Showing $S_{n}$ is not solvable for $ngeq 5$
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I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $ngeq 5$. This is the proof
Let $G$ be a group of permutations of five objects or more that
include all cyclic permutations of three elements.
$(124)(142)=e$
$(135)(153)=e$
$(123)=(124)(135)(142)(153)$
Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:Grightarrow G/H$
$f[(124)]=x$
$f[(135)]=y$
$f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$
Therefore $(123)in H$. This is true for any cyclic permutation of three
elements. Therefore, $G$ is not solvable.
I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)in H$. I would appreciate some help.
abstract-algebra group-theory galois-theory symmetric-groups
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I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $ngeq 5$. This is the proof
Let $G$ be a group of permutations of five objects or more that
include all cyclic permutations of three elements.
$(124)(142)=e$
$(135)(153)=e$
$(123)=(124)(135)(142)(153)$
Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:Grightarrow G/H$
$f[(124)]=x$
$f[(135)]=y$
$f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$
Therefore $(123)in H$. This is true for any cyclic permutation of three
elements. Therefore, $G$ is not solvable.
I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)in H$. I would appreciate some help.
abstract-algebra group-theory galois-theory symmetric-groups
$endgroup$
add a comment |
$begingroup$
I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $ngeq 5$. This is the proof
Let $G$ be a group of permutations of five objects or more that
include all cyclic permutations of three elements.
$(124)(142)=e$
$(135)(153)=e$
$(123)=(124)(135)(142)(153)$
Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:Grightarrow G/H$
$f[(124)]=x$
$f[(135)]=y$
$f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$
Therefore $(123)in H$. This is true for any cyclic permutation of three
elements. Therefore, $G$ is not solvable.
I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)in H$. I would appreciate some help.
abstract-algebra group-theory galois-theory symmetric-groups
$endgroup$
I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $ngeq 5$. This is the proof
Let $G$ be a group of permutations of five objects or more that
include all cyclic permutations of three elements.
$(124)(142)=e$
$(135)(153)=e$
$(123)=(124)(135)(142)(153)$
Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:Grightarrow G/H$
$f[(124)]=x$
$f[(135)]=y$
$f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$
Therefore $(123)in H$. This is true for any cyclic permutation of three
elements. Therefore, $G$ is not solvable.
I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)in H$. I would appreciate some help.
abstract-algebra group-theory galois-theory symmetric-groups
abstract-algebra group-theory galois-theory symmetric-groups
edited Dec 25 '18 at 19:58
Alberto Navarro
asked Dec 25 '18 at 19:52
Alberto NavarroAlberto Navarro
306
306
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1 Answer
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That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.
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Got it, thanks!
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– Alberto Navarro
Dec 26 '18 at 1:28
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By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
$endgroup$
– Alberto Navarro
Dec 26 '18 at 6:34
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Yes, that's it.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 10:54
add a comment |
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1 Answer
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1 Answer
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$begingroup$
That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.
$endgroup$
$begingroup$
Got it, thanks!
$endgroup$
– Alberto Navarro
Dec 26 '18 at 1:28
$begingroup$
By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
$endgroup$
– Alberto Navarro
Dec 26 '18 at 6:34
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 10:54
add a comment |
$begingroup$
That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.
$endgroup$
$begingroup$
Got it, thanks!
$endgroup$
– Alberto Navarro
Dec 26 '18 at 1:28
$begingroup$
By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
$endgroup$
– Alberto Navarro
Dec 26 '18 at 6:34
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 10:54
add a comment |
$begingroup$
That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.
$endgroup$
That map from $G$ onto $G/H$ is the map $gmapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $gin H$.
answered Dec 25 '18 at 19:59
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
$begingroup$
Got it, thanks!
$endgroup$
– Alberto Navarro
Dec 26 '18 at 1:28
$begingroup$
By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
$endgroup$
– Alberto Navarro
Dec 26 '18 at 6:34
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 10:54
add a comment |
$begingroup$
Got it, thanks!
$endgroup$
– Alberto Navarro
Dec 26 '18 at 1:28
$begingroup$
By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
$endgroup$
– Alberto Navarro
Dec 26 '18 at 6:34
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 10:54
$begingroup$
Got it, thanks!
$endgroup$
– Alberto Navarro
Dec 26 '18 at 1:28
$begingroup$
Got it, thanks!
$endgroup$
– Alberto Navarro
Dec 26 '18 at 1:28
$begingroup$
By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
$endgroup$
– Alberto Navarro
Dec 26 '18 at 6:34
$begingroup$
By the way, when he says "This is true for any cyclic permutation of three elements " I'm assuming it is because for $ngeq 5$ every $3-$cycle is conjugate to $(123)$ in $A_{n}$, Is this ok?
$endgroup$
– Alberto Navarro
Dec 26 '18 at 6:34
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 10:54
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 10:54
add a comment |
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