Example of subgroup of Direct Product which is not in usual form [duplicate]
This question already has an answer here:
subgroup of direct product of two groups
2 answers
I wanted to find group $Atimes B$ such that it has subgroup which is not of form $Ctimes D$ where $C < A$ and $D <B$.
Actually I think this is wrong I tried to get proof. But my friend told this is true . that it possible to have group with such prpoerty
ANy suggestion is appreciated
group-theory examples-counterexamples
asked Nov 20 at 8:24
Shubham
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marked as duplicate by J.-E. Pin, amWhy group-theory
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Nov 20 at 12:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
subgroup of direct product of two groups
2 answers
I wanted to find group $Atimes B$ such that it has subgroup which is not of form $Ctimes D$ where $C < A$ and $D <B$.
Actually I think this is wrong I tried to get proof. But my friend told this is true . that it possible to have group with such prpoerty
ANy suggestion is appreciated
group-theory examples-counterexamples
asked Nov 20 at 8:24
Shubham
1,5901519
marked as duplicate by J.-E. Pin, amWhy group-theory
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Nov 20 at 12:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
You can only say that a subgroup of $A times B$ is a subdirect product of $C times D$ where $C < A$ and $D < B$.
– J.-E. Pin
Nov 20 at 9:01
@J.-E.Pin WHat is a "subdirect product"?
– DonAntonio
Nov 20 at 10:34
@donantonio Here is a link: subdirect product
– J.-E. Pin
Nov 20 at 10:49
add a comment |
This question already has an answer here:
subgroup of direct product of two groups
2 answers
I wanted to find group $Atimes B$ such that it has subgroup which is not of form $Ctimes D$ where $C < A$ and $D <B$.
Actually I think this is wrong I tried to get proof. But my friend told this is true . that it possible to have group with such prpoerty
ANy suggestion is appreciated
group-theory examples-counterexamples
asked Nov 20 at 8:24
Shubham
1,5901519
This question already has an answer here:
subgroup of direct product of two groups
2 answers
I wanted to find group $Atimes B$ such that it has subgroup which is not of form $Ctimes D$ where $C < A$ and $D <B$.
Actually I think this is wrong I tried to get proof. But my friend told this is true . that it possible to have group with such prpoerty
ANy suggestion is appreciated
This question already has an answer here:
subgroup of direct product of two groups
2 answers
group-theory examples-counterexamples
group-theory examples-counterexamples
asked Nov 20 at 8:24
Shubham
1,5901519
asked Nov 20 at 8:24
Shubham
1,5901519
edited Nov 20 at 8:29
asked Nov 20 at 8:24
Shubham
1,5901519
asked Nov 20 at 8:24
Shubham
1,5901519
asked Nov 20 at 8:24
Shubham
1,5901519
1,5901519
marked as duplicate by J.-E. Pin, amWhy group-theory
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Nov 20 at 12:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Nov 20 at 12:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
You can only say that a subgroup of $A times B$ is a subdirect product of $C times D$ where $C < A$ and $D < B$.
– J.-E. Pin
Nov 20 at 9:01
@J.-E.Pin WHat is a "subdirect product"?
– DonAntonio
Nov 20 at 10:34
@donantonio Here is a link: subdirect product
– J.-E. Pin
Nov 20 at 10:49
add a comment |
You can only say that a subgroup of $A times B$ is a subdirect product of $C times D$ where $C < A$ and $D < B$.
– J.-E. Pin
Nov 20 at 9:01
@J.-E.Pin WHat is a "subdirect product"?
– DonAntonio
Nov 20 at 10:34
@donantonio Here is a link: subdirect product
– J.-E. Pin
Nov 20 at 10:49
You can only say that a subgroup of $A times B$ is a subdirect product of $C times D$ where $C < A$ and $D < B$.
– J.-E. Pin
Nov 20 at 9:01
You can only say that a subgroup of $A times B$ is a subdirect product of $C times D$ where $C < A$ and $D < B$.
– J.-E. Pin
Nov 20 at 9:01
@J.-E.Pin WHat is a "subdirect product"?
– DonAntonio
Nov 20 at 10:34
@J.-E.Pin WHat is a "subdirect product"?
– DonAntonio
Nov 20 at 10:34
@donantonio Here is a link: subdirect product
– J.-E. Pin
Nov 20 at 10:49
@donantonio Here is a link: subdirect product
– J.-E. Pin
Nov 20 at 10:49
add a comment |
1 Answer
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Consider $G=Bbb{Z}_2 times Bbb{Z}_2$ and let $H=langle (1,1) rangle={(0,0), (1,1)}$. Then $H leq G$ but $H$ cannot be expressed as $C times D$. Because if it could be, then $0,1 in C$ and $0,1 in D$. This will mean $C=Bbb{Z}_2$ and same for $D$. But then $C times D$ will have $4$ elements. Thus cannot be same as $H$.
answered Nov 20 at 8:29
Anurag A
25.6k12249
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider $G=Bbb{Z}_2 times Bbb{Z}_2$ and let $H=langle (1,1) rangle={(0,0), (1,1)}$. Then $H leq G$ but $H$ cannot be expressed as $C times D$. Because if it could be, then $0,1 in C$ and $0,1 in D$. This will mean $C=Bbb{Z}_2$ and same for $D$. But then $C times D$ will have $4$ elements. Thus cannot be same as $H$.
answered Nov 20 at 8:29
Anurag A
25.6k12249
add a comment |
Consider $G=Bbb{Z}_2 times Bbb{Z}_2$ and let $H=langle (1,1) rangle={(0,0), (1,1)}$. Then $H leq G$ but $H$ cannot be expressed as $C times D$. Because if it could be, then $0,1 in C$ and $0,1 in D$. This will mean $C=Bbb{Z}_2$ and same for $D$. But then $C times D$ will have $4$ elements. Thus cannot be same as $H$.
answered Nov 20 at 8:29
Anurag A
25.6k12249
add a comment |
Consider $G=Bbb{Z}_2 times Bbb{Z}_2$ and let $H=langle (1,1) rangle={(0,0), (1,1)}$. Then $H leq G$ but $H$ cannot be expressed as $C times D$. Because if it could be, then $0,1 in C$ and $0,1 in D$. This will mean $C=Bbb{Z}_2$ and same for $D$. But then $C times D$ will have $4$ elements. Thus cannot be same as $H$.
answered Nov 20 at 8:29
Anurag A
25.6k12249
Consider $G=Bbb{Z}_2 times Bbb{Z}_2$ and let $H=langle (1,1) rangle={(0,0), (1,1)}$. Then $H leq G$ but $H$ cannot be expressed as $C times D$. Because if it could be, then $0,1 in C$ and $0,1 in D$. This will mean $C=Bbb{Z}_2$ and same for $D$. But then $C times D$ will have $4$ elements. Thus cannot be same as $H$.
answered Nov 20 at 8:29
Anurag A
25.6k12249
answered Nov 20 at 8:29
Anurag A
25.6k12249
answered Nov 20 at 8:29
Anurag A
25.6k12249
answered Nov 20 at 8:29
Anurag A
25.6k12249
25.6k12249
add a comment |
add a comment |
You can only say that a subgroup of $A times B$ is a subdirect product of $C times D$ where $C < A$ and $D < B$.
– J.-E. Pin
Nov 20 at 9:01
@J.-E.Pin WHat is a "subdirect product"?
– DonAntonio
Nov 20 at 10:34
@donantonio Here is a link: subdirect product
– J.-E. Pin
Nov 20 at 10:49