Mixing
Does this property characterize abelian groups?
$begingroup$
Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?
This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.
If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.
abstract-algebra group-theory permutations
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,56822466
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?
This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.
If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.
abstract-algebra group-theory permutations
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,56822466
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?
This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.
If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.
abstract-algebra group-theory permutations
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,56822466
$endgroup$
Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?
This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.
If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,56822466
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,56822466
edited Jan 2 at 23:29
MathematicsStudent1122
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,56822466
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,56822466
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,56822466
8,56822466
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,323828
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$begingroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,323828
$endgroup$
add a comment |
$begingroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,323828
$endgroup$
add a comment |
$begingroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,323828
$endgroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,323828
answered Jan 2 at 23:32
YCorYCor
7,323828
answered Jan 2 at 23:32
YCorYCor
7,323828
answered Jan 2 at 23:32
YCorYCor
7,323828
7,323828
add a comment |
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