Express (1,…,n) as a product of 2-cycles
In Alan F. Beardon's "Algebra and geometry" he asks in an exercise to express $(1 ldots n)$ as a product of two cycles:
Show that $(1 2 3 4)=(1 4)(1 3)(1 2)$. Express $(1 2 3 4 5)$ as a product of $2$-cycles. Express $(1 2 ldots n)$ as a product of $2$-cycles.
I assume that if we are working with the set of numbers ${1,ldots,n}$ doing so is not possible. Since all of the numbers of the set will be part of the cycle $(1 ldots n)$ already. I am new in this subject, so I do not know if I am making a mistake. Thanks.
abstract-algebra group-theory finite-groups symmetric-groups permutation-cycles
edited Nov 30 '18 at 21:42
Servaes
22.4k33793
asked Nov 30 '18 at 12:17
César D. VázquezCésar D. Vázquez
1767
|
show 2 more comments
In Alan F. Beardon's "Algebra and geometry" he asks in an exercise to express $(1 ldots n)$ as a product of two cycles:
Show that $(1 2 3 4)=(1 4)(1 3)(1 2)$. Express $(1 2 3 4 5)$ as a product of $2$-cycles. Express $(1 2 ldots n)$ as a product of $2$-cycles.
I assume that if we are working with the set of numbers ${1,ldots,n}$ doing so is not possible. Since all of the numbers of the set will be part of the cycle $(1 ldots n)$ already. I am new in this subject, so I do not know if I am making a mistake. Thanks.
abstract-algebra group-theory finite-groups symmetric-groups permutation-cycles
edited Nov 30 '18 at 21:42
Servaes
22.4k33793
asked Nov 30 '18 at 12:17
César D. VázquezCésar D. Vázquez
1767
3
Note that it doesn't have to be two disjoint cycles.
– Arthur
Nov 30 '18 at 12:20
aaaaah! Great, thanks
– César D. Vázquez
Nov 30 '18 at 12:21
Are you sure it is not asking for a product of $2$-cycles? Either is possible, but the latter would be more generalizable.
– Tobias Kildetoft
Nov 30 '18 at 12:25
You were precisely right @TobiasKildetoft. But how would you do it?
– César D. Vázquez
Nov 30 '18 at 12:27
1
The problem was stated the way it was in order to helpfully guide you away from tackling the hardest part right away. Did you work out how $(1 2 3 4) = (1 4)(1 3)(1 2)$? Did you try to do a similar thing with $(1 2 3 4 5)$? What happened then?
– David K
Nov 30 '18 at 12:41
|
show 2 more comments
In Alan F. Beardon's "Algebra and geometry" he asks in an exercise to express $(1 ldots n)$ as a product of two cycles:
Show that $(1 2 3 4)=(1 4)(1 3)(1 2)$. Express $(1 2 3 4 5)$ as a product of $2$-cycles. Express $(1 2 ldots n)$ as a product of $2$-cycles.
I assume that if we are working with the set of numbers ${1,ldots,n}$ doing so is not possible. Since all of the numbers of the set will be part of the cycle $(1 ldots n)$ already. I am new in this subject, so I do not know if I am making a mistake. Thanks.
abstract-algebra group-theory finite-groups symmetric-groups permutation-cycles
edited Nov 30 '18 at 21:42
Servaes
22.4k33793
asked Nov 30 '18 at 12:17
César D. VázquezCésar D. Vázquez
1767
In Alan F. Beardon's "Algebra and geometry" he asks in an exercise to express $(1 ldots n)$ as a product of two cycles:
Show that $(1 2 3 4)=(1 4)(1 3)(1 2)$. Express $(1 2 3 4 5)$ as a product of $2$-cycles. Express $(1 2 ldots n)$ as a product of $2$-cycles.
I assume that if we are working with the set of numbers ${1,ldots,n}$ doing so is not possible. Since all of the numbers of the set will be part of the cycle $(1 ldots n)$ already. I am new in this subject, so I do not know if I am making a mistake. Thanks.
abstract-algebra group-theory finite-groups symmetric-groups permutation-cycles
abstract-algebra group-theory finite-groups symmetric-groups permutation-cycles
edited Nov 30 '18 at 21:42
Servaes
22.4k33793
asked Nov 30 '18 at 12:17
César D. VázquezCésar D. Vázquez
1767
edited Nov 30 '18 at 21:42
Servaes
22.4k33793
asked Nov 30 '18 at 12:17
César D. VázquezCésar D. Vázquez
1767
edited Nov 30 '18 at 21:42
Servaes
22.4k33793
edited Nov 30 '18 at 21:42
Servaes
22.4k33793
edited Nov 30 '18 at 21:42
Servaes
22.4k33793
22.4k33793
asked Nov 30 '18 at 12:17
César D. VázquezCésar D. Vázquez
1767
asked Nov 30 '18 at 12:17
César D. VázquezCésar D. Vázquez
1767
asked Nov 30 '18 at 12:17
César D. VázquezCésar D. Vázquez
1767
1767
3
Note that it doesn't have to be two disjoint cycles.
– Arthur
Nov 30 '18 at 12:20
aaaaah! Great, thanks
– César D. Vázquez
Nov 30 '18 at 12:21
Are you sure it is not asking for a product of $2$-cycles? Either is possible, but the latter would be more generalizable.
– Tobias Kildetoft
Nov 30 '18 at 12:25
You were precisely right @TobiasKildetoft. But how would you do it?
– César D. Vázquez
Nov 30 '18 at 12:27
1
The problem was stated the way it was in order to helpfully guide you away from tackling the hardest part right away. Did you work out how $(1 2 3 4) = (1 4)(1 3)(1 2)$? Did you try to do a similar thing with $(1 2 3 4 5)$? What happened then?
– David K
Nov 30 '18 at 12:41
|
show 2 more comments
3
Note that it doesn't have to be two disjoint cycles.
– Arthur
Nov 30 '18 at 12:20
aaaaah! Great, thanks
– César D. Vázquez
Nov 30 '18 at 12:21
Are you sure it is not asking for a product of $2$-cycles? Either is possible, but the latter would be more generalizable.
– Tobias Kildetoft
Nov 30 '18 at 12:25
You were precisely right @TobiasKildetoft. But how would you do it?
– César D. Vázquez
Nov 30 '18 at 12:27
1
The problem was stated the way it was in order to helpfully guide you away from tackling the hardest part right away. Did you work out how $(1 2 3 4) = (1 4)(1 3)(1 2)$? Did you try to do a similar thing with $(1 2 3 4 5)$? What happened then?
– David K
Nov 30 '18 at 12:41
3
3
Note that it doesn't have to be two disjoint cycles.
– Arthur
Nov 30 '18 at 12:20
Note that it doesn't have to be two disjoint cycles.
– Arthur
Nov 30 '18 at 12:20
aaaaah! Great, thanks
– César D. Vázquez
Nov 30 '18 at 12:21
aaaaah! Great, thanks
– César D. Vázquez
Nov 30 '18 at 12:21
Are you sure it is not asking for a product of $2$-cycles? Either is possible, but the latter would be more generalizable.
– Tobias Kildetoft
Nov 30 '18 at 12:25
Are you sure it is not asking for a product of $2$-cycles? Either is possible, but the latter would be more generalizable.
– Tobias Kildetoft
Nov 30 '18 at 12:25
You were precisely right @TobiasKildetoft. But how would you do it?
– César D. Vázquez
Nov 30 '18 at 12:27
You were precisely right @TobiasKildetoft. But how would you do it?
– César D. Vázquez
Nov 30 '18 at 12:27
1
1
The problem was stated the way it was in order to helpfully guide you away from tackling the hardest part right away. Did you work out how $(1 2 3 4) = (1 4)(1 3)(1 2)$? Did you try to do a similar thing with $(1 2 3 4 5)$? What happened then?
– David K
Nov 30 '18 at 12:41
The problem was stated the way it was in order to helpfully guide you away from tackling the hardest part right away. Did you work out how $(1 2 3 4) = (1 4)(1 3)(1 2)$? Did you try to do a similar thing with $(1 2 3 4 5)$? What happened then?
– David K
Nov 30 '18 at 12:41
|
show 2 more comments
1 Answer
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oldest
votes
Because $(1 n)(1 n)$ is the identity, the $n$-cycle $(1 2 ldots n)$ is the product of the two cycles $(1 n)$ and
$$(1 n)(1 2 ldots n)=(1 2 ldots n-1).$$
Induction on $n$ now also shows that $(1 2 ldots n)$ is a product of $2$-cycles.
answered Nov 30 '18 at 13:30
ServaesServaes
22.4k33793
What I've been seeing is that actually any cycle can be decomposed into a product of 2-cycles. For example, if I take (3 7 5 8 ) then, it would be equal to (3 8) (3 5) (3 7). So I will always have to fix some number within the cycle and then cyclically put in the second part of the parenthesis the next one in the cycle until I finish. Is that correct? What would be the general notation?
– César D. Vázquez
Dec 1 '18 at 16:18
1
Yes, that is correct. A proof of this is given by a minor variation on my answer; in stead of the $n$-cycle, apply the same argument to any cycle $(k_1 k_2 ldots k_n)$ with the $2$-cycle $(k_1 k_n)$. You will find that $$(k_1 k_2 ldots k_n)=(k_1 k_n)(k_1 k_{n-1})cdots(k_1 k_2).$$
– Servaes
Dec 1 '18 at 19:03
add a comment |
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Because $(1 n)(1 n)$ is the identity, the $n$-cycle $(1 2 ldots n)$ is the product of the two cycles $(1 n)$ and
$$(1 n)(1 2 ldots n)=(1 2 ldots n-1).$$
Induction on $n$ now also shows that $(1 2 ldots n)$ is a product of $2$-cycles.
answered Nov 30 '18 at 13:30
ServaesServaes
22.4k33793
What I've been seeing is that actually any cycle can be decomposed into a product of 2-cycles. For example, if I take (3 7 5 8 ) then, it would be equal to (3 8) (3 5) (3 7). So I will always have to fix some number within the cycle and then cyclically put in the second part of the parenthesis the next one in the cycle until I finish. Is that correct? What would be the general notation?
– César D. Vázquez
Dec 1 '18 at 16:18
1
Yes, that is correct. A proof of this is given by a minor variation on my answer; in stead of the $n$-cycle, apply the same argument to any cycle $(k_1 k_2 ldots k_n)$ with the $2$-cycle $(k_1 k_n)$. You will find that $$(k_1 k_2 ldots k_n)=(k_1 k_n)(k_1 k_{n-1})cdots(k_1 k_2).$$
– Servaes
Dec 1 '18 at 19:03
add a comment |
Because $(1 n)(1 n)$ is the identity, the $n$-cycle $(1 2 ldots n)$ is the product of the two cycles $(1 n)$ and
$$(1 n)(1 2 ldots n)=(1 2 ldots n-1).$$
Induction on $n$ now also shows that $(1 2 ldots n)$ is a product of $2$-cycles.
answered Nov 30 '18 at 13:30
ServaesServaes
22.4k33793
What I've been seeing is that actually any cycle can be decomposed into a product of 2-cycles. For example, if I take (3 7 5 8 ) then, it would be equal to (3 8) (3 5) (3 7). So I will always have to fix some number within the cycle and then cyclically put in the second part of the parenthesis the next one in the cycle until I finish. Is that correct? What would be the general notation?
– César D. Vázquez
Dec 1 '18 at 16:18
1
Yes, that is correct. A proof of this is given by a minor variation on my answer; in stead of the $n$-cycle, apply the same argument to any cycle $(k_1 k_2 ldots k_n)$ with the $2$-cycle $(k_1 k_n)$. You will find that $$(k_1 k_2 ldots k_n)=(k_1 k_n)(k_1 k_{n-1})cdots(k_1 k_2).$$
– Servaes
Dec 1 '18 at 19:03
add a comment |
Because $(1 n)(1 n)$ is the identity, the $n$-cycle $(1 2 ldots n)$ is the product of the two cycles $(1 n)$ and
$$(1 n)(1 2 ldots n)=(1 2 ldots n-1).$$
Induction on $n$ now also shows that $(1 2 ldots n)$ is a product of $2$-cycles.
answered Nov 30 '18 at 13:30
ServaesServaes
22.4k33793
Because $(1 n)(1 n)$ is the identity, the $n$-cycle $(1 2 ldots n)$ is the product of the two cycles $(1 n)$ and
$$(1 n)(1 2 ldots n)=(1 2 ldots n-1).$$
Induction on $n$ now also shows that $(1 2 ldots n)$ is a product of $2$-cycles.
answered Nov 30 '18 at 13:30
ServaesServaes
22.4k33793
answered Nov 30 '18 at 13:30
ServaesServaes
22.4k33793
answered Nov 30 '18 at 13:30
ServaesServaes
22.4k33793
answered Nov 30 '18 at 13:30
ServaesServaes
22.4k33793
22.4k33793
What I've been seeing is that actually any cycle can be decomposed into a product of 2-cycles. For example, if I take (3 7 5 8 ) then, it would be equal to (3 8) (3 5) (3 7). So I will always have to fix some number within the cycle and then cyclically put in the second part of the parenthesis the next one in the cycle until I finish. Is that correct? What would be the general notation?
– César D. Vázquez
Dec 1 '18 at 16:18
1
Yes, that is correct. A proof of this is given by a minor variation on my answer; in stead of the $n$-cycle, apply the same argument to any cycle $(k_1 k_2 ldots k_n)$ with the $2$-cycle $(k_1 k_n)$. You will find that $$(k_1 k_2 ldots k_n)=(k_1 k_n)(k_1 k_{n-1})cdots(k_1 k_2).$$
– Servaes
Dec 1 '18 at 19:03
add a comment |
What I've been seeing is that actually any cycle can be decomposed into a product of 2-cycles. For example, if I take (3 7 5 8 ) then, it would be equal to (3 8) (3 5) (3 7). So I will always have to fix some number within the cycle and then cyclically put in the second part of the parenthesis the next one in the cycle until I finish. Is that correct? What would be the general notation?
– César D. Vázquez
Dec 1 '18 at 16:18
1
Yes, that is correct. A proof of this is given by a minor variation on my answer; in stead of the $n$-cycle, apply the same argument to any cycle $(k_1 k_2 ldots k_n)$ with the $2$-cycle $(k_1 k_n)$. You will find that $$(k_1 k_2 ldots k_n)=(k_1 k_n)(k_1 k_{n-1})cdots(k_1 k_2).$$
– Servaes
Dec 1 '18 at 19:03
What I've been seeing is that actually any cycle can be decomposed into a product of 2-cycles. For example, if I take (3 7 5 8 ) then, it would be equal to (3 8) (3 5) (3 7). So I will always have to fix some number within the cycle and then cyclically put in the second part of the parenthesis the next one in the cycle until I finish. Is that correct? What would be the general notation?
– César D. Vázquez
Dec 1 '18 at 16:18
What I've been seeing is that actually any cycle can be decomposed into a product of 2-cycles. For example, if I take (3 7 5 8 ) then, it would be equal to (3 8) (3 5) (3 7). So I will always have to fix some number within the cycle and then cyclically put in the second part of the parenthesis the next one in the cycle until I finish. Is that correct? What would be the general notation?
– César D. Vázquez
Dec 1 '18 at 16:18
1
1
Yes, that is correct. A proof of this is given by a minor variation on my answer; in stead of the $n$-cycle, apply the same argument to any cycle $(k_1 k_2 ldots k_n)$ with the $2$-cycle $(k_1 k_n)$. You will find that $$(k_1 k_2 ldots k_n)=(k_1 k_n)(k_1 k_{n-1})cdots(k_1 k_2).$$
– Servaes
Dec 1 '18 at 19:03
Yes, that is correct. A proof of this is given by a minor variation on my answer; in stead of the $n$-cycle, apply the same argument to any cycle $(k_1 k_2 ldots k_n)$ with the $2$-cycle $(k_1 k_n)$. You will find that $$(k_1 k_2 ldots k_n)=(k_1 k_n)(k_1 k_{n-1})cdots(k_1 k_2).$$
– Servaes
Dec 1 '18 at 19:03
add a comment |
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3
Note that it doesn't have to be two disjoint cycles.
– Arthur
Nov 30 '18 at 12:20
aaaaah! Great, thanks
– César D. Vázquez
Nov 30 '18 at 12:21
Are you sure it is not asking for a product of $2$-cycles? Either is possible, but the latter would be more generalizable.
– Tobias Kildetoft
Nov 30 '18 at 12:25
You were precisely right @TobiasKildetoft. But how would you do it?
– César D. Vázquez
Nov 30 '18 at 12:27
1
The problem was stated the way it was in order to helpfully guide you away from tackling the hardest part right away. Did you work out how $(1 2 3 4) = (1 4)(1 3)(1 2)$? Did you try to do a similar thing with $(1 2 3 4 5)$? What happened then?
– David K
Nov 30 '18 at 12:41