Does there exist a nearly immaculate group not of the form $C_2^n$?

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Define a nearly immaculate group as a finite group $G$, such, that the sum of the orders of all its normal subgroups is $2|G| - 1$. It is quite obvious to see, that all groups of the form $C_2^n$ are nearly immaculate. But are there any not of the type?



One can see, that a cyclic group $C_n$ is nearly immaculate iff $n$ is almost perfect. Whether all almost perfect numbers are of the form $2^n$ or not, is an open problem. But, what’s about non-cyclic nearly immaculate groups? Do they exist?







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    up vote
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    Define a nearly immaculate group as a finite group $G$, such, that the sum of the orders of all its normal subgroups is $2|G| - 1$. It is quite obvious to see, that all groups of the form $C_2^n$ are nearly immaculate. But are there any not of the type?



    One can see, that a cyclic group $C_n$ is nearly immaculate iff $n$ is almost perfect. Whether all almost perfect numbers are of the form $2^n$ or not, is an open problem. But, what’s about non-cyclic nearly immaculate groups? Do they exist?







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      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      Define a nearly immaculate group as a finite group $G$, such, that the sum of the orders of all its normal subgroups is $2|G| - 1$. It is quite obvious to see, that all groups of the form $C_2^n$ are nearly immaculate. But are there any not of the type?



      One can see, that a cyclic group $C_n$ is nearly immaculate iff $n$ is almost perfect. Whether all almost perfect numbers are of the form $2^n$ or not, is an open problem. But, what’s about non-cyclic nearly immaculate groups? Do they exist?







      share|cite|improve this question












      Define a nearly immaculate group as a finite group $G$, such, that the sum of the orders of all its normal subgroups is $2|G| - 1$. It is quite obvious to see, that all groups of the form $C_2^n$ are nearly immaculate. But are there any not of the type?



      One can see, that a cyclic group $C_n$ is nearly immaculate iff $n$ is almost perfect. Whether all almost perfect numbers are of the form $2^n$ or not, is an open problem. But, what’s about non-cyclic nearly immaculate groups? Do they exist?









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      asked Aug 15 at 9:33









      Yanior Weg

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          I do not have a full answer, but I can show that such an example cannot be a $p$-group. It seems challanging to generalize this to nilpotent groups.



          Claim: among the (finite) $p$-groups, $Gcong C_2^n$ are the only examples.



          Proof: If $pneq 2$, then the order of every nontrivial normal subgroup is divisible by $p$, so the sum of orders is congruent to $1$ modulo $p$.
          Hence, it cannot be $2|G|-1$.
          If $p=2$ and $G=2^n$, then there exists a normal subgroup in $G$ of order $2^k$ for every $0leq kleq n$.
          As the sum of these orders is $2^n+1-1= 2|G|-1$, there must be exactly one of each.
          In particular, there is a unique maximal normal subgroup $Mtriangleleft G$, and thus any element $gnotin M$ generates $G$.






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            up vote
            5
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            I do not have a full answer, but I can show that such an example cannot be a $p$-group. It seems challanging to generalize this to nilpotent groups.



            Claim: among the (finite) $p$-groups, $Gcong C_2^n$ are the only examples.



            Proof: If $pneq 2$, then the order of every nontrivial normal subgroup is divisible by $p$, so the sum of orders is congruent to $1$ modulo $p$.
            Hence, it cannot be $2|G|-1$.
            If $p=2$ and $G=2^n$, then there exists a normal subgroup in $G$ of order $2^k$ for every $0leq kleq n$.
            As the sum of these orders is $2^n+1-1= 2|G|-1$, there must be exactly one of each.
            In particular, there is a unique maximal normal subgroup $Mtriangleleft G$, and thus any element $gnotin M$ generates $G$.






            share|cite|improve this answer


























              up vote
              5
              down vote













              I do not have a full answer, but I can show that such an example cannot be a $p$-group. It seems challanging to generalize this to nilpotent groups.



              Claim: among the (finite) $p$-groups, $Gcong C_2^n$ are the only examples.



              Proof: If $pneq 2$, then the order of every nontrivial normal subgroup is divisible by $p$, so the sum of orders is congruent to $1$ modulo $p$.
              Hence, it cannot be $2|G|-1$.
              If $p=2$ and $G=2^n$, then there exists a normal subgroup in $G$ of order $2^k$ for every $0leq kleq n$.
              As the sum of these orders is $2^n+1-1= 2|G|-1$, there must be exactly one of each.
              In particular, there is a unique maximal normal subgroup $Mtriangleleft G$, and thus any element $gnotin M$ generates $G$.






              share|cite|improve this answer
























                up vote
                5
                down vote










                up vote
                5
                down vote









                I do not have a full answer, but I can show that such an example cannot be a $p$-group. It seems challanging to generalize this to nilpotent groups.



                Claim: among the (finite) $p$-groups, $Gcong C_2^n$ are the only examples.



                Proof: If $pneq 2$, then the order of every nontrivial normal subgroup is divisible by $p$, so the sum of orders is congruent to $1$ modulo $p$.
                Hence, it cannot be $2|G|-1$.
                If $p=2$ and $G=2^n$, then there exists a normal subgroup in $G$ of order $2^k$ for every $0leq kleq n$.
                As the sum of these orders is $2^n+1-1= 2|G|-1$, there must be exactly one of each.
                In particular, there is a unique maximal normal subgroup $Mtriangleleft G$, and thus any element $gnotin M$ generates $G$.






                share|cite|improve this answer














                I do not have a full answer, but I can show that such an example cannot be a $p$-group. It seems challanging to generalize this to nilpotent groups.



                Claim: among the (finite) $p$-groups, $Gcong C_2^n$ are the only examples.



                Proof: If $pneq 2$, then the order of every nontrivial normal subgroup is divisible by $p$, so the sum of orders is congruent to $1$ modulo $p$.
                Hence, it cannot be $2|G|-1$.
                If $p=2$ and $G=2^n$, then there exists a normal subgroup in $G$ of order $2^k$ for every $0leq kleq n$.
                As the sum of these orders is $2^n+1-1= 2|G|-1$, there must be exactly one of each.
                In particular, there is a unique maximal normal subgroup $Mtriangleleft G$, and thus any element $gnotin M$ generates $G$.







                share|cite|improve this answer














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                edited Aug 15 at 10:56

























                answered Aug 15 at 9:55









                A. Pongrácz

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