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
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up vote
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down vote
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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
New contributor
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
New contributor
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
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New contributor
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asked 1 hour ago
J. Wang
262
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2 Answers
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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.
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. ThatâÂÂs 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?
â J. Wang
57 secs ago
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up vote
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Here's a counterexample . . .
Let $G=Z_n$, where $n=3cdot 5cdot 11$, regarded as a multiplicative group, and let $A,B,C$ be the subgroups of $G$ given by
$$A=langlex^3rangle,;;;B=langlex^5rangle,;;;C=langlex^11rangle$$
where $x$ is a cyclic generator of $G$.
Then we have
beginalign*
AB&=langlex^8rangle=G\[4pt]
BC&=langlex^16rangle=G\[4pt]
CA&=langlex^14rangle=G\[4pt]
endalign*
and then, since $Acap Bcap C=1$, we get
$$LargeG/(Acap Bcap C)Large=|G|=n$$
but
$$
Large(G/A)times(G/B)times(G/C)Large
=
fracn3cdot fracn5cdotfracn11
=fracn^3n=n^2 > n
$$
This is wrong: $|G/A|$ has $3$ elements, not $n/3$ elements, for instance. Indeed, in this case $Gcong (G/A)times (G/B)times (G/C)$ just by the ordinary Chinese remainder theorem for modular arithmetic.
â Eric Wofsey
6 secs ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
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.
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. ThatâÂÂs 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?
â J. Wang
57 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.
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. ThatâÂÂs 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?
â J. Wang
57 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.
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
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. ThatâÂÂs 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?
â J. Wang
57 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. ThatâÂÂs 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?
â J. Wang
57 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. ThatâÂÂs 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?
â J. Wang
57 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. ThatâÂÂs 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?
â J. Wang
57 secs ago
add a comment |Â
up vote
0
down vote
Here's a counterexample . . .
Let $G=Z_n$, where $n=3cdot 5cdot 11$, regarded as a multiplicative group, and let $A,B,C$ be the subgroups of $G$ given by
$$A=langlex^3rangle,;;;B=langlex^5rangle,;;;C=langlex^11rangle$$
where $x$ is a cyclic generator of $G$.
Then we have
beginalign*
AB&=langlex^8rangle=G\[4pt]
BC&=langlex^16rangle=G\[4pt]
CA&=langlex^14rangle=G\[4pt]
endalign*
and then, since $Acap Bcap C=1$, we get
$$LargeG/(Acap Bcap C)Large=|G|=n$$
but
$$
Large(G/A)times(G/B)times(G/C)Large
=
fracn3cdot fracn5cdotfracn11
=fracn^3n=n^2 > n
$$
This is wrong: $|G/A|$ has $3$ elements, not $n/3$ elements, for instance. Indeed, in this case $Gcong (G/A)times (G/B)times (G/C)$ just by the ordinary Chinese remainder theorem for modular arithmetic.
â Eric Wofsey
6 secs ago
add a comment |Â
up vote
0
down vote
Here's a counterexample . . .
Let $G=Z_n$, where $n=3cdot 5cdot 11$, regarded as a multiplicative group, and let $A,B,C$ be the subgroups of $G$ given by
$$A=langlex^3rangle,;;;B=langlex^5rangle,;;;C=langlex^11rangle$$
where $x$ is a cyclic generator of $G$.
Then we have
beginalign*
AB&=langlex^8rangle=G\[4pt]
BC&=langlex^16rangle=G\[4pt]
CA&=langlex^14rangle=G\[4pt]
endalign*
and then, since $Acap Bcap C=1$, we get
$$LargeG/(Acap Bcap C)Large=|G|=n$$
but
$$
Large(G/A)times(G/B)times(G/C)Large
=
fracn3cdot fracn5cdotfracn11
=fracn^3n=n^2 > n
$$
This is wrong: $|G/A|$ has $3$ elements, not $n/3$ elements, for instance. Indeed, in this case $Gcong (G/A)times (G/B)times (G/C)$ just by the ordinary Chinese remainder theorem for modular arithmetic.
â Eric Wofsey
6 secs ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here's a counterexample . . .
Let $G=Z_n$, where $n=3cdot 5cdot 11$, regarded as a multiplicative group, and let $A,B,C$ be the subgroups of $G$ given by
$$A=langlex^3rangle,;;;B=langlex^5rangle,;;;C=langlex^11rangle$$
where $x$ is a cyclic generator of $G$.
Then we have
beginalign*
AB&=langlex^8rangle=G\[4pt]
BC&=langlex^16rangle=G\[4pt]
CA&=langlex^14rangle=G\[4pt]
endalign*
and then, since $Acap Bcap C=1$, we get
$$LargeG/(Acap Bcap C)Large=|G|=n$$
but
$$
Large(G/A)times(G/B)times(G/C)Large
=
fracn3cdot fracn5cdotfracn11
=fracn^3n=n^2 > n
$$
Here's a counterexample . . .
Let $G=Z_n$, where $n=3cdot 5cdot 11$, regarded as a multiplicative group, and let $A,B,C$ be the subgroups of $G$ given by
$$A=langlex^3rangle,;;;B=langlex^5rangle,;;;C=langlex^11rangle$$
where $x$ is a cyclic generator of $G$.
Then we have
beginalign*
AB&=langlex^8rangle=G\[4pt]
BC&=langlex^16rangle=G\[4pt]
CA&=langlex^14rangle=G\[4pt]
endalign*
and then, since $Acap Bcap C=1$, we get
$$LargeG/(Acap Bcap C)Large=|G|=n$$
but
$$
Large(G/A)times(G/B)times(G/C)Large
=
fracn3cdot fracn5cdotfracn11
=fracn^3n=n^2 > n
$$
edited 27 mins ago
answered 41 mins ago
quasi
34.4k22561
34.4k22561
This is wrong: $|G/A|$ has $3$ elements, not $n/3$ elements, for instance. Indeed, in this case $Gcong (G/A)times (G/B)times (G/C)$ just by the ordinary Chinese remainder theorem for modular arithmetic.
â Eric Wofsey
6 secs ago
add a comment |Â
This is wrong: $|G/A|$ has $3$ elements, not $n/3$ elements, for instance. Indeed, in this case $Gcong (G/A)times (G/B)times (G/C)$ just by the ordinary Chinese remainder theorem for modular arithmetic.
â Eric Wofsey
6 secs ago
This is wrong: $|G/A|$ has $3$ elements, not $n/3$ elements, for instance. Indeed, in this case $Gcong (G/A)times (G/B)times (G/C)$ just by the ordinary Chinese remainder theorem for modular arithmetic.
â Eric Wofsey
6 secs ago
This is wrong: $|G/A|$ has $3$ elements, not $n/3$ elements, for instance. Indeed, in this case $Gcong (G/A)times (G/B)times (G/C)$ just by the ordinary Chinese remainder theorem for modular arithmetic.
â Eric Wofsey
6 secs ago
add a comment |Â
J. Wang is a new contributor. Be nice, and check out our Code of Conduct.
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J. Wang is a new contributor. Be nice, and check out our Code of Conduct.
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