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Asaf Karagila
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Without sitting to verify the details in full, here is a sketch of a proof:

Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of Choice in problem 10.5 (p. 149). The construction is to create two infinite sets which span isomorphic vector spaces, but have different cardinalities.

Replace "countable" by $\aleph_2$ and "finite" by $\aleph_1$. The result will satisfy $\sf DC_{\aleph_1}$, and the conclusion should still hold. Moreover, you can take the free abelian group generated by those sets instead of a vector space, and it would solve the problem.

If one prefers to do that by forcing instead of atoms, then it is also possible, although slightly more technical.

The second question seem to me answerable positively in $\sf ZF$. If the two groups are spanned by $A'$ and $B'$, then a bijection $f\colon A'\to B'$ extends uniquely to an isomorphism between $A$ and $B$ which is surjective.

Without sitting to verify the details in full, here is a sketch of a proof:

Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of Choice in problem 10.5 (p. 149). The construction is to create two infinite sets which span isomorphic vector spaces, but have different cardinalities.

Replace "countable" by $\aleph_2$ and "finite" by $\aleph_1$. The result will satisfy $\sf DC_{\aleph_1}$, and the conclusion should still hold. Moreover, you can take the free abelian group generated by those sets instead of a vector space, and it would solve the problem.

If one prefers to do that by forcing instead of atoms, then it is also possible, although slightly more technical.

The second question seem to me answerable positively in $\sf ZF$. If the two groups are spanned by $A'$ and $B'$, then a bijection $f\colon A'\to B'$ extends uniquely to an isomorphism between $A$ and $B$ which is surjective.

Without sitting to verify the details in full, here is a sketch of a proof:

Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of Choice in problem 10.5 (p. 149). The construction is to create two infinite sets which span isomorphic vector spaces, but have different cardinalities.

Replace "countable" by $\aleph_2$ and "finite" by $\aleph_1$. The result will satisfy $\sf DC_{\aleph_1}$, and the conclusion should still hold. Moreover, you can take the free abelian group generated by those sets instead of a vector space, and it would solve the problem.

If one prefers to do that by forcing instead of atoms, then it is also possible, although slightly more technical.

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Source Link
Asaf Karagila
  • 39.9k
  • 8
  • 135
  • 283

Without sitting to verify the details in full, here is a sketch of a proof:

Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of Choice in problem 10.5 (p. 149). The construction is to create two infinite sets which span isomorphic vector spaces, but have different cardinalities.

Replace "countable" by $\aleph_2$ and "finite" by $\aleph_1$. The result shouldwill satisfy $\sf DC_{\aleph_1}$, and the conclusion wouldshould still hold. Moreover, you can take the free abelian group generated by those sets instead of a vector space, and it would solve the problem.

If one prefers to do that by forcing instead of atoms, then it is also possible, although slightly more technical.

The second question seem to me answerable positively in $\sf ZF$. If the two groups are spanned by $A'$ and $B'$, then a bijection $f\colon A'\to B'$ extends uniquely to an isomorphism between $A$ and $B$ which is surjective.

Without sitting to verify the details in full, here is a sketch of a proof:

Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of Choice in problem 10.5 (p. 149). The construction is to create two infinite sets which span isomorphic vector spaces, but have different cardinalities.

Replace "countable" by $\aleph_2$ and "finite" by $\aleph_1$. The result should satisfy $\sf DC_{\aleph_1}$, and the conclusion would still hold. Moreover, you can take the free abelian group generated by those sets instead of a vector space, and it would solve the problem.

If one prefers to do that by forcing instead of atoms, then it is also possible, although slightly more technical.

The second question seem to me answerable positively in $\sf ZF$. If the two groups are spanned by $A'$ and $B'$, then a bijection $f\colon A'\to B'$ extends uniquely to an isomorphism between $A$ and $B$ which is surjective.

Without sitting to verify the details in full, here is a sketch of a proof:

Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of Choice in problem 10.5 (p. 149). The construction is to create two infinite sets which span isomorphic vector spaces, but have different cardinalities.

Replace "countable" by $\aleph_2$ and "finite" by $\aleph_1$. The result will satisfy $\sf DC_{\aleph_1}$, and the conclusion should still hold. Moreover, you can take the free abelian group generated by those sets instead of a vector space, and it would solve the problem.

If one prefers to do that by forcing instead of atoms, then it is also possible, although slightly more technical.

The second question seem to me answerable positively in $\sf ZF$. If the two groups are spanned by $A'$ and $B'$, then a bijection $f\colon A'\to B'$ extends uniquely to an isomorphism between $A$ and $B$ which is surjective.

Source Link
Asaf Karagila
  • 39.9k
  • 8
  • 135
  • 283

Without sitting to verify the details in full, here is a sketch of a proof:

Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of Choice in problem 10.5 (p. 149). The construction is to create two infinite sets which span isomorphic vector spaces, but have different cardinalities.

Replace "countable" by $\aleph_2$ and "finite" by $\aleph_1$. The result should satisfy $\sf DC_{\aleph_1}$, and the conclusion would still hold. Moreover, you can take the free abelian group generated by those sets instead of a vector space, and it would solve the problem.

If one prefers to do that by forcing instead of atoms, then it is also possible, although slightly more technical.

The second question seem to me answerable positively in $\sf ZF$. If the two groups are spanned by $A'$ and $B'$, then a bijection $f\colon A'\to B'$ extends uniquely to an isomorphism between $A$ and $B$ which is surjective.