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Let G be a group with composition series. Let $Htriangleleft G$. Then $H$ is in some of the series.

Examples of abelian group with some propertyWhen does an abelian group have a composition series?Proving that every finite group has a composition seriesAn infinite non-Abelian group with an involutive automorphism that preserves only the identity?Is the center of a p-group non-trivial?Infinite group $G$ with subgroup $H$ such that $xHx^-1 subset H$, for some $x in G$Noncyclic (infinite) group with totally ordered subgroup latticeGroup with infinite orderLet $|H|=pq$ where $q$ and $p$ are both prime. Show that $H$ is not always cyclic.A group $G$ with the following property …

1

$begingroup$

I know this is true for $G$ being finite. How about infinite group? Is this statement still valid?

I proved the finite case by induction on $|G|$ (which supposedly not valid for infinite case)

Could anyone give a hint or a counterexample?

Thanks!

group-theory infinite-groups

share|cite|improve this question

edited Mar 13 at 17:31

Tortuga

asked Mar 13 at 15:00

TortugaTortuga

35112

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I assume that you assume in the definition of composition series that successive quotient are simple (which I'd call "Jordan-Hölder series"). Anyway there's a counterexample to your claim which is a cyclic group of order 6.
  $endgroup$
  – YCor
  Mar 13 at 16:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi @YCor, Thanks for the comment! Yes, the definition I am using is exactly the one you called Jordan-H''older series. For the counterexample, could you please be more explicitly? Cyclic group of order $6$ is finite, isn't it? You mean this is not even true for finite case?
  $endgroup$
  – Tortuga
  Mar 13 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why don't you check instead of asking? You should be able to write a Jordan-Hölder series and list normal subgroups of a cyclic group of order 6.
  $endgroup$
  – YCor
  Mar 13 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was my bad. I didn't ask the question in an appropriate way. There are only 2 composition series, and one of them contain $0,2,4$ and another contain $0,3$. Have already edited the title.
  $endgroup$
  – Tortuga
  Mar 13 at 17:33
  

  
  
  
  

add a comment | 

1

$begingroup$

I know this is true for $G$ being finite. How about infinite group? Is this statement still valid?

I proved the finite case by induction on $|G|$ (which supposedly not valid for infinite case)

Could anyone give a hint or a counterexample?

Thanks!

group-theory infinite-groups

share|cite|improve this question

edited Mar 13 at 17:31

Tortuga

asked Mar 13 at 15:00

TortugaTortuga

35112

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I assume that you assume in the definition of composition series that successive quotient are simple (which I'd call "Jordan-Hölder series"). Anyway there's a counterexample to your claim which is a cyclic group of order 6.
  $endgroup$
  – YCor
  Mar 13 at 16:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi @YCor, Thanks for the comment! Yes, the definition I am using is exactly the one you called Jordan-H''older series. For the counterexample, could you please be more explicitly? Cyclic group of order $6$ is finite, isn't it? You mean this is not even true for finite case?
  $endgroup$
  – Tortuga
  Mar 13 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why don't you check instead of asking? You should be able to write a Jordan-Hölder series and list normal subgroups of a cyclic group of order 6.
  $endgroup$
  – YCor
  Mar 13 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was my bad. I didn't ask the question in an appropriate way. There are only 2 composition series, and one of them contain $0,2,4$ and another contain $0,3$. Have already edited the title.
  $endgroup$
  – Tortuga
  Mar 13 at 17:33
  

  
  
  
  

add a comment | 

$begingroup$

I know this is true for $G$ being finite. How about infinite group? Is this statement still valid?

I proved the finite case by induction on $|G|$ (which supposedly not valid for infinite case)

Could anyone give a hint or a counterexample?

Thanks!

group-theory infinite-groups

share|cite|improve this question

edited Mar 13 at 17:31

Tortuga

asked Mar 13 at 15:00

TortugaTortuga

35112

$endgroup$

share|cite|improve this question

edited Mar 13 at 17:31

Tortuga

asked Mar 13 at 15:00

TortugaTortuga

35112

share|cite|improve this question

edited Mar 13 at 17:31

Tortuga

asked Mar 13 at 15:00

TortugaTortuga

35112

asked Mar 13 at 15:00

TortugaTortuga

35112

asked Mar 13 at 15:00

TortugaTortuga

35112

1 Answer
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active

oldest

votes

1

$begingroup$

Let's say that a group $G$ is poly-simple if it admits a Jordan-Hölder series (= a finite chain from $1$ to $G$ of subgroups, each normal in the next one, with simple successive quotients). This is the smallest class of groups, containing simple groups and stable under taking extension.

Proposition: every normal or quotient subgroup of a poly-simple group is poly-simple.

Proof: let $C$ be the class of groups in which every normal subgroup is poly-simple. Clearly, every simple group is in $C$. So, it's enough to show that $C$ is closed under taking extensions. Let $G$ be a group with a normal subgroup $N$, such that $N$ and $G/N$ are in $C$. Let $H$ be a normal subgroup of $G$. Then $Hcap N$ is normal in $H$, so is poly-simple, and $H/(Hcap C)$ is isomorphic to a normal subgroup of $G/N$, namely the projection of $H$ in $G/N$. Hence $H/(Hcap C)$ is poly-simple. Stability under extensions implies that $H$ is poly-simple, finshing the proof. The proof with quotients is almost the same. $Box$

So your question has a positive answer.

Corollary: every normal subgroup of every poly-simple group lies in a Jordan-Hölder series.

share|cite|improve this answer

answered Mar 13 at 17:53

YCorYCor

8,132929

$endgroup$

add a comment | 

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1

$begingroup$

Let's say that a group $G$ is poly-simple if it admits a Jordan-Hölder series (= a finite chain from $1$ to $G$ of subgroups, each normal in the next one, with simple successive quotients). This is the smallest class of groups, containing simple groups and stable under taking extension.

Proposition: every normal or quotient subgroup of a poly-simple group is poly-simple.

Proof: let $C$ be the class of groups in which every normal subgroup is poly-simple. Clearly, every simple group is in $C$. So, it's enough to show that $C$ is closed under taking extensions. Let $G$ be a group with a normal subgroup $N$, such that $N$ and $G/N$ are in $C$. Let $H$ be a normal subgroup of $G$. Then $Hcap N$ is normal in $H$, so is poly-simple, and $H/(Hcap C)$ is isomorphic to a normal subgroup of $G/N$, namely the projection of $H$ in $G/N$. Hence $H/(Hcap C)$ is poly-simple. Stability under extensions implies that $H$ is poly-simple, finshing the proof. The proof with quotients is almost the same. $Box$

So your question has a positive answer.

Corollary: every normal subgroup of every poly-simple group lies in a Jordan-Hölder series.

share|cite|improve this answer

answered Mar 13 at 17:53

YCorYCor

8,132929

$endgroup$

add a comment | 

1

$begingroup$

Let's say that a group $G$ is poly-simple if it admits a Jordan-Hölder series (= a finite chain from $1$ to $G$ of subgroups, each normal in the next one, with simple successive quotients). This is the smallest class of groups, containing simple groups and stable under taking extension.

Proposition: every normal or quotient subgroup of a poly-simple group is poly-simple.

Proof: let $C$ be the class of groups in which every normal subgroup is poly-simple. Clearly, every simple group is in $C$. So, it's enough to show that $C$ is closed under taking extensions. Let $G$ be a group with a normal subgroup $N$, such that $N$ and $G/N$ are in $C$. Let $H$ be a normal subgroup of $G$. Then $Hcap N$ is normal in $H$, so is poly-simple, and $H/(Hcap C)$ is isomorphic to a normal subgroup of $G/N$, namely the projection of $H$ in $G/N$. Hence $H/(Hcap C)$ is poly-simple. Stability under extensions implies that $H$ is poly-simple, finshing the proof. The proof with quotients is almost the same. $Box$

So your question has a positive answer.

Corollary: every normal subgroup of every poly-simple group lies in a Jordan-Hölder series.

share|cite|improve this answer

answered Mar 13 at 17:53

YCorYCor

8,132929

$endgroup$

add a comment | 

$begingroup$

Let's say that a group $G$ is poly-simple if it admits a Jordan-Hölder series (= a finite chain from $1$ to $G$ of subgroups, each normal in the next one, with simple successive quotients). This is the smallest class of groups, containing simple groups and stable under taking extension.

Proposition: every normal or quotient subgroup of a poly-simple group is poly-simple.

Proof: let $C$ be the class of groups in which every normal subgroup is poly-simple. Clearly, every simple group is in $C$. So, it's enough to show that $C$ is closed under taking extensions. Let $G$ be a group with a normal subgroup $N$, such that $N$ and $G/N$ are in $C$. Let $H$ be a normal subgroup of $G$. Then $Hcap N$ is normal in $H$, so is poly-simple, and $H/(Hcap C)$ is isomorphic to a normal subgroup of $G/N$, namely the projection of $H$ in $G/N$. Hence $H/(Hcap C)$ is poly-simple. Stability under extensions implies that $H$ is poly-simple, finshing the proof. The proof with quotients is almost the same. $Box$

So your question has a positive answer.

Corollary: every normal subgroup of every poly-simple group lies in a Jordan-Hölder series.

share|cite|improve this answer

answered Mar 13 at 17:53

YCorYCor

8,132929

$endgroup$

share|cite|improve this answer

answered Mar 13 at 17:53

YCorYCor

8,132929

answered Mar 13 at 17:53

YCorYCor

8,132929

answered Mar 13 at 17:53

YCorYCor

8,132929