ldots,d_rb_r$. It follows from that. For infinite rank just observe that $A$ and $B$ have the same cardinality. – Derek Holt Oct 03 '19 at 16:38
  • @DerekHolt from that it follows that $rank A\leq rank B$, but where do you get equality from? In the infinite case, the same but with cardinalities – Javi Oct 03 '19 at 16:45
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    Think about it! What would the quotient be otherwise? – Derek Holt Oct 03 '19 at 16:46
  • @DerekHolt ok I see it in the finitely generated case. But I'm not sure why groups of different cardinality cannot yield $\Bbb{Z}/p$ as a quotient. I intuitively see that the difference is too huge but I'd like to see a rigorous argument, since I don't know if there is an analogue of taking coordinates with respect to a free basis. – Javi Oct 03 '19 at 20:52
  • Well $B$ is a union of $p$ sets having the same cardinality as $A$, so $B$ must have the same cardinality. – Derek Holt Oct 03 '19 at 21:39

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