Mixing
In a group we have $abc=cba$. Is it abelian?
$begingroup$
Let $G$ be a group such that for any $a,b,cne1$:
$$abc=cba$$
Is $G$ abelian?
group-theory abelian-groups
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group such that for any $a,b,cne1$:
$$abc=cba$$
Is $G$ abelian?
group-theory abelian-groups
$endgroup$
$begingroup$
Seems like a perfectly good question to me. I vote to leave open.
$endgroup$
– Brett Frankel
Feb 11 '13 at 1:11
add a comment |
$begingroup$
Let $G$ be a group such that for any $a,b,cne1$:
$$abc=cba$$
Is $G$ abelian?
group-theory abelian-groups
$endgroup$
Let $G$ be a group such that for any $a,b,cne1$:
$$abc=cba$$
Is $G$ abelian?
group-theory abelian-groups
group-theory abelian-groups
edited Feb 10 '13 at 21:31
asked Feb 9 '13 at 23:29
user59671
$begingroup$
Seems like a perfectly good question to me. I vote to leave open.
$endgroup$
– Brett Frankel
Feb 11 '13 at 1:11
add a comment |
$begingroup$
Seems like a perfectly good question to me. I vote to leave open.
$endgroup$
– Brett Frankel
Feb 11 '13 at 1:11
$begingroup$
Seems like a perfectly good question to me. I vote to leave open.
$endgroup$
– Brett Frankel
Feb 11 '13 at 1:11
$begingroup$
Seems like a perfectly good question to me. I vote to leave open.
$endgroup$
– Brett Frankel
Feb 11 '13 at 1:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The proof is easy to find by writing equations $,x = y,$ in the "relator" form $,xy' = 1,, $ for $,y' = y^{-1}.:$ Doing so below, we seek to transform the $rmcolor{#0A0}{green}$ term into the $rmcolor{#C00}{red}$ term, and the value of $rm:c:$ that produces such a transformation is clear when written this way.
$$begin{eqnarray}
abc = cba &iff&color{#0A0} a&color{#0A0} b&color{#0A0} c&color{#0A0}{ !a'b'} &color{#0A0}{!!c'} &=& ,1\
&iff& a& b& c& !!(ba)'&!!c' &=& ,1\
& Rightarrow&color{#C00} a&color{#C00} b& & &!!!!!!!color{#C00}{(ba)'} &=& ,1quadtext{for} c, =, ba\
&iff& a&b&=& b a
end{eqnarray}$$
Remark $ $ This method of matching-up equations normalized into relator form $, r = 1,$ proves quite handy when proving equations in groups presented by generators and relations. By viewing the relators as puzzle pieces, we can rotate (conjugate) them and fit them together to form a planar diagram whose boundary represents a proof of the desired equation. A basic result of combinatorial group theory implies that if an equation is a consequence of the given equations (relations), then the proof can be represented by such a "cancellation" diagram, known as a Van-Kampen diagram.
Below is a simple Van-Kampen diagram. It yields a visual "proof without words" of
$$r= x^2yxy^3 = 1, s= y^2xyx^3 = 1 Rightarrow x^7 = 1$$
The puzzle pieces are the relators $,r,s.,$ To verify the proof one simply checks that traversing the boundary of each interior region yields a relator, and that traversing the external boundary yields the sought relator, here $, x^7$ (invert the label if you traverse the edge opposite its arrow direction).
$qquad$ 
answered Feb 10 '13 at 0:25
Math GemsMath Gems
16.9k11937
$endgroup$
$begingroup$
you have to assume $abne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer.
$endgroup$
– user59671
Feb 10 '13 at 0:34
$begingroup$
Yes, that's the easy part, which most readers will have no problem with. My point was to show how to discover the hard part.
$endgroup$
– Math Gems
Feb 10 '13 at 0:37
$begingroup$
I think your diagram method is more like a street fight mathematics... unless had we a more complicated case and would it allow us to construct an algorithm out of it.
$endgroup$
– freehumorist
Jan 3 at 16:38
add a comment |
$begingroup$
Suppose $a,bin G$, $abne 1$ and $a,bne 1$. Then $1=ab(ab)^{-1}=(ab)^{-1}ba$, so $ba=ab$. Of course, if $ab=1$ then $b=a^{-1}$ and $ba=1$ as well, and finally, if one of $a,b$ is $1$, then also $ab=ba$. So the group is abelian.
answered Feb 9 '13 at 23:34
Andrés E. CaicedoAndrés E. Caicedo
65.1k8158247
$endgroup$
$begingroup$
Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:10
$begingroup$
@MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions!
$endgroup$
– user59671
Feb 10 '13 at 0:20
$begingroup$
@CutieKrait During my undergraduate studies I was math olympiad contestant a few times. Andrés' answer brought me fond memories about that, because of the style of the answer: concise and impeccable.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:59
2
$begingroup$
Congrats on getting the Populist badge off of Math Gems' answer!
$endgroup$
– Zev Chonoles
Feb 10 '13 at 22:29
add a comment |
$begingroup$
If we reformulate the question as follows:
Let $G$ be a group, and $a,b,c in G$,
Let's define two 3-cycles: $tau_1 = (abc)$ and $tau_2 = (acb)$ to identify the left$(1)$ and the right$(2)$ side of the equality given.
Then isn't it obvious that, as $tau_{1,2} in mathscr{S}_3 $ and $mathscr{S}_3$ is known not to be commutative, the only condition which would imply this equality, is $G$ being abelian?
answered Jan 3 at 16:26
freehumoristfreehumorist
173112
$endgroup$
$begingroup$
What does this have to do with commutativity of $G$?
$endgroup$
– Tobias Kildetoft
Jan 3 at 18:04
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I should have detailed: Let's introduce $sigma_1 = left ( begin{matrix} a & b & c\ a & b & c end{matrix} right ) = Id_{mathscr{S}_3}$ to represent left side of our equality; and $sigma_2 = left ( begin{matrix} a & b & c\ a & c & b end{matrix} right ) = (bc)$ to represent the right side. Can't we have then the transposition $(bc)$ is identity, thus $bc = cb$?
$endgroup$
– freehumorist
Jan 3 at 18:41
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The proof is easy to find by writing equations $,x = y,$ in the "relator" form $,xy' = 1,, $ for $,y' = y^{-1}.:$ Doing so below, we seek to transform the $rmcolor{#0A0}{green}$ term into the $rmcolor{#C00}{red}$ term, and the value of $rm:c:$ that produces such a transformation is clear when written this way.
$$begin{eqnarray}
abc = cba &iff&color{#0A0} a&color{#0A0} b&color{#0A0} c&color{#0A0}{ !a'b'} &color{#0A0}{!!c'} &=& ,1\
&iff& a& b& c& !!(ba)'&!!c' &=& ,1\
& Rightarrow&color{#C00} a&color{#C00} b& & &!!!!!!!color{#C00}{(ba)'} &=& ,1quadtext{for} c, =, ba\
&iff& a&b&=& b a
end{eqnarray}$$
Remark $ $ This method of matching-up equations normalized into relator form $, r = 1,$ proves quite handy when proving equations in groups presented by generators and relations. By viewing the relators as puzzle pieces, we can rotate (conjugate) them and fit them together to form a planar diagram whose boundary represents a proof of the desired equation. A basic result of combinatorial group theory implies that if an equation is a consequence of the given equations (relations), then the proof can be represented by such a "cancellation" diagram, known as a Van-Kampen diagram.
Below is a simple Van-Kampen diagram. It yields a visual "proof without words" of
$$r= x^2yxy^3 = 1, s= y^2xyx^3 = 1 Rightarrow x^7 = 1$$
The puzzle pieces are the relators $,r,s.,$ To verify the proof one simply checks that traversing the boundary of each interior region yields a relator, and that traversing the external boundary yields the sought relator, here $, x^7$ (invert the label if you traverse the edge opposite its arrow direction).
$qquad$
answered Feb 10 '13 at 0:25
Math GemsMath Gems
16.9k11937
$endgroup$
$begingroup$
you have to assume $abne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer.
$endgroup$
– user59671
Feb 10 '13 at 0:34
$begingroup$
Yes, that's the easy part, which most readers will have no problem with. My point was to show how to discover the hard part.
$endgroup$
– Math Gems
Feb 10 '13 at 0:37
$begingroup$
I think your diagram method is more like a street fight mathematics... unless had we a more complicated case and would it allow us to construct an algorithm out of it.
$endgroup$
– freehumorist
Jan 3 at 16:38
add a comment |
$begingroup$
The proof is easy to find by writing equations $,x = y,$ in the "relator" form $,xy' = 1,, $ for $,y' = y^{-1}.:$ Doing so below, we seek to transform the $rmcolor{#0A0}{green}$ term into the $rmcolor{#C00}{red}$ term, and the value of $rm:c:$ that produces such a transformation is clear when written this way.
$$begin{eqnarray}
abc = cba &iff&color{#0A0} a&color{#0A0} b&color{#0A0} c&color{#0A0}{ !a'b'} &color{#0A0}{!!c'} &=& ,1\
&iff& a& b& c& !!(ba)'&!!c' &=& ,1\
& Rightarrow&color{#C00} a&color{#C00} b& & &!!!!!!!color{#C00}{(ba)'} &=& ,1quadtext{for} c, =, ba\
&iff& a&b&=& b a
end{eqnarray}$$
Remark $ $ This method of matching-up equations normalized into relator form $, r = 1,$ proves quite handy when proving equations in groups presented by generators and relations. By viewing the relators as puzzle pieces, we can rotate (conjugate) them and fit them together to form a planar diagram whose boundary represents a proof of the desired equation. A basic result of combinatorial group theory implies that if an equation is a consequence of the given equations (relations), then the proof can be represented by such a "cancellation" diagram, known as a Van-Kampen diagram.
Below is a simple Van-Kampen diagram. It yields a visual "proof without words" of
$$r= x^2yxy^3 = 1, s= y^2xyx^3 = 1 Rightarrow x^7 = 1$$
The puzzle pieces are the relators $,r,s.,$ To verify the proof one simply checks that traversing the boundary of each interior region yields a relator, and that traversing the external boundary yields the sought relator, here $, x^7$ (invert the label if you traverse the edge opposite its arrow direction).
$qquad$
answered Feb 10 '13 at 0:25
Math GemsMath Gems
16.9k11937
$endgroup$
$begingroup$
you have to assume $abne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer.
$endgroup$
– user59671
Feb 10 '13 at 0:34
$begingroup$
Yes, that's the easy part, which most readers will have no problem with. My point was to show how to discover the hard part.
$endgroup$
– Math Gems
Feb 10 '13 at 0:37
$begingroup$
I think your diagram method is more like a street fight mathematics... unless had we a more complicated case and would it allow us to construct an algorithm out of it.
$endgroup$
– freehumorist
Jan 3 at 16:38
add a comment |
$begingroup$
The proof is easy to find by writing equations $,x = y,$ in the "relator" form $,xy' = 1,, $ for $,y' = y^{-1}.:$ Doing so below, we seek to transform the $rmcolor{#0A0}{green}$ term into the $rmcolor{#C00}{red}$ term, and the value of $rm:c:$ that produces such a transformation is clear when written this way.
$$begin{eqnarray}
abc = cba &iff&color{#0A0} a&color{#0A0} b&color{#0A0} c&color{#0A0}{ !a'b'} &color{#0A0}{!!c'} &=& ,1\
&iff& a& b& c& !!(ba)'&!!c' &=& ,1\
& Rightarrow&color{#C00} a&color{#C00} b& & &!!!!!!!color{#C00}{(ba)'} &=& ,1quadtext{for} c, =, ba\
&iff& a&b&=& b a
end{eqnarray}$$
Remark $ $ This method of matching-up equations normalized into relator form $, r = 1,$ proves quite handy when proving equations in groups presented by generators and relations. By viewing the relators as puzzle pieces, we can rotate (conjugate) them and fit them together to form a planar diagram whose boundary represents a proof of the desired equation. A basic result of combinatorial group theory implies that if an equation is a consequence of the given equations (relations), then the proof can be represented by such a "cancellation" diagram, known as a Van-Kampen diagram.
Below is a simple Van-Kampen diagram. It yields a visual "proof without words" of
$$r= x^2yxy^3 = 1, s= y^2xyx^3 = 1 Rightarrow x^7 = 1$$
The puzzle pieces are the relators $,r,s.,$ To verify the proof one simply checks that traversing the boundary of each interior region yields a relator, and that traversing the external boundary yields the sought relator, here $, x^7$ (invert the label if you traverse the edge opposite its arrow direction).
$qquad$
answered Feb 10 '13 at 0:25
Math GemsMath Gems
16.9k11937
$endgroup$
The proof is easy to find by writing equations $,x = y,$ in the "relator" form $,xy' = 1,, $ for $,y' = y^{-1}.:$ Doing so below, we seek to transform the $rmcolor{#0A0}{green}$ term into the $rmcolor{#C00}{red}$ term, and the value of $rm:c:$ that produces such a transformation is clear when written this way.
$$begin{eqnarray}
abc = cba &iff&color{#0A0} a&color{#0A0} b&color{#0A0} c&color{#0A0}{ !a'b'} &color{#0A0}{!!c'} &=& ,1\
&iff& a& b& c& !!(ba)'&!!c' &=& ,1\
& Rightarrow&color{#C00} a&color{#C00} b& & &!!!!!!!color{#C00}{(ba)'} &=& ,1quadtext{for} c, =, ba\
&iff& a&b&=& b a
end{eqnarray}$$
Remark $ $ This method of matching-up equations normalized into relator form $, r = 1,$ proves quite handy when proving equations in groups presented by generators and relations. By viewing the relators as puzzle pieces, we can rotate (conjugate) them and fit them together to form a planar diagram whose boundary represents a proof of the desired equation. A basic result of combinatorial group theory implies that if an equation is a consequence of the given equations (relations), then the proof can be represented by such a "cancellation" diagram, known as a Van-Kampen diagram.
Below is a simple Van-Kampen diagram. It yields a visual "proof without words" of
$$r= x^2yxy^3 = 1, s= y^2xyx^3 = 1 Rightarrow x^7 = 1$$
The puzzle pieces are the relators $,r,s.,$ To verify the proof one simply checks that traversing the boundary of each interior region yields a relator, and that traversing the external boundary yields the sought relator, here $, x^7$ (invert the label if you traverse the edge opposite its arrow direction).
$qquad$
answered Feb 10 '13 at 0:25
Math GemsMath Gems
16.9k11937
edited Feb 10 '13 at 2:39
answered Feb 10 '13 at 0:25
Math GemsMath Gems
16.9k11937
answered Feb 10 '13 at 0:25
Math GemsMath Gems
16.9k11937
answered Feb 10 '13 at 0:25
Math GemsMath Gems
16.9k11937
16.9k11937
$begingroup$
you have to assume $abne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer.
$endgroup$
– user59671
Feb 10 '13 at 0:34
$begingroup$
Yes, that's the easy part, which most readers will have no problem with. My point was to show how to discover the hard part.
$endgroup$
– Math Gems
Feb 10 '13 at 0:37
$begingroup$
I think your diagram method is more like a street fight mathematics... unless had we a more complicated case and would it allow us to construct an algorithm out of it.
$endgroup$
– freehumorist
Jan 3 at 16:38
add a comment |
$begingroup$
you have to assume $abne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer.
$endgroup$
– user59671
Feb 10 '13 at 0:34
$begingroup$
Yes, that's the easy part, which most readers will have no problem with. My point was to show how to discover the hard part.
$endgroup$
– Math Gems
Feb 10 '13 at 0:37
$begingroup$
I think your diagram method is more like a street fight mathematics... unless had we a more complicated case and would it allow us to construct an algorithm out of it.
$endgroup$
– freehumorist
Jan 3 at 16:38
$begingroup$
you have to assume $abne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer.
$endgroup$
– user59671
Feb 10 '13 at 0:34
$begingroup$
you have to assume $abne 1$ (if $ab=1$ clearly $ba=ab$). It's a more constructive answer.
$endgroup$
– user59671
Feb 10 '13 at 0:34
$begingroup$
Yes, that's the easy part, which most readers will have no problem with. My point was to show how to discover the hard part.
$endgroup$
– Math Gems
Feb 10 '13 at 0:37
$begingroup$
Yes, that's the easy part, which most readers will have no problem with. My point was to show how to discover the hard part.
$endgroup$
– Math Gems
Feb 10 '13 at 0:37
$begingroup$
I think your diagram method is more like a street fight mathematics... unless had we a more complicated case and would it allow us to construct an algorithm out of it.
$endgroup$
– freehumorist
Jan 3 at 16:38
$begingroup$
I think your diagram method is more like a street fight mathematics... unless had we a more complicated case and would it allow us to construct an algorithm out of it.
$endgroup$
– freehumorist
Jan 3 at 16:38
add a comment |
$begingroup$
Suppose $a,bin G$, $abne 1$ and $a,bne 1$. Then $1=ab(ab)^{-1}=(ab)^{-1}ba$, so $ba=ab$. Of course, if $ab=1$ then $b=a^{-1}$ and $ba=1$ as well, and finally, if one of $a,b$ is $1$, then also $ab=ba$. So the group is abelian.
answered Feb 9 '13 at 23:34
Andrés E. CaicedoAndrés E. Caicedo
65.1k8158247
$endgroup$
$begingroup$
Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:10
$begingroup$
@MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions!
$endgroup$
– user59671
Feb 10 '13 at 0:20
$begingroup$
@CutieKrait During my undergraduate studies I was math olympiad contestant a few times. Andrés' answer brought me fond memories about that, because of the style of the answer: concise and impeccable.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:59
2
$begingroup$
Congrats on getting the Populist badge off of Math Gems' answer!
$endgroup$
– Zev Chonoles
Feb 10 '13 at 22:29
add a comment |
$begingroup$
Suppose $a,bin G$, $abne 1$ and $a,bne 1$. Then $1=ab(ab)^{-1}=(ab)^{-1}ba$, so $ba=ab$. Of course, if $ab=1$ then $b=a^{-1}$ and $ba=1$ as well, and finally, if one of $a,b$ is $1$, then also $ab=ba$. So the group is abelian.
answered Feb 9 '13 at 23:34
Andrés E. CaicedoAndrés E. Caicedo
65.1k8158247
$endgroup$
$begingroup$
Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:10
$begingroup$
@MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions!
$endgroup$
– user59671
Feb 10 '13 at 0:20
$begingroup$
@CutieKrait During my undergraduate studies I was math olympiad contestant a few times. Andrés' answer brought me fond memories about that, because of the style of the answer: concise and impeccable.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:59
2
$begingroup$
Congrats on getting the Populist badge off of Math Gems' answer!
$endgroup$
– Zev Chonoles
Feb 10 '13 at 22:29
add a comment |
$begingroup$
Suppose $a,bin G$, $abne 1$ and $a,bne 1$. Then $1=ab(ab)^{-1}=(ab)^{-1}ba$, so $ba=ab$. Of course, if $ab=1$ then $b=a^{-1}$ and $ba=1$ as well, and finally, if one of $a,b$ is $1$, then also $ab=ba$. So the group is abelian.
answered Feb 9 '13 at 23:34
Andrés E. CaicedoAndrés E. Caicedo
65.1k8158247
$endgroup$
Suppose $a,bin G$, $abne 1$ and $a,bne 1$. Then $1=ab(ab)^{-1}=(ab)^{-1}ba$, so $ba=ab$. Of course, if $ab=1$ then $b=a^{-1}$ and $ba=1$ as well, and finally, if one of $a,b$ is $1$, then also $ab=ba$. So the group is abelian.
answered Feb 9 '13 at 23:34
Andrés E. CaicedoAndrés E. Caicedo
65.1k8158247
answered Feb 9 '13 at 23:34
Andrés E. CaicedoAndrés E. Caicedo
65.1k8158247
answered Feb 9 '13 at 23:34
Andrés E. CaicedoAndrés E. Caicedo
65.1k8158247
answered Feb 9 '13 at 23:34
Andrés E. CaicedoAndrés E. Caicedo
65.1k8158247
65.1k8158247
$begingroup$
Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:10
$begingroup$
@MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions!
$endgroup$
– user59671
Feb 10 '13 at 0:20
$begingroup$
@CutieKrait During my undergraduate studies I was math olympiad contestant a few times. Andrés' answer brought me fond memories about that, because of the style of the answer: concise and impeccable.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:59
2
$begingroup$
Congrats on getting the Populist badge off of Math Gems' answer!
$endgroup$
– Zev Chonoles
Feb 10 '13 at 22:29
add a comment |
$begingroup$
Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:10
$begingroup$
@MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions!
$endgroup$
– user59671
Feb 10 '13 at 0:20
$begingroup$
@CutieKrait During my undergraduate studies I was math olympiad contestant a few times. Andrés' answer brought me fond memories about that, because of the style of the answer: concise and impeccable.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:59
2
$begingroup$
Congrats on getting the Populist badge off of Math Gems' answer!
$endgroup$
– Zev Chonoles
Feb 10 '13 at 22:29
$begingroup$
Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:10
$begingroup$
Hmmm, an answer from a seasoned math olympiads excontestant, brilliant!.
$endgroup$
– Matemáticos Chibchas
Feb 10 '13 at 0:10
$begingroup$
@MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions!
$endgroup$
– user59671
Feb 10 '13 at 0:20
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@MatemáticosChibchas: Was that an olympiad question?! I just guessed it may be true! And even there are more generalized questions!
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– user59671
Feb 10 '13 at 0:20
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@CutieKrait During my undergraduate studies I was math olympiad contestant a few times. Andrés' answer brought me fond memories about that, because of the style of the answer: concise and impeccable.
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– Matemáticos Chibchas
Feb 10 '13 at 0:59
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@CutieKrait During my undergraduate studies I was math olympiad contestant a few times. Andrés' answer brought me fond memories about that, because of the style of the answer: concise and impeccable.
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– Matemáticos Chibchas
Feb 10 '13 at 0:59
2
2
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Congrats on getting the Populist badge off of Math Gems' answer!
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– Zev Chonoles
Feb 10 '13 at 22:29
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Congrats on getting the Populist badge off of Math Gems' answer!
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– Zev Chonoles
Feb 10 '13 at 22:29
add a comment |
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If we reformulate the question as follows:
Let $G$ be a group, and $a,b,c in G$,
Let's define two 3-cycles: $tau_1 = (abc)$ and $tau_2 = (acb)$ to identify the left$(1)$ and the right$(2)$ side of the equality given.
Then isn't it obvious that, as $tau_{1,2} in mathscr{S}_3 $ and $mathscr{S}_3$ is known not to be commutative, the only condition which would imply this equality, is $G$ being abelian?
answered Jan 3 at 16:26
freehumoristfreehumorist
173112
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What does this have to do with commutativity of $G$?
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– Tobias Kildetoft
Jan 3 at 18:04
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I should have detailed: Let's introduce $sigma_1 = left ( begin{matrix} a & b & c\ a & b & c end{matrix} right ) = Id_{mathscr{S}_3}$ to represent left side of our equality; and $sigma_2 = left ( begin{matrix} a & b & c\ a & c & b end{matrix} right ) = (bc)$ to represent the right side. Can't we have then the transposition $(bc)$ is identity, thus $bc = cb$?
$endgroup$
– freehumorist
Jan 3 at 18:41
add a comment |
$begingroup$
If we reformulate the question as follows:
Let $G$ be a group, and $a,b,c in G$,
Let's define two 3-cycles: $tau_1 = (abc)$ and $tau_2 = (acb)$ to identify the left$(1)$ and the right$(2)$ side of the equality given.
Then isn't it obvious that, as $tau_{1,2} in mathscr{S}_3 $ and $mathscr{S}_3$ is known not to be commutative, the only condition which would imply this equality, is $G$ being abelian?
answered Jan 3 at 16:26
freehumoristfreehumorist
173112
$endgroup$
$begingroup$
What does this have to do with commutativity of $G$?
$endgroup$
– Tobias Kildetoft
Jan 3 at 18:04
$begingroup$
I should have detailed: Let's introduce $sigma_1 = left ( begin{matrix} a & b & c\ a & b & c end{matrix} right ) = Id_{mathscr{S}_3}$ to represent left side of our equality; and $sigma_2 = left ( begin{matrix} a & b & c\ a & c & b end{matrix} right ) = (bc)$ to represent the right side. Can't we have then the transposition $(bc)$ is identity, thus $bc = cb$?
$endgroup$
– freehumorist
Jan 3 at 18:41
add a comment |
$begingroup$
If we reformulate the question as follows:
Let $G$ be a group, and $a,b,c in G$,
Let's define two 3-cycles: $tau_1 = (abc)$ and $tau_2 = (acb)$ to identify the left$(1)$ and the right$(2)$ side of the equality given.
Then isn't it obvious that, as $tau_{1,2} in mathscr{S}_3 $ and $mathscr{S}_3$ is known not to be commutative, the only condition which would imply this equality, is $G$ being abelian?
answered Jan 3 at 16:26
freehumoristfreehumorist
173112
$endgroup$
If we reformulate the question as follows:
Let $G$ be a group, and $a,b,c in G$,
Let's define two 3-cycles: $tau_1 = (abc)$ and $tau_2 = (acb)$ to identify the left$(1)$ and the right$(2)$ side of the equality given.
Then isn't it obvious that, as $tau_{1,2} in mathscr{S}_3 $ and $mathscr{S}_3$ is known not to be commutative, the only condition which would imply this equality, is $G$ being abelian?
answered Jan 3 at 16:26
freehumoristfreehumorist
173112
edited Jan 3 at 18:00
answered Jan 3 at 16:26
freehumoristfreehumorist
173112
answered Jan 3 at 16:26
freehumoristfreehumorist
173112
answered Jan 3 at 16:26
freehumoristfreehumorist
173112
173112
$begingroup$
What does this have to do with commutativity of $G$?
$endgroup$
– Tobias Kildetoft
Jan 3 at 18:04
$begingroup$
I should have detailed: Let's introduce $sigma_1 = left ( begin{matrix} a & b & c\ a & b & c end{matrix} right ) = Id_{mathscr{S}_3}$ to represent left side of our equality; and $sigma_2 = left ( begin{matrix} a & b & c\ a & c & b end{matrix} right ) = (bc)$ to represent the right side. Can't we have then the transposition $(bc)$ is identity, thus $bc = cb$?
$endgroup$
– freehumorist
Jan 3 at 18:41
add a comment |
$begingroup$
What does this have to do with commutativity of $G$?
$endgroup$
– Tobias Kildetoft
Jan 3 at 18:04
$begingroup$
I should have detailed: Let's introduce $sigma_1 = left ( begin{matrix} a & b & c\ a & b & c end{matrix} right ) = Id_{mathscr{S}_3}$ to represent left side of our equality; and $sigma_2 = left ( begin{matrix} a & b & c\ a & c & b end{matrix} right ) = (bc)$ to represent the right side. Can't we have then the transposition $(bc)$ is identity, thus $bc = cb$?
$endgroup$
– freehumorist
Jan 3 at 18:41
$begingroup$
What does this have to do with commutativity of $G$?
$endgroup$
– Tobias Kildetoft
Jan 3 at 18:04
$begingroup$
What does this have to do with commutativity of $G$?
$endgroup$
– Tobias Kildetoft
Jan 3 at 18:04
$begingroup$
I should have detailed: Let's introduce $sigma_1 = left ( begin{matrix} a & b & c\ a & b & c end{matrix} right ) = Id_{mathscr{S}_3}$ to represent left side of our equality; and $sigma_2 = left ( begin{matrix} a & b & c\ a & c & b end{matrix} right ) = (bc)$ to represent the right side. Can't we have then the transposition $(bc)$ is identity, thus $bc = cb$?
$endgroup$
– freehumorist
Jan 3 at 18:41
$begingroup$
I should have detailed: Let's introduce $sigma_1 = left ( begin{matrix} a & b & c\ a & b & c end{matrix} right ) = Id_{mathscr{S}_3}$ to represent left side of our equality; and $sigma_2 = left ( begin{matrix} a & b & c\ a & c & b end{matrix} right ) = (bc)$ to represent the right side. Can't we have then the transposition $(bc)$ is identity, thus $bc = cb$?
$endgroup$
– freehumorist
Jan 3 at 18:41
add a comment |
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Seems like a perfectly good question to me. I vote to leave open.
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– Brett Frankel
Feb 11 '13 at 1:11