Subgroups of $G^n$, where $G$ is a $p$-group











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Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.




Are all subgroups of $G^n$ isomorphic to $H_1 times dotsm times H_n$ for some subgroups $H_1, dots, H_n$
of $G$?




Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here.
This question and its answer, which relies on Goursat's lemma might also be relevant.



P.S. I am especially interested in the case $p = 2$.










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  • Have you run any computer checks for small values of $|G|$ and $n$?
    – the_fox
    Nov 22 at 12:25










  • I tried a few examples on GAP without finding any counterexample so far.
    – J.-E. Pin
    Nov 22 at 12:34










  • I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
    – the_fox
    Nov 22 at 12:41










  • I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
    – Derek Holt
    Nov 22 at 12:49






  • 1




    I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
    – Derek Holt
    Nov 22 at 12:55















up vote
3
down vote

favorite












Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.




Are all subgroups of $G^n$ isomorphic to $H_1 times dotsm times H_n$ for some subgroups $H_1, dots, H_n$
of $G$?




Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here.
This question and its answer, which relies on Goursat's lemma might also be relevant.



P.S. I am especially interested in the case $p = 2$.










share|cite|improve this question
























  • Have you run any computer checks for small values of $|G|$ and $n$?
    – the_fox
    Nov 22 at 12:25










  • I tried a few examples on GAP without finding any counterexample so far.
    – J.-E. Pin
    Nov 22 at 12:34










  • I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
    – the_fox
    Nov 22 at 12:41










  • I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
    – Derek Holt
    Nov 22 at 12:49






  • 1




    I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
    – Derek Holt
    Nov 22 at 12:55













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.




Are all subgroups of $G^n$ isomorphic to $H_1 times dotsm times H_n$ for some subgroups $H_1, dots, H_n$
of $G$?




Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here.
This question and its answer, which relies on Goursat's lemma might also be relevant.



P.S. I am especially interested in the case $p = 2$.










share|cite|improve this question















Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.




Are all subgroups of $G^n$ isomorphic to $H_1 times dotsm times H_n$ for some subgroups $H_1, dots, H_n$
of $G$?




Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here.
This question and its answer, which relies on Goursat's lemma might also be relevant.



P.S. I am especially interested in the case $p = 2$.







group-theory finite-groups p-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 12:48









Derek Holt

52.3k53570




52.3k53570










asked Nov 22 at 12:12









J.-E. Pin

18.3k21754




18.3k21754












  • Have you run any computer checks for small values of $|G|$ and $n$?
    – the_fox
    Nov 22 at 12:25










  • I tried a few examples on GAP without finding any counterexample so far.
    – J.-E. Pin
    Nov 22 at 12:34










  • I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
    – the_fox
    Nov 22 at 12:41










  • I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
    – Derek Holt
    Nov 22 at 12:49






  • 1




    I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
    – Derek Holt
    Nov 22 at 12:55


















  • Have you run any computer checks for small values of $|G|$ and $n$?
    – the_fox
    Nov 22 at 12:25










  • I tried a few examples on GAP without finding any counterexample so far.
    – J.-E. Pin
    Nov 22 at 12:34










  • I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
    – the_fox
    Nov 22 at 12:41










  • I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
    – Derek Holt
    Nov 22 at 12:49






  • 1




    I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
    – Derek Holt
    Nov 22 at 12:55
















Have you run any computer checks for small values of $|G|$ and $n$?
– the_fox
Nov 22 at 12:25




Have you run any computer checks for small values of $|G|$ and $n$?
– the_fox
Nov 22 at 12:25












I tried a few examples on GAP without finding any counterexample so far.
– J.-E. Pin
Nov 22 at 12:34




I tried a few examples on GAP without finding any counterexample so far.
– J.-E. Pin
Nov 22 at 12:34












I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
– the_fox
Nov 22 at 12:41




I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
– the_fox
Nov 22 at 12:41












I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
– Derek Holt
Nov 22 at 12:49




I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
– Derek Holt
Nov 22 at 12:49




1




1




I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
– Derek Holt
Nov 22 at 12:55




I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
– Derek Holt
Nov 22 at 12:55










1 Answer
1






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5
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Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.



If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.



In fact $H$ is indecomposable.






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    1 Answer
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    accepted










    Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.



    If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.



    In fact $H$ is indecomposable.






    share|cite|improve this answer

























      up vote
      5
      down vote



      accepted










      Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.



      If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.



      In fact $H$ is indecomposable.






      share|cite|improve this answer























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.



        If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.



        In fact $H$ is indecomposable.






        share|cite|improve this answer












        Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.



        If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.



        In fact $H$ is indecomposable.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 14:10









        Derek Holt

        52.3k53570




        52.3k53570






























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