Can I determine a $p$-group by the number of subgroups












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Let $G$ and $H$ be finite $p$-groups of order $p^n$ where $p$ is a prime. Assume that for every $1leq ileq n$, the number of subgroups of $G$ and $H$ of order $p^i$ are the same. Can I deduce that $G$ and $H$ are isomorphic?



It seems like it works for $n=1,2,3$ but these cases are more or less trivial. For $n>3$, examples become too complicated for me to investigate.










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    $begingroup$


    Let $G$ and $H$ be finite $p$-groups of order $p^n$ where $p$ is a prime. Assume that for every $1leq ileq n$, the number of subgroups of $G$ and $H$ of order $p^i$ are the same. Can I deduce that $G$ and $H$ are isomorphic?



    It seems like it works for $n=1,2,3$ but these cases are more or less trivial. For $n>3$, examples become too complicated for me to investigate.










    share|cite|improve this question









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      $begingroup$


      Let $G$ and $H$ be finite $p$-groups of order $p^n$ where $p$ is a prime. Assume that for every $1leq ileq n$, the number of subgroups of $G$ and $H$ of order $p^i$ are the same. Can I deduce that $G$ and $H$ are isomorphic?



      It seems like it works for $n=1,2,3$ but these cases are more or less trivial. For $n>3$, examples become too complicated for me to investigate.










      share|cite|improve this question









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      Let $G$ and $H$ be finite $p$-groups of order $p^n$ where $p$ is a prime. Assume that for every $1leq ileq n$, the number of subgroups of $G$ and $H$ of order $p^i$ are the same. Can I deduce that $G$ and $H$ are isomorphic?



      It seems like it works for $n=1,2,3$ but these cases are more or less trivial. For $n>3$, examples become too complicated for me to investigate.







      abstract-algebra group-theory finite-groups






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      asked Jan 20 at 1:40









      LeventLevent

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          No of course you cannot deduce that? Why would expect to? (Counterexample: $mathtt{SmallGroup}(16,i)$ for $i=2$ and $4$; or for $i=5$ and $6$.)






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            $begingroup$

            No of course you cannot deduce that? Why would expect to? (Counterexample: $mathtt{SmallGroup}(16,i)$ for $i=2$ and $4$; or for $i=5$ and $6$.)






            share|cite|improve this answer









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              $begingroup$

              No of course you cannot deduce that? Why would expect to? (Counterexample: $mathtt{SmallGroup}(16,i)$ for $i=2$ and $4$; or for $i=5$ and $6$.)






              share|cite|improve this answer









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                $begingroup$

                No of course you cannot deduce that? Why would expect to? (Counterexample: $mathtt{SmallGroup}(16,i)$ for $i=2$ and $4$; or for $i=5$ and $6$.)






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                No of course you cannot deduce that? Why would expect to? (Counterexample: $mathtt{SmallGroup}(16,i)$ for $i=2$ and $4$; or for $i=5$ and $6$.)







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                answered Jan 20 at 8:12









                Derek HoltDerek Holt

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