Mixing
A combinatorial property of uncountable groups
Clash Royale CLAN TAG#URR8PPP
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Let $A,B$ be two uncountable sets in a group $G$ such that for any elements $x,yin G$ the intersection $xAcap yB$ is finite. Let $Phi:Gto 2^G$ be a function assigning to each element $xin G$ some finite set $Phi(x)subset G$.
Question. Is it true that there exist elements $x,yin G$ and $ain AsetminusPhi(x)$ and $bin BsetminusPhi(y)$ such that $xa=yb$?
Comment. The answer is affirmative if $ab=ba$ for any $ain A$ and $bin B$. In this case we can choose two elements $ain AsetminusPhi(b)$ and $bin Bsetminus Phi(a)$ and put $x=b$ and $y=a$.
gr.group-theory set-theory additive-combinatorics infinite-combinatorics ramsey-theory
asked 5 hours ago
Taras Banakh
14.2k12882
add a comment |Â
up vote
7
down vote
favorite
Let $A,B$ be two uncountable sets in a group $G$ such that for any elements $x,yin G$ the intersection $xAcap yB$ is finite. Let $Phi:Gto 2^G$ be a function assigning to each element $xin G$ some finite set $Phi(x)subset G$.
Question. Is it true that there exist elements $x,yin G$ and $ain AsetminusPhi(x)$ and $bin BsetminusPhi(y)$ such that $xa=yb$?
Comment. The answer is affirmative if $ab=ba$ for any $ain A$ and $bin B$. In this case we can choose two elements $ain AsetminusPhi(b)$ and $bin Bsetminus Phi(a)$ and put $x=b$ and $y=a$.
gr.group-theory set-theory additive-combinatorics infinite-combinatorics ramsey-theory
asked 5 hours ago
Taras Banakh
14.2k12882
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Let $A,B$ be two uncountable sets in a group $G$ such that for any elements $x,yin G$ the intersection $xAcap yB$ is finite. Let $Phi:Gto 2^G$ be a function assigning to each element $xin G$ some finite set $Phi(x)subset G$.
Question. Is it true that there exist elements $x,yin G$ and $ain AsetminusPhi(x)$ and $bin BsetminusPhi(y)$ such that $xa=yb$?
Comment. The answer is affirmative if $ab=ba$ for any $ain A$ and $bin B$. In this case we can choose two elements $ain AsetminusPhi(b)$ and $bin Bsetminus Phi(a)$ and put $x=b$ and $y=a$.
gr.group-theory set-theory additive-combinatorics infinite-combinatorics ramsey-theory
asked 5 hours ago
Taras Banakh
14.2k12882
Let $A,B$ be two uncountable sets in a group $G$ such that for any elements $x,yin G$ the intersection $xAcap yB$ is finite. Let $Phi:Gto 2^G$ be a function assigning to each element $xin G$ some finite set $Phi(x)subset G$.
Question. Is it true that there exist elements $x,yin G$ and $ain AsetminusPhi(x)$ and $bin BsetminusPhi(y)$ such that $xa=yb$?
Comment. The answer is affirmative if $ab=ba$ for any $ain A$ and $bin B$. In this case we can choose two elements $ain AsetminusPhi(b)$ and $bin Bsetminus Phi(a)$ and put $x=b$ and $y=a$.
gr.group-theory set-theory additive-combinatorics infinite-combinatorics ramsey-theory
gr.group-theory set-theory additive-combinatorics infinite-combinatorics ramsey-theory
asked 5 hours ago
Taras Banakh
14.2k12882
asked 5 hours ago
Taras Banakh
14.2k12882
edited 3 hours ago
asked 5 hours ago
Taras Banakh
14.2k12882
asked 5 hours ago
Taras Banakh
14.2k12882
asked 5 hours ago
Taras Banakh
14.2k12882
14.2k12882
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
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up vote
2
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Unfortunately (for my further plans) this question has negative answer. Just take any two disjoint uncountable sets $A,B$ and consider the free group $G$ over the union $Acup B$. Let $Phi:Gto 2^G$ be the function assigning to each $gin G$ the set of letters in the irreducible representation of $g$.
Now assume that there exist elements $x,yin G$ and $ain GsetminusPhi(x)$ and $bin Gsetminus Phi(y)$ such that $xa=yb$. Since $anotinPhi(x)$ the letter $a$ does not appear in the irreducible representation of $x$ and hence $Phi(y)=Phi(xab^-1)=Phi(x)cupa,b$, which contradicts the choice of $bnotin Phi(y)$.
By the way, my purpose was to resolve Question 2.2 from this survey of Protasov. This question asks if the countability of a group $G$ is equivalent to the normality of the finitary ballean on $G$. It is known that a commutative or free group is countable if and only if its finitary ballean is normal.
answered 1 hour ago
Taras Banakh
14.2k12882
1
Just to mention, in this case, $xAcap yB$ is at most a singleton for every pair $(x,y)$, namely if $x^-1y=a_0b_0^-1$ for some $(a_0,b_0)in Atimes B$, this singleton is equal to $xa_0(=yb_0)$; otherwise it's empty. In general, this argument holds as soon as $langle Aranglecap langle Brangle=1$.
– YCor
1 hour ago
Do you have a restatement of Protasov's question using the language of infinite combinatorics?
– YCor
49 mins ago
@YCor Yes, and I am thinking of posing this as a separate problem.
– Taras Banakh
42 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Unfortunately (for my further plans) this question has negative answer. Just take any two disjoint uncountable sets $A,B$ and consider the free group $G$ over the union $Acup B$. Let $Phi:Gto 2^G$ be the function assigning to each $gin G$ the set of letters in the irreducible representation of $g$.
Now assume that there exist elements $x,yin G$ and $ain GsetminusPhi(x)$ and $bin Gsetminus Phi(y)$ such that $xa=yb$. Since $anotinPhi(x)$ the letter $a$ does not appear in the irreducible representation of $x$ and hence $Phi(y)=Phi(xab^-1)=Phi(x)cupa,b$, which contradicts the choice of $bnotin Phi(y)$.
By the way, my purpose was to resolve Question 2.2 from this survey of Protasov. This question asks if the countability of a group $G$ is equivalent to the normality of the finitary ballean on $G$. It is known that a commutative or free group is countable if and only if its finitary ballean is normal.
answered 1 hour ago
Taras Banakh
14.2k12882
1
Just to mention, in this case, $xAcap yB$ is at most a singleton for every pair $(x,y)$, namely if $x^-1y=a_0b_0^-1$ for some $(a_0,b_0)in Atimes B$, this singleton is equal to $xa_0(=yb_0)$; otherwise it's empty. In general, this argument holds as soon as $langle Aranglecap langle Brangle=1$.
– YCor
1 hour ago
Do you have a restatement of Protasov's question using the language of infinite combinatorics?
– YCor
49 mins ago
@YCor Yes, and I am thinking of posing this as a separate problem.
– Taras Banakh
42 mins ago
add a comment |Â
up vote
2
down vote
Unfortunately (for my further plans) this question has negative answer. Just take any two disjoint uncountable sets $A,B$ and consider the free group $G$ over the union $Acup B$. Let $Phi:Gto 2^G$ be the function assigning to each $gin G$ the set of letters in the irreducible representation of $g$.
Now assume that there exist elements $x,yin G$ and $ain GsetminusPhi(x)$ and $bin Gsetminus Phi(y)$ such that $xa=yb$. Since $anotinPhi(x)$ the letter $a$ does not appear in the irreducible representation of $x$ and hence $Phi(y)=Phi(xab^-1)=Phi(x)cupa,b$, which contradicts the choice of $bnotin Phi(y)$.
By the way, my purpose was to resolve Question 2.2 from this survey of Protasov. This question asks if the countability of a group $G$ is equivalent to the normality of the finitary ballean on $G$. It is known that a commutative or free group is countable if and only if its finitary ballean is normal.
answered 1 hour ago
Taras Banakh
14.2k12882
1
Just to mention, in this case, $xAcap yB$ is at most a singleton for every pair $(x,y)$, namely if $x^-1y=a_0b_0^-1$ for some $(a_0,b_0)in Atimes B$, this singleton is equal to $xa_0(=yb_0)$; otherwise it's empty. In general, this argument holds as soon as $langle Aranglecap langle Brangle=1$.
– YCor
1 hour ago
Do you have a restatement of Protasov's question using the language of infinite combinatorics?
– YCor
49 mins ago
@YCor Yes, and I am thinking of posing this as a separate problem.
– Taras Banakh
42 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Unfortunately (for my further plans) this question has negative answer. Just take any two disjoint uncountable sets $A,B$ and consider the free group $G$ over the union $Acup B$. Let $Phi:Gto 2^G$ be the function assigning to each $gin G$ the set of letters in the irreducible representation of $g$.
Now assume that there exist elements $x,yin G$ and $ain GsetminusPhi(x)$ and $bin Gsetminus Phi(y)$ such that $xa=yb$. Since $anotinPhi(x)$ the letter $a$ does not appear in the irreducible representation of $x$ and hence $Phi(y)=Phi(xab^-1)=Phi(x)cupa,b$, which contradicts the choice of $bnotin Phi(y)$.
By the way, my purpose was to resolve Question 2.2 from this survey of Protasov. This question asks if the countability of a group $G$ is equivalent to the normality of the finitary ballean on $G$. It is known that a commutative or free group is countable if and only if its finitary ballean is normal.
answered 1 hour ago
Taras Banakh
14.2k12882
Unfortunately (for my further plans) this question has negative answer. Just take any two disjoint uncountable sets $A,B$ and consider the free group $G$ over the union $Acup B$. Let $Phi:Gto 2^G$ be the function assigning to each $gin G$ the set of letters in the irreducible representation of $g$.
Now assume that there exist elements $x,yin G$ and $ain GsetminusPhi(x)$ and $bin Gsetminus Phi(y)$ such that $xa=yb$. Since $anotinPhi(x)$ the letter $a$ does not appear in the irreducible representation of $x$ and hence $Phi(y)=Phi(xab^-1)=Phi(x)cupa,b$, which contradicts the choice of $bnotin Phi(y)$.
By the way, my purpose was to resolve Question 2.2 from this survey of Protasov. This question asks if the countability of a group $G$ is equivalent to the normality of the finitary ballean on $G$. It is known that a commutative or free group is countable if and only if its finitary ballean is normal.
answered 1 hour ago
Taras Banakh
14.2k12882
edited 1 hour ago
answered 1 hour ago
Taras Banakh
14.2k12882
answered 1 hour ago
Taras Banakh
14.2k12882
answered 1 hour ago
Taras Banakh
14.2k12882
14.2k12882
1
Just to mention, in this case, $xAcap yB$ is at most a singleton for every pair $(x,y)$, namely if $x^-1y=a_0b_0^-1$ for some $(a_0,b_0)in Atimes B$, this singleton is equal to $xa_0(=yb_0)$; otherwise it's empty. In general, this argument holds as soon as $langle Aranglecap langle Brangle=1$.
– YCor
1 hour ago
Do you have a restatement of Protasov's question using the language of infinite combinatorics?
– YCor
49 mins ago
@YCor Yes, and I am thinking of posing this as a separate problem.
– Taras Banakh
42 mins ago
add a comment |Â
1
Just to mention, in this case, $xAcap yB$ is at most a singleton for every pair $(x,y)$, namely if $x^-1y=a_0b_0^-1$ for some $(a_0,b_0)in Atimes B$, this singleton is equal to $xa_0(=yb_0)$; otherwise it's empty. In general, this argument holds as soon as $langle Aranglecap langle Brangle=1$.
– YCor
1 hour ago
Do you have a restatement of Protasov's question using the language of infinite combinatorics?
– YCor
49 mins ago
@YCor Yes, and I am thinking of posing this as a separate problem.
– Taras Banakh
42 mins ago
1
1
Just to mention, in this case, $xAcap yB$ is at most a singleton for every pair $(x,y)$, namely if $x^-1y=a_0b_0^-1$ for some $(a_0,b_0)in Atimes B$, this singleton is equal to $xa_0(=yb_0)$; otherwise it's empty. In general, this argument holds as soon as $langle Aranglecap langle Brangle=1$.
– YCor
1 hour ago
Just to mention, in this case, $xAcap yB$ is at most a singleton for every pair $(x,y)$, namely if $x^-1y=a_0b_0^-1$ for some $(a_0,b_0)in Atimes B$, this singleton is equal to $xa_0(=yb_0)$; otherwise it's empty. In general, this argument holds as soon as $langle Aranglecap langle Brangle=1$.
– YCor
1 hour ago
Do you have a restatement of Protasov's question using the language of infinite combinatorics?
– YCor
49 mins ago
Do you have a restatement of Protasov's question using the language of infinite combinatorics?
– YCor
49 mins ago
@YCor Yes, and I am thinking of posing this as a separate problem.
– Taras Banakh
42 mins ago
@YCor Yes, and I am thinking of posing this as a separate problem.
– Taras Banakh
42 mins ago
add a comment |Â
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