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Permutation group products (23)(12)(34)=(1243)?
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I just wanted to ask why it makes sense that $(23)(12)(34)=(1243)$. Note I'm going from right to left.
I'm trying to understand the concepts trying to find a true "method" to follow in all cases. My reasoning at the moment is: $1 mapsto 2$ but since it doesn't close we must then leave the cycle open right (I'm assuming this is the case for when a cycle doesn't close unless when it returns to itself). Then we have $(12$. To proceed we consider where $2$ goes to. Then we see that $2 mapsto 3, 3 mapsto 4$. I'm getting the idea that because we see that $3$ much like a "path" to $4$ then the mapping is then $2 mapsto 4$, giving us $(124$. Then since we know $4mapsto 3$ we just close with $(1243)$. To me however it feels more accurate to see that I have $1 mapsto 2, 2mapsto 3$. Then we see that we have $(13$. Then since we went through all cycles once we then start at 2 since it hasn't closed then we go from $2 mapsto 3, 3 mapsto 4$, then we would have $(1324)$. What am I not seeing? I'm trying to imagine functions where maybe each cycle is its own function and sends to another cycle but I never seem to find anything consistent.
Thanks in advance for any of your clarifications.
group-theory permutations symmetric-groups permutation-cycles
asked 4 hours ago
dls
786
add a comment |Â
up vote
1
down vote
favorite
I just wanted to ask why it makes sense that $(23)(12)(34)=(1243)$. Note I'm going from right to left.
I'm trying to understand the concepts trying to find a true "method" to follow in all cases. My reasoning at the moment is: $1 mapsto 2$ but since it doesn't close we must then leave the cycle open right (I'm assuming this is the case for when a cycle doesn't close unless when it returns to itself). Then we have $(12$. To proceed we consider where $2$ goes to. Then we see that $2 mapsto 3, 3 mapsto 4$. I'm getting the idea that because we see that $3$ much like a "path" to $4$ then the mapping is then $2 mapsto 4$, giving us $(124$. Then since we know $4mapsto 3$ we just close with $(1243)$. To me however it feels more accurate to see that I have $1 mapsto 2, 2mapsto 3$. Then we see that we have $(13$. Then since we went through all cycles once we then start at 2 since it hasn't closed then we go from $2 mapsto 3, 3 mapsto 4$, then we would have $(1324)$. What am I not seeing? I'm trying to imagine functions where maybe each cycle is its own function and sends to another cycle but I never seem to find anything consistent.
Thanks in advance for any of your clarifications.
group-theory permutations symmetric-groups permutation-cycles
asked 4 hours ago
dls
786
1
I also get $(1342)$. Could $(1243)$ be a typo?
– Arthur
4 hours ago
@Arthur Okay thanks, maybe I plugged it into WolframAlpha incorrectly.
– dls
4 hours ago
Wolfram Alpha is weird. Usually, $(2 3)(1 2)$ means you apply $(1 2)$, then you apply $(2 3)$. However, Wolfram Alpha does it so that $(2 3)(1 2)$ means you apply $(2 3)$ then $(1 2)$. Thus, they get the inverse of what it would be normally, which is $(1 2 4 3)$.
– Noble Mushtak
4 hours ago
@NobleMushtak yeah that's what I'm having some difficulty with as well. They perform the operation from left to right, which seems to be normal in some cases but not really how we're learning it in class so it a bit troublesome
– dls
4 hours ago
2
Unfortunately, this is something you kind of have to deal with in abstract algebra because there are two different conventions. Most people use right to left like function composition, but you always have to make sure you check what convention the article/CAS you are reading uses because some people use left to right.
– Noble Mushtak
4 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I just wanted to ask why it makes sense that $(23)(12)(34)=(1243)$. Note I'm going from right to left.
I'm trying to understand the concepts trying to find a true "method" to follow in all cases. My reasoning at the moment is: $1 mapsto 2$ but since it doesn't close we must then leave the cycle open right (I'm assuming this is the case for when a cycle doesn't close unless when it returns to itself). Then we have $(12$. To proceed we consider where $2$ goes to. Then we see that $2 mapsto 3, 3 mapsto 4$. I'm getting the idea that because we see that $3$ much like a "path" to $4$ then the mapping is then $2 mapsto 4$, giving us $(124$. Then since we know $4mapsto 3$ we just close with $(1243)$. To me however it feels more accurate to see that I have $1 mapsto 2, 2mapsto 3$. Then we see that we have $(13$. Then since we went through all cycles once we then start at 2 since it hasn't closed then we go from $2 mapsto 3, 3 mapsto 4$, then we would have $(1324)$. What am I not seeing? I'm trying to imagine functions where maybe each cycle is its own function and sends to another cycle but I never seem to find anything consistent.
Thanks in advance for any of your clarifications.
group-theory permutations symmetric-groups permutation-cycles
asked 4 hours ago
dls
786
I just wanted to ask why it makes sense that $(23)(12)(34)=(1243)$. Note I'm going from right to left.
I'm trying to understand the concepts trying to find a true "method" to follow in all cases. My reasoning at the moment is: $1 mapsto 2$ but since it doesn't close we must then leave the cycle open right (I'm assuming this is the case for when a cycle doesn't close unless when it returns to itself). Then we have $(12$. To proceed we consider where $2$ goes to. Then we see that $2 mapsto 3, 3 mapsto 4$. I'm getting the idea that because we see that $3$ much like a "path" to $4$ then the mapping is then $2 mapsto 4$, giving us $(124$. Then since we know $4mapsto 3$ we just close with $(1243)$. To me however it feels more accurate to see that I have $1 mapsto 2, 2mapsto 3$. Then we see that we have $(13$. Then since we went through all cycles once we then start at 2 since it hasn't closed then we go from $2 mapsto 3, 3 mapsto 4$, then we would have $(1324)$. What am I not seeing? I'm trying to imagine functions where maybe each cycle is its own function and sends to another cycle but I never seem to find anything consistent.
Thanks in advance for any of your clarifications.
group-theory permutations symmetric-groups permutation-cycles
group-theory permutations symmetric-groups permutation-cycles
asked 4 hours ago
dls
786
asked 4 hours ago
dls
786
asked 4 hours ago
dls
786
asked 4 hours ago
dls
786
asked 4 hours ago
dls
786
786
1
I also get $(1342)$. Could $(1243)$ be a typo?
– Arthur
4 hours ago
@Arthur Okay thanks, maybe I plugged it into WolframAlpha incorrectly.
– dls
4 hours ago
Wolfram Alpha is weird. Usually, $(2 3)(1 2)$ means you apply $(1 2)$, then you apply $(2 3)$. However, Wolfram Alpha does it so that $(2 3)(1 2)$ means you apply $(2 3)$ then $(1 2)$. Thus, they get the inverse of what it would be normally, which is $(1 2 4 3)$.
– Noble Mushtak
4 hours ago
@NobleMushtak yeah that's what I'm having some difficulty with as well. They perform the operation from left to right, which seems to be normal in some cases but not really how we're learning it in class so it a bit troublesome
– dls
4 hours ago
2
Unfortunately, this is something you kind of have to deal with in abstract algebra because there are two different conventions. Most people use right to left like function composition, but you always have to make sure you check what convention the article/CAS you are reading uses because some people use left to right.
– Noble Mushtak
4 hours ago
add a comment |Â
1
I also get $(1342)$. Could $(1243)$ be a typo?
– Arthur
4 hours ago
@Arthur Okay thanks, maybe I plugged it into WolframAlpha incorrectly.
– dls
4 hours ago
Wolfram Alpha is weird. Usually, $(2 3)(1 2)$ means you apply $(1 2)$, then you apply $(2 3)$. However, Wolfram Alpha does it so that $(2 3)(1 2)$ means you apply $(2 3)$ then $(1 2)$. Thus, they get the inverse of what it would be normally, which is $(1 2 4 3)$.
– Noble Mushtak
4 hours ago
@NobleMushtak yeah that's what I'm having some difficulty with as well. They perform the operation from left to right, which seems to be normal in some cases but not really how we're learning it in class so it a bit troublesome
– dls
4 hours ago
2
Unfortunately, this is something you kind of have to deal with in abstract algebra because there are two different conventions. Most people use right to left like function composition, but you always have to make sure you check what convention the article/CAS you are reading uses because some people use left to right.
– Noble Mushtak
4 hours ago
1
1
I also get $(1342)$. Could $(1243)$ be a typo?
– Arthur
4 hours ago
I also get $(1342)$. Could $(1243)$ be a typo?
– Arthur
4 hours ago
@Arthur Okay thanks, maybe I plugged it into WolframAlpha incorrectly.
– dls
4 hours ago
@Arthur Okay thanks, maybe I plugged it into WolframAlpha incorrectly.
– dls
4 hours ago
Wolfram Alpha is weird. Usually, $(2 3)(1 2)$ means you apply $(1 2)$, then you apply $(2 3)$. However, Wolfram Alpha does it so that $(2 3)(1 2)$ means you apply $(2 3)$ then $(1 2)$. Thus, they get the inverse of what it would be normally, which is $(1 2 4 3)$.
– Noble Mushtak
4 hours ago
Wolfram Alpha is weird. Usually, $(2 3)(1 2)$ means you apply $(1 2)$, then you apply $(2 3)$. However, Wolfram Alpha does it so that $(2 3)(1 2)$ means you apply $(2 3)$ then $(1 2)$. Thus, they get the inverse of what it would be normally, which is $(1 2 4 3)$.
– Noble Mushtak
4 hours ago
@NobleMushtak yeah that's what I'm having some difficulty with as well. They perform the operation from left to right, which seems to be normal in some cases but not really how we're learning it in class so it a bit troublesome
– dls
4 hours ago
@NobleMushtak yeah that's what I'm having some difficulty with as well. They perform the operation from left to right, which seems to be normal in some cases but not really how we're learning it in class so it a bit troublesome
– dls
4 hours ago
2
2
Unfortunately, this is something you kind of have to deal with in abstract algebra because there are two different conventions. Most people use right to left like function composition, but you always have to make sure you check what convention the article/CAS you are reading uses because some people use left to right.
– Noble Mushtak
4 hours ago
Unfortunately, this is something you kind of have to deal with in abstract algebra because there are two different conventions. Most people use right to left like function composition, but you always have to make sure you check what convention the article/CAS you are reading uses because some people use left to right.
– Noble Mushtak
4 hours ago
add a comment |Â
4 Answers
4
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oldest
votes
up vote
2
down vote
accepted
I would also get $(1342)$ if I were to calculate that product, following basically the same argument you did.
If you got $(1243)$ from WolframAlpha, note what it says in the input interpretation field:
the product $sigmatau$ represents the permutation $tau(sigma(cdot))$
meaning they apply permutations from left to right (first $sigma$, then $tau$). In that light, their calculations are entirely correct, but they do not get the same answer as you do because they interpret the product of permutations differently.
At least they are nice enough to tell you what convention they use.
answered 4 hours ago
Arthur
105k799182
Thanks for that clarification! I guess I'm tired of computing these cycle products and messed up plugging it in. Wish I knew more programming to just compute these for me haha
– dls
4 hours ago
add a comment |Â
up vote
1
down vote
For me, cycle notation is always confusing when multiplying, so I always try to write the whole permutation out. I will also go right to left, so we start with $(34)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 1 & 2 & 4 & 3endmatrixright)$$
Then, we apply $(12)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 2 & 1 & 4 & 3endmatrixright)$$
Finally, we apply $(23)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 3 & 1 & 4 & 2endmatrixright)$$
Thus, $1 to 3 to 4 to 2 to 1$, so we get the cycle $(1342)$. This is the same answer you got, so I hope this alternative method gives you more confidence that your reasoning is correct!
answered 4 hours ago
Noble Mushtak
12.6k1632
Thanks for the reminder, I do feel more confident with the array form as well haha
– dls
4 hours ago
add a comment |Â
up vote
1
down vote
You want to clarify $$(23)(12)(34)=(1243)$$
We need to move from left to right through these permutations.
Well, let us start with $1$
$1$ does not move under $(2,3)$ but it goes to $2$ under $(1,2)$ and does not move under $(3,4)$.
As the result, we have $$ 1to 2$$
Now let us see what happens to $2$
Well $2$ goes to $3$ under $(2,3)$ and $3$ does not move under $(1,2)$ and maps to $4$ under $(3,4)$
As the result $$ 2to 4$$
Let us see what happens to $4$
Well , $4$ does not move under $(2,3)$ and $(1,2)$ but maps to $3$ under $(3,4)$.
As the result $$ 4to 3$$
Let us see what happens to $3$
Well , $3$ moves to $2$ under $(2,3)$ and $2$ moves to
$1$ under $(1,2)$ and $1$ does not move under $(3,4)$.
As the result $$ 3to 1$$
The final result is $$ (23)(12)(34)=(1243)$$
Which is what we want to prove.
answered 4 hours ago
Mohammad Riazi-Kermani
35.5k41855
Thanks! I was doing this from right to left and it is nice to be shown the other way too :)
– dls
3 hours ago
@dls I do not know why they do that but they do it anyway. Thanks for your comment.
– Mohammad Riazi-Kermani
2 hours ago
add a comment |Â
up vote
-1
down vote
Just plug in numbers and see where they end up when you finish going through (23)(12)(34) a single time.
Plug in 1: It first gets sent to 2 by (12), then that 2 gets sent to 3 by (23), so we end at 3. Hence, 1 -> 3, so we start with (13 . [sic]
Plug in 3: It goes to 4 via (34), but the 4 doesn’t get sent anywhere after that (34). Hence, 3 -> 4, and we now have (134 . [sic]
Plug in 4: It goes to 3 via (34), and that 3 goes to 2 by (23). Hence, 4 -> 2, and we have (1342 . [sic]
Plug in 2: It goes to 1 via (12), and that 1 doesn’t go anywhere else. Hence, 2 -> 1. Since 1 is the first entry of (1342 , this means we have closed the loop—literally, we close the loop and get (1342) (by adding a parenthesis :P).
answered 4 hours ago
Junhugie
644
1
That's exactly what dls has already done, if you read the question post.
– Arthur
4 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I would also get $(1342)$ if I were to calculate that product, following basically the same argument you did.
If you got $(1243)$ from WolframAlpha, note what it says in the input interpretation field:
the product $sigmatau$ represents the permutation $tau(sigma(cdot))$
meaning they apply permutations from left to right (first $sigma$, then $tau$). In that light, their calculations are entirely correct, but they do not get the same answer as you do because they interpret the product of permutations differently.
At least they are nice enough to tell you what convention they use.
answered 4 hours ago
Arthur
105k799182
Thanks for that clarification! I guess I'm tired of computing these cycle products and messed up plugging it in. Wish I knew more programming to just compute these for me haha
– dls
4 hours ago
add a comment |Â
up vote
2
down vote
accepted
I would also get $(1342)$ if I were to calculate that product, following basically the same argument you did.
If you got $(1243)$ from WolframAlpha, note what it says in the input interpretation field:
the product $sigmatau$ represents the permutation $tau(sigma(cdot))$
meaning they apply permutations from left to right (first $sigma$, then $tau$). In that light, their calculations are entirely correct, but they do not get the same answer as you do because they interpret the product of permutations differently.
At least they are nice enough to tell you what convention they use.
answered 4 hours ago
Arthur
105k799182
Thanks for that clarification! I guess I'm tired of computing these cycle products and messed up plugging it in. Wish I knew more programming to just compute these for me haha
– dls
4 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I would also get $(1342)$ if I were to calculate that product, following basically the same argument you did.
If you got $(1243)$ from WolframAlpha, note what it says in the input interpretation field:
the product $sigmatau$ represents the permutation $tau(sigma(cdot))$
meaning they apply permutations from left to right (first $sigma$, then $tau$). In that light, their calculations are entirely correct, but they do not get the same answer as you do because they interpret the product of permutations differently.
At least they are nice enough to tell you what convention they use.
answered 4 hours ago
Arthur
105k799182
I would also get $(1342)$ if I were to calculate that product, following basically the same argument you did.
If you got $(1243)$ from WolframAlpha, note what it says in the input interpretation field:
the product $sigmatau$ represents the permutation $tau(sigma(cdot))$
meaning they apply permutations from left to right (first $sigma$, then $tau$). In that light, their calculations are entirely correct, but they do not get the same answer as you do because they interpret the product of permutations differently.
At least they are nice enough to tell you what convention they use.
answered 4 hours ago
Arthur
105k799182
edited 4 hours ago
answered 4 hours ago
Arthur
105k799182
answered 4 hours ago
Arthur
105k799182
answered 4 hours ago
Arthur
105k799182
105k799182
Thanks for that clarification! I guess I'm tired of computing these cycle products and messed up plugging it in. Wish I knew more programming to just compute these for me haha
– dls
4 hours ago
add a comment |Â
Thanks for that clarification! I guess I'm tired of computing these cycle products and messed up plugging it in. Wish I knew more programming to just compute these for me haha
– dls
4 hours ago
Thanks for that clarification! I guess I'm tired of computing these cycle products and messed up plugging it in. Wish I knew more programming to just compute these for me haha
– dls
4 hours ago
Thanks for that clarification! I guess I'm tired of computing these cycle products and messed up plugging it in. Wish I knew more programming to just compute these for me haha
– dls
4 hours ago
add a comment |Â
up vote
1
down vote
For me, cycle notation is always confusing when multiplying, so I always try to write the whole permutation out. I will also go right to left, so we start with $(34)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 1 & 2 & 4 & 3endmatrixright)$$
Then, we apply $(12)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 2 & 1 & 4 & 3endmatrixright)$$
Finally, we apply $(23)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 3 & 1 & 4 & 2endmatrixright)$$
Thus, $1 to 3 to 4 to 2 to 1$, so we get the cycle $(1342)$. This is the same answer you got, so I hope this alternative method gives you more confidence that your reasoning is correct!
answered 4 hours ago
Noble Mushtak
12.6k1632
Thanks for the reminder, I do feel more confident with the array form as well haha
– dls
4 hours ago
add a comment |Â
up vote
1
down vote
For me, cycle notation is always confusing when multiplying, so I always try to write the whole permutation out. I will also go right to left, so we start with $(34)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 1 & 2 & 4 & 3endmatrixright)$$
Then, we apply $(12)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 2 & 1 & 4 & 3endmatrixright)$$
Finally, we apply $(23)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 3 & 1 & 4 & 2endmatrixright)$$
Thus, $1 to 3 to 4 to 2 to 1$, so we get the cycle $(1342)$. This is the same answer you got, so I hope this alternative method gives you more confidence that your reasoning is correct!
answered 4 hours ago
Noble Mushtak
12.6k1632
Thanks for the reminder, I do feel more confident with the array form as well haha
– dls
4 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For me, cycle notation is always confusing when multiplying, so I always try to write the whole permutation out. I will also go right to left, so we start with $(34)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 1 & 2 & 4 & 3endmatrixright)$$
Then, we apply $(12)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 2 & 1 & 4 & 3endmatrixright)$$
Finally, we apply $(23)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 3 & 1 & 4 & 2endmatrixright)$$
Thus, $1 to 3 to 4 to 2 to 1$, so we get the cycle $(1342)$. This is the same answer you got, so I hope this alternative method gives you more confidence that your reasoning is correct!
answered 4 hours ago
Noble Mushtak
12.6k1632
For me, cycle notation is always confusing when multiplying, so I always try to write the whole permutation out. I will also go right to left, so we start with $(34)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 1 & 2 & 4 & 3endmatrixright)$$
Then, we apply $(12)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 2 & 1 & 4 & 3endmatrixright)$$
Finally, we apply $(23)$:
$$left(beginmatrix 1 & 2 & 3 & 4 \ 3 & 1 & 4 & 2endmatrixright)$$
Thus, $1 to 3 to 4 to 2 to 1$, so we get the cycle $(1342)$. This is the same answer you got, so I hope this alternative method gives you more confidence that your reasoning is correct!
answered 4 hours ago
Noble Mushtak
12.6k1632
answered 4 hours ago
Noble Mushtak
12.6k1632
answered 4 hours ago
Noble Mushtak
12.6k1632
answered 4 hours ago
Noble Mushtak
12.6k1632
12.6k1632
Thanks for the reminder, I do feel more confident with the array form as well haha
– dls
4 hours ago
add a comment |Â
Thanks for the reminder, I do feel more confident with the array form as well haha
– dls
4 hours ago
Thanks for the reminder, I do feel more confident with the array form as well haha
– dls
4 hours ago
Thanks for the reminder, I do feel more confident with the array form as well haha
– dls
4 hours ago
add a comment |Â
up vote
1
down vote
You want to clarify $$(23)(12)(34)=(1243)$$
We need to move from left to right through these permutations.
Well, let us start with $1$
$1$ does not move under $(2,3)$ but it goes to $2$ under $(1,2)$ and does not move under $(3,4)$.
As the result, we have $$ 1to 2$$
Now let us see what happens to $2$
Well $2$ goes to $3$ under $(2,3)$ and $3$ does not move under $(1,2)$ and maps to $4$ under $(3,4)$
As the result $$ 2to 4$$
Let us see what happens to $4$
Well , $4$ does not move under $(2,3)$ and $(1,2)$ but maps to $3$ under $(3,4)$.
As the result $$ 4to 3$$
Let us see what happens to $3$
Well , $3$ moves to $2$ under $(2,3)$ and $2$ moves to
$1$ under $(1,2)$ and $1$ does not move under $(3,4)$.
As the result $$ 3to 1$$
The final result is $$ (23)(12)(34)=(1243)$$
Which is what we want to prove.
answered 4 hours ago
Mohammad Riazi-Kermani
35.5k41855
Thanks! I was doing this from right to left and it is nice to be shown the other way too :)
– dls
3 hours ago
@dls I do not know why they do that but they do it anyway. Thanks for your comment.
– Mohammad Riazi-Kermani
2 hours ago
add a comment |Â
up vote
1
down vote
You want to clarify $$(23)(12)(34)=(1243)$$
We need to move from left to right through these permutations.
Well, let us start with $1$
$1$ does not move under $(2,3)$ but it goes to $2$ under $(1,2)$ and does not move under $(3,4)$.
As the result, we have $$ 1to 2$$
Now let us see what happens to $2$
Well $2$ goes to $3$ under $(2,3)$ and $3$ does not move under $(1,2)$ and maps to $4$ under $(3,4)$
As the result $$ 2to 4$$
Let us see what happens to $4$
Well , $4$ does not move under $(2,3)$ and $(1,2)$ but maps to $3$ under $(3,4)$.
As the result $$ 4to 3$$
Let us see what happens to $3$
Well , $3$ moves to $2$ under $(2,3)$ and $2$ moves to
$1$ under $(1,2)$ and $1$ does not move under $(3,4)$.
As the result $$ 3to 1$$
The final result is $$ (23)(12)(34)=(1243)$$
Which is what we want to prove.
answered 4 hours ago
Mohammad Riazi-Kermani
35.5k41855
Thanks! I was doing this from right to left and it is nice to be shown the other way too :)
– dls
3 hours ago
@dls I do not know why they do that but they do it anyway. Thanks for your comment.
– Mohammad Riazi-Kermani
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You want to clarify $$(23)(12)(34)=(1243)$$
We need to move from left to right through these permutations.
Well, let us start with $1$
$1$ does not move under $(2,3)$ but it goes to $2$ under $(1,2)$ and does not move under $(3,4)$.
As the result, we have $$ 1to 2$$
Now let us see what happens to $2$
Well $2$ goes to $3$ under $(2,3)$ and $3$ does not move under $(1,2)$ and maps to $4$ under $(3,4)$
As the result $$ 2to 4$$
Let us see what happens to $4$
Well , $4$ does not move under $(2,3)$ and $(1,2)$ but maps to $3$ under $(3,4)$.
As the result $$ 4to 3$$
Let us see what happens to $3$
Well , $3$ moves to $2$ under $(2,3)$ and $2$ moves to
$1$ under $(1,2)$ and $1$ does not move under $(3,4)$.
As the result $$ 3to 1$$
The final result is $$ (23)(12)(34)=(1243)$$
Which is what we want to prove.
answered 4 hours ago
Mohammad Riazi-Kermani
35.5k41855
You want to clarify $$(23)(12)(34)=(1243)$$
We need to move from left to right through these permutations.
Well, let us start with $1$
$1$ does not move under $(2,3)$ but it goes to $2$ under $(1,2)$ and does not move under $(3,4)$.
As the result, we have $$ 1to 2$$
Now let us see what happens to $2$
Well $2$ goes to $3$ under $(2,3)$ and $3$ does not move under $(1,2)$ and maps to $4$ under $(3,4)$
As the result $$ 2to 4$$
Let us see what happens to $4$
Well , $4$ does not move under $(2,3)$ and $(1,2)$ but maps to $3$ under $(3,4)$.
As the result $$ 4to 3$$
Let us see what happens to $3$
Well , $3$ moves to $2$ under $(2,3)$ and $2$ moves to
$1$ under $(1,2)$ and $1$ does not move under $(3,4)$.
As the result $$ 3to 1$$
The final result is $$ (23)(12)(34)=(1243)$$
Which is what we want to prove.
answered 4 hours ago
Mohammad Riazi-Kermani
35.5k41855
answered 4 hours ago
Mohammad Riazi-Kermani
35.5k41855
answered 4 hours ago
Mohammad Riazi-Kermani
35.5k41855
answered 4 hours ago
Mohammad Riazi-Kermani
35.5k41855
35.5k41855
Thanks! I was doing this from right to left and it is nice to be shown the other way too :)
– dls
3 hours ago
@dls I do not know why they do that but they do it anyway. Thanks for your comment.
– Mohammad Riazi-Kermani
2 hours ago
add a comment |Â
Thanks! I was doing this from right to left and it is nice to be shown the other way too :)
– dls
3 hours ago
@dls I do not know why they do that but they do it anyway. Thanks for your comment.
– Mohammad Riazi-Kermani
2 hours ago
Thanks! I was doing this from right to left and it is nice to be shown the other way too :)
– dls
3 hours ago
Thanks! I was doing this from right to left and it is nice to be shown the other way too :)
– dls
3 hours ago
@dls I do not know why they do that but they do it anyway. Thanks for your comment.
– Mohammad Riazi-Kermani
2 hours ago
@dls I do not know why they do that but they do it anyway. Thanks for your comment.
– Mohammad Riazi-Kermani
2 hours ago
add a comment |Â
up vote
-1
down vote
Just plug in numbers and see where they end up when you finish going through (23)(12)(34) a single time.
Plug in 1: It first gets sent to 2 by (12), then that 2 gets sent to 3 by (23), so we end at 3. Hence, 1 -> 3, so we start with (13 . [sic]
Plug in 3: It goes to 4 via (34), but the 4 doesn’t get sent anywhere after that (34). Hence, 3 -> 4, and we now have (134 . [sic]
Plug in 4: It goes to 3 via (34), and that 3 goes to 2 by (23). Hence, 4 -> 2, and we have (1342 . [sic]
Plug in 2: It goes to 1 via (12), and that 1 doesn’t go anywhere else. Hence, 2 -> 1. Since 1 is the first entry of (1342 , this means we have closed the loop—literally, we close the loop and get (1342) (by adding a parenthesis :P).
answered 4 hours ago
Junhugie
644
1
That's exactly what dls has already done, if you read the question post.
– Arthur
4 hours ago
add a comment |Â
up vote
-1
down vote
Just plug in numbers and see where they end up when you finish going through (23)(12)(34) a single time.
Plug in 1: It first gets sent to 2 by (12), then that 2 gets sent to 3 by (23), so we end at 3. Hence, 1 -> 3, so we start with (13 . [sic]
Plug in 3: It goes to 4 via (34), but the 4 doesn’t get sent anywhere after that (34). Hence, 3 -> 4, and we now have (134 . [sic]
Plug in 4: It goes to 3 via (34), and that 3 goes to 2 by (23). Hence, 4 -> 2, and we have (1342 . [sic]
Plug in 2: It goes to 1 via (12), and that 1 doesn’t go anywhere else. Hence, 2 -> 1. Since 1 is the first entry of (1342 , this means we have closed the loop—literally, we close the loop and get (1342) (by adding a parenthesis :P).
answered 4 hours ago
Junhugie
644
1
That's exactly what dls has already done, if you read the question post.
– Arthur
4 hours ago
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Just plug in numbers and see where they end up when you finish going through (23)(12)(34) a single time.
Plug in 1: It first gets sent to 2 by (12), then that 2 gets sent to 3 by (23), so we end at 3. Hence, 1 -> 3, so we start with (13 . [sic]
Plug in 3: It goes to 4 via (34), but the 4 doesn’t get sent anywhere after that (34). Hence, 3 -> 4, and we now have (134 . [sic]
Plug in 4: It goes to 3 via (34), and that 3 goes to 2 by (23). Hence, 4 -> 2, and we have (1342 . [sic]
Plug in 2: It goes to 1 via (12), and that 1 doesn’t go anywhere else. Hence, 2 -> 1. Since 1 is the first entry of (1342 , this means we have closed the loop—literally, we close the loop and get (1342) (by adding a parenthesis :P).
answered 4 hours ago
Junhugie
644
Just plug in numbers and see where they end up when you finish going through (23)(12)(34) a single time.
Plug in 1: It first gets sent to 2 by (12), then that 2 gets sent to 3 by (23), so we end at 3. Hence, 1 -> 3, so we start with (13 . [sic]
Plug in 3: It goes to 4 via (34), but the 4 doesn’t get sent anywhere after that (34). Hence, 3 -> 4, and we now have (134 . [sic]
Plug in 4: It goes to 3 via (34), and that 3 goes to 2 by (23). Hence, 4 -> 2, and we have (1342 . [sic]
Plug in 2: It goes to 1 via (12), and that 1 doesn’t go anywhere else. Hence, 2 -> 1. Since 1 is the first entry of (1342 , this means we have closed the loop—literally, we close the loop and get (1342) (by adding a parenthesis :P).
answered 4 hours ago
Junhugie
644
answered 4 hours ago
Junhugie
644
answered 4 hours ago
Junhugie
644
answered 4 hours ago
Junhugie
644
644
1
That's exactly what dls has already done, if you read the question post.
– Arthur
4 hours ago
add a comment |Â
1
That's exactly what dls has already done, if you read the question post.
– Arthur
4 hours ago
1
1
That's exactly what dls has already done, if you read the question post.
– Arthur
4 hours ago
That's exactly what dls has already done, if you read the question post.
– Arthur
4 hours ago
add a comment |Â
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1
I also get $(1342)$. Could $(1243)$ be a typo?
– Arthur
4 hours ago
@Arthur Okay thanks, maybe I plugged it into WolframAlpha incorrectly.
– dls
4 hours ago
Wolfram Alpha is weird. Usually, $(2 3)(1 2)$ means you apply $(1 2)$, then you apply $(2 3)$. However, Wolfram Alpha does it so that $(2 3)(1 2)$ means you apply $(2 3)$ then $(1 2)$. Thus, they get the inverse of what it would be normally, which is $(1 2 4 3)$.
– Noble Mushtak
4 hours ago
@NobleMushtak yeah that's what I'm having some difficulty with as well. They perform the operation from left to right, which seems to be normal in some cases but not really how we're learning it in class so it a bit troublesome
– dls
4 hours ago
2
Unfortunately, this is something you kind of have to deal with in abstract algebra because there are two different conventions. Most people use right to left like function composition, but you always have to make sure you check what convention the article/CAS you are reading uses because some people use left to right.
– Noble Mushtak
4 hours ago