Show that $tau (18632)(47) tau^{-1} = (12345)(67)$.
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Example 4.7. of Aluffi's Algebra says
In $S_8$, $(18632)(47)$ and $(12345)(67)$ must be conjugate since they have the same type. So there there exist $tau$ such that $tau (18632)(47) tau^{-1} = (12345)(67)$ for $tau = (285)(364)$.
If $tau (18632)(47) tau^{-1} = (12345)(67)$ means $tau (18632)(47) = (12345)(67) tau$ so for RHS applying $tau$ to $(text{id})$ then $(12345)(67)$ results in $(128)(3765)$; and for LHS applying $(18632)(47)$ to $(text{id})$ then $tau$ results in $(152)(3847)$ which indeed are not equal! Which part am I doing wrong?
Edit. $tau$ is suggested by the book:
abstract-algebra permutations permutation-cycles
asked Nov 18 at 11:32
72D
50916
add a comment |
up vote
0
down vote
favorite
Example 4.7. of Aluffi's Algebra says
In $S_8$, $(18632)(47)$ and $(12345)(67)$ must be conjugate since they have the same type. So there there exist $tau$ such that $tau (18632)(47) tau^{-1} = (12345)(67)$ for $tau = (285)(364)$.
If $tau (18632)(47) tau^{-1} = (12345)(67)$ means $tau (18632)(47) = (12345)(67) tau$ so for RHS applying $tau$ to $(text{id})$ then $(12345)(67)$ results in $(128)(3765)$; and for LHS applying $(18632)(47)$ to $(text{id})$ then $tau$ results in $(152)(3847)$ which indeed are not equal! Which part am I doing wrong?
Edit. $tau$ is suggested by the book:
abstract-algebra permutations permutation-cycles
asked Nov 18 at 11:32
72D
50916
1
I get $tau =(8 2 5)(6 3 4)$
– mathnoob
Nov 18 at 12:04
@mathnoob, I edited it.
– 72D
Nov 18 at 12:10
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Example 4.7. of Aluffi's Algebra says
In $S_8$, $(18632)(47)$ and $(12345)(67)$ must be conjugate since they have the same type. So there there exist $tau$ such that $tau (18632)(47) tau^{-1} = (12345)(67)$ for $tau = (285)(364)$.
If $tau (18632)(47) tau^{-1} = (12345)(67)$ means $tau (18632)(47) = (12345)(67) tau$ so for RHS applying $tau$ to $(text{id})$ then $(12345)(67)$ results in $(128)(3765)$; and for LHS applying $(18632)(47)$ to $(text{id})$ then $tau$ results in $(152)(3847)$ which indeed are not equal! Which part am I doing wrong?
Edit. $tau$ is suggested by the book:
abstract-algebra permutations permutation-cycles
asked Nov 18 at 11:32
72D
50916
Example 4.7. of Aluffi's Algebra says
In $S_8$, $(18632)(47)$ and $(12345)(67)$ must be conjugate since they have the same type. So there there exist $tau$ such that $tau (18632)(47) tau^{-1} = (12345)(67)$ for $tau = (285)(364)$.
If $tau (18632)(47) tau^{-1} = (12345)(67)$ means $tau (18632)(47) = (12345)(67) tau$ so for RHS applying $tau$ to $(text{id})$ then $(12345)(67)$ results in $(128)(3765)$; and for LHS applying $(18632)(47)$ to $(text{id})$ then $tau$ results in $(152)(3847)$ which indeed are not equal! Which part am I doing wrong?
Edit. $tau$ is suggested by the book:
abstract-algebra permutations permutation-cycles
abstract-algebra permutations permutation-cycles
asked Nov 18 at 11:32
72D
50916
asked Nov 18 at 11:32
72D
50916
edited Nov 18 at 12:09
asked Nov 18 at 11:32
72D
50916
asked Nov 18 at 11:32
72D
50916
asked Nov 18 at 11:32
72D
50916
50916
1
I get $tau =(8 2 5)(6 3 4)$
– mathnoob
Nov 18 at 12:04
@mathnoob, I edited it.
– 72D
Nov 18 at 12:10
add a comment |
1
I get $tau =(8 2 5)(6 3 4)$
– mathnoob
Nov 18 at 12:04
@mathnoob, I edited it.
– 72D
Nov 18 at 12:10
1
1
I get $tau =(8 2 5)(6 3 4)$
– mathnoob
Nov 18 at 12:04
I get $tau =(8 2 5)(6 3 4)$
– mathnoob
Nov 18 at 12:04
@mathnoob, I edited it.
– 72D
Nov 18 at 12:10
@mathnoob, I edited it.
– 72D
Nov 18 at 12:10
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Maybe it is a printing mistake for $tau$ , I think $tau$ in the book is actually suppose to be $tau^{-1}$ and that would work. because in general: If $sigma=(i_1^1...i_{t(1)}^1)(i_1^2...i_{t(2)}^2)...(i_1^r...i_{t(r)}^r)$ and $rho=(j_1^1...j_{t(1)}^1)(j_1^2...j_{t(2)}^2)...(j_1^r...j_{t(r)}^r)$, Then define $tau$ by $tau(i_b^a)=j_b^a$, then $tausigmatau^{-1}=rho$.
answered Nov 18 at 12:17
mathnoob
1,133116
I think the book meant actions start from left to right! (?)
– 72D
Nov 18 at 12:26
Ah, you are right! sorry, that make sence now! But then still, $tau(18632)(47)=(12345)(67)tau =(1 8 5)(2 6 7 4)$. Also, sorry What did you mean by apply to (id)?
– mathnoob
Nov 18 at 12:38
By $(text{id})$ I meant the original set of ordered 1-2-...-8 that no $sigma$ acted yet; i.e. the first line of $tau$ in picture. Also the book defines acting $sigma$ from right to left not like operators or functions. A bit absurd!
– 72D
Nov 18 at 13:55
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Maybe it is a printing mistake for $tau$ , I think $tau$ in the book is actually suppose to be $tau^{-1}$ and that would work. because in general: If $sigma=(i_1^1...i_{t(1)}^1)(i_1^2...i_{t(2)}^2)...(i_1^r...i_{t(r)}^r)$ and $rho=(j_1^1...j_{t(1)}^1)(j_1^2...j_{t(2)}^2)...(j_1^r...j_{t(r)}^r)$, Then define $tau$ by $tau(i_b^a)=j_b^a$, then $tausigmatau^{-1}=rho$.
answered Nov 18 at 12:17
mathnoob
1,133116
I think the book meant actions start from left to right! (?)
– 72D
Nov 18 at 12:26
Ah, you are right! sorry, that make sence now! But then still, $tau(18632)(47)=(12345)(67)tau =(1 8 5)(2 6 7 4)$. Also, sorry What did you mean by apply to (id)?
– mathnoob
Nov 18 at 12:38
By $(text{id})$ I meant the original set of ordered 1-2-...-8 that no $sigma$ acted yet; i.e. the first line of $tau$ in picture. Also the book defines acting $sigma$ from right to left not like operators or functions. A bit absurd!
– 72D
Nov 18 at 13:55
add a comment |
up vote
0
down vote
accepted
Maybe it is a printing mistake for $tau$ , I think $tau$ in the book is actually suppose to be $tau^{-1}$ and that would work. because in general: If $sigma=(i_1^1...i_{t(1)}^1)(i_1^2...i_{t(2)}^2)...(i_1^r...i_{t(r)}^r)$ and $rho=(j_1^1...j_{t(1)}^1)(j_1^2...j_{t(2)}^2)...(j_1^r...j_{t(r)}^r)$, Then define $tau$ by $tau(i_b^a)=j_b^a$, then $tausigmatau^{-1}=rho$.
answered Nov 18 at 12:17
mathnoob
1,133116
I think the book meant actions start from left to right! (?)
– 72D
Nov 18 at 12:26
Ah, you are right! sorry, that make sence now! But then still, $tau(18632)(47)=(12345)(67)tau =(1 8 5)(2 6 7 4)$. Also, sorry What did you mean by apply to (id)?
– mathnoob
Nov 18 at 12:38
By $(text{id})$ I meant the original set of ordered 1-2-...-8 that no $sigma$ acted yet; i.e. the first line of $tau$ in picture. Also the book defines acting $sigma$ from right to left not like operators or functions. A bit absurd!
– 72D
Nov 18 at 13:55
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Maybe it is a printing mistake for $tau$ , I think $tau$ in the book is actually suppose to be $tau^{-1}$ and that would work. because in general: If $sigma=(i_1^1...i_{t(1)}^1)(i_1^2...i_{t(2)}^2)...(i_1^r...i_{t(r)}^r)$ and $rho=(j_1^1...j_{t(1)}^1)(j_1^2...j_{t(2)}^2)...(j_1^r...j_{t(r)}^r)$, Then define $tau$ by $tau(i_b^a)=j_b^a$, then $tausigmatau^{-1}=rho$.
answered Nov 18 at 12:17
mathnoob
1,133116
Maybe it is a printing mistake for $tau$ , I think $tau$ in the book is actually suppose to be $tau^{-1}$ and that would work. because in general: If $sigma=(i_1^1...i_{t(1)}^1)(i_1^2...i_{t(2)}^2)...(i_1^r...i_{t(r)}^r)$ and $rho=(j_1^1...j_{t(1)}^1)(j_1^2...j_{t(2)}^2)...(j_1^r...j_{t(r)}^r)$, Then define $tau$ by $tau(i_b^a)=j_b^a$, then $tausigmatau^{-1}=rho$.
answered Nov 18 at 12:17
mathnoob
1,133116
answered Nov 18 at 12:17
mathnoob
1,133116
answered Nov 18 at 12:17
mathnoob
1,133116
answered Nov 18 at 12:17
mathnoob
1,133116
1,133116
I think the book meant actions start from left to right! (?)
– 72D
Nov 18 at 12:26
Ah, you are right! sorry, that make sence now! But then still, $tau(18632)(47)=(12345)(67)tau =(1 8 5)(2 6 7 4)$. Also, sorry What did you mean by apply to (id)?
– mathnoob
Nov 18 at 12:38
By $(text{id})$ I meant the original set of ordered 1-2-...-8 that no $sigma$ acted yet; i.e. the first line of $tau$ in picture. Also the book defines acting $sigma$ from right to left not like operators or functions. A bit absurd!
– 72D
Nov 18 at 13:55
add a comment |
I think the book meant actions start from left to right! (?)
– 72D
Nov 18 at 12:26
Ah, you are right! sorry, that make sence now! But then still, $tau(18632)(47)=(12345)(67)tau =(1 8 5)(2 6 7 4)$. Also, sorry What did you mean by apply to (id)?
– mathnoob
Nov 18 at 12:38
By $(text{id})$ I meant the original set of ordered 1-2-...-8 that no $sigma$ acted yet; i.e. the first line of $tau$ in picture. Also the book defines acting $sigma$ from right to left not like operators or functions. A bit absurd!
– 72D
Nov 18 at 13:55
I think the book meant actions start from left to right! (?)
– 72D
Nov 18 at 12:26
I think the book meant actions start from left to right! (?)
– 72D
Nov 18 at 12:26
Ah, you are right! sorry, that make sence now! But then still, $tau(18632)(47)=(12345)(67)tau =(1 8 5)(2 6 7 4)$. Also, sorry What did you mean by apply to (id)?
– mathnoob
Nov 18 at 12:38
Ah, you are right! sorry, that make sence now! But then still, $tau(18632)(47)=(12345)(67)tau =(1 8 5)(2 6 7 4)$. Also, sorry What did you mean by apply to (id)?
– mathnoob
Nov 18 at 12:38
By $(text{id})$ I meant the original set of ordered 1-2-...-8 that no $sigma$ acted yet; i.e. the first line of $tau$ in picture. Also the book defines acting $sigma$ from right to left not like operators or functions. A bit absurd!
– 72D
Nov 18 at 13:55
By $(text{id})$ I meant the original set of ordered 1-2-...-8 that no $sigma$ acted yet; i.e. the first line of $tau$ in picture. Also the book defines acting $sigma$ from right to left not like operators or functions. A bit absurd!
– 72D
Nov 18 at 13:55
add a comment |
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I get $tau =(8 2 5)(6 3 4)$
– mathnoob
Nov 18 at 12:04
@mathnoob, I edited it.
– 72D
Nov 18 at 12:10