Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.
$begingroup$
Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.
I was thinking that you could rewrite $sigma=g_1cdots g_k$ with $g_i$ disjoint cycles and $tau=h_1cdots h_l$ with $h_i$ disjoint cycles. But I don't know what to do next, as $g_i$ and $h_j$ doesn't have to be disjoint.
Edit: Could I prove it like this ?
We can write $σ$ in disjoint cycles: $σ=σ_1...σ_r$ with lenghts $l_1,...l_r$. So you get: $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$.
Which gives:
begin{align*}
τστ^{-1}&=τσ_1...σ_rτ^{-1}\
&=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\
&=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\
&=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\
end{align*}
Therefore $σ$ and $τστ^{-1}$ have the same cycle type. So $τσ$ and $στ$ have the same cycle type.
abstract-algebra group-theory finite-groups permutations
$endgroup$
add a comment |
$begingroup$
Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.
I was thinking that you could rewrite $sigma=g_1cdots g_k$ with $g_i$ disjoint cycles and $tau=h_1cdots h_l$ with $h_i$ disjoint cycles. But I don't know what to do next, as $g_i$ and $h_j$ doesn't have to be disjoint.
Edit: Could I prove it like this ?
We can write $σ$ in disjoint cycles: $σ=σ_1...σ_r$ with lenghts $l_1,...l_r$. So you get: $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$.
Which gives:
begin{align*}
τστ^{-1}&=τσ_1...σ_rτ^{-1}\
&=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\
&=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\
&=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\
end{align*}
Therefore $σ$ and $τστ^{-1}$ have the same cycle type. So $τσ$ and $στ$ have the same cycle type.
abstract-algebra group-theory finite-groups permutations
$endgroup$
add a comment |
$begingroup$
Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.
I was thinking that you could rewrite $sigma=g_1cdots g_k$ with $g_i$ disjoint cycles and $tau=h_1cdots h_l$ with $h_i$ disjoint cycles. But I don't know what to do next, as $g_i$ and $h_j$ doesn't have to be disjoint.
Edit: Could I prove it like this ?
We can write $σ$ in disjoint cycles: $σ=σ_1...σ_r$ with lenghts $l_1,...l_r$. So you get: $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$.
Which gives:
begin{align*}
τστ^{-1}&=τσ_1...σ_rτ^{-1}\
&=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\
&=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\
&=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\
end{align*}
Therefore $σ$ and $τστ^{-1}$ have the same cycle type. So $τσ$ and $στ$ have the same cycle type.
abstract-algebra group-theory finite-groups permutations
$endgroup$
Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.
I was thinking that you could rewrite $sigma=g_1cdots g_k$ with $g_i$ disjoint cycles and $tau=h_1cdots h_l$ with $h_i$ disjoint cycles. But I don't know what to do next, as $g_i$ and $h_j$ doesn't have to be disjoint.
Edit: Could I prove it like this ?
We can write $σ$ in disjoint cycles: $σ=σ_1...σ_r$ with lenghts $l_1,...l_r$. So you get: $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$.
Which gives:
begin{align*}
τστ^{-1}&=τσ_1...σ_rτ^{-1}\
&=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\
&=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\
&=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\
end{align*}
Therefore $σ$ and $τστ^{-1}$ have the same cycle type. So $τσ$ and $στ$ have the same cycle type.
abstract-algebra group-theory finite-groups permutations
abstract-algebra group-theory finite-groups permutations
edited Dec 24 '18 at 21:16
Shaun
9,268113684
9,268113684
asked Mar 19 '13 at 15:51
90intuition90intuition
1,13331230
1,13331230
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Hint: they are conjugate elements of the group.
$endgroup$
5
$begingroup$
My book introduces conjugates in the next chapter.
$endgroup$
– 90intuition
Mar 19 '13 at 15:59
add a comment |
$begingroup$
First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.
$endgroup$
$begingroup$
This one did it ! Thanks
$endgroup$
– 90intuition
Mar 19 '13 at 20:09
$begingroup$
Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
I hope that this is what you meant.
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
$endgroup$
– Shaun
Dec 24 '18 at 21:11
add a comment |
$begingroup$
First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.
Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.
If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?
$endgroup$
$begingroup$
Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
$endgroup$
– whacka
Aug 2 '15 at 21:06
add a comment |
$begingroup$
In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.
$endgroup$
$begingroup$
Nice! $quad ddot smile quad +1 quad$
$endgroup$
– amWhy
Mar 20 '13 at 0:15
add a comment |
$begingroup$
Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.
Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$
Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: they are conjugate elements of the group.
$endgroup$
5
$begingroup$
My book introduces conjugates in the next chapter.
$endgroup$
– 90intuition
Mar 19 '13 at 15:59
add a comment |
$begingroup$
Hint: they are conjugate elements of the group.
$endgroup$
5
$begingroup$
My book introduces conjugates in the next chapter.
$endgroup$
– 90intuition
Mar 19 '13 at 15:59
add a comment |
$begingroup$
Hint: they are conjugate elements of the group.
$endgroup$
Hint: they are conjugate elements of the group.
answered Mar 19 '13 at 15:57
Robert IsraelRobert Israel
324k23214468
324k23214468
5
$begingroup$
My book introduces conjugates in the next chapter.
$endgroup$
– 90intuition
Mar 19 '13 at 15:59
add a comment |
5
$begingroup$
My book introduces conjugates in the next chapter.
$endgroup$
– 90intuition
Mar 19 '13 at 15:59
5
5
$begingroup$
My book introduces conjugates in the next chapter.
$endgroup$
– 90intuition
Mar 19 '13 at 15:59
$begingroup$
My book introduces conjugates in the next chapter.
$endgroup$
– 90intuition
Mar 19 '13 at 15:59
add a comment |
$begingroup$
First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.
$endgroup$
$begingroup$
This one did it ! Thanks
$endgroup$
– 90intuition
Mar 19 '13 at 20:09
$begingroup$
Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
I hope that this is what you meant.
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
$endgroup$
– Shaun
Dec 24 '18 at 21:11
add a comment |
$begingroup$
First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.
$endgroup$
$begingroup$
This one did it ! Thanks
$endgroup$
– 90intuition
Mar 19 '13 at 20:09
$begingroup$
Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
I hope that this is what you meant.
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
$endgroup$
– Shaun
Dec 24 '18 at 21:11
add a comment |
$begingroup$
First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.
$endgroup$
First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.
answered Mar 19 '13 at 16:03
Santiago CanezSantiago Canez
2,07111113
2,07111113
$begingroup$
This one did it ! Thanks
$endgroup$
– 90intuition
Mar 19 '13 at 20:09
$begingroup$
Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
I hope that this is what you meant.
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
$endgroup$
– Shaun
Dec 24 '18 at 21:11
add a comment |
$begingroup$
This one did it ! Thanks
$endgroup$
– 90intuition
Mar 19 '13 at 20:09
$begingroup$
Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
I hope that this is what you meant.
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
$endgroup$
– Shaun
Dec 24 '18 at 21:11
$begingroup$
This one did it ! Thanks
$endgroup$
– 90intuition
Mar 19 '13 at 20:09
$begingroup$
This one did it ! Thanks
$endgroup$
– 90intuition
Mar 19 '13 at 20:09
$begingroup$
Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
I hope that this is what you meant.
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
I hope that this is what you meant.
$endgroup$
– 90intuition
Mar 19 '13 at 20:38
$begingroup$
Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
$endgroup$
– Shaun
Dec 24 '18 at 21:11
$begingroup$
Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
$endgroup$
– Shaun
Dec 24 '18 at 21:11
add a comment |
$begingroup$
First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.
Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.
If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?
$endgroup$
$begingroup$
Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
$endgroup$
– whacka
Aug 2 '15 at 21:06
add a comment |
$begingroup$
First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.
Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.
If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?
$endgroup$
$begingroup$
Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
$endgroup$
– whacka
Aug 2 '15 at 21:06
add a comment |
$begingroup$
First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.
Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.
If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?
$endgroup$
First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.
Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.
If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?
edited Mar 19 '13 at 18:23
answered Mar 19 '13 at 16:51
NECingNECing
3,30231628
3,30231628
$begingroup$
Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
$endgroup$
– whacka
Aug 2 '15 at 21:06
add a comment |
$begingroup$
Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
$endgroup$
– whacka
Aug 2 '15 at 21:06
$begingroup$
Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
$endgroup$
– whacka
Aug 2 '15 at 21:06
$begingroup$
Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
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– whacka
Aug 2 '15 at 21:06
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In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.
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Nice! $quad ddot smile quad +1 quad$
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– amWhy
Mar 20 '13 at 0:15
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In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.
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Nice! $quad ddot smile quad +1 quad$
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– amWhy
Mar 20 '13 at 0:15
add a comment |
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In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.
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In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.
answered Mar 19 '13 at 16:51
mrsmrs
1
1
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Nice! $quad ddot smile quad +1 quad$
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– amWhy
Mar 20 '13 at 0:15
add a comment |
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Nice! $quad ddot smile quad +1 quad$
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– amWhy
Mar 20 '13 at 0:15
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Nice! $quad ddot smile quad +1 quad$
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– amWhy
Mar 20 '13 at 0:15
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Nice! $quad ddot smile quad +1 quad$
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– amWhy
Mar 20 '13 at 0:15
add a comment |
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Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.
Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$
Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.
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add a comment |
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Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.
Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$
Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.
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add a comment |
$begingroup$
Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.
Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$
Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.
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Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.
Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$
Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.
answered Mar 19 '13 at 20:04
P..P..
13.4k22348
13.4k22348
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