Mixing
Given a finite group, does this equation involving group's order, a partition of it and centralizers' orders...
$begingroup$
Let $G$ be a finite group with $|G|=n$. In the case $Z(G)=lbrace e rbrace$, the following equation seems to hold:
begin{equation}
n(n-1)=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2+sum_{a in G setminus lbrace e rbrace}(|C_G(a)|-1)+sum_{a in G setminus lbrace e rbrace}|C_G(a)|^2([G:C_G(a)]-1)
end{equation}
where $C_G(a)$ is the centralizer of $a$ in $G$ and $lambda(n-1)$ is a partition of the integer $n-1$.
Before anyone wastes time going through my proof's sketch, can some more trained eye than mine acknowledge at a glance whether above equation is:
- trivial (e.g. an identity);
- correct and well-known;
- clearly wrong.
group-theory number-theory
asked Jan 12 at 14:07
LucaLuca
19719
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group with $|G|=n$. In the case $Z(G)=lbrace e rbrace$, the following equation seems to hold:
begin{equation}
n(n-1)=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2+sum_{a in G setminus lbrace e rbrace}(|C_G(a)|-1)+sum_{a in G setminus lbrace e rbrace}|C_G(a)|^2([G:C_G(a)]-1)
end{equation}
where $C_G(a)$ is the centralizer of $a$ in $G$ and $lambda(n-1)$ is a partition of the integer $n-1$.
Before anyone wastes time going through my proof's sketch, can some more trained eye than mine acknowledge at a glance whether above equation is:
- trivial (e.g. an identity);
- correct and well-known;
- clearly wrong.
group-theory number-theory
asked Jan 12 at 14:07
LucaLuca
19719
$endgroup$
2
$begingroup$
With $G=S_3$, I am getting $42$ for the third term in the sum, which is already bigger than $n(n-1)=30$.
$endgroup$
– Derek Holt
Jan 12 at 14:27
$begingroup$
Thank you, I'll double check my proof's sketch.
$endgroup$
– Luca
Jan 12 at 14:36
1
$begingroup$
Surely that can't be true for all partitions, since different partitions should give you different results for the first term...
$endgroup$
– Arnaud D.
Jan 12 at 14:50
$begingroup$
@Arnaud D. Not really, you're right; but indeed I was (and still am) hoping such a partition to be "characteristic" of $G$ itself.
$endgroup$
– Luca
Jan 13 at 10:00
$begingroup$
@Derek Holt. Although I'm still quite confident on 1st and 2nd term, you're right: the 3rd one is completely wrong and I'm working to thoroughly revise it.
$endgroup$
– Luca
Jan 13 at 10:01
add a comment |
$begingroup$
Let $G$ be a finite group with $|G|=n$. In the case $Z(G)=lbrace e rbrace$, the following equation seems to hold:
begin{equation}
n(n-1)=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2+sum_{a in G setminus lbrace e rbrace}(|C_G(a)|-1)+sum_{a in G setminus lbrace e rbrace}|C_G(a)|^2([G:C_G(a)]-1)
end{equation}
where $C_G(a)$ is the centralizer of $a$ in $G$ and $lambda(n-1)$ is a partition of the integer $n-1$.
Before anyone wastes time going through my proof's sketch, can some more trained eye than mine acknowledge at a glance whether above equation is:
- trivial (e.g. an identity);
- correct and well-known;
- clearly wrong.
group-theory number-theory
asked Jan 12 at 14:07
LucaLuca
19719
$endgroup$
Let $G$ be a finite group with $|G|=n$. In the case $Z(G)=lbrace e rbrace$, the following equation seems to hold:
begin{equation}
n(n-1)=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2+sum_{a in G setminus lbrace e rbrace}(|C_G(a)|-1)+sum_{a in G setminus lbrace e rbrace}|C_G(a)|^2([G:C_G(a)]-1)
end{equation}
where $C_G(a)$ is the centralizer of $a$ in $G$ and $lambda(n-1)$ is a partition of the integer $n-1$.
Before anyone wastes time going through my proof's sketch, can some more trained eye than mine acknowledge at a glance whether above equation is:
- trivial (e.g. an identity);
- correct and well-known;
- clearly wrong.
group-theory number-theory
group-theory number-theory
asked Jan 12 at 14:07
LucaLuca
19719
asked Jan 12 at 14:07
LucaLuca
19719
asked Jan 12 at 14:07
LucaLuca
19719
asked Jan 12 at 14:07
LucaLuca
19719
asked Jan 12 at 14:07
LucaLuca
19719
19719
2
$begingroup$
With $G=S_3$, I am getting $42$ for the third term in the sum, which is already bigger than $n(n-1)=30$.
$endgroup$
– Derek Holt
Jan 12 at 14:27
$begingroup$
Thank you, I'll double check my proof's sketch.
$endgroup$
– Luca
Jan 12 at 14:36
1
$begingroup$
Surely that can't be true for all partitions, since different partitions should give you different results for the first term...
$endgroup$
– Arnaud D.
Jan 12 at 14:50
$begingroup$
@Arnaud D. Not really, you're right; but indeed I was (and still am) hoping such a partition to be "characteristic" of $G$ itself.
$endgroup$
– Luca
Jan 13 at 10:00
$begingroup$
@Derek Holt. Although I'm still quite confident on 1st and 2nd term, you're right: the 3rd one is completely wrong and I'm working to thoroughly revise it.
$endgroup$
– Luca
Jan 13 at 10:01
add a comment |
2
$begingroup$
With $G=S_3$, I am getting $42$ for the third term in the sum, which is already bigger than $n(n-1)=30$.
$endgroup$
– Derek Holt
Jan 12 at 14:27
$begingroup$
Thank you, I'll double check my proof's sketch.
$endgroup$
– Luca
Jan 12 at 14:36
1
$begingroup$
Surely that can't be true for all partitions, since different partitions should give you different results for the first term...
$endgroup$
– Arnaud D.
Jan 12 at 14:50
$begingroup$
@Arnaud D. Not really, you're right; but indeed I was (and still am) hoping such a partition to be "characteristic" of $G$ itself.
$endgroup$
– Luca
Jan 13 at 10:00
$begingroup$
@Derek Holt. Although I'm still quite confident on 1st and 2nd term, you're right: the 3rd one is completely wrong and I'm working to thoroughly revise it.
$endgroup$
– Luca
Jan 13 at 10:01
2
2
$begingroup$
With $G=S_3$, I am getting $42$ for the third term in the sum, which is already bigger than $n(n-1)=30$.
$endgroup$
– Derek Holt
Jan 12 at 14:27
$begingroup$
With $G=S_3$, I am getting $42$ for the third term in the sum, which is already bigger than $n(n-1)=30$.
$endgroup$
– Derek Holt
Jan 12 at 14:27
$begingroup$
Thank you, I'll double check my proof's sketch.
$endgroup$
– Luca
Jan 12 at 14:36
$begingroup$
Thank you, I'll double check my proof's sketch.
$endgroup$
– Luca
Jan 12 at 14:36
1
1
$begingroup$
Surely that can't be true for all partitions, since different partitions should give you different results for the first term...
$endgroup$
– Arnaud D.
Jan 12 at 14:50
$begingroup$
Surely that can't be true for all partitions, since different partitions should give you different results for the first term...
$endgroup$
– Arnaud D.
Jan 12 at 14:50
$begingroup$
@Arnaud D. Not really, you're right; but indeed I was (and still am) hoping such a partition to be "characteristic" of $G$ itself.
$endgroup$
– Luca
Jan 13 at 10:00
$begingroup$
@Arnaud D. Not really, you're right; but indeed I was (and still am) hoping such a partition to be "characteristic" of $G$ itself.
$endgroup$
– Luca
Jan 13 at 10:00
$begingroup$
@Derek Holt. Although I'm still quite confident on 1st and 2nd term, you're right: the 3rd one is completely wrong and I'm working to thoroughly revise it.
$endgroup$
– Luca
Jan 13 at 10:01
$begingroup$
@Derek Holt. Although I'm still quite confident on 1st and 2nd term, you're right: the 3rd one is completely wrong and I'm working to thoroughly revise it.
$endgroup$
– Luca
Jan 13 at 10:01
add a comment |
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$begingroup$
With $G=S_3$, I am getting $42$ for the third term in the sum, which is already bigger than $n(n-1)=30$.
$endgroup$
– Derek Holt
Jan 12 at 14:27
$begingroup$
Thank you, I'll double check my proof's sketch.
$endgroup$
– Luca
Jan 12 at 14:36
1
$begingroup$
Surely that can't be true for all partitions, since different partitions should give you different results for the first term...
$endgroup$
– Arnaud D.
Jan 12 at 14:50
$begingroup$
@Arnaud D. Not really, you're right; but indeed I was (and still am) hoping such a partition to be "characteristic" of $G$ itself.
$endgroup$
– Luca
Jan 13 at 10:00
$begingroup$
@Derek Holt. Although I'm still quite confident on 1st and 2nd term, you're right: the 3rd one is completely wrong and I'm working to thoroughly revise it.
$endgroup$
– Luca
Jan 13 at 10:01