compute $aba^{−1}$
0
$begingroup$
With the usual notations, compute $aba^{−1}$
in $S_5$ and express it as the
product of disjoint cycles, where
$a = (1 2 3)(4 5)$ and $b = (2 3)(1 4).$
My attempt : $ab$ = $(1345)$ and $a^{-1} = (321)(54)$
now now i got $aba^{-1} = (12)(35)$
is its correct ?
Any hint/ solution will be appreciated
abstract-algebra
asked Jan 9 at 14:15
jasminejasmine
1,941420
$endgroup$
add a comment |
0
$begingroup$
With the usual notations, compute $aba^{−1}$
in $S_5$ and express it as the
product of disjoint cycles, where
$a = (1 2 3)(4 5)$ and $b = (2 3)(1 4).$
My attempt : $ab$ = $(1345)$ and $a^{-1} = (321)(54)$
now now i got $aba^{-1} = (12)(35)$
is its correct ?
Any hint/ solution will be appreciated
abstract-algebra
asked Jan 9 at 14:15
jasminejasmine
1,941420
$endgroup$
add a comment |
0
0
0
$begingroup$
With the usual notations, compute $aba^{−1}$
in $S_5$ and express it as the
product of disjoint cycles, where
$a = (1 2 3)(4 5)$ and $b = (2 3)(1 4).$
My attempt : $ab$ = $(1345)$ and $a^{-1} = (321)(54)$
now now i got $aba^{-1} = (12)(35)$
is its correct ?
Any hint/ solution will be appreciated
abstract-algebra
asked Jan 9 at 14:15
jasminejasmine
1,941420
$endgroup$
With the usual notations, compute $aba^{−1}$
in $S_5$ and express it as the
product of disjoint cycles, where
$a = (1 2 3)(4 5)$ and $b = (2 3)(1 4).$
My attempt : $ab$ = $(1345)$ and $a^{-1} = (321)(54)$
now now i got $aba^{-1} = (12)(35)$
is its correct ?
Any hint/ solution will be appreciated
abstract-algebra
abstract-algebra
asked Jan 9 at 14:15
jasminejasmine
1,941420
asked Jan 9 at 14:15
jasminejasmine
1,941420
asked Jan 9 at 14:15
jasminejasmine
1,941420
asked Jan 9 at 14:15
jasminejasmine
1,941420
asked Jan 9 at 14:15
jasminejasmine
1,941420
1,941420
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$
answered Jan 9 at 14:18
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
$endgroup$
– jasmine
Jan 9 at 14:24
1
$begingroup$
From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:25
$begingroup$
But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
$endgroup$
– jasmine
Jan 9 at 14:26
1
$begingroup$
I am using the standard notation here.
$endgroup$
– José Carlos Santos
Jan 9 at 14:27
1
$begingroup$
How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:36
|
show 2 more comments
1
$begingroup$
Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$
answered Jan 9 at 14:18
A. BailleulA. Bailleul
1587
$endgroup$
$begingroup$
how $a(2) =3$?..
$endgroup$
– jasmine
Jan 9 at 14:22
1
$begingroup$
You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
$endgroup$
– A. Bailleul
Jan 9 at 14:25
add a comment |
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2 Answers
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2 Answers
2
active
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active
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2
$begingroup$
No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$
answered Jan 9 at 14:18
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
$endgroup$
– jasmine
Jan 9 at 14:24
1
$begingroup$
From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:25
$begingroup$
But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
$endgroup$
– jasmine
Jan 9 at 14:26
1
$begingroup$
I am using the standard notation here.
$endgroup$
– José Carlos Santos
Jan 9 at 14:27
1
$begingroup$
How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:36
|
show 2 more comments
2
$begingroup$
No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$
answered Jan 9 at 14:18
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
$endgroup$
– jasmine
Jan 9 at 14:24
1
$begingroup$
From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:25
$begingroup$
But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
$endgroup$
– jasmine
Jan 9 at 14:26
1
$begingroup$
I am using the standard notation here.
$endgroup$
– José Carlos Santos
Jan 9 at 14:27
1
$begingroup$
How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:36
|
show 2 more comments
2
2
2
$begingroup$
No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$
answered Jan 9 at 14:18
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$
answered Jan 9 at 14:18
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 9 at 14:18
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 9 at 14:18
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 9 at 14:18
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
$begingroup$
..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
$endgroup$
– jasmine
Jan 9 at 14:24
1
$begingroup$
From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:25
$begingroup$
But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
$endgroup$
– jasmine
Jan 9 at 14:26
1
$begingroup$
I am using the standard notation here.
$endgroup$
– José Carlos Santos
Jan 9 at 14:27
1
$begingroup$
How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:36
|
show 2 more comments
$begingroup$
..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
$endgroup$
– jasmine
Jan 9 at 14:24
1
$begingroup$
From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:25
$begingroup$
But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
$endgroup$
– jasmine
Jan 9 at 14:26
1
$begingroup$
I am using the standard notation here.
$endgroup$
– José Carlos Santos
Jan 9 at 14:27
1
$begingroup$
How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:36
$begingroup$
..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
$endgroup$
– jasmine
Jan 9 at 14:24
$begingroup$
..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
$endgroup$
– jasmine
Jan 9 at 14:24
1
1
$begingroup$
From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:25
$begingroup$
From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:25
$begingroup$
But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
$endgroup$
– jasmine
Jan 9 at 14:26
$begingroup$
But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
$endgroup$
– jasmine
Jan 9 at 14:26
1
1
$begingroup$
I am using the standard notation here.
$endgroup$
– José Carlos Santos
Jan 9 at 14:27
$begingroup$
I am using the standard notation here.
$endgroup$
– José Carlos Santos
Jan 9 at 14:27
1
1
$begingroup$
How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:36
$begingroup$
How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
$endgroup$
– José Carlos Santos
Jan 9 at 14:36
|
show 2 more comments
1
$begingroup$
Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$
answered Jan 9 at 14:18
A. BailleulA. Bailleul
1587
$endgroup$
$begingroup$
how $a(2) =3$?..
$endgroup$
– jasmine
Jan 9 at 14:22
1
$begingroup$
You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
$endgroup$
– A. Bailleul
Jan 9 at 14:25
add a comment |
1
$begingroup$
Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$
answered Jan 9 at 14:18
A. BailleulA. Bailleul
1587
$endgroup$
$begingroup$
how $a(2) =3$?..
$endgroup$
– jasmine
Jan 9 at 14:22
1
$begingroup$
You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
$endgroup$
– A. Bailleul
Jan 9 at 14:25
add a comment |
1
1
1
$begingroup$
Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$
answered Jan 9 at 14:18
A. BailleulA. Bailleul
1587
$endgroup$
Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$
answered Jan 9 at 14:18
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:18
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:18
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:18
A. BailleulA. Bailleul
1587
1587
$begingroup$
how $a(2) =3$?..
$endgroup$
– jasmine
Jan 9 at 14:22
1
$begingroup$
You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
$endgroup$
– A. Bailleul
Jan 9 at 14:25
add a comment |
$begingroup$
how $a(2) =3$?..
$endgroup$
– jasmine
Jan 9 at 14:22
1
$begingroup$
You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
$endgroup$
– A. Bailleul
Jan 9 at 14:25
$begingroup$
how $a(2) =3$?..
$endgroup$
– jasmine
Jan 9 at 14:22
$begingroup$
how $a(2) =3$?..
$endgroup$
– jasmine
Jan 9 at 14:22
1
1
$begingroup$
You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
$endgroup$
– A. Bailleul
Jan 9 at 14:25
$begingroup$
You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
$endgroup$
– A. Bailleul
Jan 9 at 14:25
add a comment |
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