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Prove that if $G$ is an infinite abelian group all of whose proper subgroups are finite then $G cong C_p^∞$ for some prime $p.$

What can we say of a group all of whose proper subgroups are abelian?Give an example of a noncyclic Abelian group all of whose proper subgroups are cyclic.Nonabelian $p$-groups all of whose proper subgroups are abelian.Prove that, if all maximal subgroups of a finite group are abelian, at least one of maximal subgroups is normalA finite group with the property that all of its proper subgroups are abelianIf G is a group such that all of its proper subgroups are abelian, then G itself must be abelianProblem about finite group whose proper subgroups are abelian.Locally graded group with all proper subgroups abelianfinite non-abelian groups with all proper subgroups cyclicDoes there exist an infinite non-abelian group such that all of its proper subgroups become cyclic?

0

$begingroup$

Prove that if $G$ is an infinite abelian group all of whose proper subgroups are finite then $G cong C_p^∞$ for some prime $p.$

Is $C_p^∞ times C_q^∞ cong C_p^∞$ for $p neq q$?

Can anyone help me out on this one?

Thanks.

abstract-algebra group-theory

share|cite|improve this question

edited Mar 28 at 17:45

Shaun

10.1k113685

asked Mar 28 at 15:51

Mathematical MushroomMathematical Mushroom

1068

$endgroup$

add a comment | 

0

$begingroup$

Prove that if $G$ is an infinite abelian group all of whose proper subgroups are finite then $G cong C_p^∞$ for some prime $p.$

Is $C_p^∞ times C_q^∞ cong C_p^∞$ for $p neq q$?

Can anyone help me out on this one?

Thanks.

abstract-algebra group-theory

share|cite|improve this question

edited Mar 28 at 17:45

Shaun

10.1k113685

asked Mar 28 at 15:51

Mathematical MushroomMathematical Mushroom

1068

$endgroup$

add a comment | 

$begingroup$

Prove that if $G$ is an infinite abelian group all of whose proper subgroups are finite then $G cong C_p^∞$ for some prime $p.$

Is $C_p^∞ times C_q^∞ cong C_p^∞$ for $p neq q$?

Can anyone help me out on this one?

Thanks.

abstract-algebra group-theory

share|cite|improve this question

edited Mar 28 at 17:45

Shaun

10.1k113685

asked Mar 28 at 15:51

Mathematical MushroomMathematical Mushroom

1068

$endgroup$

share|cite|improve this question

edited Mar 28 at 17:45

Shaun

10.1k113685

asked Mar 28 at 15:51

Mathematical MushroomMathematical Mushroom

1068

share|cite|improve this question

edited Mar 28 at 17:45

Shaun

10.1k113685

asked Mar 28 at 15:51

Mathematical MushroomMathematical Mushroom

1068

edited Mar 28 at 17:45

Shaun

10.1k113685

edited Mar 28 at 17:45

Shaun

10.1k113685

asked Mar 28 at 15:51

Mathematical MushroomMathematical Mushroom

1068

asked Mar 28 at 15:51

Mathematical MushroomMathematical Mushroom

1068

1 Answer
1

active

oldest

votes

0

$begingroup$

See if you can find a problem with the argument:

You've already deduced G is a torsion abelian group. And thus is a direct product of p-groups:

$$Gcong textDr_pin X G_p $$

The set of primes over which the direct product is taken has to just consist of p. Otherwise G would have an infinite subgroup isomorphic to $C_q^infty$.

The map $Grightarrow pG$ is a homomorphism with kernel $G[p]$. Furthermore, both $pG$ and $G[p]$ are subgroups, and so one of them has to be finite. We can just look at $pG$. It has to either be $G$, or be a finite subgroup. $pG=gin G$ can't be finite because $G$ is infinite (and also that would mean $G[p]=xin G$ is infinite, which can't happen).
Thus, $G=pG$, so $G$ is p-divisible.
Then the structure theorem for divisible groups says we can write $G$ as a product of $C_p^infty$'s (there are no $mathbbQ$'s here since we have a p-group). Again invoking the fact that we can't have an infinite subgroup isomorphic to $C_q^infty$ for some $qneq p$, we must have $Gcong C_p^infty$ for a single $p$.

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 16:33

Wyatt KuehsterWyatt Kuehster

737

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yo man!! Yeah, i figured this one out actually, well actually I got to the end where it says use the map $G rightarrow pG$ but I haven't figured out how to use that clue. Other than that, i'm having trouble with certain parts of #5 >.<
  $endgroup$
  – Mathematical Mushroom
  Mar 28 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The map is not a isomorphism because $G$ is gaurentee'd to have an element of order p by cauchy's theorem and so the kernel is non-empty. The quotient is isomorphic to the image because the map is onto. I think that the point is that the kernel is a proper subgroup by hypothesis and we are supposed to use that fact to find the isomorphism we are looking for but I still don't see it.
  $endgroup$
  – Mathematical Mushroom
  Mar 29 at 10:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, I was wrong. It is not an isomorphism. But we can use that fact that $G/G[p]cong pG$ to show that $G=pG$, so it's p-divisible for all primes p.
  $endgroup$
  – Wyatt Kuehster
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the argument that $G = pG$ needs a little more justification.
  $endgroup$
  – hunter
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is the trickiest part for me as well. Let's try this: Case 1: G=pG. We hope this is the case because then G is divisible and we're done. Case 2: pG<G. In this case $G[p]$ has to be infinite, so it has to be $G$. But $G[p]$ is just the p-primary component of $G$, whose torsion group will be isomorphic to a direct product of copies of $C_p^infty$. Then we proceed as before, deducing that it has to be for a single prime p.
  $endgroup$
  – Wyatt Kuehster
  2 days ago
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

See if you can find a problem with the argument:

You've already deduced G is a torsion abelian group. And thus is a direct product of p-groups:

$$Gcong textDr_pin X G_p $$

The set of primes over which the direct product is taken has to just consist of p. Otherwise G would have an infinite subgroup isomorphic to $C_q^infty$.

The map $Grightarrow pG$ is a homomorphism with kernel $G[p]$. Furthermore, both $pG$ and $G[p]$ are subgroups, and so one of them has to be finite. We can just look at $pG$. It has to either be $G$, or be a finite subgroup. $pG=gin G$ can't be finite because $G$ is infinite (and also that would mean $G[p]=xin G$ is infinite, which can't happen).
Thus, $G=pG$, so $G$ is p-divisible.
Then the structure theorem for divisible groups says we can write $G$ as a product of $C_p^infty$'s (there are no $mathbbQ$'s here since we have a p-group). Again invoking the fact that we can't have an infinite subgroup isomorphic to $C_q^infty$ for some $qneq p$, we must have $Gcong C_p^infty$ for a single $p$.

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 16:33

Wyatt KuehsterWyatt Kuehster

737

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yo man!! Yeah, i figured this one out actually, well actually I got to the end where it says use the map $G rightarrow pG$ but I haven't figured out how to use that clue. Other than that, i'm having trouble with certain parts of #5 >.<
  $endgroup$
  – Mathematical Mushroom
  Mar 28 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The map is not a isomorphism because $G$ is gaurentee'd to have an element of order p by cauchy's theorem and so the kernel is non-empty. The quotient is isomorphic to the image because the map is onto. I think that the point is that the kernel is a proper subgroup by hypothesis and we are supposed to use that fact to find the isomorphism we are looking for but I still don't see it.
  $endgroup$
  – Mathematical Mushroom
  Mar 29 at 10:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, I was wrong. It is not an isomorphism. But we can use that fact that $G/G[p]cong pG$ to show that $G=pG$, so it's p-divisible for all primes p.
  $endgroup$
  – Wyatt Kuehster
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the argument that $G = pG$ needs a little more justification.
  $endgroup$
  – hunter
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is the trickiest part for me as well. Let's try this: Case 1: G=pG. We hope this is the case because then G is divisible and we're done. Case 2: pG<G. In this case $G[p]$ has to be infinite, so it has to be $G$. But $G[p]$ is just the p-primary component of $G$, whose torsion group will be isomorphic to a direct product of copies of $C_p^infty$. Then we proceed as before, deducing that it has to be for a single prime p.
  $endgroup$
  – Wyatt Kuehster
  2 days ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

See if you can find a problem with the argument:

You've already deduced G is a torsion abelian group. And thus is a direct product of p-groups:

$$Gcong textDr_pin X G_p $$

The set of primes over which the direct product is taken has to just consist of p. Otherwise G would have an infinite subgroup isomorphic to $C_q^infty$.

The map $Grightarrow pG$ is a homomorphism with kernel $G[p]$. Furthermore, both $pG$ and $G[p]$ are subgroups, and so one of them has to be finite. We can just look at $pG$. It has to either be $G$, or be a finite subgroup. $pG=gin G$ can't be finite because $G$ is infinite (and also that would mean $G[p]=xin G$ is infinite, which can't happen).
Thus, $G=pG$, so $G$ is p-divisible.
Then the structure theorem for divisible groups says we can write $G$ as a product of $C_p^infty$'s (there are no $mathbbQ$'s here since we have a p-group). Again invoking the fact that we can't have an infinite subgroup isomorphic to $C_q^infty$ for some $qneq p$, we must have $Gcong C_p^infty$ for a single $p$.

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 16:33

Wyatt KuehsterWyatt Kuehster

737

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yo man!! Yeah, i figured this one out actually, well actually I got to the end where it says use the map $G rightarrow pG$ but I haven't figured out how to use that clue. Other than that, i'm having trouble with certain parts of #5 >.<
  $endgroup$
  – Mathematical Mushroom
  Mar 28 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The map is not a isomorphism because $G$ is gaurentee'd to have an element of order p by cauchy's theorem and so the kernel is non-empty. The quotient is isomorphic to the image because the map is onto. I think that the point is that the kernel is a proper subgroup by hypothesis and we are supposed to use that fact to find the isomorphism we are looking for but I still don't see it.
  $endgroup$
  – Mathematical Mushroom
  Mar 29 at 10:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, I was wrong. It is not an isomorphism. But we can use that fact that $G/G[p]cong pG$ to show that $G=pG$, so it's p-divisible for all primes p.
  $endgroup$
  – Wyatt Kuehster
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the argument that $G = pG$ needs a little more justification.
  $endgroup$
  – hunter
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is the trickiest part for me as well. Let's try this: Case 1: G=pG. We hope this is the case because then G is divisible and we're done. Case 2: pG<G. In this case $G[p]$ has to be infinite, so it has to be $G$. But $G[p]$ is just the p-primary component of $G$, whose torsion group will be isomorphic to a direct product of copies of $C_p^infty$. Then we proceed as before, deducing that it has to be for a single prime p.
  $endgroup$
  – Wyatt Kuehster
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

See if you can find a problem with the argument:

You've already deduced G is a torsion abelian group. And thus is a direct product of p-groups:

$$Gcong textDr_pin X G_p $$

The set of primes over which the direct product is taken has to just consist of p. Otherwise G would have an infinite subgroup isomorphic to $C_q^infty$.

The map $Grightarrow pG$ is a homomorphism with kernel $G[p]$. Furthermore, both $pG$ and $G[p]$ are subgroups, and so one of them has to be finite. We can just look at $pG$. It has to either be $G$, or be a finite subgroup. $pG=gin G$ can't be finite because $G$ is infinite (and also that would mean $G[p]=xin G$ is infinite, which can't happen).
Thus, $G=pG$, so $G$ is p-divisible.
Then the structure theorem for divisible groups says we can write $G$ as a product of $C_p^infty$'s (there are no $mathbbQ$'s here since we have a p-group). Again invoking the fact that we can't have an infinite subgroup isomorphic to $C_q^infty$ for some $qneq p$, we must have $Gcong C_p^infty$ for a single $p$.

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 16:33

Wyatt KuehsterWyatt Kuehster

737

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 16:33

Wyatt KuehsterWyatt Kuehster

737

answered Mar 28 at 16:33

Wyatt KuehsterWyatt Kuehster

737

answered Mar 28 at 16:33

Wyatt KuehsterWyatt Kuehster

737