Mixing
Given a finite group, does this equation involving group's order, a partition of it and centralizers' orders...
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After failing this attempt, I've revised my proof sketch and I've come to the following version of the equation in the title.
So, let $G$ be a finite group, say $G=lbrace e,a_1,dots,a_{n-1} rbrace$, and $C_G(a_i)$ denote the centralizer of $a_i$ in $G$. Then, in the case $Z(G)=lbrace e rbrace$, the following equation seems to hold:
begin{equation}
n(n-1)=sum_{j=1}^{n-1}delta(Lambda)_j|C_G(a_j)|
end{equation}
where $delta(Lambda)$ is a $(n-1)$-term partition of the integer $Lambda:=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$, being $lambda(n-1)$ a partition of the integer $n-1$.
E.g., for $G=S_3$, if we take $lambda(5)=(2,3)$ and $delta(13)=(2,2,3,3,3)$, we get:
begin{equation}
6cdot5 = 2^2+3^2+2cdot(3-1)+2cdot(3-1)+3cdot(2-1)+3cdot(2-1)+3cdot(2-1)
end{equation}
As I'm not very skilled on symmetric groups, can someone check whether above formula works also for $S_4$, etc. or some other group with trivial center?
Proof.
Let's define the collections $F_a:=(varphi_g(a))_{gin G}$, where $varphi_g(a):=g^{-1}ag$, and denote by $tilde F_a$ the set composed of the distinct elements of $F_a$. We have:
$forall a in G, a in tilde F_a$; then, $tilde F_a ne emptyset$;
$forall a in G, |F_a|=n$;
$tilde F_e=lbrace e rbrace$;
$Z(G)=lbrace e rbrace Rightarrow |tilde F_a|>1, forall a in G, a ne e$;
$F_a=(varphi_{g}(a))_{g in C_G(a)} cup (varphi_{g}(a))_{g in ∁_G(C_G(a))}=(a,a,...,a) cup (varphi_{g}(a))_{g in ∁_G(C_G(a))}$, with $|(a,a,...,a)|=|C_G(a)|$;- given $a,b in G, b ne a$, it is $tilde F_b=tilde F_a Leftrightarrow b in tilde F_a$;
- the collection $(varphi_{g}(a))_{g in ∁_G(C_G(a))}$ contains as many copies of each element $b in tilde F_a setminus lbrace a rbrace$ as is the cardinality of the set $tilde{mathcal{F}}_b^{(a)}:=lbrace g in complement_G(C_G(a))|varphi_g(a)=b rbrace$, minus $1$.
We may also think of a $n times n$ "matrix" such that, given $a in G$, the "$a$-th" row is made of the elements of the set $lbrace varphi_a(g)|g in G rbrace$ and the "$a$-th" column is made of the elements of the collection $F_a=(varphi_g(a))_{g in G}$. According to this "picture", we have that:
- for 3 and 4, there's one and only one column with all the entries equal to each other (all $e$);
- for 5, the "$a$-th" column contains as many copies of $a$ as the order of $C_G(a)$, minus $1$;
- for 6 and 7, there are $|tilde F_a|$ columns with the same, differently permuted, $|tilde F_a|$ distinct elements (included the "index" $a$); each of these columns has as many copies of each element $b$ of $tilde F_a setminus lbrace a rbrace$ as is the integer $|tilde{mathcal{F}}_b^{(a)}|-1$.
Now, $tilde{mathcal{F}}_b^{(a)}=emptyset Leftrightarrow b notin tilde F_a$, and then $b in tilde F_a Rightarrow tilde{mathcal{F}}_b^{(a)} ne emptyset Rightarrow |tilde{mathcal{F}}_b^{(a)}|=|C_G(a)|$ (for the last implication, see here).
By summing on all the columns, we have that a partition $lambda(n-1)$ of the integer $n-1$ exists such that:
begin{align}
n^2& = n+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}sum_{b in tilde F_a setminus lbrace a rbrace}(|tilde{mathcal{F}}_b^{(a)}|-1)=\&=n+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}|tilde F_a setminus lbrace a rbrace|(|C_G(a)|-1)=\&=n+sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2+sum_{a in G setminus lbrace e rbrace}(|tilde F_a setminus lbrace a rbrace|+1)(|C_G(a)|-1)
end{align}
But, $$sum_{a in G setminus lbrace e rbrace}(|tilde F_a setminus lbrace a rbrace|+1)=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
so, if we define $delta_a:=|tilde F_a setminus lbrace a rbrace|+1$, we get:
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}delta_a(|C_G(a)|-1)
end{equation}
with $$sum_{a in G setminus lbrace e rbrace}delta_a=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
and then, for a given labelling of the elements of $G$,
begin{equation}
n(n-1)=sum_{j=1}^{n-1}delta(Lambda)_j|C_G(a_j)|
end{equation}
where $delta(Lambda)$ is a $(n-1)$-term partition of the integer $Lambda:=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$.
Addendum
As shown, each $a in G$, $a ne e$, gives rise to the collection $F_a$ made of:
$|tilde F_a|$ distinct elements of $G$, included $a$;
$|C_G(a)|-1$ copies of $a$;
$|mathcal{F}_b^{(a)}|-1=|C_G(a)|-1$ copies of $b$, for each $b in tilde F_a setminus lbrace a rbrace$.
Then,
begin{align}
n=&|tilde F_a|+|C_G(a)|-1+(|C_G(a)|-1)(|tilde F_a|-1)=\=&|tilde F_a|+|C_G(a)|-1+|C_G(a)||tilde F_a|-|C_G(a)|-|tilde F_a|+1=\=&|C_G(a)||tilde F_a|
end{align}
so $|tilde F_a|=[G:C_G(a)]$, and finally $$sum_{a in G setminus lbrace e rbrace}[G:C_G(a)]=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
For a finite group of order $n$ and trivial center, a partition $lambda(n-1)$ of $n-1$ with no "1" summands exists, such that the sum of the indexes of the centralizers of all the elements of the group (but the unit) is equal to the sum of the squares of the summands of $lambda(n-1)$.
combinatorics group-theory number-theory
asked Jan 13 at 22:21
LucaLuca
21419
$endgroup$
add a comment |
$begingroup$
After failing this attempt, I've revised my proof sketch and I've come to the following version of the equation in the title.
So, let $G$ be a finite group, say $G=lbrace e,a_1,dots,a_{n-1} rbrace$, and $C_G(a_i)$ denote the centralizer of $a_i$ in $G$. Then, in the case $Z(G)=lbrace e rbrace$, the following equation seems to hold:
begin{equation}
n(n-1)=sum_{j=1}^{n-1}delta(Lambda)_j|C_G(a_j)|
end{equation}
where $delta(Lambda)$ is a $(n-1)$-term partition of the integer $Lambda:=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$, being $lambda(n-1)$ a partition of the integer $n-1$.
E.g., for $G=S_3$, if we take $lambda(5)=(2,3)$ and $delta(13)=(2,2,3,3,3)$, we get:
begin{equation}
6cdot5 = 2^2+3^2+2cdot(3-1)+2cdot(3-1)+3cdot(2-1)+3cdot(2-1)+3cdot(2-1)
end{equation}
As I'm not very skilled on symmetric groups, can someone check whether above formula works also for $S_4$, etc. or some other group with trivial center?
Proof.
Let's define the collections $F_a:=(varphi_g(a))_{gin G}$, where $varphi_g(a):=g^{-1}ag$, and denote by $tilde F_a$ the set composed of the distinct elements of $F_a$. We have:
$forall a in G, a in tilde F_a$; then, $tilde F_a ne emptyset$;
$forall a in G, |F_a|=n$;
$tilde F_e=lbrace e rbrace$;
$Z(G)=lbrace e rbrace Rightarrow |tilde F_a|>1, forall a in G, a ne e$;
$F_a=(varphi_{g}(a))_{g in C_G(a)} cup (varphi_{g}(a))_{g in ∁_G(C_G(a))}=(a,a,...,a) cup (varphi_{g}(a))_{g in ∁_G(C_G(a))}$, with $|(a,a,...,a)|=|C_G(a)|$;- given $a,b in G, b ne a$, it is $tilde F_b=tilde F_a Leftrightarrow b in tilde F_a$;
- the collection $(varphi_{g}(a))_{g in ∁_G(C_G(a))}$ contains as many copies of each element $b in tilde F_a setminus lbrace a rbrace$ as is the cardinality of the set $tilde{mathcal{F}}_b^{(a)}:=lbrace g in complement_G(C_G(a))|varphi_g(a)=b rbrace$, minus $1$.
We may also think of a $n times n$ "matrix" such that, given $a in G$, the "$a$-th" row is made of the elements of the set $lbrace varphi_a(g)|g in G rbrace$ and the "$a$-th" column is made of the elements of the collection $F_a=(varphi_g(a))_{g in G}$. According to this "picture", we have that:
- for 3 and 4, there's one and only one column with all the entries equal to each other (all $e$);
- for 5, the "$a$-th" column contains as many copies of $a$ as the order of $C_G(a)$, minus $1$;
- for 6 and 7, there are $|tilde F_a|$ columns with the same, differently permuted, $|tilde F_a|$ distinct elements (included the "index" $a$); each of these columns has as many copies of each element $b$ of $tilde F_a setminus lbrace a rbrace$ as is the integer $|tilde{mathcal{F}}_b^{(a)}|-1$.
Now, $tilde{mathcal{F}}_b^{(a)}=emptyset Leftrightarrow b notin tilde F_a$, and then $b in tilde F_a Rightarrow tilde{mathcal{F}}_b^{(a)} ne emptyset Rightarrow |tilde{mathcal{F}}_b^{(a)}|=|C_G(a)|$ (for the last implication, see here).
By summing on all the columns, we have that a partition $lambda(n-1)$ of the integer $n-1$ exists such that:
begin{align}
n^2& = n+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}sum_{b in tilde F_a setminus lbrace a rbrace}(|tilde{mathcal{F}}_b^{(a)}|-1)=\&=n+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}|tilde F_a setminus lbrace a rbrace|(|C_G(a)|-1)=\&=n+sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2+sum_{a in G setminus lbrace e rbrace}(|tilde F_a setminus lbrace a rbrace|+1)(|C_G(a)|-1)
end{align}
But, $$sum_{a in G setminus lbrace e rbrace}(|tilde F_a setminus lbrace a rbrace|+1)=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
so, if we define $delta_a:=|tilde F_a setminus lbrace a rbrace|+1$, we get:
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}delta_a(|C_G(a)|-1)
end{equation}
with $$sum_{a in G setminus lbrace e rbrace}delta_a=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
and then, for a given labelling of the elements of $G$,
begin{equation}
n(n-1)=sum_{j=1}^{n-1}delta(Lambda)_j|C_G(a_j)|
end{equation}
where $delta(Lambda)$ is a $(n-1)$-term partition of the integer $Lambda:=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$.
Addendum
As shown, each $a in G$, $a ne e$, gives rise to the collection $F_a$ made of:
$|tilde F_a|$ distinct elements of $G$, included $a$;
$|C_G(a)|-1$ copies of $a$;
$|mathcal{F}_b^{(a)}|-1=|C_G(a)|-1$ copies of $b$, for each $b in tilde F_a setminus lbrace a rbrace$.
Then,
begin{align}
n=&|tilde F_a|+|C_G(a)|-1+(|C_G(a)|-1)(|tilde F_a|-1)=\=&|tilde F_a|+|C_G(a)|-1+|C_G(a)||tilde F_a|-|C_G(a)|-|tilde F_a|+1=\=&|C_G(a)||tilde F_a|
end{align}
so $|tilde F_a|=[G:C_G(a)]$, and finally $$sum_{a in G setminus lbrace e rbrace}[G:C_G(a)]=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
For a finite group of order $n$ and trivial center, a partition $lambda(n-1)$ of $n-1$ with no "1" summands exists, such that the sum of the indexes of the centralizers of all the elements of the group (but the unit) is equal to the sum of the squares of the summands of $lambda(n-1)$.
combinatorics group-theory number-theory
asked Jan 13 at 22:21
LucaLuca
21419
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I don't see how something like this could possibly work, as the sum on the right depends on the ordering of the elements of $G$. For example, even for $S_3$ and $lambda(5)=(2,3)$, this fails if we order the elements of $S_3$ differently. (For example, if $a_1$ is an involution rather than an element of order $3$.)
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– verret
Jan 14 at 3:06
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Wouldn't in that case the equation hold with, e.g., the partition $delta(13)=(3,2,3,3,2)$ ?
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– Luca
Jan 14 at 8:07
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Do you have any clue how to find a partition that will work? If not, then it seems hard computationally to check whether it holds for a given group (and it could also hold for trivial reasons, simply because there are so many partitions). Moreover, if not, then it seems hard to use the result, since we wouldn't know what the partition is.
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– verret
Jan 14 at 9:02
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If I'm not mistaken, my proof simply states that such partitions $lambda$ and $delta$ do exist, no more than that. If it were correct, I'm more inclined to think that, as you argued, it is simply for trivial reasons. So, I think I'd better to post the sketch of the proof for your review.
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– Luca
Jan 14 at 9:11
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Edited with my proof attempt.
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– Luca
Jan 14 at 12:08
add a comment |
$begingroup$
After failing this attempt, I've revised my proof sketch and I've come to the following version of the equation in the title.
So, let $G$ be a finite group, say $G=lbrace e,a_1,dots,a_{n-1} rbrace$, and $C_G(a_i)$ denote the centralizer of $a_i$ in $G$. Then, in the case $Z(G)=lbrace e rbrace$, the following equation seems to hold:
begin{equation}
n(n-1)=sum_{j=1}^{n-1}delta(Lambda)_j|C_G(a_j)|
end{equation}
where $delta(Lambda)$ is a $(n-1)$-term partition of the integer $Lambda:=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$, being $lambda(n-1)$ a partition of the integer $n-1$.
E.g., for $G=S_3$, if we take $lambda(5)=(2,3)$ and $delta(13)=(2,2,3,3,3)$, we get:
begin{equation}
6cdot5 = 2^2+3^2+2cdot(3-1)+2cdot(3-1)+3cdot(2-1)+3cdot(2-1)+3cdot(2-1)
end{equation}
As I'm not very skilled on symmetric groups, can someone check whether above formula works also for $S_4$, etc. or some other group with trivial center?
Proof.
Let's define the collections $F_a:=(varphi_g(a))_{gin G}$, where $varphi_g(a):=g^{-1}ag$, and denote by $tilde F_a$ the set composed of the distinct elements of $F_a$. We have:
$forall a in G, a in tilde F_a$; then, $tilde F_a ne emptyset$;
$forall a in G, |F_a|=n$;
$tilde F_e=lbrace e rbrace$;
$Z(G)=lbrace e rbrace Rightarrow |tilde F_a|>1, forall a in G, a ne e$;
$F_a=(varphi_{g}(a))_{g in C_G(a)} cup (varphi_{g}(a))_{g in ∁_G(C_G(a))}=(a,a,...,a) cup (varphi_{g}(a))_{g in ∁_G(C_G(a))}$, with $|(a,a,...,a)|=|C_G(a)|$;- given $a,b in G, b ne a$, it is $tilde F_b=tilde F_a Leftrightarrow b in tilde F_a$;
- the collection $(varphi_{g}(a))_{g in ∁_G(C_G(a))}$ contains as many copies of each element $b in tilde F_a setminus lbrace a rbrace$ as is the cardinality of the set $tilde{mathcal{F}}_b^{(a)}:=lbrace g in complement_G(C_G(a))|varphi_g(a)=b rbrace$, minus $1$.
We may also think of a $n times n$ "matrix" such that, given $a in G$, the "$a$-th" row is made of the elements of the set $lbrace varphi_a(g)|g in G rbrace$ and the "$a$-th" column is made of the elements of the collection $F_a=(varphi_g(a))_{g in G}$. According to this "picture", we have that:
- for 3 and 4, there's one and only one column with all the entries equal to each other (all $e$);
- for 5, the "$a$-th" column contains as many copies of $a$ as the order of $C_G(a)$, minus $1$;
- for 6 and 7, there are $|tilde F_a|$ columns with the same, differently permuted, $|tilde F_a|$ distinct elements (included the "index" $a$); each of these columns has as many copies of each element $b$ of $tilde F_a setminus lbrace a rbrace$ as is the integer $|tilde{mathcal{F}}_b^{(a)}|-1$.
Now, $tilde{mathcal{F}}_b^{(a)}=emptyset Leftrightarrow b notin tilde F_a$, and then $b in tilde F_a Rightarrow tilde{mathcal{F}}_b^{(a)} ne emptyset Rightarrow |tilde{mathcal{F}}_b^{(a)}|=|C_G(a)|$ (for the last implication, see here).
By summing on all the columns, we have that a partition $lambda(n-1)$ of the integer $n-1$ exists such that:
begin{align}
n^2& = n+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}sum_{b in tilde F_a setminus lbrace a rbrace}(|tilde{mathcal{F}}_b^{(a)}|-1)=\&=n+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}|tilde F_a setminus lbrace a rbrace|(|C_G(a)|-1)=\&=n+sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2+sum_{a in G setminus lbrace e rbrace}(|tilde F_a setminus lbrace a rbrace|+1)(|C_G(a)|-1)
end{align}
But, $$sum_{a in G setminus lbrace e rbrace}(|tilde F_a setminus lbrace a rbrace|+1)=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
so, if we define $delta_a:=|tilde F_a setminus lbrace a rbrace|+1$, we get:
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}delta_a(|C_G(a)|-1)
end{equation}
with $$sum_{a in G setminus lbrace e rbrace}delta_a=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
and then, for a given labelling of the elements of $G$,
begin{equation}
n(n-1)=sum_{j=1}^{n-1}delta(Lambda)_j|C_G(a_j)|
end{equation}
where $delta(Lambda)$ is a $(n-1)$-term partition of the integer $Lambda:=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$.
Addendum
As shown, each $a in G$, $a ne e$, gives rise to the collection $F_a$ made of:
$|tilde F_a|$ distinct elements of $G$, included $a$;
$|C_G(a)|-1$ copies of $a$;
$|mathcal{F}_b^{(a)}|-1=|C_G(a)|-1$ copies of $b$, for each $b in tilde F_a setminus lbrace a rbrace$.
Then,
begin{align}
n=&|tilde F_a|+|C_G(a)|-1+(|C_G(a)|-1)(|tilde F_a|-1)=\=&|tilde F_a|+|C_G(a)|-1+|C_G(a)||tilde F_a|-|C_G(a)|-|tilde F_a|+1=\=&|C_G(a)||tilde F_a|
end{align}
so $|tilde F_a|=[G:C_G(a)]$, and finally $$sum_{a in G setminus lbrace e rbrace}[G:C_G(a)]=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
For a finite group of order $n$ and trivial center, a partition $lambda(n-1)$ of $n-1$ with no "1" summands exists, such that the sum of the indexes of the centralizers of all the elements of the group (but the unit) is equal to the sum of the squares of the summands of $lambda(n-1)$.
combinatorics group-theory number-theory
asked Jan 13 at 22:21
LucaLuca
21419
$endgroup$
After failing this attempt, I've revised my proof sketch and I've come to the following version of the equation in the title.
So, let $G$ be a finite group, say $G=lbrace e,a_1,dots,a_{n-1} rbrace$, and $C_G(a_i)$ denote the centralizer of $a_i$ in $G$. Then, in the case $Z(G)=lbrace e rbrace$, the following equation seems to hold:
begin{equation}
n(n-1)=sum_{j=1}^{n-1}delta(Lambda)_j|C_G(a_j)|
end{equation}
where $delta(Lambda)$ is a $(n-1)$-term partition of the integer $Lambda:=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$, being $lambda(n-1)$ a partition of the integer $n-1$.
E.g., for $G=S_3$, if we take $lambda(5)=(2,3)$ and $delta(13)=(2,2,3,3,3)$, we get:
begin{equation}
6cdot5 = 2^2+3^2+2cdot(3-1)+2cdot(3-1)+3cdot(2-1)+3cdot(2-1)+3cdot(2-1)
end{equation}
As I'm not very skilled on symmetric groups, can someone check whether above formula works also for $S_4$, etc. or some other group with trivial center?
Proof.
Let's define the collections $F_a:=(varphi_g(a))_{gin G}$, where $varphi_g(a):=g^{-1}ag$, and denote by $tilde F_a$ the set composed of the distinct elements of $F_a$. We have:
$forall a in G, a in tilde F_a$; then, $tilde F_a ne emptyset$;
$forall a in G, |F_a|=n$;
$tilde F_e=lbrace e rbrace$;
$Z(G)=lbrace e rbrace Rightarrow |tilde F_a|>1, forall a in G, a ne e$;
$F_a=(varphi_{g}(a))_{g in C_G(a)} cup (varphi_{g}(a))_{g in ∁_G(C_G(a))}=(a,a,...,a) cup (varphi_{g}(a))_{g in ∁_G(C_G(a))}$, with $|(a,a,...,a)|=|C_G(a)|$;- given $a,b in G, b ne a$, it is $tilde F_b=tilde F_a Leftrightarrow b in tilde F_a$;
- the collection $(varphi_{g}(a))_{g in ∁_G(C_G(a))}$ contains as many copies of each element $b in tilde F_a setminus lbrace a rbrace$ as is the cardinality of the set $tilde{mathcal{F}}_b^{(a)}:=lbrace g in complement_G(C_G(a))|varphi_g(a)=b rbrace$, minus $1$.
We may also think of a $n times n$ "matrix" such that, given $a in G$, the "$a$-th" row is made of the elements of the set $lbrace varphi_a(g)|g in G rbrace$ and the "$a$-th" column is made of the elements of the collection $F_a=(varphi_g(a))_{g in G}$. According to this "picture", we have that:
- for 3 and 4, there's one and only one column with all the entries equal to each other (all $e$);
- for 5, the "$a$-th" column contains as many copies of $a$ as the order of $C_G(a)$, minus $1$;
- for 6 and 7, there are $|tilde F_a|$ columns with the same, differently permuted, $|tilde F_a|$ distinct elements (included the "index" $a$); each of these columns has as many copies of each element $b$ of $tilde F_a setminus lbrace a rbrace$ as is the integer $|tilde{mathcal{F}}_b^{(a)}|-1$.
Now, $tilde{mathcal{F}}_b^{(a)}=emptyset Leftrightarrow b notin tilde F_a$, and then $b in tilde F_a Rightarrow tilde{mathcal{F}}_b^{(a)} ne emptyset Rightarrow |tilde{mathcal{F}}_b^{(a)}|=|C_G(a)|$ (for the last implication, see here).
By summing on all the columns, we have that a partition $lambda(n-1)$ of the integer $n-1$ exists such that:
begin{align}
n^2& = n+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}sum_{b in tilde F_a setminus lbrace a rbrace}(|tilde{mathcal{F}}_b^{(a)}|-1)=\&=n+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}|tilde F_a setminus lbrace a rbrace|(|C_G(a)|-1)=\&=n+sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2+sum_{a in G setminus lbrace e rbrace}(|tilde F_a setminus lbrace a rbrace|+1)(|C_G(a)|-1)
end{align}
But, $$sum_{a in G setminus lbrace e rbrace}(|tilde F_a setminus lbrace a rbrace|+1)=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
so, if we define $delta_a:=|tilde F_a setminus lbrace a rbrace|+1$, we get:
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}delta_a(|C_G(a)|-1)
end{equation}
with $$sum_{a in G setminus lbrace e rbrace}delta_a=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
and then, for a given labelling of the elements of $G$,
begin{equation}
n(n-1)=sum_{j=1}^{n-1}delta(Lambda)_j|C_G(a_j)|
end{equation}
where $delta(Lambda)$ is a $(n-1)$-term partition of the integer $Lambda:=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$.
Addendum
As shown, each $a in G$, $a ne e$, gives rise to the collection $F_a$ made of:
$|tilde F_a|$ distinct elements of $G$, included $a$;
$|C_G(a)|-1$ copies of $a$;
$|mathcal{F}_b^{(a)}|-1=|C_G(a)|-1$ copies of $b$, for each $b in tilde F_a setminus lbrace a rbrace$.
Then,
begin{align}
n=&|tilde F_a|+|C_G(a)|-1+(|C_G(a)|-1)(|tilde F_a|-1)=\=&|tilde F_a|+|C_G(a)|-1+|C_G(a)||tilde F_a|-|C_G(a)|-|tilde F_a|+1=\=&|C_G(a)||tilde F_a|
end{align}
so $|tilde F_a|=[G:C_G(a)]$, and finally $$sum_{a in G setminus lbrace e rbrace}[G:C_G(a)]=sum_{substack{lambda_i in lambda(n-1)\ lambda_i>1}}lambda_i^2$$
For a finite group of order $n$ and trivial center, a partition $lambda(n-1)$ of $n-1$ with no "1" summands exists, such that the sum of the indexes of the centralizers of all the elements of the group (but the unit) is equal to the sum of the squares of the summands of $lambda(n-1)$.
combinatorics group-theory number-theory
combinatorics group-theory number-theory
asked Jan 13 at 22:21
LucaLuca
21419
asked Jan 13 at 22:21
LucaLuca
21419
edited Jan 17 at 21:17
Luca
asked Jan 13 at 22:21
LucaLuca
21419
asked Jan 13 at 22:21
LucaLuca
21419
asked Jan 13 at 22:21
LucaLuca
21419
21419
$begingroup$
I don't see how something like this could possibly work, as the sum on the right depends on the ordering of the elements of $G$. For example, even for $S_3$ and $lambda(5)=(2,3)$, this fails if we order the elements of $S_3$ differently. (For example, if $a_1$ is an involution rather than an element of order $3$.)
$endgroup$
– verret
Jan 14 at 3:06
$begingroup$
Wouldn't in that case the equation hold with, e.g., the partition $delta(13)=(3,2,3,3,2)$ ?
$endgroup$
– Luca
Jan 14 at 8:07
$begingroup$
Do you have any clue how to find a partition that will work? If not, then it seems hard computationally to check whether it holds for a given group (and it could also hold for trivial reasons, simply because there are so many partitions). Moreover, if not, then it seems hard to use the result, since we wouldn't know what the partition is.
$endgroup$
– verret
Jan 14 at 9:02
$begingroup$
If I'm not mistaken, my proof simply states that such partitions $lambda$ and $delta$ do exist, no more than that. If it were correct, I'm more inclined to think that, as you argued, it is simply for trivial reasons. So, I think I'd better to post the sketch of the proof for your review.
$endgroup$
– Luca
Jan 14 at 9:11
$begingroup$
Edited with my proof attempt.
$endgroup$
– Luca
Jan 14 at 12:08
add a comment |
$begingroup$
I don't see how something like this could possibly work, as the sum on the right depends on the ordering of the elements of $G$. For example, even for $S_3$ and $lambda(5)=(2,3)$, this fails if we order the elements of $S_3$ differently. (For example, if $a_1$ is an involution rather than an element of order $3$.)
$endgroup$
– verret
Jan 14 at 3:06
$begingroup$
Wouldn't in that case the equation hold with, e.g., the partition $delta(13)=(3,2,3,3,2)$ ?
$endgroup$
– Luca
Jan 14 at 8:07
$begingroup$
Do you have any clue how to find a partition that will work? If not, then it seems hard computationally to check whether it holds for a given group (and it could also hold for trivial reasons, simply because there are so many partitions). Moreover, if not, then it seems hard to use the result, since we wouldn't know what the partition is.
$endgroup$
– verret
Jan 14 at 9:02
$begingroup$
If I'm not mistaken, my proof simply states that such partitions $lambda$ and $delta$ do exist, no more than that. If it were correct, I'm more inclined to think that, as you argued, it is simply for trivial reasons. So, I think I'd better to post the sketch of the proof for your review.
$endgroup$
– Luca
Jan 14 at 9:11
$begingroup$
Edited with my proof attempt.
$endgroup$
– Luca
Jan 14 at 12:08
$begingroup$
I don't see how something like this could possibly work, as the sum on the right depends on the ordering of the elements of $G$. For example, even for $S_3$ and $lambda(5)=(2,3)$, this fails if we order the elements of $S_3$ differently. (For example, if $a_1$ is an involution rather than an element of order $3$.)
$endgroup$
– verret
Jan 14 at 3:06
$begingroup$
I don't see how something like this could possibly work, as the sum on the right depends on the ordering of the elements of $G$. For example, even for $S_3$ and $lambda(5)=(2,3)$, this fails if we order the elements of $S_3$ differently. (For example, if $a_1$ is an involution rather than an element of order $3$.)
$endgroup$
– verret
Jan 14 at 3:06
$begingroup$
Wouldn't in that case the equation hold with, e.g., the partition $delta(13)=(3,2,3,3,2)$ ?
$endgroup$
– Luca
Jan 14 at 8:07
$begingroup$
Wouldn't in that case the equation hold with, e.g., the partition $delta(13)=(3,2,3,3,2)$ ?
$endgroup$
– Luca
Jan 14 at 8:07
$begingroup$
Do you have any clue how to find a partition that will work? If not, then it seems hard computationally to check whether it holds for a given group (and it could also hold for trivial reasons, simply because there are so many partitions). Moreover, if not, then it seems hard to use the result, since we wouldn't know what the partition is.
$endgroup$
– verret
Jan 14 at 9:02
$begingroup$
Do you have any clue how to find a partition that will work? If not, then it seems hard computationally to check whether it holds for a given group (and it could also hold for trivial reasons, simply because there are so many partitions). Moreover, if not, then it seems hard to use the result, since we wouldn't know what the partition is.
$endgroup$
– verret
Jan 14 at 9:02
$begingroup$
If I'm not mistaken, my proof simply states that such partitions $lambda$ and $delta$ do exist, no more than that. If it were correct, I'm more inclined to think that, as you argued, it is simply for trivial reasons. So, I think I'd better to post the sketch of the proof for your review.
$endgroup$
– Luca
Jan 14 at 9:11
$begingroup$
If I'm not mistaken, my proof simply states that such partitions $lambda$ and $delta$ do exist, no more than that. If it were correct, I'm more inclined to think that, as you argued, it is simply for trivial reasons. So, I think I'd better to post the sketch of the proof for your review.
$endgroup$
– Luca
Jan 14 at 9:11
$begingroup$
Edited with my proof attempt.
$endgroup$
– Luca
Jan 14 at 12:08
$begingroup$
Edited with my proof attempt.
$endgroup$
– Luca
Jan 14 at 12:08
add a comment |
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$begingroup$
I don't see how something like this could possibly work, as the sum on the right depends on the ordering of the elements of $G$. For example, even for $S_3$ and $lambda(5)=(2,3)$, this fails if we order the elements of $S_3$ differently. (For example, if $a_1$ is an involution rather than an element of order $3$.)
$endgroup$
– verret
Jan 14 at 3:06
$begingroup$
Wouldn't in that case the equation hold with, e.g., the partition $delta(13)=(3,2,3,3,2)$ ?
$endgroup$
– Luca
Jan 14 at 8:07
$begingroup$
Do you have any clue how to find a partition that will work? If not, then it seems hard computationally to check whether it holds for a given group (and it could also hold for trivial reasons, simply because there are so many partitions). Moreover, if not, then it seems hard to use the result, since we wouldn't know what the partition is.
$endgroup$
– verret
Jan 14 at 9:02
$begingroup$
If I'm not mistaken, my proof simply states that such partitions $lambda$ and $delta$ do exist, no more than that. If it were correct, I'm more inclined to think that, as you argued, it is simply for trivial reasons. So, I think I'd better to post the sketch of the proof for your review.
$endgroup$
– Luca
Jan 14 at 9:11
$begingroup$
Edited with my proof attempt.
$endgroup$
– Luca
Jan 14 at 12:08