Generating Symmetric Group with Transpositions
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There is a theorem in our book which I'm trying to prove, It says:
Theorem: Number of Transpositions which generate $mathbb{S}n$ can not be lower than $n-1$.
I've already proved that $mathbb{S}n$ can be generated by $n-1$ Transpositions, but I don't have any Idea about to show it can not be lower than that.
combinatorics symmetric-groups finitely-generated involutions
edited Dec 28 '18 at 5:24
Shaun
9,241113684
asked Dec 2 '17 at 14:39
SajadSajad
103
$endgroup$
add a comment |
$begingroup$
There is a theorem in our book which I'm trying to prove, It says:
Theorem: Number of Transpositions which generate $mathbb{S}n$ can not be lower than $n-1$.
I've already proved that $mathbb{S}n$ can be generated by $n-1$ Transpositions, but I don't have any Idea about to show it can not be lower than that.
combinatorics symmetric-groups finitely-generated involutions
edited Dec 28 '18 at 5:24
Shaun
9,241113684
asked Dec 2 '17 at 14:39
SajadSajad
103
$endgroup$
1
$begingroup$
If you know graph theory then this is clear. Consider a graph whose set of vertices is ${1,ldots,n}$. We define two vertices $ineq j$ to be adjacent iff $(i j)in X$, the generating set. Now this graph is necessarily connected, since $S_n$ is transitive. This forces the graph to have at least $n-1$ edges.
$endgroup$
– Colescu
Dec 2 '17 at 14:53
$begingroup$
That was awesome. Thank you
$endgroup$
– Sajad
Dec 2 '17 at 14:57
add a comment |
$begingroup$
There is a theorem in our book which I'm trying to prove, It says:
Theorem: Number of Transpositions which generate $mathbb{S}n$ can not be lower than $n-1$.
I've already proved that $mathbb{S}n$ can be generated by $n-1$ Transpositions, but I don't have any Idea about to show it can not be lower than that.
combinatorics symmetric-groups finitely-generated involutions
edited Dec 28 '18 at 5:24
Shaun
9,241113684
asked Dec 2 '17 at 14:39
SajadSajad
103
$endgroup$
There is a theorem in our book which I'm trying to prove, It says:
Theorem: Number of Transpositions which generate $mathbb{S}n$ can not be lower than $n-1$.
I've already proved that $mathbb{S}n$ can be generated by $n-1$ Transpositions, but I don't have any Idea about to show it can not be lower than that.
combinatorics symmetric-groups finitely-generated involutions
combinatorics symmetric-groups finitely-generated involutions
edited Dec 28 '18 at 5:24
Shaun
9,241113684
asked Dec 2 '17 at 14:39
SajadSajad
103
edited Dec 28 '18 at 5:24
Shaun
9,241113684
asked Dec 2 '17 at 14:39
SajadSajad
103
edited Dec 28 '18 at 5:24
Shaun
9,241113684
edited Dec 28 '18 at 5:24
Shaun
9,241113684
edited Dec 28 '18 at 5:24
Shaun
9,241113684
9,241113684
asked Dec 2 '17 at 14:39
SajadSajad
103
asked Dec 2 '17 at 14:39
SajadSajad
103
asked Dec 2 '17 at 14:39
SajadSajad
103
103
1
$begingroup$
If you know graph theory then this is clear. Consider a graph whose set of vertices is ${1,ldots,n}$. We define two vertices $ineq j$ to be adjacent iff $(i j)in X$, the generating set. Now this graph is necessarily connected, since $S_n$ is transitive. This forces the graph to have at least $n-1$ edges.
$endgroup$
– Colescu
Dec 2 '17 at 14:53
$begingroup$
That was awesome. Thank you
$endgroup$
– Sajad
Dec 2 '17 at 14:57
add a comment |
1
$begingroup$
If you know graph theory then this is clear. Consider a graph whose set of vertices is ${1,ldots,n}$. We define two vertices $ineq j$ to be adjacent iff $(i j)in X$, the generating set. Now this graph is necessarily connected, since $S_n$ is transitive. This forces the graph to have at least $n-1$ edges.
$endgroup$
– Colescu
Dec 2 '17 at 14:53
$begingroup$
That was awesome. Thank you
$endgroup$
– Sajad
Dec 2 '17 at 14:57
1
1
$begingroup$
If you know graph theory then this is clear. Consider a graph whose set of vertices is ${1,ldots,n}$. We define two vertices $ineq j$ to be adjacent iff $(i j)in X$, the generating set. Now this graph is necessarily connected, since $S_n$ is transitive. This forces the graph to have at least $n-1$ edges.
$endgroup$
– Colescu
Dec 2 '17 at 14:53
$begingroup$
If you know graph theory then this is clear. Consider a graph whose set of vertices is ${1,ldots,n}$. We define two vertices $ineq j$ to be adjacent iff $(i j)in X$, the generating set. Now this graph is necessarily connected, since $S_n$ is transitive. This forces the graph to have at least $n-1$ edges.
$endgroup$
– Colescu
Dec 2 '17 at 14:53
$begingroup$
That was awesome. Thank you
$endgroup$
– Sajad
Dec 2 '17 at 14:57
$begingroup$
That was awesome. Thank you
$endgroup$
– Sajad
Dec 2 '17 at 14:57
add a comment |
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1
$begingroup$
If you know graph theory then this is clear. Consider a graph whose set of vertices is ${1,ldots,n}$. We define two vertices $ineq j$ to be adjacent iff $(i j)in X$, the generating set. Now this graph is necessarily connected, since $S_n$ is transitive. This forces the graph to have at least $n-1$ edges.
$endgroup$
– Colescu
Dec 2 '17 at 14:53
$begingroup$
That was awesome. Thank you
$endgroup$
– Sajad
Dec 2 '17 at 14:57