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Examples of smooth manifolds admitting inbetween one and a continuum of complex structures

Alternative definitions of “internal direct product”

Example of a group $G$ such that $G = N_1cdots N_n$ and $N_i cap N_j = e$ for all $i neq j$ but $G$ is not the internal direct product of them.Finding $G$ with normal $H_1,ldots,H_n$ such that $G=H_1cdots H_n$ and $H_icap H_j=e$ for $ineq j$, but $Gnotcong H_1timescdotstimes H_n$Number of non isomorphic groups of order $122$, My attempt through Sylow theory.Show the map from the product of disjoint subgroups of G to G is an isomorphismUnderstanding the proof of finite abelian group G as the internal direct product of p-groups.If $G$ is isomorphic to $H_1timescdotstimes H_n$, $H_i$ subgroup of $G$, then *is* $G$ the internal direct product of $H_1cdots H_n$?Prove or disprove: The equivalence of $forall i in 1,dots,nquad |H_1dots H_n|=|H_1|dots|H_n|$Comparing definitions of internal direct productDirect product for minimal normal subgroupEquivalence of Internal & External Direct ProductDefinition of subgroup generated by a set

1

$begingroup$

Let be $H_1,cdots,H_n$ subgroups of the group $G$, then we say that $G$ is an internal direct product of $H_1,cdots,H_n$ iff

1. 
   $H_jtriangleleft G$ for $1le jle n$, and


2. 
   $G=H_1H_2cdots H_n$, and


3. 
   $H_kcap H_1H_2cdots hatH_kcdots H_n=e$.

Can property (3) be replaced with the following property?

4. 
   $H_jcap H_k=e$ for all $jne k$.

I proved 1+2+3 imply 1+2+3'. Now I have to prove the converse.

By contradiction suppose that there is a $k$ and a $h_kin H_k$ with $h_kne e$ and $h_k=h_1cdots h_k-1h_k+1cdots h_n$.

I dont't need necessarily a proof. To know that the 1+2+3' is a rightfully alternative would suffice.

EDIT It seams that it is not correct according to this post.

Could it be an alternative in case of an abelian group?

abstract-algebra direct-product

share|cite|improve this question

edited Mar 28 at 21:07

Aaron Lenz

asked Mar 28 at 21:01

Aaron LenzAaron Lenz

5610

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The group in the example in the post you link to and the group in my answer are both abelian.
  $endgroup$
  – Rob Arthan
  Mar 28 at 21:12
  

  
  
  
  

add a comment | 

1

$begingroup$

Let be $H_1,cdots,H_n$ subgroups of the group $G$, then we say that $G$ is an internal direct product of $H_1,cdots,H_n$ iff

1. 
   $H_jtriangleleft G$ for $1le jle n$, and


2. 
   $G=H_1H_2cdots H_n$, and


3. 
   $H_kcap H_1H_2cdots hatH_kcdots H_n=e$.

Can property (3) be replaced with the following property?

4. 
   $H_jcap H_k=e$ for all $jne k$.

I proved 1+2+3 imply 1+2+3'. Now I have to prove the converse.

By contradiction suppose that there is a $k$ and a $h_kin H_k$ with $h_kne e$ and $h_k=h_1cdots h_k-1h_k+1cdots h_n$.

I dont't need necessarily a proof. To know that the 1+2+3' is a rightfully alternative would suffice.

EDIT It seams that it is not correct according to this post.

Could it be an alternative in case of an abelian group?

abstract-algebra direct-product

share|cite|improve this question

edited Mar 28 at 21:07

Aaron Lenz

asked Mar 28 at 21:01

Aaron LenzAaron Lenz

5610

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The group in the example in the post you link to and the group in my answer are both abelian.
  $endgroup$
  – Rob Arthan
  Mar 28 at 21:12
  

  
  
  
  

add a comment | 

$begingroup$

Let be $H_1,cdots,H_n$ subgroups of the group $G$, then we say that $G$ is an internal direct product of $H_1,cdots,H_n$ iff

1. 
   $H_jtriangleleft G$ for $1le jle n$, and


2. 
   $G=H_1H_2cdots H_n$, and


3. 
   $H_kcap H_1H_2cdots hatH_kcdots H_n=e$.

Can property (3) be replaced with the following property?

4. 
   $H_jcap H_k=e$ for all $jne k$.

I proved 1+2+3 imply 1+2+3'. Now I have to prove the converse.

By contradiction suppose that there is a $k$ and a $h_kin H_k$ with $h_kne e$ and $h_k=h_1cdots h_k-1h_k+1cdots h_n$.

I dont't need necessarily a proof. To know that the 1+2+3' is a rightfully alternative would suffice.

EDIT It seams that it is not correct according to this post.

Could it be an alternative in case of an abelian group?

abstract-algebra direct-product

share|cite|improve this question

edited Mar 28 at 21:07

Aaron Lenz

asked Mar 28 at 21:01

Aaron LenzAaron Lenz

5610

$endgroup$

share|cite|improve this question

edited Mar 28 at 21:07

Aaron Lenz

asked Mar 28 at 21:01

Aaron LenzAaron Lenz

5610

share|cite|improve this question

edited Mar 28 at 21:07

Aaron Lenz

asked Mar 28 at 21:01

Aaron LenzAaron Lenz

5610

asked Mar 28 at 21:01

Aaron LenzAaron Lenz

5610

asked Mar 28 at 21:01

Aaron LenzAaron Lenz

5610

1 Answer
1

active

oldest

votes

1

$begingroup$

Your property (4) is strictly weaker than (3). For example, take $G = BbbZtimes BbbZ$ (pairs of integers) under pointwise addition and let us write the group operation as $+$: $(a, b) + (c, d) = (a + c, b + d)$. Now take $n = 3$ and take $H_1$, $H_2$ and $H_3$ to be the subgroups generated by $(1, 0)$, $(0, 1)$ and $(1, 1)$ respectively. Then your properties (1), (2) and (4) hold, but (3) does not: $H_3 subseteq H_1 + H_2$.

share|cite|improve this answer

answered Mar 28 at 21:10

Rob ArthanRob Arthan

29.6k42967

$endgroup$

add a comment | 

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1

$begingroup$

Your property (4) is strictly weaker than (3). For example, take $G = BbbZtimes BbbZ$ (pairs of integers) under pointwise addition and let us write the group operation as $+$: $(a, b) + (c, d) = (a + c, b + d)$. Now take $n = 3$ and take $H_1$, $H_2$ and $H_3$ to be the subgroups generated by $(1, 0)$, $(0, 1)$ and $(1, 1)$ respectively. Then your properties (1), (2) and (4) hold, but (3) does not: $H_3 subseteq H_1 + H_2$.

share|cite|improve this answer

answered Mar 28 at 21:10

Rob ArthanRob Arthan

29.6k42967

$endgroup$

add a comment | 

1

$begingroup$

Your property (4) is strictly weaker than (3). For example, take $G = BbbZtimes BbbZ$ (pairs of integers) under pointwise addition and let us write the group operation as $+$: $(a, b) + (c, d) = (a + c, b + d)$. Now take $n = 3$ and take $H_1$, $H_2$ and $H_3$ to be the subgroups generated by $(1, 0)$, $(0, 1)$ and $(1, 1)$ respectively. Then your properties (1), (2) and (4) hold, but (3) does not: $H_3 subseteq H_1 + H_2$.

share|cite|improve this answer

answered Mar 28 at 21:10

Rob ArthanRob Arthan

29.6k42967

$endgroup$

add a comment | 

$begingroup$

Your property (4) is strictly weaker than (3). For example, take $G = BbbZtimes BbbZ$ (pairs of integers) under pointwise addition and let us write the group operation as $+$: $(a, b) + (c, d) = (a + c, b + d)$. Now take $n = 3$ and take $H_1$, $H_2$ and $H_3$ to be the subgroups generated by $(1, 0)$, $(0, 1)$ and $(1, 1)$ respectively. Then your properties (1), (2) and (4) hold, but (3) does not: $H_3 subseteq H_1 + H_2$.

share|cite|improve this answer

answered Mar 28 at 21:10

Rob ArthanRob Arthan

29.6k42967

$endgroup$

share|cite|improve this answer

answered Mar 28 at 21:10

Rob ArthanRob Arthan

29.6k42967

answered Mar 28 at 21:10

Rob ArthanRob Arthan

29.6k42967

answered Mar 28 at 21:10

Rob ArthanRob Arthan

29.6k42967