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!










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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!










    share|cite|improve this question







    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.





















      up vote
      5
      down vote

      favorite
      1









      up vote
      5
      down vote

      favorite
      1






      1





      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!










      share|cite|improve this question







      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






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      J. Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 1 hour ago









      J. Wang

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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.






          share|cite|improve this answer




















          • 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
            57 secs ago

















          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
          $$
          Large(G/A)times(G/B)times(G/C)Large
          =
          fracn3cdot fracn5cdotfracn11
          =fracn^3n=n^2 > n
          $$






          share|cite|improve this answer






















          • 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.
            – Eric Wofsey
            6 secs ago










          Your Answer




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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

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          active

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          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.






          share|cite|improve this answer




















          • 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
            57 secs ago














          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.






          share|cite|improve this answer




















          • 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
            57 secs ago












          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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
            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. 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
            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. 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
          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. 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
          57 secs ago










          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
          $$
          Large(G/A)times(G/B)times(G/C)Large
          =
          fracn3cdot fracn5cdotfracn11
          =fracn^3n=n^2 > n
          $$






          share|cite|improve this answer






















          • 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.
            – Eric Wofsey
            6 secs ago














          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
          $$
          Large(G/A)times(G/B)times(G/C)Large
          =
          fracn3cdot fracn5cdotfracn11
          =fracn^3n=n^2 > n
          $$






          share|cite|improve this answer






















          • 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.
            – Eric Wofsey
            6 secs ago












          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
          $$
          Large(G/A)times(G/B)times(G/C)Large
          =
          fracn3cdot fracn5cdotfracn11
          =fracn^3n=n^2 > n
          $$






          share|cite|improve this answer














          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
          $$
          Large(G/A)times(G/B)times(G/C)Large
          =
          fracn3cdot fracn5cdotfracn11
          =fracn^3n=n^2 > n
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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.
            – 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.
            – 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.
          – 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.
          – Eric Wofsey
          6 secs ago










          J. Wang is a new contributor. Be nice, and check out our Code of Conduct.









           

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