Existence of abelian group which has no “square-root” but whose “cube” has a “square-root”
4
$begingroup$
Does there exist an abelian group $G$ such that $G ncong H times H$ for every abelian group $H$ but $G times G times G cong K times K$ for some abelian group $K$ ?
Also see Existence of topological space which has no "square-root" but whose "cube" has a "square-root"
group-theory category-theory abelian-groups group-isomorphism group-homomorphism
asked Dec 27 '18 at 1:05
user521337user521337
1,1861417
$endgroup$
add a comment |
4
$begingroup$
Does there exist an abelian group $G$ such that $G ncong H times H$ for every abelian group $H$ but $G times G times G cong K times K$ for some abelian group $K$ ?
Also see Existence of topological space which has no "square-root" but whose "cube" has a "square-root"
group-theory category-theory abelian-groups group-isomorphism group-homomorphism
asked Dec 27 '18 at 1:05
user521337user521337
1,1861417
$endgroup$
1
$begingroup$
Structure theorem for finitely generated abelian groups clearly shows that such a group cannot be finitely generated.
$endgroup$
– Crostul
Dec 27 '18 at 1:15
4
$begingroup$
@RobArthan: no ... I do mean what I wrote ...
$endgroup$
– user521337
Dec 27 '18 at 1:38
1
$begingroup$
@Crostul: I'm not so sure about the torsion part ...
$endgroup$
– user521337
Dec 27 '18 at 1:39
$begingroup$
OK. I just wanted to check that you didn't mean "some". "any" would be more idiomatic than "every" in this context,
$endgroup$
– Rob Arthan
Dec 27 '18 at 23:42
add a comment |
4
4
4
4
$begingroup$
Does there exist an abelian group $G$ such that $G ncong H times H$ for every abelian group $H$ but $G times G times G cong K times K$ for some abelian group $K$ ?
Also see Existence of topological space which has no "square-root" but whose "cube" has a "square-root"
group-theory category-theory abelian-groups group-isomorphism group-homomorphism
asked Dec 27 '18 at 1:05
user521337user521337
1,1861417
$endgroup$
Does there exist an abelian group $G$ such that $G ncong H times H$ for every abelian group $H$ but $G times G times G cong K times K$ for some abelian group $K$ ?
Also see Existence of topological space which has no "square-root" but whose "cube" has a "square-root"
group-theory category-theory abelian-groups group-isomorphism group-homomorphism
group-theory category-theory abelian-groups group-isomorphism group-homomorphism
asked Dec 27 '18 at 1:05
user521337user521337
1,1861417
asked Dec 27 '18 at 1:05
user521337user521337
1,1861417
asked Dec 27 '18 at 1:05
user521337user521337
1,1861417
asked Dec 27 '18 at 1:05
user521337user521337
1,1861417
asked Dec 27 '18 at 1:05
user521337user521337
1,1861417
1,1861417
1
$begingroup$
Structure theorem for finitely generated abelian groups clearly shows that such a group cannot be finitely generated.
$endgroup$
– Crostul
Dec 27 '18 at 1:15
4
$begingroup$
@RobArthan: no ... I do mean what I wrote ...
$endgroup$
– user521337
Dec 27 '18 at 1:38
1
$begingroup$
@Crostul: I'm not so sure about the torsion part ...
$endgroup$
– user521337
Dec 27 '18 at 1:39
$begingroup$
OK. I just wanted to check that you didn't mean "some". "any" would be more idiomatic than "every" in this context,
$endgroup$
– Rob Arthan
Dec 27 '18 at 23:42
add a comment |
1
$begingroup$
Structure theorem for finitely generated abelian groups clearly shows that such a group cannot be finitely generated.
$endgroup$
– Crostul
Dec 27 '18 at 1:15
4
$begingroup$
@RobArthan: no ... I do mean what I wrote ...
$endgroup$
– user521337
Dec 27 '18 at 1:38
1
$begingroup$
@Crostul: I'm not so sure about the torsion part ...
$endgroup$
– user521337
Dec 27 '18 at 1:39
$begingroup$
OK. I just wanted to check that you didn't mean "some". "any" would be more idiomatic than "every" in this context,
$endgroup$
– Rob Arthan
Dec 27 '18 at 23:42
1
1
$begingroup$
Structure theorem for finitely generated abelian groups clearly shows that such a group cannot be finitely generated.
$endgroup$
– Crostul
Dec 27 '18 at 1:15
$begingroup$
Structure theorem for finitely generated abelian groups clearly shows that such a group cannot be finitely generated.
$endgroup$
– Crostul
Dec 27 '18 at 1:15
4
4
$begingroup$
@RobArthan: no ... I do mean what I wrote ...
$endgroup$
– user521337
Dec 27 '18 at 1:38
$begingroup$
@RobArthan: no ... I do mean what I wrote ...
$endgroup$
– user521337
Dec 27 '18 at 1:38
1
1
$begingroup$
@Crostul: I'm not so sure about the torsion part ...
$endgroup$
– user521337
Dec 27 '18 at 1:39
$begingroup$
@Crostul: I'm not so sure about the torsion part ...
$endgroup$
– user521337
Dec 27 '18 at 1:39
$begingroup$
OK. I just wanted to check that you didn't mean "some". "any" would be more idiomatic than "every" in this context,
$endgroup$
– Rob Arthan
Dec 27 '18 at 23:42
$begingroup$
OK. I just wanted to check that you didn't mean "some". "any" would be more idiomatic than "every" in this context,
$endgroup$
– Rob Arthan
Dec 27 '18 at 23:42
add a comment |
0
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1
$begingroup$
Structure theorem for finitely generated abelian groups clearly shows that such a group cannot be finitely generated.
$endgroup$
– Crostul
Dec 27 '18 at 1:15
4
$begingroup$
@RobArthan: no ... I do mean what I wrote ...
$endgroup$
– user521337
Dec 27 '18 at 1:38
1
$begingroup$
@Crostul: I'm not so sure about the torsion part ...
$endgroup$
– user521337
Dec 27 '18 at 1:39
$begingroup$
OK. I just wanted to check that you didn't mean "some". "any" would be more idiomatic than "every" in this context,
$endgroup$
– Rob Arthan
Dec 27 '18 at 23:42