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Validation for a conjecture about Chinese Remainder Theorem for groups
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I was wondering if the following statement is true:
Let $G$ be a group with normal subgroups $H_1,H_2,...H_n$. Suppose $H_iH_j=G$ for all $ineq j$. Then $G/H_1cap H_2...cap H_ncong G/H_1 times...times G/H_n.$
There is a similar question here, which is the case when $n=2$. There is also a similar question here, but the conditions involve the index of a subgroup. I want to get rid of such condition so that the conclusion is applicable to some other situations.
But when I tried to proceed proof by induction to get the conclusion with arbitrary $n$, things became not so approachable.
If you think this is false, please give a counter-example. If you think this is true, please share your ideas of the proof. Thank you!
group-theory normal-subgroups conjectures chinese-remainder-theorem
asked 1 hour ago
J. Wang
262
New contributor
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
5
down vote
favorite
I was wondering if the following statement is true:
Let $G$ be a group with normal subgroups $H_1,H_2,...H_n$. Suppose $H_iH_j=G$ for all $ineq j$. Then $G/H_1cap H_2...cap H_ncong G/H_1 times...times G/H_n.$
There is a similar question here, which is the case when $n=2$. There is also a similar question here, but the conditions involve the index of a subgroup. I want to get rid of such condition so that the conclusion is applicable to some other situations.
But when I tried to proceed proof by induction to get the conclusion with arbitrary $n$, things became not so approachable.
If you think this is false, please give a counter-example. If you think this is true, please share your ideas of the proof. Thank you!
group-theory normal-subgroups conjectures chinese-remainder-theorem
asked 1 hour ago
J. Wang
262
New contributor
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I was wondering if the following statement is true:
Let $G$ be a group with normal subgroups $H_1,H_2,...H_n$. Suppose $H_iH_j=G$ for all $ineq j$. Then $G/H_1cap H_2...cap H_ncong G/H_1 times...times G/H_n.$
There is a similar question here, which is the case when $n=2$. There is also a similar question here, but the conditions involve the index of a subgroup. I want to get rid of such condition so that the conclusion is applicable to some other situations.
But when I tried to proceed proof by induction to get the conclusion with arbitrary $n$, things became not so approachable.
If you think this is false, please give a counter-example. If you think this is true, please share your ideas of the proof. Thank you!
group-theory normal-subgroups conjectures chinese-remainder-theorem
asked 1 hour ago
J. Wang
262
New contributor
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I was wondering if the following statement is true:
Let $G$ be a group with normal subgroups $H_1,H_2,...H_n$. Suppose $H_iH_j=G$ for all $ineq j$. Then $G/H_1cap H_2...cap H_ncong G/H_1 times...times G/H_n.$
There is a similar question here, which is the case when $n=2$. There is also a similar question here, but the conditions involve the index of a subgroup. I want to get rid of such condition so that the conclusion is applicable to some other situations.
But when I tried to proceed proof by induction to get the conclusion with arbitrary $n$, things became not so approachable.
If you think this is false, please give a counter-example. If you think this is true, please share your ideas of the proof. Thank you!
group-theory normal-subgroups conjectures chinese-remainder-theorem
group-theory normal-subgroups conjectures chinese-remainder-theorem
asked 1 hour ago
J. Wang
262
New contributor
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
J. Wang
262
New contributor
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
J. Wang
262
New contributor
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
J. Wang
262
asked 1 hour ago
J. Wang
262
262
New contributor
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
3
down vote
This is false. For instance, let $G=(mathbbZ/2mathbbZ)^2$ and let $H_1,H_2,H_3subset G$ be the three subgroups of order $2$. Then $H_iH_j=G$ for all $ineq j$ but $G/(H_1cap H_2cap H_3)cong G$ has $4$ elements while $G/H_1times G/H_2times G/H_3cong(mathbbZ/2mathbbZ)^3$ has $8$ elements.
answered 1 hour ago
Eric Wofsey
170k12198316
Thank you. This example is clear. I will appreciate it if you can explain more why things fail to work for n>2. My intuitive understanding is that the information of a group is fully contained in the first two subgroup in the condition. So when we add one more, we count things redundantly. That’s why you and the other answer argue through the cardinality. However, is there a more in-depth reason why the statement fails? And how should we modify the condition to make it work?
– J. Wang
3 mins ago
I think you're asking the wrong question. The right question is why it does work for $n=2$. There are various proofs you can give in that case and you can see where they use the fact that $n=2$.
– Eric Wofsey
50 secs ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is false. For instance, let $G=(mathbbZ/2mathbbZ)^2$ and let $H_1,H_2,H_3subset G$ be the three subgroups of order $2$. Then $H_iH_j=G$ for all $ineq j$ but $G/(H_1cap H_2cap H_3)cong G$ has $4$ elements while $G/H_1times G/H_2times G/H_3cong(mathbbZ/2mathbbZ)^3$ has $8$ elements.
answered 1 hour ago
Eric Wofsey
170k12198316
Thank you. This example is clear. I will appreciate it if you can explain more why things fail to work for n>2. My intuitive understanding is that the information of a group is fully contained in the first two subgroup in the condition. So when we add one more, we count things redundantly. That’s why you and the other answer argue through the cardinality. However, is there a more in-depth reason why the statement fails? And how should we modify the condition to make it work?
– J. Wang
3 mins ago
I think you're asking the wrong question. The right question is why it does work for $n=2$. There are various proofs you can give in that case and you can see where they use the fact that $n=2$.
– Eric Wofsey
50 secs ago
add a comment |Â
up vote
3
down vote
This is false. For instance, let $G=(mathbbZ/2mathbbZ)^2$ and let $H_1,H_2,H_3subset G$ be the three subgroups of order $2$. Then $H_iH_j=G$ for all $ineq j$ but $G/(H_1cap H_2cap H_3)cong G$ has $4$ elements while $G/H_1times G/H_2times G/H_3cong(mathbbZ/2mathbbZ)^3$ has $8$ elements.
answered 1 hour ago
Eric Wofsey
170k12198316
Thank you. This example is clear. I will appreciate it if you can explain more why things fail to work for n>2. My intuitive understanding is that the information of a group is fully contained in the first two subgroup in the condition. So when we add one more, we count things redundantly. That’s why you and the other answer argue through the cardinality. However, is there a more in-depth reason why the statement fails? And how should we modify the condition to make it work?
– J. Wang
3 mins ago
I think you're asking the wrong question. The right question is why it does work for $n=2$. There are various proofs you can give in that case and you can see where they use the fact that $n=2$.
– Eric Wofsey
50 secs ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is false. For instance, let $G=(mathbbZ/2mathbbZ)^2$ and let $H_1,H_2,H_3subset G$ be the three subgroups of order $2$. Then $H_iH_j=G$ for all $ineq j$ but $G/(H_1cap H_2cap H_3)cong G$ has $4$ elements while $G/H_1times G/H_2times G/H_3cong(mathbbZ/2mathbbZ)^3$ has $8$ elements.
answered 1 hour ago
Eric Wofsey
170k12198316
This is false. For instance, let $G=(mathbbZ/2mathbbZ)^2$ and let $H_1,H_2,H_3subset G$ be the three subgroups of order $2$. Then $H_iH_j=G$ for all $ineq j$ but $G/(H_1cap H_2cap H_3)cong G$ has $4$ elements while $G/H_1times G/H_2times G/H_3cong(mathbbZ/2mathbbZ)^3$ has $8$ elements.
answered 1 hour ago
Eric Wofsey
170k12198316
answered 1 hour ago
Eric Wofsey
170k12198316
answered 1 hour ago
Eric Wofsey
170k12198316
answered 1 hour ago
Eric Wofsey
170k12198316
170k12198316
Thank you. This example is clear. I will appreciate it if you can explain more why things fail to work for n>2. My intuitive understanding is that the information of a group is fully contained in the first two subgroup in the condition. So when we add one more, we count things redundantly. That’s why you and the other answer argue through the cardinality. However, is there a more in-depth reason why the statement fails? And how should we modify the condition to make it work?
– J. Wang
3 mins ago
I think you're asking the wrong question. The right question is why it does work for $n=2$. There are various proofs you can give in that case and you can see where they use the fact that $n=2$.
– Eric Wofsey
50 secs ago
add a comment |Â
Thank you. This example is clear. I will appreciate it if you can explain more why things fail to work for n>2. My intuitive understanding is that the information of a group is fully contained in the first two subgroup in the condition. So when we add one more, we count things redundantly. That’s why you and the other answer argue through the cardinality. However, is there a more in-depth reason why the statement fails? And how should we modify the condition to make it work?
– J. Wang
3 mins ago
I think you're asking the wrong question. The right question is why it does work for $n=2$. There are various proofs you can give in that case and you can see where they use the fact that $n=2$.
– Eric Wofsey
50 secs ago
Thank you. This example is clear. I will appreciate it if you can explain more why things fail to work for n>2. My intuitive understanding is that the information of a group is fully contained in the first two subgroup in the condition. So when we add one more, we count things redundantly. That’s why you and the other answer argue through the cardinality. However, is there a more in-depth reason why the statement fails? And how should we modify the condition to make it work?
– J. Wang
3 mins ago
Thank you. This example is clear. I will appreciate it if you can explain more why things fail to work for n>2. My intuitive understanding is that the information of a group is fully contained in the first two subgroup in the condition. So when we add one more, we count things redundantly. That’s why you and the other answer argue through the cardinality. However, is there a more in-depth reason why the statement fails? And how should we modify the condition to make it work?
– J. Wang
3 mins ago
I think you're asking the wrong question. The right question is why it does work for $n=2$. There are various proofs you can give in that case and you can see where they use the fact that $n=2$.
– Eric Wofsey
50 secs ago
I think you're asking the wrong question. The right question is why it does work for $n=2$. There are various proofs you can give in that case and you can see where they use the fact that $n=2$.
– Eric Wofsey
50 secs ago
add a comment |Â
J. Wang is a new contributor. Be nice, and check out our Code of Conduct.
J. Wang is a new contributor. Be nice, and check out our Code of Conduct.
J. Wang is a new contributor. Be nice, and check out our Code of Conduct.
J. Wang is a new contributor. Be nice, and check out our Code of Conduct.
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