Finite-by-torsion-free abelian groups (or compact abelian groups with finitely many components)












5












$begingroup$


Here's a question I should know the answer to but don't:




Suppose $1to F to G to G/F to 1$ is a short exact sequence of abelian groups with $F$ finite and $G/F$ torsion-free. Must the sequence split?




This is not true if you merely assume that $F$ is torsion. A counterexample is given by YCor here: https://mathoverflow.net/a/314536/20598.



Equivalently, suppose $G$ is a compact abelian group with finitely many components. Then does $G_0 to G to G/G_0$ split? This is not true without assuming there are finitely many components, as YCor's example shows, and it's also not true for nonabelian groups, as Max's answer here shows: https://math.stackexchange.com/a/954539/23805 (though I think it's true whenever $G/G_0$ is cyclic).






gr.group-theory abelian-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    The answer is yes. I don't remember right now where to find a reference.
    $endgroup$
    – YCor
    Jan 12 at 11:55






  • 1




    $begingroup$
    In fact, you only need the torsion subgroup has bounded order. See the (currently broken, but I’ll fix it as soon as I’ve posted this) link in mathoverflow.net/questions/60525/…
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 12:10






  • 1




    $begingroup$
    @JeremyRickard great! "bounder order" (used in the post you link) is awkward, it should be "of bounded exponent".
    $endgroup$
    – YCor
    Jan 12 at 12:15










  • $begingroup$
    @YCor I agree, but I decided to stick with the terminology of the reference.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 12:24
















5












$begingroup$


Here's a question I should know the answer to but don't:




Suppose $1to F to G to G/F to 1$ is a short exact sequence of abelian groups with $F$ finite and $G/F$ torsion-free. Must the sequence split?




This is not true if you merely assume that $F$ is torsion. A counterexample is given by YCor here: https://mathoverflow.net/a/314536/20598.



Equivalently, suppose $G$ is a compact abelian group with finitely many components. Then does $G_0 to G to G/G_0$ split? This is not true without assuming there are finitely many components, as YCor's example shows, and it's also not true for nonabelian groups, as Max's answer here shows: https://math.stackexchange.com/a/954539/23805 (though I think it's true whenever $G/G_0$ is cyclic).






gr.group-theory abelian-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    The answer is yes. I don't remember right now where to find a reference.
    $endgroup$
    – YCor
    Jan 12 at 11:55






  • 1




    $begingroup$
    In fact, you only need the torsion subgroup has bounded order. See the (currently broken, but I’ll fix it as soon as I’ve posted this) link in mathoverflow.net/questions/60525/…
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 12:10






  • 1




    $begingroup$
    @JeremyRickard great! "bounder order" (used in the post you link) is awkward, it should be "of bounded exponent".
    $endgroup$
    – YCor
    Jan 12 at 12:15










  • $begingroup$
    @YCor I agree, but I decided to stick with the terminology of the reference.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 12:24














5












5








5


2



$begingroup$


Here's a question I should know the answer to but don't:




Suppose $1to F to G to G/F to 1$ is a short exact sequence of abelian groups with $F$ finite and $G/F$ torsion-free. Must the sequence split?




This is not true if you merely assume that $F$ is torsion. A counterexample is given by YCor here: https://mathoverflow.net/a/314536/20598.



Equivalently, suppose $G$ is a compact abelian group with finitely many components. Then does $G_0 to G to G/G_0$ split? This is not true without assuming there are finitely many components, as YCor's example shows, and it's also not true for nonabelian groups, as Max's answer here shows: https://math.stackexchange.com/a/954539/23805 (though I think it's true whenever $G/G_0$ is cyclic).






gr.group-theory abelian-groups







share|cite|improve this question











$endgroup$




Here's a question I should know the answer to but don't:




Suppose $1to F to G to G/F to 1$ is a short exact sequence of abelian groups with $F$ finite and $G/F$ torsion-free. Must the sequence split?




This is not true if you merely assume that $F$ is torsion. A counterexample is given by YCor here: https://mathoverflow.net/a/314536/20598.



Equivalently, suppose $G$ is a compact abelian group with finitely many components. Then does $G_0 to G to G/G_0$ split? This is not true without assuming there are finitely many components, as YCor's example shows, and it's also not true for nonabelian groups, as Max's answer here shows: https://math.stackexchange.com/a/954539/23805 (though I think it's true whenever $G/G_0$ is cyclic).






gr.group-theory abelian-groups




gr.group-theory abelian-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 7:30







Sean Eberhard

















asked Jan 12 at 11:33









Sean EberhardSean Eberhard

3,5911431




3,5911431












  • $begingroup$
    The answer is yes. I don't remember right now where to find a reference.
    $endgroup$
    – YCor
    Jan 12 at 11:55






  • 1




    $begingroup$
    In fact, you only need the torsion subgroup has bounded order. See the (currently broken, but I’ll fix it as soon as I’ve posted this) link in mathoverflow.net/questions/60525/…
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 12:10






  • 1




    $begingroup$
    @JeremyRickard great! "bounder order" (used in the post you link) is awkward, it should be "of bounded exponent".
    $endgroup$
    – YCor
    Jan 12 at 12:15










  • $begingroup$
    @YCor I agree, but I decided to stick with the terminology of the reference.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 12:24


















  • $begingroup$
    The answer is yes. I don't remember right now where to find a reference.
    $endgroup$
    – YCor
    Jan 12 at 11:55






  • 1




    $begingroup$
    In fact, you only need the torsion subgroup has bounded order. See the (currently broken, but I’ll fix it as soon as I’ve posted this) link in mathoverflow.net/questions/60525/…
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 12:10






  • 1




    $begingroup$
    @JeremyRickard great! "bounder order" (used in the post you link) is awkward, it should be "of bounded exponent".
    $endgroup$
    – YCor
    Jan 12 at 12:15










  • $begingroup$
    @YCor I agree, but I decided to stick with the terminology of the reference.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 12:24
















$begingroup$
The answer is yes. I don't remember right now where to find a reference.
$endgroup$
– YCor
Jan 12 at 11:55




$begingroup$
The answer is yes. I don't remember right now where to find a reference.
$endgroup$
– YCor
Jan 12 at 11:55




1




1




$begingroup$
In fact, you only need the torsion subgroup has bounded order. See the (currently broken, but I’ll fix it as soon as I’ve posted this) link in mathoverflow.net/questions/60525/…
$endgroup$
– Jeremy Rickard
Jan 12 at 12:10




$begingroup$
In fact, you only need the torsion subgroup has bounded order. See the (currently broken, but I’ll fix it as soon as I’ve posted this) link in mathoverflow.net/questions/60525/…
$endgroup$
– Jeremy Rickard
Jan 12 at 12:10




1




1




$begingroup$
@JeremyRickard great! "bounder order" (used in the post you link) is awkward, it should be "of bounded exponent".
$endgroup$
– YCor
Jan 12 at 12:15




$begingroup$
@JeremyRickard great! "bounder order" (used in the post you link) is awkward, it should be "of bounded exponent".
$endgroup$
– YCor
Jan 12 at 12:15












$begingroup$
@YCor I agree, but I decided to stick with the terminology of the reference.
$endgroup$
– Jeremy Rickard
Jan 12 at 12:24




$begingroup$
@YCor I agree, but I decided to stick with the terminology of the reference.
$endgroup$
– Jeremy Rickard
Jan 12 at 12:24










1 Answer
1






active

oldest

votes


















12












$begingroup$

Here’s a quick homological proof.



Suppose $F$ is finite and $H$ torsion free. Then $Fcongtext{Hom}(F,mathbb{Q}/mathbb{Z})$, so
$$text{Ext}^1(H,F)congtext{Ext}^1left(H,text{Hom}(F,mathbb{Q}/mathbb{Z})right)
congtext{Hom}left(text{Tor}_1(H,F),mathbb{Q}/mathbb{Z}right),
$$
which is zero since torsion free abelian groups are flat.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Is the last isomorphism obvious? I mean, I could probably easily prove it, but...
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 13:06






  • 2




    $begingroup$
    @მამუკაჯიბლაძე $text{Hom}(-,text{Hom}(F,mathbb{Q}/mathbb{Z}))congtext{Hom}(-otimes F,mathbb{Q}/mathbb{Z})$. I’m just applying the first right derived functor of each side to $H$.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 14:01












  • $begingroup$
    Very slick. Nice.
    $endgroup$
    – Sean Eberhard
    Jan 12 at 14:14






  • 1




    $begingroup$
    ...and you get $text{Tor}$ on the right since $text{Hom}(-,mathbb Q/mathbb Z)$ is exact. Fine.
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 19:34













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1 Answer
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1 Answer
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oldest

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12












$begingroup$

Here’s a quick homological proof.



Suppose $F$ is finite and $H$ torsion free. Then $Fcongtext{Hom}(F,mathbb{Q}/mathbb{Z})$, so
$$text{Ext}^1(H,F)congtext{Ext}^1left(H,text{Hom}(F,mathbb{Q}/mathbb{Z})right)
congtext{Hom}left(text{Tor}_1(H,F),mathbb{Q}/mathbb{Z}right),
$$
which is zero since torsion free abelian groups are flat.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Is the last isomorphism obvious? I mean, I could probably easily prove it, but...
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 13:06






  • 2




    $begingroup$
    @მამუკაჯიბლაძე $text{Hom}(-,text{Hom}(F,mathbb{Q}/mathbb{Z}))congtext{Hom}(-otimes F,mathbb{Q}/mathbb{Z})$. I’m just applying the first right derived functor of each side to $H$.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 14:01












  • $begingroup$
    Very slick. Nice.
    $endgroup$
    – Sean Eberhard
    Jan 12 at 14:14






  • 1




    $begingroup$
    ...and you get $text{Tor}$ on the right since $text{Hom}(-,mathbb Q/mathbb Z)$ is exact. Fine.
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 19:34


















12












$begingroup$

Here’s a quick homological proof.



Suppose $F$ is finite and $H$ torsion free. Then $Fcongtext{Hom}(F,mathbb{Q}/mathbb{Z})$, so
$$text{Ext}^1(H,F)congtext{Ext}^1left(H,text{Hom}(F,mathbb{Q}/mathbb{Z})right)
congtext{Hom}left(text{Tor}_1(H,F),mathbb{Q}/mathbb{Z}right),
$$
which is zero since torsion free abelian groups are flat.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Is the last isomorphism obvious? I mean, I could probably easily prove it, but...
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 13:06






  • 2




    $begingroup$
    @მამუკაჯიბლაძე $text{Hom}(-,text{Hom}(F,mathbb{Q}/mathbb{Z}))congtext{Hom}(-otimes F,mathbb{Q}/mathbb{Z})$. I’m just applying the first right derived functor of each side to $H$.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 14:01












  • $begingroup$
    Very slick. Nice.
    $endgroup$
    – Sean Eberhard
    Jan 12 at 14:14






  • 1




    $begingroup$
    ...and you get $text{Tor}$ on the right since $text{Hom}(-,mathbb Q/mathbb Z)$ is exact. Fine.
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 19:34
















12












12








12





$begingroup$

Here’s a quick homological proof.



Suppose $F$ is finite and $H$ torsion free. Then $Fcongtext{Hom}(F,mathbb{Q}/mathbb{Z})$, so
$$text{Ext}^1(H,F)congtext{Ext}^1left(H,text{Hom}(F,mathbb{Q}/mathbb{Z})right)
congtext{Hom}left(text{Tor}_1(H,F),mathbb{Q}/mathbb{Z}right),
$$
which is zero since torsion free abelian groups are flat.






share|cite|improve this answer









$endgroup$



Here’s a quick homological proof.



Suppose $F$ is finite and $H$ torsion free. Then $Fcongtext{Hom}(F,mathbb{Q}/mathbb{Z})$, so
$$text{Ext}^1(H,F)congtext{Ext}^1left(H,text{Hom}(F,mathbb{Q}/mathbb{Z})right)
congtext{Hom}left(text{Tor}_1(H,F),mathbb{Q}/mathbb{Z}right),
$$
which is zero since torsion free abelian groups are flat.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 12:22









Jeremy RickardJeremy Rickard

21.1k15797




21.1k15797








  • 1




    $begingroup$
    Is the last isomorphism obvious? I mean, I could probably easily prove it, but...
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 13:06






  • 2




    $begingroup$
    @მამუკაჯიბლაძე $text{Hom}(-,text{Hom}(F,mathbb{Q}/mathbb{Z}))congtext{Hom}(-otimes F,mathbb{Q}/mathbb{Z})$. I’m just applying the first right derived functor of each side to $H$.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 14:01












  • $begingroup$
    Very slick. Nice.
    $endgroup$
    – Sean Eberhard
    Jan 12 at 14:14






  • 1




    $begingroup$
    ...and you get $text{Tor}$ on the right since $text{Hom}(-,mathbb Q/mathbb Z)$ is exact. Fine.
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 19:34
















  • 1




    $begingroup$
    Is the last isomorphism obvious? I mean, I could probably easily prove it, but...
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 13:06






  • 2




    $begingroup$
    @მამუკაჯიბლაძე $text{Hom}(-,text{Hom}(F,mathbb{Q}/mathbb{Z}))congtext{Hom}(-otimes F,mathbb{Q}/mathbb{Z})$. I’m just applying the first right derived functor of each side to $H$.
    $endgroup$
    – Jeremy Rickard
    Jan 12 at 14:01












  • $begingroup$
    Very slick. Nice.
    $endgroup$
    – Sean Eberhard
    Jan 12 at 14:14






  • 1




    $begingroup$
    ...and you get $text{Tor}$ on the right since $text{Hom}(-,mathbb Q/mathbb Z)$ is exact. Fine.
    $endgroup$
    – მამუკა ჯიბლაძე
    Jan 12 at 19:34










1




1




$begingroup$
Is the last isomorphism obvious? I mean, I could probably easily prove it, but...
$endgroup$
– მამუკა ჯიბლაძე
Jan 12 at 13:06




$begingroup$
Is the last isomorphism obvious? I mean, I could probably easily prove it, but...
$endgroup$
– მამუკა ჯიბლაძე
Jan 12 at 13:06




2




2




$begingroup$
@მამუკაჯიბლაძე $text{Hom}(-,text{Hom}(F,mathbb{Q}/mathbb{Z}))congtext{Hom}(-otimes F,mathbb{Q}/mathbb{Z})$. I’m just applying the first right derived functor of each side to $H$.
$endgroup$
– Jeremy Rickard
Jan 12 at 14:01






$begingroup$
@მამუკაჯიბლაძე $text{Hom}(-,text{Hom}(F,mathbb{Q}/mathbb{Z}))congtext{Hom}(-otimes F,mathbb{Q}/mathbb{Z})$. I’m just applying the first right derived functor of each side to $H$.
$endgroup$
– Jeremy Rickard
Jan 12 at 14:01














$begingroup$
Very slick. Nice.
$endgroup$
– Sean Eberhard
Jan 12 at 14:14




$begingroup$
Very slick. Nice.
$endgroup$
– Sean Eberhard
Jan 12 at 14:14




1




1




$begingroup$
...and you get $text{Tor}$ on the right since $text{Hom}(-,mathbb Q/mathbb Z)$ is exact. Fine.
$endgroup$
– მამუკა ჯიბლაძე
Jan 12 at 19:34






$begingroup$
...and you get $text{Tor}$ on the right since $text{Hom}(-,mathbb Q/mathbb Z)$ is exact. Fine.
$endgroup$
– მამუკა ჯიბლაძე
Jan 12 at 19:34




















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