Mixing
Zero divisors in complex group algebras of residually finite groups
Clash Royale CLAN TAG#URR8PPP
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Conjecture. There exists a function $f:mathbbN rightarrow mathbbN$ such that if $alpha$ and $beta$ are non-zero elements of the complex group algebra $mathbbC[G]$ of a finite group $G$ such that $1in textsupp(alpha) cap textsupp(beta)$, $|textsupp(alpha)|leq |textsupp(beta)|$ and $alpha cdot beta =0$, then $textexpbig (langle textsupp(alpha) rangle big) leq f(|textsupp(beta)|)$, where $textexp(H)$ denotes the exponent of a finite group $H$ and $textsupp(gamma)$ for an element $gamma=sum_gin G gamma_g ; g in mathbbC[G]$ denotes the set $ gin G ;$.
Motivation. If the conjecture is true, then the support of any zero divisor of the complex group algebra of any residually finite group generates a finite subgroup.
Proof. Let $alpha$ be a non-zero element of $mathbbC[G]$ for some residually finite group $G$ such that $alpha cdot beta =0$ for
some non-zero $beta in mathbbC[G]$. Let $H=langle textsupp(alpha) rangle$. By Zelmanov's celebrated result on restricted Burnside problem, it is enough to show that the exponent of $H$ is finite. Since $H$ is finitely generated residually finite, there exists a descending series $H=N_1geq N_2 geq cdots$ of normal subgroups $H$ of finite index such that $cap_iinmathbbN N_i=1$. Note that there exists $kinmathbbN$ such that $baralpha cdot barbeta=0$ in $mathbbC[H/N_i]$ for all $igeq k$ and $|textsupp(barbeta)|=|textsupp(beta)|=:t$, where $bar$ is the natural ring epimorphism from $mathbbC[H]$ onto $mathbbC[H/N_i]$. Now if the above conjecture is true then $textexp(H/N_i)leq f(t)$ and so $textexp(H)$ is finite. This completes the proof.
If the above proof and conjecture are true then the complex group algebras of torsion-free residually finite groups have no zero divisor.
gr.group-theory
asked 1 hour ago
Alireza Abdollahi
2,4941123
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up vote
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Conjecture. There exists a function $f:mathbbN rightarrow mathbbN$ such that if $alpha$ and $beta$ are non-zero elements of the complex group algebra $mathbbC[G]$ of a finite group $G$ such that $1in textsupp(alpha) cap textsupp(beta)$, $|textsupp(alpha)|leq |textsupp(beta)|$ and $alpha cdot beta =0$, then $textexpbig (langle textsupp(alpha) rangle big) leq f(|textsupp(beta)|)$, where $textexp(H)$ denotes the exponent of a finite group $H$ and $textsupp(gamma)$ for an element $gamma=sum_gin G gamma_g ; g in mathbbC[G]$ denotes the set $ gin G ;$.
Motivation. If the conjecture is true, then the support of any zero divisor of the complex group algebra of any residually finite group generates a finite subgroup.
Proof. Let $alpha$ be a non-zero element of $mathbbC[G]$ for some residually finite group $G$ such that $alpha cdot beta =0$ for
some non-zero $beta in mathbbC[G]$. Let $H=langle textsupp(alpha) rangle$. By Zelmanov's celebrated result on restricted Burnside problem, it is enough to show that the exponent of $H$ is finite. Since $H$ is finitely generated residually finite, there exists a descending series $H=N_1geq N_2 geq cdots$ of normal subgroups $H$ of finite index such that $cap_iinmathbbN N_i=1$. Note that there exists $kinmathbbN$ such that $baralpha cdot barbeta=0$ in $mathbbC[H/N_i]$ for all $igeq k$ and $|textsupp(barbeta)|=|textsupp(beta)|=:t$, where $bar$ is the natural ring epimorphism from $mathbbC[H]$ onto $mathbbC[H/N_i]$. Now if the above conjecture is true then $textexp(H/N_i)leq f(t)$ and so $textexp(H)$ is finite. This completes the proof.
If the above proof and conjecture are true then the complex group algebras of torsion-free residually finite groups have no zero divisor.
gr.group-theory
asked 1 hour ago
Alireza Abdollahi
2,4941123
There are finitely generated, infinite groups with finite exponent. So I think the argument is not quite complete, but for applications to torsion-free groups, this should not be a problem.
– Steffen Kionke
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Conjecture. There exists a function $f:mathbbN rightarrow mathbbN$ such that if $alpha$ and $beta$ are non-zero elements of the complex group algebra $mathbbC[G]$ of a finite group $G$ such that $1in textsupp(alpha) cap textsupp(beta)$, $|textsupp(alpha)|leq |textsupp(beta)|$ and $alpha cdot beta =0$, then $textexpbig (langle textsupp(alpha) rangle big) leq f(|textsupp(beta)|)$, where $textexp(H)$ denotes the exponent of a finite group $H$ and $textsupp(gamma)$ for an element $gamma=sum_gin G gamma_g ; g in mathbbC[G]$ denotes the set $ gin G ;$.
Motivation. If the conjecture is true, then the support of any zero divisor of the complex group algebra of any residually finite group generates a finite subgroup.
Proof. Let $alpha$ be a non-zero element of $mathbbC[G]$ for some residually finite group $G$ such that $alpha cdot beta =0$ for
some non-zero $beta in mathbbC[G]$. Let $H=langle textsupp(alpha) rangle$. By Zelmanov's celebrated result on restricted Burnside problem, it is enough to show that the exponent of $H$ is finite. Since $H$ is finitely generated residually finite, there exists a descending series $H=N_1geq N_2 geq cdots$ of normal subgroups $H$ of finite index such that $cap_iinmathbbN N_i=1$. Note that there exists $kinmathbbN$ such that $baralpha cdot barbeta=0$ in $mathbbC[H/N_i]$ for all $igeq k$ and $|textsupp(barbeta)|=|textsupp(beta)|=:t$, where $bar$ is the natural ring epimorphism from $mathbbC[H]$ onto $mathbbC[H/N_i]$. Now if the above conjecture is true then $textexp(H/N_i)leq f(t)$ and so $textexp(H)$ is finite. This completes the proof.
If the above proof and conjecture are true then the complex group algebras of torsion-free residually finite groups have no zero divisor.
gr.group-theory
asked 1 hour ago
Alireza Abdollahi
2,4941123
Conjecture. There exists a function $f:mathbbN rightarrow mathbbN$ such that if $alpha$ and $beta$ are non-zero elements of the complex group algebra $mathbbC[G]$ of a finite group $G$ such that $1in textsupp(alpha) cap textsupp(beta)$, $|textsupp(alpha)|leq |textsupp(beta)|$ and $alpha cdot beta =0$, then $textexpbig (langle textsupp(alpha) rangle big) leq f(|textsupp(beta)|)$, where $textexp(H)$ denotes the exponent of a finite group $H$ and $textsupp(gamma)$ for an element $gamma=sum_gin G gamma_g ; g in mathbbC[G]$ denotes the set $ gin G ;$.
Motivation. If the conjecture is true, then the support of any zero divisor of the complex group algebra of any residually finite group generates a finite subgroup.
Proof. Let $alpha$ be a non-zero element of $mathbbC[G]$ for some residually finite group $G$ such that $alpha cdot beta =0$ for
some non-zero $beta in mathbbC[G]$. Let $H=langle textsupp(alpha) rangle$. By Zelmanov's celebrated result on restricted Burnside problem, it is enough to show that the exponent of $H$ is finite. Since $H$ is finitely generated residually finite, there exists a descending series $H=N_1geq N_2 geq cdots$ of normal subgroups $H$ of finite index such that $cap_iinmathbbN N_i=1$. Note that there exists $kinmathbbN$ such that $baralpha cdot barbeta=0$ in $mathbbC[H/N_i]$ for all $igeq k$ and $|textsupp(barbeta)|=|textsupp(beta)|=:t$, where $bar$ is the natural ring epimorphism from $mathbbC[H]$ onto $mathbbC[H/N_i]$. Now if the above conjecture is true then $textexp(H/N_i)leq f(t)$ and so $textexp(H)$ is finite. This completes the proof.
If the above proof and conjecture are true then the complex group algebras of torsion-free residually finite groups have no zero divisor.
gr.group-theory
gr.group-theory
asked 1 hour ago
Alireza Abdollahi
2,4941123
asked 1 hour ago
Alireza Abdollahi
2,4941123
asked 1 hour ago
Alireza Abdollahi
2,4941123
asked 1 hour ago
Alireza Abdollahi
2,4941123
asked 1 hour ago
Alireza Abdollahi
2,4941123
2,4941123
There are finitely generated, infinite groups with finite exponent. So I think the argument is not quite complete, but for applications to torsion-free groups, this should not be a problem.
– Steffen Kionke
1 hour ago
add a comment |Â
There are finitely generated, infinite groups with finite exponent. So I think the argument is not quite complete, but for applications to torsion-free groups, this should not be a problem.
– Steffen Kionke
1 hour ago
There are finitely generated, infinite groups with finite exponent. So I think the argument is not quite complete, but for applications to torsion-free groups, this should not be a problem.
– Steffen Kionke
1 hour ago
There are finitely generated, infinite groups with finite exponent. So I think the argument is not quite complete, but for applications to torsion-free groups, this should not be a problem.
– Steffen Kionke
1 hour ago
add a comment |Â
1 Answer
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up vote
3
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The conjecture does not hold.
Let $m geq 1$ be an arbitrary integer.
Let $G = langle x,y | x^m, y^5 , [x,y]rangle$ be the finite abelian group isomorphic to a product of two cyclic groups $C_m times C_5$.
Let $alpha = xy +y -x -1$ and let $beta = y^4 +y^3 +y^2 +y +1$ in $mathbbZ[G]$.
Then $alpha beta = (x+1)(y-1)beta= 0$.
The support of $beta$ has $5$ elements and the support of $alpha$ has $4$ elements. Moreover, the support of $alpha$ generates the group $G$, which has exponent $mathrmexp(G) geq m$.
answered 51 mins ago
Steffen Kionke
26617
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The conjecture does not hold.
Let $m geq 1$ be an arbitrary integer.
Let $G = langle x,y | x^m, y^5 , [x,y]rangle$ be the finite abelian group isomorphic to a product of two cyclic groups $C_m times C_5$.
Let $alpha = xy +y -x -1$ and let $beta = y^4 +y^3 +y^2 +y +1$ in $mathbbZ[G]$.
Then $alpha beta = (x+1)(y-1)beta= 0$.
The support of $beta$ has $5$ elements and the support of $alpha$ has $4$ elements. Moreover, the support of $alpha$ generates the group $G$, which has exponent $mathrmexp(G) geq m$.
answered 51 mins ago
Steffen Kionke
26617
add a comment |Â
up vote
3
down vote
The conjecture does not hold.
Let $m geq 1$ be an arbitrary integer.
Let $G = langle x,y | x^m, y^5 , [x,y]rangle$ be the finite abelian group isomorphic to a product of two cyclic groups $C_m times C_5$.
Let $alpha = xy +y -x -1$ and let $beta = y^4 +y^3 +y^2 +y +1$ in $mathbbZ[G]$.
Then $alpha beta = (x+1)(y-1)beta= 0$.
The support of $beta$ has $5$ elements and the support of $alpha$ has $4$ elements. Moreover, the support of $alpha$ generates the group $G$, which has exponent $mathrmexp(G) geq m$.
answered 51 mins ago
Steffen Kionke
26617
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The conjecture does not hold.
Let $m geq 1$ be an arbitrary integer.
Let $G = langle x,y | x^m, y^5 , [x,y]rangle$ be the finite abelian group isomorphic to a product of two cyclic groups $C_m times C_5$.
Let $alpha = xy +y -x -1$ and let $beta = y^4 +y^3 +y^2 +y +1$ in $mathbbZ[G]$.
Then $alpha beta = (x+1)(y-1)beta= 0$.
The support of $beta$ has $5$ elements and the support of $alpha$ has $4$ elements. Moreover, the support of $alpha$ generates the group $G$, which has exponent $mathrmexp(G) geq m$.
answered 51 mins ago
Steffen Kionke
26617
The conjecture does not hold.
Let $m geq 1$ be an arbitrary integer.
Let $G = langle x,y | x^m, y^5 , [x,y]rangle$ be the finite abelian group isomorphic to a product of two cyclic groups $C_m times C_5$.
Let $alpha = xy +y -x -1$ and let $beta = y^4 +y^3 +y^2 +y +1$ in $mathbbZ[G]$.
Then $alpha beta = (x+1)(y-1)beta= 0$.
The support of $beta$ has $5$ elements and the support of $alpha$ has $4$ elements. Moreover, the support of $alpha$ generates the group $G$, which has exponent $mathrmexp(G) geq m$.
answered 51 mins ago
Steffen Kionke
26617
answered 51 mins ago
Steffen Kionke
26617
answered 51 mins ago
Steffen Kionke
26617
answered 51 mins ago
Steffen Kionke
26617
26617
add a comment |Â
add a comment |Â
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There are finitely generated, infinite groups with finite exponent. So I think the argument is not quite complete, but for applications to torsion-free groups, this should not be a problem.
– Steffen Kionke
1 hour ago