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, whatâÂÂs about non-cyclic nearly immaculate groups? Do they exist?
abstract-algebra group-theory finite-groups normal-subgroups
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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, whatâÂÂs about non-cyclic nearly immaculate groups? Do they exist?
abstract-algebra group-theory finite-groups normal-subgroups
add a comment |Â
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, whatâÂÂs about non-cyclic nearly immaculate groups? Do they exist?
abstract-algebra group-theory finite-groups normal-subgroups
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, whatâÂÂs about non-cyclic nearly immaculate groups? Do they exist?
abstract-algebra group-theory finite-groups normal-subgroups
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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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $G$.
add a comment |Â
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 $G$.
add a comment |Â
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 $G$.
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 $G$.
edited Aug 15 at 10:56
answered Aug 15 at 9:55
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A. Pongrácz
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3,797625
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